Electromagnetic
Fields


Sergei A. Schelkunoff


COLUMBIA UNIVERSITY


BLAISDELL PUBLISHING COMPANY
New York . London




First Edition, 1963
@ Copyright 1963, by Blaisdell Publishing Company,
A Division of Ginn and Company
All rights reserved under International and Pan-American Copyright Con-
ventions. Published in Ne\v York, Toronto, and London by Blaisdell
Publishing Company.
Library of Congress Catalog Card Number: 63-8925
l\lanufactured in the United States of America

To my wife
Jean Kennedy Schelkunoff

Preface
This text has been prepared for a sequence of two basic courses in
electromah'l1etic field theory. One course can be based upon the fIrst
five chapters, the main prerequisites for which are (1) a course in
general physics, including sections on electricity and magnetism, and
(2) calculus. With this background. students can master these chap-
ters in four semester-hours provided they work hard.
The first part of the book emphasizes the physical aspects of fields
rather than mathematical manipulation. It is for this reason that only
rather elementary college mathematics is required. More advanced
mathematics such as vector analysis, functions of a complex variable,
partial differential equations, and special functions are certainly
essential in more advanced courses for those who wish to specialize
in field theory; but in a basic introductory course this mathematics
is unnecessary and would only divert students' attention from the
essen tial characteristics of fields. There is real danger that the use of
advanced mathematics in a basic in troductory course would encourage
pencil pushing at the expense of thinking. The object is to learn to
express the physical concepts as simply as possible. Why seek to kill
a fly with a 16-inch gun when a fly swatter is available?
There is another danger in relying on too advanced mathematics in
an introductory course on flelds. Mathematics suggests rigor, and
students may get an erroneous idea that rigor assures truth in the
domain of physics. What is even worse, they may come to believe
that mere algebraic manipulation of symbols constitutes rigor and
assures the correctness of the results. The faith in such manipulation
can grow to such an extent that students may accept results which
are obviously wrong from the physical point of vic,v, without looking
over their solutions in search of an error. Students should be en-
couraged to develop a habit of sound even more than of rigorous,
thinking.
For ready reference, Chapter 1 contains a fairly detailed review of
fundamental field concepts. The main purpose, ho,vever, is to show
. .
Vl1

...
V1l1
Preface
the interrelation bet,veen static and time-varying fIelds. This inter-
relation is the basis for many approximate methods to follow. In
Chapter 2 we obtain the fields of basic sources. The results are im-
portant in themselves; they also illustrate fundamental field concepts.
In addition we develop approximate techniques for handling the
"almost static fIelds." These tcchniques enable us to solve many
important fIcld problems not amenable to exact analysis. They also
illustrate the physical nature of :\laxwell's equations. Chapter 3
in troduces the ideas of dissipation. storage. and transfer of energy in
fields. Thcse ideas are then applied to approximate analysis of cavity
resonators and to representations of physical circuit elements by net-
works of ideal circuit elements. Chaptcr 4 is devotcd to fundamentals
of wave propagation in transmission lines and to approximate analysis
of certain \vaveguides. In Chapter 5 we consider ,vaves guided by an
infInitely thin semi-infinite wire. The resulting formulas are then used
to analyze waves guided by coaxial cones. by diverging cones, and by
diverging ,vires. Finally, the same formulas enable us to obtain the
field of an electric current element or an oscillating electric charge.
The remainder of the chapter is devoted to some basic applications.
The remaining chapters have been prepared for a more advanced
basic course in fields. Here, more mathematical preparation is de-
sirable. 1"'hus for Chapter 6. dealing \vith normal modes of field
distribution and ,,"ave propagation. it would be helpful if the student
had some kno,vledge of the method of separation of variables. In
Chapter 7 scattering by small objects is considered, and, for the most
part. only elementary mathematics is needed. In Chapter 8 we treat
coupled oscillations and derive the equivalent networks for certain
continuous structures. In Chapter 9 the basic ideas for developing
generalized telegraphist's equations are explained and illustrated by
simple examples. For the last two chapters some knowledge of Fourier
series is essen tial.
Problems have been designed to develop the ideas and methods in
the text still further. For this reason quite a few problems contain
suggestions for their solution, and all problems are supplied with
anS'\lCrs.
I have deeply appreciated the comments and suggestions made by
Professors William H. Huggins and John R. \Vhinnery who read the
first draft of this book. I also wish to thank lr. Paul R. Karmel for
his help \vith reading the typed copies of the manuscript.
S. A. S.
COlUl1zbia Uni'versity
January 27, 1963

Contents
1. Basic Concepts and Equations
1.0 Introduction 1
1.1 Force, mass, work, energy 3
1.2 Electric charge, electric field, magnetic field 5
1.3 Electric in tensity E 8
1.4 Electric lines of force 10
1.5 Electromotive force 11
1.6 Electric current I and its density j 13
1.7 Ohm's law and conductivity q 14
1.8 Dissipation of energy 16
1.9 Tubes of electric current 16
1.10 The electric field of a poin t source of stcady electric curren t 17
1.11 A dipole source of curren t or a current elemcn t 18
1.12 Charge distribution in conductors 21
1.13 Faraday's law of electrostatic induction 23
1.14 Electric flux or displacemen t density jj 25
1.15 Relations between jj and E 28
1.16 Electric dipolc 29
1.%17 Magnetic flux density f3 30
1.18 The second aspect of B and the Faraday-11axwelllaw 32
1.19 1Iagnetic intensity jj 37
1.20 Operational definition of jj 39
1.21 l\1agnetomotive force U and Ampere's law 41
1.22 l\Jagnetic fields of a point source and a double source of
current in an infinite conducting medium 42
1.23 Displacement current and the Ampere-Maxwell law 42
1.24 l\Jagnetic field of a moving charge 46
.
IX

x
Contents
1.25 The force bctwecn two moving charged particles and
bet\veen t\VO current elements 49
1.26 Summary of field equations 50
1.27 Boundary conditions 55
1.28 Discon tinuities 58
1.29 Step-by-step calculation of electromagnetic fields 60
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20
3.0
3.1
3.2
2. Static and Almost Static Fields
Introduction
Potential
Calcula tion of electric fIelds
Calculation of charge distributions
l\lctal sphere in uniform electric field
Dielectric sphere in uniform electric field
Proximity effect
IVlagnctic scalar potential
l\lagnetic vector poten tial
Straight uniform current filaments
Circulating current
Comparison of electric and magnetic fields
Images
Tubes of flo'v and cquipotcntial surfaces
Properties of fields in the large
Resistance and conductance coefficients
Poten tial and capacitance coefficicn ts
Inductance cocflicients
l'ransmission lines
Coaxial transmission lines
Limitations of the step-by-step method of
sclf-consistcn t fields
62
62
66
72
76
80
82
85
86
89
91
94
102
107
113
114
117
117
118
124
calculating
126
3. Energy Storage, Dissipation, and Transfer
Introduction
Energy conversion and fio\v
Distribution of magnetic energy
127
128
130

Contents
.
Xl
3.3 Distribution of electric energy 131
3.4 Oscillations in a cavity 132
3.5 Damping constant 136
3.6 Equivalent circuit for parallel wires shorted at the far end 137
3.7 Equivalent circuit for parallel wires open at the far end 139
3.8 Equivalent circuit for a parallel plate capacitor 140
3.9 Equivalent circuit for a slotted toroidal conductor 141
3.10 Use of equivalent circuits 142
3.11 Higher modes of oscillation 144
3.12 Comparison of strengths of electric and magnetic fields 146
3.13 The meaning of "slowly varying field" 147
3.14 Electric networks 148
4. Waves
4.0 Introduction 150
4.1 lax\vell's laws of interaction between time-harmonic
electric and magnetic fields 150
4.2 Equations for time-harmonic fields in transmission lines 151
4.3 Field in the interior of a metal cylinder 155
4.4 Step-by-step solution of transmission equations 156
4.5 Equivalent circuits 159
4.6 Differential equations 160
4.7 Characteristic impedance 161
4.8 Propagation constant 162
4.9 Phase velocity, wavelength 162
4.10 Transfer of power 163
4.11 Attenuation constant 163
4.12 Reflection 164
4.13 Input impedance 165
4.14 Standing waves 165
4.15 l\tlodes of oscillation 166
4.16 Propagation in highly dissipative media and skin effect 167
4.17 Nonuniform transmission lines 169
4.18 Image parameters 170
4.19 Waves in hollow tubes 171

..
XII
Contents
5.
Spherical Waves
5.0 Introduction 177
5.1 Jaxwell's equations for circularly symmetric fields 177
5.2 V\'aves on semi-infinite wire 180
5.3 Waves bet\veen coaxial cones 182
5.4 \Vaves between coaxial cylinders 184
5.5 \Vaves between parallel planes 185
5.6 \Vaves guided by thin diverging cones 186
5.7 Waves guided by parallel wires 187
5.8 Waves generated by an electric current element 188
5.9 \\Taves above perfectly conducting planes 190
5.10 Radiation 191
5.11 In tcrference and directive radiation 194
5.12 Current distribution in thin wires 195
5.13 Short antenna 198
5.14 Half-\vave antenna 200
5.15 Retarded potentials 202
6. Normal Modes
6.0 Introduction 205
6.1 Direct cur.rent in conducting plates 206
6.2 Direct current in stratified plates 213
6.3 Direct current in expanding plates 217
6.4 Direct current in bent plates 220
6.5 Electric current filament, between perfectly conducting
parallel planes 222
6.6 Point charge inside a hollow metal tube of rectangular
cross section 224
6.7 Normal modes of field distribution and \vave propagation 227
6.8 Transverse magnetic (T1vI) waves between perfectly con-
ducting parallel planes 227
6.9 Transverse electric (TE) waves between perfectly con-
ducting parallel planes 229
6.10 Waves in perfectly conducting rectangular wavesguides 230
6.11 Natural oscillations in metal cavities 232

Contents
Xll1
6.12 Attenuation
6.13 Damping constant
6.14 Waveguides and cavities of general shapes
6.15 Excitation of guided waves
232
234
235
236
7. Reflection and Scattering
7.0 Introduction 239
7.1 Reflection at a junction of two transmission lines 239
7.2 Reflection from a discontinuity in a transmission line 241
7.3 Reflection of plane waves at normal incidence 243
7.4 Reflection of plane waves at oblique incidence 245
7.5 Waves at grazing incidence over imperfect ground 250
7.6 Wave antenna 250
7.7 Scattering by a discontinuity in a transmission line 252
7.8 Scattering by a small perfectly conducting sphere in free
space 254
7.9 Scattering by a small perfectly conducting sphere above a
perfectly conducting plane 255
7.10 Scattering by a long rectangular loop 256
7.11 Scattering by a short wire 258
7.12 Scattering by a half-wave wire 259
7.13 Scattering by a half-wave receiving antenna 262
7 .14 Scattering in waveguides 263
8. Coupled Oscillations
8.0 Introduction 268
8.1 Oscillations in two coupled circuits 268
8.2 Beats 269
8.3 Concentrated coupling between sections of transmission
lines 271
8.4 An equivalent network for a shorted section of a uniform
nondissipative transmission line and its admittance in
terms of resonant frequencies 272
8.5 Another equivalent network for a shorted section of a
uniform nondissipative transmission line and its im-
pedance in terms of antircsonant frequencies 276

xiv
Contents
8.6 Equivalent networks for nonuniform transmission lines 280
8.7 Lagrange's equations in circuit theory 283
9. Generalized Telegraphist's Equations
9.0 Introduction 285
9.1 Coupled transmission lines 285
9.2 Weak coupling 287
9.3 Directional coupling 289
9.4 Waves in stratified media between perfectly conducting
parallel planes 292
9.5 Waves in completely nonhomogeneous media between
perfectly conducting planes 294
9.6 Waves between uniformly bent planes 300
9.7 Waves betwecn imperfectly conducting parallel planes 302
9.8 Generalized coordinates 304
A.I
A.2
A.3
A.4
A.5
A.6
A.?
Appendix I
Coordinate Systems and Vectors
Coordinatcs systems and vector components
Transformation of coordinates
Elements of length, area, and volume
Gradient
Circulation of a vector and curl of a vector
Flux of a vector and divergencc
Laplacian
305
306
307
308
309
310
311
Appendix II
Appendix III
Problems
List of Symbols
Index
Maxwell's Equations
Laplace's Equation
312
314
315
407
409

Electromagnetic Fields

1
Basic Concepts and Equations
1.0 Introduction
1"he interaction between electric and magnetic fields is at the root
of electromagnetic wave propagation in free space-on which depend
radio communication, radar, light, noise from outer space, etc. It is
at the root of propagation in waveguides, in linear accelerators, in
cyclotrons, etc. And it is at the root of wave propagation in cables
and overhead transmission lines used in telephone and telegraph
communication. The interaction between electric and magnetic
fields is responsible for the behavior of physical electric circuit ele-
ments, electric circuits, and networks at low and at high frequencies.
In electric circuit and network theories one studies the behavior of
mathematical models which are made up of ideal resistors, inductors,
and capacitors. Such ideal circuit elements do not exist in nature.
Physical resistors, inductors, and capacitors may be approximated
by their ideal counterparts in a restricted frequency range. These
physical elements may be approximated much better and in a more
extended frequency range by equivalent networks of ideal elements.
In certain situations they may be represented exactly by appropriate
equivalent networks. Knowledge of field theory is essential for making
reliable approximations and equivalent representations even in
ordinary electric networks at low frequencies, not to mention micro-
wave networks. The objective of this book is to provide a physical
and mathematical background needed for understanding electric and
magnetic fields, the interaction between them as expressed by 1\1ax-
well's equations, and the most important consequences of this inter-
action. Modern physical research and engineering applications require
such a knowledge.
Our knowledge of electric and magnetic fields is derived from
circumstantial evidence and is based on interpretation of this evi-
dence. We are nearest to" seeing" a field when we perform the follow-
ing experiment. If we scatter iron filings on a sheet of paper at random
and bring a bar magnet under the sheet. the filings will rearrange
1

2
1/erlr()l11l1gnetic Jie/tls
themselves into an ordered pattern along lines diverging from the
vicinity of one end of the magnet and converging to the other end.
Apparently some sort of invisible force operates in the space around
the nlagnct.
'fhe needle of a magnetic compass aligns itself in the north-south
direction. If deflected, the needle tends to return to the original
position. If prevented from returning to the original position by a
spring, it will exert a force on the spring, either stretching or com-
pressing it. l'hus we speak of the earth's magnetic field (of force).
On a dry summer day it is not uncommon to receive a shock on
touching a doorknob, hear an accompanying crackling noise and
sometimes see a spark. Familiar to many are demonstrations showing
that amber rubbed with fur acquires the power of attracting light
objects such as pith balls and paper. A few experiments \vould
suffice to sho\\l' that the field of force around a piece of amber and fur,
an electric field, has different properties from the field surrounding a
bar magnet, even though there are similarities.
It is \vithin the province of physics to present systematically the
facts and to formulate the basic concepts relatcd to electricity and
magnetism. It is assumed that the reader is familiar with them.
1"'0 ensure that he understands the terms and the symbols as they are
used in this book, we shall summarize and illustrate the principal
conclusions, definitions, and equations vvhich are relevant to field
theory. In this chapter the order of presentation of basic field con-
cepts, illustrations, and point of view is intended to hclp the student
form mental pictures by an analogy between intangible quantitic5
such as the clectric flux density tJ and the "displacement current
density" aIJjat and more tangible quantities such as the density of
electric current in conductors. l'he student is advised to become
thoroughly familiar with the concept of "displacemen t curren t"
which is the cornerstone of 1\Jaxwell's field theory and its applications.
1\n intuitativc understanding of displacement current will enable
the student to analyze fields and \vavcs under various conditions at
least qualitatively and often semi-quantitatively.
Special attention should be given to the interrelation bet\veen
static and time-varying fields. 'fhis interrelation is the basis for
approximation methods developed in subsequent chapters. In order to
stress this idea \ve have abandoned thc conventional grouping of sub-
ject ma tter in to "electrostatics." H magnctosta tics," etc. Instead,
the grouping is arranged to emphasize key ideas, analogies, methods
of analysis, and methods of approximation.

Basic concl'pts and eql1a.tion
3
1.1 Force, mass, work, energy
'ATc derive our initial ideas about force from experience with pushing
and pulling, and then extend them to include invisible "forces"
such as those of gravitation, electric attraction, and repulsion, etc.
'fhe key ideas are expressed in Newton's laws of motion:
1. A body at rest will remain at rest, and a body in motion
will continue to move in the same direction and with the
same speed, that is, with the same velocity, unless acted
upon by some external force.
2. Whenever a force acts upon a body, it produces in the
motion of the body an acceleration which is proportional
to the force acting and is in the same direction as the force,
and is inversely proportional to the mass of the body
acted upon.
Quantitatively the second law is expressed as
dv -
m-=F
dt
(1.1)
in all coherent units, that is, in units based on independent and
arbitrarily chosen" fundamental" units of length, time, mass, and
electric charge (or electric current). In this book we use the MKSC
(meter-kilogram-second-coulomb) system of units in which the
"meter is the unit of length, the kilogram is the unit of mass .t1't, and
second is the unit of time t, the nteter per second is the unit of velocity
V, and the meter-kilogram per second per second, ntherwise known as
the 1lewton, is the unit of force F. The unit of electric charge, the
coulomb, will be defined approximately in the next section. Since it
is casier to measure accurately electric current, the legal standard
is a unit of electric current, the ampere, and the coulomb is defined
precisely as the alnpere-second. Hence the MKSC system is usually
called the MKSA system.
\\fl'hen motion is in a straight line, equation (1.1) may be written as
dv
11t - = F
dt '
(1.1)
where the speed v is the magnitude of the velocity vector v and F
is the magnitude of the force vector F. 1\1ultiplying this equation by

4
Eleclronlagl1ctic fields
the differential ds of the distance traveled by the mass in time dt
and integrating, \ve obtain
f ' dv J'
11l - ds = F ds,
,dt ,
o 0
fV mv dv = f' F ds,
V o '0
mv2 - 1Jlvij = r F ds.
'0
( 1.2)
The quantity on the right of equation (1.2) is called the work done
by the force in moving the body through the distance s - so. 1'1he
quantity 1n,P/2 is called the kinetic energy of the body. 'rhus the
equation states that the work done equals the incrcase in kinetic
energy.
If a body is lifted against the force of gravity 'Ing (\vhere the ac-
celeration of gravity g equals 9.8 m/sec 2 , more or less, depending
on the locality) to the height h, the \vork done is l1zgh. It is said that
the body has acquired a potential e1lerKY 11ZKIz. If this body is allo\vcd
to fall back. the potential energy \vill be converted to kinetic energy,
which in turn \vill be converted upon impact \vith the earth into heat
and dissipated. 'l'he l\lKSA unit of \vork and energy is the nd)ton-
l1zeter. called the joule. _
1\Iore generally the \vork If V done by a force 1 1 ' on a body moving
in an arbitrary curvilinear path AB is the line integral of the scalar
product
w = f P.dS = f F, ds = f F cas (ft, dS) ds, (1.3)
AB An An
\vhcre F, is the component of ft tangential to the curve and (F, dS)
is the angle bctween thc vector ft and the <Ii ff cren tial displacemen t
vector dS.
According to  e\vton '5 theory of gravitation, amply justified
by its applications, any t\VO material particles attract each other
along the line joining them \\'ith a force inversely proportional to the
square of the distance bet\veen them and directly proportional to
their n1asses:
F = - kg (1111111'2/r2) .
( 1.4)
'rhe negative sign is included to indicate that the force acts in the
direction of decreasing r. The gravitational constant kg equals 6.67 X

Basic concepts and equations
5
10- 11 meter 3 per kilogram-sccond 2 . It can be shown mathematically
that this equation applies to uniformly dense spherical shells if ,.
is the distance between their centers. It is also possible to determine
for any body its center of mass such that equation (1.4) applies as if
the entire mass of each body were concentrated in its center of mass
(or center of gravity).
Thus in the space around a material body a force operates on any
other material body, and we may say that a mass is surrounded by a
gravitatio11,al field.
1.2 Electric charge, electric field, magnetic field
Historically two kinds of electric charge, positive and ,negative, each
conceived as some sort of invisible fluid, were postulated to explain
primitive experiments with wax and ebonite rubbed with fur and
with bodies brought in contact with them. Simple experiments
suffice to demonstrate that charges of the same sign repel each other,
those of opposite signs attract, and that equal charges of opposite
signs can neutralize each other as far as external action is concerned.
Coulomb's law of force between charged particles and bodies is
analogous to Newton's law of gravitation
F = ke(qlq2/r2) ,
(1.5)
where ql and q2 are the electric charges on the particles and the
force acts along the line joining the particles.
Recent studies indicate that matter and electricity consist of a
relatively small number of elementary particles. Particles that are of
particular interest to us are electrically neutral particles such as
hydrogen atoms, positive electric particles called protons, and
negative electric particles called electrons. The mass of a hydrogen
atom is 1.67 X 10- 27 kg and the force of attraction between t\VO such
atoms is given by equation (1.4). A hydrogen atom can be split into a
proton and an electron. The forces existing between the various
particles are summarized in Figure 1.1. The force between a proton
and a hydrogen atom is nearly the same as the force between two
hydrogen atoms. The force between an electron and a hydrogen atom
is smaller by a factor 1844. This is consistent with an assumption that
these forces are gravitational and that the mass of an electron is
1/1844 times the mass of a proton; that is, the mass of an electron is
9.1 X .10-31 kg. A proton and an electron also attract each other with
a force inversely proportional to the square of the distance but this
force is larger than that given by equation (1.4) by a factor 4 X 10 42 .

6
Electromagnetic fields
e
e
It
FIGURE 1.1 A diagram illustrating the directions and, on a greatly
compressed scale, the relative magnitudes of: (1) the gravitational forces
between an electrically neutral hydrogen atom h and a proton p, an'
electron e, or another hydrogen atom; and (2) the electric forces be-
tween protons and electrons.
Two protons repel each other with the same force. In the same manner,
two electrons repel each other with this force. Thus we assume the
existece of two new kinds of "mass" called electric charge, positive
for the proton and negative for the electron. The signs are chosen to
agree with those assigned in early macroscopic experiments in which
the electric particles were separated by friction.
No electric charge whose absolute magnitude is smaller than
that of a proton or of an electron has ever been observed. Hence this
electric charge would be a natural choice for the" unit charge." How-
ever, it is too small for ordinary purposes. The practical (MKSA) unit
of electric charge, the coulomb, has been defined in relation to the
practical unit of electric current, the ampere. In the past the ampere
was defined in relation to certain electro-chemical phenomena;
at presen t it is defined in relation to forces existing between two
long parallel wires carrying electric currents. The charge of a proton
turns out to be 1.6 X 10- 19 coulomb and the charge of an electron
-1.6 X 10- 19 coulomb. Thus approximately (5/8) 10 19 electrons,
would constitute one coulomb of negative charge. The electric constant
ke in equation (1.5) equals approximately 9 X 10' joule-meters per
coulomb per coulomb.
There is also an electrically neutral particle, the neutron, whose
mass is nearly equal to that of a proton. In the diagram 1.1 we
can substitute neutrons for hydrogen atoms to illustrate the differ-
ence between gravitational and electric forces.
So far ,no magnetic particles have been discovered. Primitive

Basic concepts and equations
7
experiments show that two thin magnets exert forces on each other
consistent with the following assumptions.
1. Each end of one magnet exerts a force on each end of the
other.
2. These forces are inversely proportional to the square of
the distance between respective pairs of ends.
3. One end of one magnet repels one end of the second and
attracts the other with the same force if the distances are
the same.
4. rrhe north-seeking ends repel each other; in the same
manner sou th-seeking ends repel. The north-seeking and
south-seeking ends attract each other.
At this stage the evidence points to the existence of opposite "mag-
netic poles" at opposite ends of each thin magnet. Arbitrarily, the
north-seeking pole was named "positive" and the south-seeking
"negative." Although magnets in common with all material bodies are
composed of atoms and hence contain positive and negative electric
particles, they are normally electrically neutral and exert no force
on an external electric charge because the numbers of opposite par-
ticles are the same and the particles are close together. Thus, it ap-
pears that there exist "magnetic charges" different from electric
charges. However, if we cut a magnet in half, we find that both
halves are magnets, each with two poles.
Further experiments have shown that while a stationary electric
charge exerts no force on a magnet, an electric current, that is, a
moving charge, acts on a magnet and the action is directly propor-
tional to the magnitude of the current. Two wires carrying electric
currents also act on each other unless they are perpendicular to each
other. Two solenoids, that is, long and thin coils carrying direct
electric currents, act on each other and on thin magnets as magnets
act on each other. One end of a solenoid is north-seeking and the other
south-seeking. If the direction of the current is reversed, the polarity
is reversed. On the basis of such evidence it has been concluded that
"magnetic" action is a property of a moving electric charge and that
the action of a permanen t magnet is due to atomic circulating curren ts.
To summarize:
1. The force between stationary electric particles is given by
equation (1.5) and we say that an electric charge is sur-
rounded by an electric field. The charge is said to be the
source of this field.
2. The force between moving electric particles has two COffi-

8
Jlectro1nag11etic fields
ponents: One is given by equation (1.5) and the other
depends not only on their charges but also on their ve-
locities and the angle between directions of motion. Thus
it is said that moving charges create a magnetic .field super-
imposed on the electric field. In the case of permanent
magnets, no external electric field has been observed (ex-
cept, of course, when they are deliberately charged).
Even though electric charge is granular in nature, we shall consider
it a continuous fluid because we shall deal with very large numbers
of charged particles which are extremely close. They will be confined
to material bodies, conductors, and dielectrics, where their individuali-
ties will be lost. It would be quite different in electron streams, where
some particles might overtake the others, or fall behind them. The
" smoothing" assumption is permissible in electromagnetic field
theory wherein we are not concerned with noise phenomena which
are attributable to random movements of discrete particles.
1.3 Electric intensity E
The force F on a stationary electric charge q at a point P of a given
electric field is proportional to the charge. The ratio of this force
to the" test charge,"
E = Flq,
(1.6)
may be taken as a measure of the strength of the field. Various terms
are used to denote vector E: electric field strength, electric field in-
tensity, or simply electric intensity at a point P.
It is essential to note that when a test charge is introduced into an
actual electric field for measuring purposes, i t may disturb the
positions of the charged particles producing the field. The above
ratio will then be the electric intensity of the altered field. It will be
the intensity of the original field in either of the following cases.
1. The sources of the field are held fixed.
2. Point P is so far away from the sources that the test charge
does not affect their positions.
3. The test charge is so small that it does not affect the
positions of the sources.
The MKSA unit of electric intensity is one newton per coulomb
or, as we shall presently see, the volt per meter.
From Coulomb's law, equation (1.5), we find that the electric

Basic concepts and eq uations
9
intensity of the field produced by a point charge q has only the
radial component (assuming that the center of coordinate system is
at the point charge)
Er = k,!}/r2.
If q is positive, E points away from the charge; otherwise it points
toward the charge as shown in Figure 1.2 (a, b) .
The electric intensity of any given distribution of charged particles.
\p
\
\
\
\
+qe
/
/
/
/
/
/
/
/ a
p
I
I
-q'
,
,
,
,
+q
-q
(a)
(b)
(c)
/ 
/ ,
/ ,
,
,
,
,
,
+q
(e)
+2q
/
//
"
,/
/"
/
,,/
fI'
+q
,
,
,
\
\
,
,
\
/ ,
/ ,
/ ,
/ ,
/ ,
/ ,
/ ,
//0' ,
/
/
/
/
/
/
/
/
,
,
,
,
,
,
,
,
(d)
+q
+q
(f)
\
,
\
\
\
,
.
+q
FIGURE 1.2 Illustrations of vectorial addition of electric intensities.
may be found by adding vectorially the electric intensities produced
by individual particles. Thus, at any point in the plane perpendicular
to and bisecting the line joining two equal and opposite charges,
the electric intensity is perpendicular to the plane. See Figure 1.2 (c).
Its magnitude equals that given by the above last equation times 2
cos lX. If the charges are of the same sign, E is in the plane [Figure
1.2 (d) ] and the multiplying factor is 2 sin a.

10
Electro111agllett"c fields
In general vectorial addition becomes complicated [Figure 1.2 (e, f) ]
since the magnitudes of the vectors to be added depend on the
magnitudes of the corresponding charges and on the distances in-
volved. In such a case \VC can add the corresponding Cartesian
components of all vectors and then obtain the magnitude and direc-
tion of the resultant.
1.4 Electric lines of force
A n electric line of force is a line tangential to the electric vector at
every point (see Figure 1.3). Such lines are useful for visual repre-
sentation of fields. If drawn properly, they show not only the direction
of the electric intensity but its relative magnitude as well. Electric
lines of a single charged particle are radial. For a positive particle
they start from the particle and go to infinity, as shown in Figure
1.3 (a). For a negative particle they start from infinity and end on the
particle. If we draw a certain number of lines, starting them uni-
(a)
(b)
FIGURE 1.3 Electric lines of force: (a) depicting the field produced by a point
charge; (b) depicting the field produced by tU.,lO equal but opposite charges.
formly around the charge, we note that they are more dense where
the field is strong than \vhere the field is weak. The numbers of lines
issuing from differen t particles may be taken proportional to their
charges to represent correctly the relative field strengths in the
vicini ties of the particles. The lines should issue from each particle
uniformly in all directions; but as we proceed on each line, always
in the direction of the E vector, we find that these lines diverge or
converge unevenly and exhibit the relative strengths of the field at
various points.

Basic concepts and equations
11
J4'igure 1.3 (b) sho\vs two equal and opposite charges \vhcrc all lines
issuing from the positive charge converge to the negative charge.
If the positive charge is t\vice as large. only one half of its lines
terminate on the negative charge; the other half go to infinity.
1.5 Electromotive force
"fhe elcctro1110tive force, or the 'voltage, along a given path AB
(Figure 1.4) is the line in tcgral of the tangen tial componen t of elec-
tric intensity
VAll = f E, ds = f BodS. (1.7)
AU AIJ
It is clear that V BA = - V A/J.
Suppose that a charged particle q is carried along i1 B. 1 ultiplying
equation (1.7) by q, \ve have
qV AB = f qB'dS. (1.8)
AB
Here q E is the force acting on q and the line in tcgral is the \vork
done by this force. Hence if E docs not vary \vith lime, the voltage
B
A
FIGl'RE 1.4 A 11 illustration of equation (1.7) 'li.'h ich defines the
elce/ronlothle force (the "7 1 oltaJ!.e") bet7.leen t'u..'o points, "I and /3, along
(l J.;l1'ell (unle.
If All is the \,'ork done by the 1ield I)cr unit charge carried along the
path lIB. If l varies \vith tinlc but the transit time of the particle is
so short tha t 1 has not changed appreciably in this time. V All is still
suhstan tially the \\'ork done by the licld per unit charge carried along
J113. Other\\"isc lf All is just the line integral of 11 \\'hich plays an
importan t role in electromagnetic theory.

12
1/('(/ r0l11l11:" c/ ie Jiclt! s
Subsequently. we shall find that in the case of elcctrotatic Jields it
is permissible to speak of a 'voltagc bctu./ccn tu'o points because the
voltage is independent of the path joining the t\VO points. };'urthcr-
more, this is often approximately true even for time-variable fields.
Since q V JaB is in the nature of work or energy, its unit is the joule.
Therefore, the unit of electromotive force is one joule per coulomb,
which is called the volt. From equation (1.7) it is clear that the
unit of electric intensity may be called the volt/meter as well as the
newton/coulomb.
Consider Figure 1.5 which illustrates two parallel coaxial circular
plates, equally and oppositely charged, with holes at their centers.
Let an electron (whose charge is -e) enter the left hole with a speed
-+
-+
E
A 
va -+ B 1-'1
 0 0 ...
... z
- e -+ -e
-+
-+
FIGURE 1.5 A n electron r110t 1 inK through an electric field bel' ween equally but
oppositel')' charged parallel plates.
Vo. What is its speed t'l when it has passed through the right hole?
Equation (1.1), Ne\vton's equation of motion, gives
dv
1n - = -eEz.
dt '
where Ez is the electric intensity along the axis and nl is the mass of
the electron. From equation (1.2) \ve find
mvi - mv = - f eE.dz = -eV AB = eV BA .
AU
Hence
Vl = [2 (e/1n) V BA + VJ1/2,

Basic concepts and equa.tions
13
\\here elnz 1.76 X 1011 coulonlbs per kilogram. If V UA 1000
volts and l'U O. then 'i'l = 1.87 X 10 7 m/sec. 'rhi speed is about 6%
of the speed of light and is small enough to justify the use of the
"rest mass" of the electron in the above equation. If V BA is so large
that 'l becomes an appreciable fraction of the velocity of light. the
relativistic effects should be included.
'I'he voltage bet\veen the terminals of a dry cell is of the order of
one volt. "fhe voltages between electric po\vcr lines, brought into
homes, and ground are usually about 110 to 115 volts (" effective"
or mean square values since these voltages are alternating). The
electric intensity of strong sunlight at the surface of the earth is
713 volts/meter (effective).
1.6 Electric current I and its density J.
In some media, notably metals, there are many easily movable
electrons. Such media are called conductors. As long as there is an
electric field in the interior of a cond uctor, electric charge will move.
1'his flow of charge is called electric current. The movements of
individual electrons are erratic; but on the average there will be a
drift in the direction of the electric intensity E. There are so many
electrons and they are so close together, that it is convenient to
think of the moving charge as a fluid in motion. The positive direction
of electric current is the direction opposite to that in which the
electrons are drifting. 'This con ven tion was adopted long before
electrons were discovered when electric current was thought to be a
flo\v of positive charge. It is convenient to maintain this fiction in
order to avoid a\vkward statements. Thus we shall think of electric
curren t as flow of positive charge even though in reality it is the
t10\V of negative charge in the opposite direction.
"fhe electric curren t I passing through a given surface is thus
defined as the time rate of f10\V of electric charge
I = dqjdt,
( 1.9)
\vhcrc dq is the charge crossing tc surface in time tit.
'fhe density of electric current], is defined as the limit
J = lim(ljS)rnllx, as as  o.
(1.1 0)
I*Icre D.I is the curren t passing through an clemen t of area  S.
'I'herc \vill be a particular orientation of this area, nanlcly one per-
pendicular to the lines of current flow, for which AI is maxinlum.

14
l/('rtrOl1l(l gllC/ ic Jieltls
For this orientation l/AS is the magnitude of the average current
density, and its Iin1it as A) approaches zero is the Inagnitudc J
of the current density at the point in question. 'l'he direction of the
vector is the direction of flo\v a t the point.
For any other orientation of the elementary area
I = ] (S) cos(], ii),
(1.11)
where (j, ii) is the angle between j and the normal 11, to the area
S. If the vector elcment of area is defined by
E:S = (S)ii,
(1.12)
wc can write equation (1.11) as follo\vs:
- 
I = J.S.
(1.13)
The current passing through any given area can then be expressed in
.
varIOUS ways as
I = f J.dS = f J cos (J, ii) dS = f I n dS,
(1.14)
where] n is the component of J in the direction normal to dS.
The 1\,lKSA unit of current is the cOli/onzb per second, called the
al1zpere. The unit of curren t density is the (11n pere per square rneier.
The current passing through a IOO-watt incandescent lamp is
about 10/11 ampere (effective). If the current in a wire one milli-
meter square is one ampere, the current density is lOll amp/m 2 .
1.7 Ohm's law and conductivity u
In metals and some other conducting media the current density is
proportional to thc electric in tcnsi ty E
- -
J = uE.
(1.15)
l"he coefficient of proportionality u is called the conductiL'ity of the
medium.
'rhis form of "Ohm's l"a\v" is deduced from experiments \vith
homogeneous conductors of uniform cross section (14"igurc 1.6). It
is found that the current I is directly proportional to the voltage
l ' bet\\.cen the ends and the area S of the cross section, and inversely
proportional to the length I,
/ = uVSjl,
(1.16)

Basic concepts and equations
15
",
",
/
E

A I
---------------
/'" 1 
FIGURE 1.6 lUus/ratin?, an expert@rnental t'erification of Ohm's la'll.!.
where the coefficient of proportionality depends on the substance
from which the conductor is made. Since 1/ S = J and V Il = E, we
find that equation (1.16) is consistent with equation (1.15). 'fhe
consequences of equation (1.15) under other varying conditions
have been found to agree with measurements.
The ratio G = flV in equation (1.16) is called the co, fuctance
of the conductor and its reciprocal R = V II the resistance. 'rhus
for a conductor of uniform cross section
G = uSll,
R = tluS.
( 1.17)
'The unit of resistance is the volt per ampere, called the ohm. The
unit of conductance, the an'lpere per volt, is called the mho. Hence,
the unit of conductivity is the mho per 'meter.
Table 1.1 exhibits the wide range of conductivities of various
substances
TABLE 1.1
Substance
Conductivity
copper
aluminum
1ron, pure
carbon (incandescent lamps)
sea \vater
soil
sand
quartz
5.8 X 10 7
3.5 X 10 7
10 7
2.5 X 10 4
5
0.015
0.002
8.3 X 10- 13
Very feeble electric fields can maintain strong currents in metals.
Quartz, on the other hand, is almost an ideal dielectric (noncon-
ductor) .

16
1lcctro11t111.11e/ic fields
1.8 Dissipation of energy
\\.hen electric charge is moving in a conducting mcdiun1 it Inovcs
in rcp()nse to a force. 'fhis force is doing ,vork. Consider an clenlcnt
of voluIlle (.1S) s) \vith the elenlcnt of length s in the direction
of lines of flo,\. and the elenlent of area S at right angles to thcnl.
In accordance ,vith equation (1.8) the ".ork done by the field on the
charge q moving through the distance s is (Es) :1q. 'fhc ,vork
donc per unit time is (Es) q, t1t) EJ dSS. lIenee, tbe ,\.ork
done per unit time per unit volume. that is. the dissipated po,vcr
per unit volume, is
p1v
l
E.T.
(1.18)
l'he principle of conserva tion of cnergy demands tha t this ".ork appear
as sonlC foml of energy. EX})crience sho\vs that it appears as heat.
I-Ica t is generated ,vhenever an electric current passes through a con-
(Iucting medium, and it represents the energy consunlcd in nlaintain-
ing the electric current. It is said that the latter energy is disipatcd
in heat. Dissipation of energy is distributed throughout the entire
volume ,vhcrc therc is elcctric currcnt.
In nlcdia obcying Ohm's la\\T equation 1.18) nlay also be ,vritten
as
PlV
I
uE'2 = J2 u.
1.19)
,
I:quation (1.18) is n10re general and it applies even \vhen J is a
nonlinear function of E. In nonisotropic nlcdia the directions of
E and J arc not the :-lame and nly that conlponent of Ji is doing
\vork \vhich is in the direction of ..T. lIenee, for such media
I;) = E.J.
( 1.20)
1.9 Tubes of electric current
"rubes of flo,v of electric charge or tubes of electric current arc regions
bounded by lines of Ilo\v. If the current is steady, there can be no
accunlulation of charge any,vhere in the nlcdiunl since such an
accunlulation \vould develop a tinle-variablc electric Jield. 'rhus
the sanlC current passes through every cross cction of a tube of 11o\v,
Figure 1.7
f J1, dS I = f J 2 ,(152.
(1.21 )
Also the total current passing through any closed surface equals zero

Basic concepts and equations
17
f j.ers = f JndS
o.
( 1.22)
In later sections \ve shall develop the concept of a "tube of flo\v"
still further in connection \vith less tangible field phenomena. It is
useful for visualizing abstract mathematical relations.
FIGL'RE 1.7 A tube of electric current and /71..'0 of its cross sections.
1.10 The electric field of a point source of steady electric current
1"he law of conservation of electric charge requires that if charge
is steadily streaming out of a point it must be supplied to this point
at the same rate. Also if charge is converging on a point, it must
00
,
I
I
1 1
,,/
 /,, 
FIGURE 1.8 A point source of electric current.

18
J.;/lytr0l11l1 gncl ic fields
enlanatc fron1 that point. In 14"igurc 1.8 \ve have a semi-infinite wire
insulated fronl the surrounding conducting medium except at the
end from \vhich the curren t I in the \\'ire escapes and spreads ou t\\'ard
to infinity. If the medium is homogeneous and isotropic, the lines of
flo\v will be radial and the curren t density J r will depend only on
the distance r from the end of the wire. Hence from equation (1.22),
we fInd
47rT 2 J r - I = 0
so that in the medium outside the wire
J r = I/47r1'2, Er = lrlO' = I/47rur 2 .
( 1.23)
1"he wire carrying current to the point source need not be straight.
The electric field will be the same; but the magnetic field will be
altered.
1.11 A dipole source of current or a current element
Equation (1.23) is similar to the equation for the electric intensity
of a charged particle. Hence, for two or more point sources of current,
we can calculate j and E by a method suggested in Section 1.3.
FIGURE 1.9 4 current elenlel1t cr)1zsistill?t of a short wire insulated
ironl the surrounding nlcdiunl e:\"cept at the ends A and B. Current
e111erges fronz B 'into the nlediunl and C01lverges to A.
Consider Figure 1.9 which shows a thin wire of length l, insulated

Basic concepts and equation
19
from the surrounding medium except at its ends A and B. Let an
electric current I be driven from ./1 to B. l'his current will emerge
from B into the surrounding medium and converge to A. On account
of circular symmetry about the axis AB we need consider only the
components Jz parallel to AB and J p perpendicular to AB.
Referring to Figure 1.10 and using equation (1.23), we find that
z
p
B
-r;
I It
L
P(p,z)
A
FIGURE 1.10 A diagram assisting in the calculation of the field of an electric
current elenzent.
the p and z components of J ar,e
I sin 0 1
J p =
47rri
I sin O 2
4 2
7rr2
( 1.24)
and
1 cos 0 1
Jz=
47rri
1 cos O 2
4 2 '
7rr2
where
rl = (r 2 - Ir cos 0 +l2) 1/2,
72 = (r 2 + Ir cos 0 + il 2 ) 1/2,
(1.25 )

20
1/('(tr()nl(lf!.l1rti( firlds
rl sin 0 1 = '2 sin O'J, r sin () = p,
Tl cos 0 1 = Z - (l/2), r2 cos ()2 = z + (l/2), r cos () = z.
lquations (1.24) become much simpler when l/r approaches
zero either because l approaches zero or because r increases in-
definitely. In the first instance we shall have a diPole source of cur-
rent. or a currel'zt element. In the second instance we can say that at
large distances the field of any double source of current may be ap-
proximated by the field of a dipole source. To obtain the simplifIed
expressions, we shall expand 11 and T2 in equations (1.25) in power
series of l/r and neglect the terms of order (1/r)2 and higher. For this
the binomial series
n(n - 1)
(1 + u) n = 1 + nu + tt 2 + ...
2!
 1 + tl1t as Zl  0
is needed. Thus
"1/" = [1 - (t/T) COS 8 + 1(l/r)2JI/2
 1 - (l/2r) cos 8.
Hence
rl  r - l cos 0,
and similarly
r2  r + l cos O.
'rherefore
"2 - rl  1 cos 0
'[his approximation is quite obvious from the geometric picture.
\Vc now write equations (1.24) as follows:
I rl sin (h [r2 sin (}2
J p = ,
47rr1 47rr
1Tl cos 0 1
Jz=
47rr
I r 2 cos ()2
47rr
By taking advantage of the last two rows of equations (1.25), we
obtain
J p = IT sin () ( _ ) = [r(r - ri) sin 0
47r ri, 47rrr
[r("2 - "1) (r + "2r1 + r1) sin 8
-
47rTir

Basic concepts and equations
21
As llr approaches zero, the first-order effect is given by the difference
T2 - rl = I cos o. In the remaining expressions Tl and r2 may be
replaced by r.
Similarly,
J. = : Ci - : ) -  Ci + : )
which can be analyzed as above.
Thus we find the field of a current element of motnent Il
3Il sin 0 cos 0
J p =
41rr 3
Il(3 C05 2 0 - 1)
J" = .
41rr 3
1'his is also an approximate field for any length l when (liT) is small
compared with unity.
In spherical coordinates (see Appendix I) we have
J r = J"cosO+JpsinO,
J 8 = - J z si n () + J p cos 0,
and the field of the current element becomes
,
21rr 3
Il sin (J
J 8 =
4.".,3 '
(1.26 )
It cas 0
J r =
where r is the distance from the current element and 0 is the angle
between its axis and the radius.
While the above direct calculation of the current density is straight-
forward it is laborious and \vould involve more work for a larger
number of sources. In Chapter 2 we shall introduce the concept of
paten tial which simplifies such calculations and is useful in many
other ways.
1.12 Charge distribution in conductors
Conducting bodies are normally electrically neutral. '!'hey con tain
equal numbers of protons and electrons so distributed that their
forces on an external charge cancel and there is no external field.
Also. on the average there is no internal field. If a quantity of electrons
is removed from a body. the body becomes positively charged. It is'

22
Electronlagnetic fields
negatively charged when there is an excess of electrons. If electrons
are introduced in to a conductor, the forces of repulsion will disperse
them. :They will keep moving as long as there is an e]ectric intensity
inside the conductor and a tangential component of E on the surface.
The normal component of E will try to pull them out but unless it is
extremely strong, the electrons will stay on the surface. A static state
is reached when the field inside the conductor and its tangential com-
ponent on the surface vanish. Thus in a static state the electric field is
normal to the surface of any conductor. For example, on a metal
sphere the electrons are distributed uniformly and the field is radial
[igure 1.11 (a) J.

t
t
/
/
"


"
/
'x
t
t
(a)
(b)
FIGURE 1.11 On a conducting sphere the excess electrons (or their deficiency)
are distributed uniformly at the surface.
The above argument was based on the excess of electrons. The
same argument applies where there is a deficiency. The protons will
pull the electrons until the positive field inside the conductor dis-
appears so that the deficiency of electrons will exist only on the sur-
face, and there the final distribution will be such that the tangential
componen t of E is zero [Figure 1.11 (b) J.
If a neutral conductor is introduced in an electric field, the free
electrons are displaced, Figure 1.12 (a), in such a way that the
electric intensity due to their displacement within, the conducting
body is equal and opposite to the original or "impressed" field. Also
the tangential component of the electric field due to the displaced

Basic concepts and equations
23
charge is equal and opposite to the impressed tangential component.
'I'his phcnonlcnon is kn()\vn as" electrostatic induction."
If \ve in troduce a positive chargc (electron defIciency) equal
to tha t displaced by the field in l;'igure 1.12 ( a), thc total displaced
charge i then positive and is distributed nl0re densely on one end of
the conductor [I"igurc 1.12 ( b) J.
+
Et +
+
+
Et +
+
+
(a)
+
(b)
Et
+
+
+
+
(c)
FIGURE 1.12 The disPlacernent of charge under the influence of an electric
field: (a) on a nell/ral sphere in an isotroPic nlediunz, (b) on a positively charged
sphere, (c) on a nelltral sphere in a nonisotropic nlediunl.
If the mcdiulTI outside a mctal sphere is crystalline, the electrons
arc uually displaced in some direction other than that of i1 [I;'igure
1.12 (c) ]. except \vhen ilis along a "principal" axis of the medium.
1.13 Faraday's law of electrostatic induction
I'araday ciiscovcrcd that if a charge q is enclosed by a neutral metal
sphere, an equal charge of the same sign appears on the external
surface of the sphere. He found that the externallicld is symmetric

24
EleclrOtllagnclic fields
\vhether the sphere is concentric with the enclosed charge or not
(Figure 1.13). Also, if the charge on the external surface is removed
from the sphere by momentarily grounding it, a charge equal and
opposi te to the enclosed charge will be left on the sphere. 1"his is
true regardless of the dielectric (nonconducting) medium surrounding
the charge q.
+
...-
t+
t+
+
+
----
+
+
+
+,,/
+
(b)
(a)
FIC;URE 1.13 Illustrating Faraday's experinlents 1.;)£111 disPlacetncnt of charge.
For his experiments, he fonned each sphere from t\VO hemispheres
so that he could easily enclose a charged body suspended from an
insulating string. An electric field was detected with a test charge.
The equality of t\VO charges of opposite signs can be easily established
by letting them combine and neutralize each other. 1"he equality of
like charges can be established by letting them combine with equal
charges of opposite sign. And it is always easy to obtain equal and
opposite charges.
};araday's observations arc explained in vic\v of present-day
knowledge about free electrons in conductors. 1"he enclosed charge
will either repel or attract the electrons, depending on its sign.
Hence the like charge \vill always be on the external surface and the
opposite charge on the internal surface. When the displacement of
charge has taken place, there \vill be no ficld in the metallic shell.
Hence the charge on the external surface is free to distribute itself
according to its own forces. .
rrhe fact that the displaced chargc is equal to the enclosed charge
can be predicted from Coulomb's law. Conversely, Coulornb's law
can be derived frol1t Faraday's law of electrostatic induction.

I
E
I
E
Baic conCl'pts and equations
I
E
25
(b)
'\.
-q
Cd)
FIGURE 1.14 Illustrating the definition of the density of electric displacenlent
at a point -in an electric field.
x 6
'\. q
- 6.q
(a)
I
E
1.14 Electric flux or displacement density jj
The quantity of electric charge displaced on a conductor under the
influence of a given electric intensity E depends on the area of the
conductor exposed to the field, on its shape, and on the surrounding
medium. The effect of area can be established by experiments with
thin metal plates. Figure 1.14 shows a solid metal plate (a), two thin
metal plates connected with a wire (b), two separate metal plates
in contact with each other, each having an insulating handle (c),
and two metal plates connected through a device (ballistic galva-
nometer) capable of measuring the time integral of electric current
I (t) passing through it or to total charge q that has passed through
it (d). If the device is connected at time t = 0 and if at t = T an
cssen tially static condition has been reached (the in terval happens
to be extremely short), then
T
I!.q = 1 I(t) dt. (1.27)
o
]"'his quantity is illustrated in Figure 1.15 by the area under the
curve represcn ting I (t) as a function of time. In all cases the charge
moves until the field in the metal plates (a) and (c) and between
ry
X
(c)

26
E/ectroruuglletic fields
ICt)
t
FIGURE 1.15 GraPhical representation of the tinze integral of electric current
by the area under the curve representing the current I(t) as a function of time t.
the metal plates (b) and (d) is reduced to zero by the displaced
charge. In case (b) the platcs can be disconnectcd while still in the
field. In (c) the plates can be separated while still in the fIeld. In
either case the displaced charge is trapped and can be measured.
Likewise it can be measurcd in the case (d) by closing the s\vitch
connecting originally neutral plates to the device measuring the
time integral of electric current. In the latter case an ammeter may
be used to measure the time rate of change of the displaced charge if
E is varying with time, (slowly enough).
With these experimcnts it is possible to establish the following:
1. The displaced charge q depends on the orientation of the
test plate.
2. q is proportional to the area tiS of the test platc.
3. When the test plate is perpendicular to a certain unit
vector d, the positive charge displaced in the direction d
(and the equal negative charge in the opposite direction)
is maximum
4. 'fhe charge displaced in any other orientation given by
the unit normal ii equals this maximum displaced carge
multiplied by the cosine of the angle bet\veen ii and d.
5. 1he displaced charge depends not only on the electric

13asic concept sand eq ua tions
27
intensity E but also on the medium. In pure \vater, for
instance, the displaced charge \vould be about eighty times
f15 large as in air.
In vic\v of the fIrst four experimental results we conclude that for
a complete quantitative description of an electric field we need
another vector jj whose magnitude D is the maximum displaced
charge per unit area
D = lim(gmax/ S), as S  0, (1.28)
and whose direction is the direction in which the displaced charge is
.
maXImum.
For any other direction ii the charge q displaced across a plate of
area S is
fig = jj. = D cos (D, 1i)S = DnS. (1.29)
The unit of jj is the coulomb per square meter.
'rher is no complete agreemen [ on the modern name for the
vector D. 'rhe classical name is "displacemen t." Another extensively
used term is "electric flux density," which suggests the physical fact
that D expresses clectric charge displaced per unit area. We are in
favor of calling jj electric displaceme1tt density at a given point of
the field, thus departing from classical tradition. The advantage of
using this term becomes clear when we consider the time rate of
ichange of D, aD/at which is expressed in amperes per square meter.
In computing magnetic fields produced by electric currents, the quan-
tity aD/at must be added to the density j of the conduction current,
and the surface integral of aD/at must be added to the true electric
current. For this reason laxwell introduced the concept of disp0ce-
11unt current. By follo\ving the classical tradition of calling aD/at
"displacement current." we are adding this" current" to the density
of conduction current. This is an awkward use of words. 'rhus the
follo\ving tcrminology is more appropriate.
1. 'rhe vector /) js the (electric) displacement density.
2. '[he vcctor alJ/at is the density of displacement current.
3. 'I'he surface integral
'11 = f b.dS = f Dn dS
(1.30)
is the electric displace1Jlellt through the surface of in tcgra-
tion (or electric flux) .
4. 'rhe tinlC rate of change of the electric displacement
through a surface is the (electric) displacCl1lent current

28
Jlec/ronlaglle/ic fields
through the surface
a 'I' a f - -. a f
It! = = D.dS = - IJ n dS.
at at at
1.15 Relations between Jj and E
In free space and in many media jj is proportional to E,
D = EE.
(1.31 )
Such media are called isotropic. 'I'he coefficient of proportionality E
is called the dielectric constal'zt of the medium.
In crystalline media where jj and E usually have differen t direc-
tions, the Cartesian components of jj are linear functions of the
Cartesian components of E.
'['he unit of E is the coulomb per volt per meter. l'he coulomb per volt
is the unit of capacitance and is called the farad. Hence the unit of E
is the farad per 1neter, the same unit as fOf capacitance per unit length.
In the case of a poin t charge Of a charged metal sphere in an
isotropic medium both E and /J are radial. From Faraday's exper-
iment in \vhich the metal sphere is concentric with the enclosed
. charge and docs not alter the geometry of the original field, we have
41rr 2 Dr = q,
Dr = q /41rr 2 .
( 1.32)
From this and equation (1.31) we find
Er = q/47rEr 2
( 1.33)
for a point charge and for an isolated metal sphere.
'I'he dielectric constan t of free space is denoted by Eo and its
value is
EO = 8.854 X 10- 12  (1/361r) lO- 9 farad/meter.
( 1.34)
'[he ratio
Er = E/t\)
(1.35 )
is the relative dielectric constant of the medium.

Basic concepts and equations
29
'l'ahle 1.2 gives relative dielectric constant5 for a few media.
TABLE 1.2
l\{cdia
Relative
dielectric constant,;
Quartz
Sand
])ry soil
\Vet soil
Sea \vater
f r = 4.5
Er = 10
Er = 10
Er = 30
Er = 78
In the case of soil and sand, the dielectric constan is will vary of
course depending on the sample; the above values are indicative of
the order of magnitude only.
1.16 Electric dipole
A pair of equal and opposite charges q and -q (Figure 1.16) when
separated by a small distance l constitute an electric dipole of 11l011lent
qt. Ideally. q is infinitely large and l infinitely small in such a way
that the moment ql is flnite. Equations (1.23), (1.32), and (1.33)
z
P(r,e, cp)
q
o
-q
FHiURE 1.16 1111 electric dipole alon? the z axis at the origin of the
spherical coordhulle s')'stenz.

30
llectromagnetic fields
exhibit the analogy between the current I and the charge q as sources
of flelds in infinite conducting media on the one hand, and in infi-
nite dielectric media on the other. Hence, by analogy with equation
(1.26), we have at point P
ql cos 8
Dr = ,
21rr 3
ql sin 8
D 6 =
41rT 3
(1.36 )
ql cos 8
Er = ,
27rEr 3
ql sin 8
Es = .
47rEr 3
1.17 Magnetic flux density B
For a quantitative description of electric fields in conducting media,
we have introduced two vectors: the electric intensity E acting on
electric charge and the resulting current density J. These are ob-
viously different physical quantities. For nonconducting (dielectric)
media we have also defined two quantities: the electric intensity E
and the electric displacement density D. From our operational defi-
nitions of D and J, we will prove that aD/at, the density of displace-
ment current, is related to J.
Although the operational definitions of E and D are different, iJ
can be defined in terms of E on the basis of present-day knowledge
about the constitution of matter. Defining tJ in terms of E requires
a knowledge of appropriate physical theories and considerable
mathematical detail. However, such a definition loses sight of the
macroscopic meaning of D, especially in vacuum. Similarly j can be
defined in terms of E and microscopic properties of matter. In effect
these definitions are equivalent to a theoretical calculation of the
relative dielectric constant Er and the conductivity (1. In the approach
we have adopted, Er and q are experimental constants.
In the case of magnetic fields it is also convenient to define tw
quantities for each point of the field: the magnetic flux density B
and the magnetic intensity [I. There are two physical aspects to B.
One is particularly important in particle physics and electronics and
the other is more prominent in field theory and its applications.
In particle dynamics f3 is defmed in terms f the force acting on a
moving charged particle and thus is analogous to E. In 1axwell's
e_quations governil.2g the behavior of fields jj is analogous to E and
B is analogous to D.
First we shall define i3 from the point of view of particle dynamics.
In a pure magnetic field (around a permanent magnet, for instance)

Basic concepts and cq ua tions
31
no force is exerted on a stationary charge q. \Vhen the charge is nlOV-
ing. the force on it is proportional to the product of its speed v and
the charge q. 'rhc force depends also on a direction of motion. 'fhcre
arc t\VO opposite directions such that no force is exerted on a charge
1110ving in these lirections. A nlagnctic needle aligns itself along these
directions. TJct b be a unit vector along this line. in the direction
from the south-seeking to the north-seeking end. \\Thcn the charge
is llloving in any other direction, the force j; m acting on it is perpen-
dicular to this direction and to the unit vector b. See Figure 1.17 (a).
Fm Fm F-qE
v v v
(a) (b) (c)
FIGURE 1.17 A 11, illustration of the force acting on a moving electric charge.
The magnitude of this force is proportional to the product of qv and
the sine of the angle bctween the velocity v and the vector b. 'fhe
coellicicn t of proportionality is defined as the magnitule B f the
1nagnetic flux density at the point occupied by q so that B = Bb.
In other words. in a pure magnetic field the force on a moving
charge is [l"\igurc 1.17 (b) ]
- -
l'm = qv X B. (1.37)
\Vhen there is an electric field as \vell, this equation gives the differ-
ence bet\\Tcen the total force j; acting on the particle and the force
qE \vhich \vould have been exerted on the particle if it were sta-
tionary [Figure 1.17 (c) J. lIenee. the total force is
1:' = q l + qv X i3.
( 1.38)
In light of this operational dcfinition of /3 the nan1C t1zagnctic jlux
density appears to be ill-suited. It might have been nlore proper to
call it "magnetic fIeld strength" or "nlagnetic in tensity." On the
other hand, 1nagnetic flux density i well suited when \\e consider the

32
llrctronulllctic firlds
second aspect of B. ,vhich is of primary importance in elcctronlagnetic
field theory.
l\lthough from this poin t of vic\v i3 and E arc analogou. their
physical dimensions arc different and they are measured in different
units. It is possible to define another quantity
GeR.
\vhere c is the speed of light in vacuum, \vhich has the same (limen-
sions as E and, therefore, \vould be nleasurcd in volts per meter.
lly introducing this quantity in equation (1.38) we have
F = qE + q(vjc) X G.
l"his quantity G would have certain advantages over i3 in particle
dynamics. "Gaussian field strength" might be a suitable name for it,
to distinguish it frm the magnetic flux density 13 and from the
magnetic intensity II.
1.18 The second aspect of B and the Faraday-Maxwell law
In accordance with equation (1.37), the free electrons in a conducting
\vire moving in a magnetic tield \vill experience a force and will be
displaced in such a way that the electric in tensity
Ei = V X B
( 1.39)
induced in the wire by the motion of the wire in the magnetic field is
annihilated by the electric intensity of the displaced charge. Note
that Ei exists only in the 1noving wire and is not due to any charge.
l"he displaced charge, on the other hand, creates a distributed electric
fIcld. 'fhe moving wire is a simple kind of" electric generator" and is a
prototype of practical electric generators at lo\v frequencies.
Consider Figure 1.18(aL in which the wire ]JIV, its velocity V,
and magnetic flux density B are mutually perpendicular. Assume that
B is pointing out of the paper and does not vary either in time or in
space. Then the induced electric intensity is along the \vire, poin ling
do\vn\\Tard. 'fhe induced voltage V.I.V causes an equal and opposite
voltage l'.\',,,t due to the displaccd charge
l'sM = VI.N = Blv.
( 1.40)
\\'herc l is the length of the wire. If ;.\1 N is sliding along statiunary
parallel ,vires connected to a voltmeter, the induced voltage can be
measured. If these leads are connected to a long-period ballistic

Basic concepts and equations
33
.11
v
.
B +  Ei
/" 
+
r N
(a)
Al
-

v
It'
t C2J
 is V NM l
N + Q 
p
v1t.v
x = Xo
x = x
(b)
FIGUHE 1.18 A wire moving in a magnetic field.
galvanomcter as shown in Figure 1.18(b), the ti1neilltegral of the
induced 'oltage can be measured. 1"his time integral is
/ V;UJ dt = / Blv dt = 1% Bl dx ,
  
\vhere JfN is at x = Xo when t = to and at x = x \vhen t = t. Note
tha t the time in tcgral is the same regardless of the speed of the wire.
Since \Vc have assumed that B docs not dcpend on x,
/ V;I.V dt = Bl(x - xo)
to
VtThcre S is the area S\vcpt by the wire.
1"hc quantity
= BS
,
(1.41 )
cp = BS
( 1.42)

34
Electrot1zagnetic fields
is called the 11lagnetic flux passing through the area of the rectangle
JflVQPM. l'he time integral of the induced voltage is seen to equal
the magnetic flux" cut" by the moving ,vireo
Suppose now that the magnetic field is static but that B varies
from point to point. The induced voltage '\vill be the line integral
of the induced in tensity
VfN = f Ei ds = [ Bv ds
J./ N 0
and its time integral will be
/ VlN dt = {'1' B ds dx,
to %0 0
where s is taken along the wire. Since the integrand on the right is
now by definition (1.42) the magnetic flux through an element of
area dS = ds dx, the integral itself is the magnetic flux through
MNQPM which is cut by the moving wire
<I> = J B dS.
Let us now remove the remaining restrictions and show that
f' Vl lN dt = <1>,
to
(1.43)
where
<I> = J Bn dS = J jj.tTS
(1.44)
even when Band MN are not perpendicular to v (Figure 1.19).
Vector i3 can be resolved in two components: En normal to the plane
defined by MN and v; B p parallel to the plane. The force exerted
by the latter on charge in the moving wire is perpendicular to the
wire and contributes nothing to the voltage along the_ wire. The
induced intensity due to the normal component is v X Bn sin {}. Its
magnitude is vBn sin {}. Hence
VfN = f vBn sin tJ ds,
MN
and
f' VN dt = f f Bn sin tJ ds dx,
to NQ MN

Basic concepts and eq ua tions
35

c::ZJ
FIGURE 1.19 A wire moving in a magnetic field.
where ds is an clement of length along MN. Since sin {} ds dx equals
the area d S of an clemen tary parallelogram, the integral on the right
is the maJ{netic flux. equation (1.44), crossing the area J.rINQPM and
which is ell t by }J lV in time t - to. The induced voltage is the time rate
with u,lzich 11lagnetic Jlux is cut by the 1noving wire.
Let us now calculate the line integral
f E; ds
of the induced electric intensity round a conducting loop moving
with a velocity v, illustrated in Figure 1.20. Assume that the inte-
gration is in the counterclockwise {lirection when the positive direc-
tion for the magnetic flux through the loop is chosen toward the
reader. Considering two clements on the opposite sides of the loop,
v
)l
FIGURE 1.20 A conductini!, loop nlovinK in a nlagnet-ic field.

36
Electro1nagnetic fields
AB and CD, we note that Vll equals the rate with which the flux is
leaving the loop while V1c equals the rate \vith which the flux is
etztering the loop so that the net contribution VB + VD = VB -
V tc to the counterclockwise voltage equals the time rate of decrease in
the magnetic flux cl>, linked with the loop. Thus
f . dcfJ
E; ds = -
dt
( 1.45)
is tbe voltage induced in the loop if the loop is moving in the field of a
stationary magnet. Since motion is relative, equation (1.45) gives
the voltage induced in a sta1tionary loop by a moving magnet. This is
indeed the case. A moving magnet exerts a force on any stationary
charge. Thus when the magnetic field at various points is varying
with time, there will be generated an electric field and its line integral
round a closed curve will be
f a <I> a f
E, ds = -- = -- Bn dS
at at '
(1.46 )
where the surface integration on the right may be extended over
any surface, the edge of which is the closed curve of integration
on the left. The partial derivative is used here because i3 is now a
function of independent space and time variables. This equation was
V(t)
t
FIGURE 1.21 The time integral of the voltage is shown by the area
under the curve. The voltage as a function of tirne is represented by ll(t).

Basic concepts and equations
37
formulated by l\Iaxwell to express experimental results obtained by
Faraday. \Ve shall call it the Faraday-Maxwell Law.
From equations (1.43) and (1.44) we observe that the unit of
magnetic flux is the volt-second. This unit is called the weber. Hence
the unit of magnetic flux density f3 is the volt-second per square meter
or the weber per square meter. This unit is too large for practical mag-
netic fields. A convenient subunit is the gauss equal to 10- 4 weberjm 2
or one weber per square hectometer.
Referring to equations (1.43) and (1.44), we see that with the aid
of a ballistic galvanometer we can measure the magnetic flux linked
with a small loop by measuring the time integral (Figure 1.21) of the
voltage induced in the loop. Hence we can measure the flux density,
that is, th flux per unit area of the loop. l"'his will give the com-
ponent of B normal to the area. We can also determine the particular
orientation of the loop for which the flux density is maximum. This
\vill give the magnitude B of magnetic flux density vector, and the
normal to the loop will gi'ye the direction. This is often used as the
operational definition of B because it is easier to apply in practice
than the one given in the preceding section.
1.19 Magnetic intensity H
It is an experimental fact that in free space and in some other homo-
· gencous media the vector i3 in the magnetic field around a straight
filament of electric current I is tangential to circles coaxial with the
filament as represented in -"igurc 1.22 (a). At distances small com-
pared to the length of the fIlament the magnitude of this vector
varies inversely as the distance p from the axis of the filament. This
suggests that for an infInitely long filament Bfj> = B would vary
inversely as p for all values of p. Thus the product of Bq, and the
length of the circumference is independent of p. It is an experimental
fact that tlzis product is proportional to the current
27rpBt/J = JJ.I.
( 1.47)
The coefficien t of proportionali ty JJ. is called the perl1teability of the
medium. 1'he permeability of vacuum is denoted by JJ.o and its magni-
tude is
volt-second henry
J.l.o = 47rlO- 7 or (1.48)
meter-ampere meter
In such media the line integral of B along any closed curve en-

38
Elec/ronlaglletic fields
(a)
(b)
F'I<iURE 1.22 (a) The 111aglletic lines of force around a straight
current jilanlCllt; (b) ./1 ssistinf.: in the exPlanation of Ampere's law.
circling I is proportional to !, Figure 1.22 (b),
f jj.ds = f B. ds = /-II.
( 1.49)
'l'hat this is so for any broken line made up of segments of radii

Basic concepts and Lluations
39
and circles coaxial with the filament is obvious. 'I'he radial segments
contribute nothing to the integral and the contributions from the
circular arcs depend only on the subtcnded angles dcp. Any curve in a
plane normal to the current filament is the limit of such a broken
curve and equation (1.49) applies to this limiting curve too. We can
also argue that the component of f3 tangential to the curve is B, =
BiP cos fJ while the clement of length of ds = p dcp/cos fJ. Hence, the
product is B, ds = Btj,p del>. From equation (1.47) we find BiPP =
p,I j27r and the in tegral of dcp around a closed curve encircling the fila-
ment is 27r. Hence, equation (1.49).
A similar argument shows that equation (1.49) applies to all
curves encircling I and not only to curves in planes normal to I.
For those closed curves which do not encircle I,
f B. ds = O.
( 1.50)
We now introduce another vector, the magnetic intensity
- -
II = B/p,
(1.51 )
which enables us to write equation (1.49) in the form
f H. ds = I.
( 1.52)
'fhe vector 11 is related directly to the current generating the mag-
netic field and we might conjecture that equation (1.52) applies to
nonhomogeneous media and to nonisotropic media in which i3 is not
tangential to circles coaxial \vith the current. But in order to verify
equation (1.52) experinzentally for such media we need an operational
definition of magnetic intensity f1 wich is independent of the
definition of the magnetic flux density B.
1.20 Operational definition of H
Consider a solelloid that is, a closely wound coil carrying curren t I
[F"igure 1.23 (a)] in air. Suppoe that the solenoid is straight and
long compared to its diameter. Experiments show that the magnetic
field inside such a solenoid is uniform except when the measure-
ments are made close to the winding where there are gaps in the
current or near the ends of the solenoid. If the solenoid closely
approximates a continuous sheet of circulating current [Figure
1.23 (b)], a nearly uniform field is obtained. Experiments further

40
l/ectrOnl(l[!.lleti( fields
H
.....------------....
.., -
J
t
1
, ",
'-__ H ___/
--- - - -. -- ....
(a)
H
EJ
)))})
(b)
FIGURE 1.23 A ssisting tOn an operational defi1uOtion of nlagl1etic intensity 11.
show that a magnetic needle tcnds to align itself parallel to the
axis of the solenoid, and that the torque on the magnet depends only
on the circulating current C per unit length of the solenoid. l'his torque
is independcn t of the length of the solenoid, of the shape of the cross
section, of the area of the cross section. of the number of turns per
unit length N, and of the actual current Ill} in the wire as long as
C = N I w is kept constan t.
We now define the magni tude 11 of the 1nag1tetic intensity inside
the solenoid as the circulating current per unit length,
II == C = Nlu..
( 1.53)
1'he direction of II is taken to be the direction of the force on the
north-seeking end of the magnet, that is, parallel to the axis of the
solenoid in the direction of the advance of a righ t-hande screw
turned in the direction of circulating current. l"hc unit of II is the
unit of current per unit length, that is. the anzpere per 1Jleter. 'rhus
the magnetic intensity II is directly related to the electric current
,!hich generates the l1eld in a laboratory. Any device for measuring
II can be calihrated by using definition (1.53).
Conceptually, we always relate the magnetic intensity at any
point of any magnetic field to the field inside a solenoid. For example,
in a solid (\vhich may bc nonisotropic), \ve imagine a thin tunnel
with a solenoid in it. 1"'herc is a circulating current C per unit length

Basic concepts and equations
41
of the tunnel for which the conlponcnt of the magnetic field in the
direction of the tunnel will be reduced to zero no matter ho\v this
field is being detected. By definition, the II component of the field
in the solid in the direction of the tunnel is equal and opposite to jj
generated by the solenoid. For any solid, there is a direction of the
tunnel for which the generated II will be maximum. rrhus we obtain
both the magnitude and the direction of /j in the medium.
When i3 and II arc defined independently, the relation bet\veen
them may be obtained either experimentally or from the physical
theory of matter. Thus it is found that in some media they are
proportional as in equation (1.51) for all values of ii. In other media
the proportionality holds only for small values of II and the equation
becomes nonlinear for the large values. Yet in other media, the
Cartesian components ofB are linear functions of the Cartesian
components of jj.
In a solenoid with 10 turns/em and with I equal to one ampere,
II = 1000 amp/meter. With an air-core B = 47r10- 4 weber/m 2 or
47r gauss. 1"'his is a rather weak flux density. In the earth's magnetic
field B varies from 0.3 gauss at the equator to 0.6 gauss at a pole.
In strong sunlight at the surface of the earth H = 1.89 amp/m
(effective), and B = 0.022 gauss. l'lux density of one ,veber/m2,
or 10 000 gauss, is strong. "fhe density of 100 000 gauss is very strong.
1.21 Magnetomotive force U and Ampere's law
'fhe line integral of the magnetic intensity along a curve AB is called
the nl{lgneto1noti'e force along A B,
VAll = f 11.ds = f Il. ds.
(1.54)
Experiments indicate that under all circumstances involving closed
steady currents, the magnetomotive force round any closed curve
equals the total current [linked with the curve
f H.ds = I = f JndS.
( 1.55)
In this equation the curve of integration on the left is the edge of
the surface of integration on the right. It is assumed that, \vhcn the
handle of a right-handed screw is turned in the direction of integration
round the edge, chosen as positive, the scrc\V advances in the positive
direction of the normal to the surface of integration. J'lzis is 11npere's
Law.

42
Electromagnetic fields
1.22 Magnetic fields of a point source a,nd a double source of current
in an infinite conducting medium
In Section 1.10 we obtained the current density of the field of a point
source in an infinite conducting medium. See Figure 1.8. Assume now
that the point source is at the origin of the coordinate system (Ap-
pendix I, ];'igures 1 and 2) and that the insulated ,vire carrying the
curren t to the source is along the positive z axis. On accoun t of
circular symmetry the magnetic lines arc circles coaxial with the z
axis and the magnetic intensity IItjJ is independent of the angle .
Applying equation (1.55) to a typical circle of latitude of radius
p = r sin () and to the spherical cap of radius r which surmounts this
circle and noti.ng that the radial current density outside the wire is
given by equation (1.23), we have
f 8f2r
27rr sin () 1/  = - I + J r r 2 sin 0 d dO
o 0
- - [ + {[" (I j47r) sin 0 drjJ dO
o 0
- -(1 + cos (})/.
Therefore,
1(1 + cos 0)
41rr sin (}
Similarly, we can calculate the magnetic intensity of a current
clement of moment It, situated along the z axis at the origin, from J r
given by equation (1.26). lhus
f 8f2r It cos () sin () dcf> dO It sin 2 ()
21rrsin(}II= =
o 0 27rr 2r
fliP =
( 1.56)
and
It sin 0
I/ =
47rr 2
( 1.57)
1.23 Displacement current and the Ampere-Maxwell law
Steady electric currents can exist only in closed conducting circuits
and there is no ambiguity in equation (1.55). On the other hand,
the conducting paths do not have to be closed to permit currents

Basic concepts and equations
43
varying \vith time. If a neutral wire is brought into an electric field,
the two ends will become oppositely charged (Figure 1.24) and
temporary electric currents must have existed during the period of
displacement of charge. If the wire is in an electric field varying \vith
time, the flo\v of charge in it will be taking place continuously.
l\loving charges gecratc a magnetic field. In this case, however,
the line integral of 11 cannot be given by equation (1.55) as it stands
since the closed curve of integration is not" linked" with an open
wire. A moving charged particle generates a magnetic field and again
equation (1.55) can not be applied as it stands. l\laxwcll in troduced
a concept of displace1nent current, which when added to the conduc-
tion current constitutes the total current. The total current is con-
tinuous in space so that equation (1.55) can be applied without
ambiguity. Equation (1.55), modified to apply to this total current,
represents correctly what happens in time-varying fields.
E
...
+
+++
o
FIGURE 1.24 A wire in an electric field.
, Displacement" current" is not a real current in the sense of moving
charge. To explain the meaning of this concept we shall imagine
a fc\v hypothetical siInplc situations even though it would be im-
possible to realize them in a laboratory. Subsequently, we can con-
sider some situations that can be realized. Suppose that at the
instant t = 0 ,ve begin to drive steady electric current I in a semi-
infinite wire, Figure 1.25, in a nonconducting medium, to\vard the
end O. Electric charge
q = [ I dt = It
o
( 1.58)
will start accumulating at point O. Equations (1.32) and (1.33) give
q It
E ----
r- -
47rEr 2 47rEr 2 '
( 1.59)
Dr =....!L - 
47rr 2 47rr 2 ·

44
llect rOnlU?on ct ieft el ds
'rhc electric field is increasing with time and, hence, the force driving
the current \viII have to increase with time. l'he time derivative of
the radial displacemcn t density
aIJ, J
at 41rr 2
( 1.60)
is seen to be constant and its value is given by the same expression
as the conduction current density J, [equation (1.23) ] for the case
when the medium is conducting. The conduction current in the wire
is now "con tin ued" as the radial displace1nent current of density
aDr/at in the medium outside. \Ve now modify equation (1.55) \vhich
gives the magnctomotive force ar')und a closed curve by including
this displacement current density. Thus we obtain the Ampere-
}.,{ GX"'dJelllaw,
f Il, ds = f (In + az n ) dS.
(1.61 )
Ampere is given the credit for equation (1.55), which is correct
when the conduction current is steady and closed. l\Jaxwell is given
the credit for introducing displacement current and thus making
the equation applicable to time-varying curren ts in either closed or
open conductors.
Using equation (1.60), we now find an expression for the magnetic
intensity II t/J,
Ilq, =
1(1 + cos 8)
411"r sin 8
( 1.62)
of the field generated by a semi-infinite uniform current filament
in a dielectric medium.'"fhis expression is, of course, identical with
expression (1.56) for the same filament in a conducting medium.
14'rom equa tion (1.36) for an electric dipole of momen t ql, 've obtain
the density of the displacemcn t curren t at poin t P in the field of the
dipole, Figure 1.26,
aD8
-=
at
fit sin ()
47rr 3
( 1.63)
al)r ql cos ()
at 27rr 3
Compare these equations ''lith equations (1.25) for the current
density of a double source in a conducting medium, from \vhich
equation (1.57) for the magnetic field ,vas obtained. i\ comparison
shows that the magnetic intensity of a current clement of nlonlent

Basic concepts and equations
45
00
I
I
I
I
l
,p
I
r/
I
I
I
FIGURE 1.25 A semi-infinite direct-current filament in a dielectric
,nedium and an accumulated charKe q = It! dt.
Il = (dq/dt)l in a dielectric medium is
It sin 8
H4J =
(1.64)
41rT 2
The concept of displacement current in vacuum is less tangible
than the concept of electric current in a conducting medium where
charged material particles are in motion. The most tangible aspect
of displacement current is: If two parallel conducting Plates are placed
in a tinte-varying electric field and connected to an a11zmeter [Figure
1.14 (d) ] there will be an indication, of current on the al1uneter which
can be accepted as the measure of the displace1nent current in the space
occupied by the plates when" the plates are removed. Similarly, the elec-
tric intensity E is detected only where an electric charge is introduced
into the field. In material media a part of displacenlent current, the
polarization current, consists of motion of bound electrons; but the
remainder is just as intangible as displacement current in vacuum.
By analogy, with tubes of true current flow, ,vhich arc illustrated in
Figure 1.7, Section 1.9. we can think of tubes of displacement current
and_ tubes of displaccmen t as regions bounded by lines tangen tial
to D. '[he tubes start on positive charges and end on negative charges
or at infinity, or they start at infinity and end on negative charges.
1"hcy may also be completely closed tubes, like doughnuts. The dis-
placement and the displacement current through every cross section
of a given tube arc the saIne. The total displacement (or electric flux)
through any surface enclosing a. charge q equals q. The displacel1zent

46
l/('ctrol1Ulllct ie field,,,
current through tlzis surface is q and thus is equal to the til1zC rate 7.oith
which tlte clza rge is entering the '(}olum"e eneloscd by the surface. If no
charge is enclosed. the tubes of displacement enter the volunlc and
then leave it. rrhe net displacement through such a surface is zero.
I)isplacement currents in one direction are either very feeble or of
very short duration. For example, if E increases at the rate of 1
volt/m/sec, the displacement current density in vacuum is J) =
EoE == 8.854 X 10- 12 amp/m 2 = 8.854 micro-microamp/m 2 . If E
increases at the rate of one volt per micro-microsecond. tJ = 8.854
amp/m 2 . This rate of increase cannot be maintained steadily for more
than a small fraction of a second since the electric generator would
have to develop tremendous internal forces to drive the charge which
creates the field.
On the other hand, if the field is alternating, then E = Ea sin 27rft
and I) = Eo27rfEa COS27rft so that for high frequencies tJ may be
fairly substantial. If f = 1010 cycles/see and Ea 100 volts/m, the
amplitude of tJ is 50.5 amp/m2.
Displacement currents are important even at low frequencies
because they can flow across large areas; conduction currents, how-
ever, are usually confined to relatively thin \vires. 1he great dispartity
in densities mayor may not be compensated by an equally great
disparity in the areas across \vhich the currents flo\v.
1.24 Magnetic field of a moving charge
We shall now calculate the magnetic intensity of a charged particle
moving vvith the speed v. 1vfagnetic lines are circles coaxial with the
line of motion. Consider Figure 1.27 which illustrates a typical mag-
netic line of radius p in the plane perpendicular to the line of motion
at distance z above the particle. Applying equation (1.61) and noting
that there is only a displacement current through the spherical cap of
radius r, bounded by the magnetic line, we have
II q, =  /27rp =  /27rr sin 0,
\vhere  is the time derivative of the displacemen t '11. Since
f 8f21"
\}I - (q/47rr 2 )r 2 sin () dc/> do = q(l - cos 0).
o 0
we have
. 1 · .
\}1 = zq sIn () 0,
\vhere () is the tinlC rate of change in 0 as the particle is moving

Basic concepts and equations
47
P(,,(),cP)
FIGURE 1.26 A n electric current element in a dielectric medium.
upward with the speed v = -z. Since
cot 8 = zip,
we find
- csc 2 8.()
zip = -vip,
and
() = (v sin 2 8) I p = (v sin 8) /r.
By substituting in the expression for  and then in II cp, we obtain
q'"u sin 8
II q, =. ( 1.65)
47rr 2
Comparing this with II q, for an electric curren t clemen t of moment
fl. equation (1.64), \ve note that. as far as the magnetic field is con-
cerned, the charged particle q moving with the specd v is equivalcn t to
an electric curren t clemen t of nlon1cn t /1 qv. 'I'his is hardly surpris-
ing since the movenlcnt of the particle may be represented by super-
inlposing a pair of charges, q and - g, an infinitesimal distance 1 apart,
on the charge g. See l.'igurc 1.28. 'fhe curren t in this dipole is I = q / T,
\vhcre T is the tinlC required for the charge to move a distance l.
lIenee the nl0111cnt It = qilT = qv. Neither J nor I are defined sepa-
rately; only thcir product has a meaning. On the other hand, the
currcnt density J has a perfectly definite meaning. Suppose that the
curren t I of the elcmen t is distribu tca over an infinitesimal volume

48
1le(tro11laglletic fields
z
q
FHIl1RE 1.27 Assistil1J!. in the calculation of the l11a.Knetic field
f!.cnera led by (/ I1tot'i 11 J!. (Ita rf!.c.
Sl, where I is in the direction of current and S is at right angles to it.
Then the momcnt of the current clement is j St. l"his equals the
moment of the moving charge qv = pSlv \vhere p is the density of
charge. Hence
] = pv.
( 1.66)
In deriving equation (1.65) it was assumed that only displace-
ment current is crossing the pherical cap surmounting the magnetic
line. 'J"his is true as long as the particle itself is not crossing the cap.
If it is, \ve should include the true current density given by equation
(1.66) in equation (1.61). l"he end result is the samc.
Equation (1.66) gives the density of COJl'i)cction currcnt at any place
of a stream of nl0ving charged particles. "l'his density is constan t
over the volume occupied by each particle and equals zero else\vherc.
Frequently \ve are intercsted only in the average value of con-
vection curren t in a stream. 'rhis is certainly the case in conducting
media ,vhere the average value i proportional to the electric in-
tensity, and the convection current is usually called the conduction
current. [See eq ua tion (1.15). ]

Basic concept and ('quatiolls
49
1.25 The force between two moving charged particles and between
two current elements
It is possible now to obtain the force one moving charged particle,
ql, exerts on another particle q2. First of all there is a force due to the
electric field of the first particle. Then there is a force due to the
magnetic field. Vector notation is convenient here. Let '12 be the vector
v
- 
, ,-----------------...
-
(+q)
+q
(a)
-q
@
+q
ll=qv

.
+q
(b)
FIGURE 1.28 (a) A moving electric charge; (b) an equivalent superposition of
an electric current element on a stationary charge.
from the first particle to the second as shown in I"igure 1.29. 'Then the
electric in tensity of the fIrst particle at the second is
E 1 -
qf r l2
47rtri2.
( 1.67)
VI
q2
ql
FIGURE 1.29 T'wo "loving point charges.
'fhe magnetic field is given by equation (1.65) and may be written as
ql"ih X '12
III =
47rr12
- }J.q 1 fh X ;12
HI = p,II = .
47rr2
( 1.68)
'J'hc total force FI'!. exerted by the tirst particle on the second is no\v
obtained from these equations and equation (1.38)
1';12 = q 2 E 1 + qi V 2 X B 1
or

50
J/('c/rO"lagl1rtir jirld s
qlq'2 r I2 p.qlq.i v 2 X (VI X ;I,.J
1'I2 = +
4 r :I 4 r:' .
7rf I 7r 12
In the case of t\VO current elements of nl0mcnts /11 1 and /.)'2, the
force due to the magnetic field of the first element on the second is
- -
P = J.lfdl2 X (ll X '12)
12 , m 4 :s .
7rr12
In addition there will be four electric forces between the end charges
of the clemen ts.
(1 J)9)
( 1.70)
1.26 Summary of field equations
In the preceding sections we defined and illustrated five field quan-
tities: the electric intensity E, the density of conduction current
J cond , the electric displacemen t densi ty jj, the magnetic in tensity II,
a.!ld the magnetic flux density (or magnetic" displacement" density)
13. In isotropic media these quantities are related as follows:
J concl = (J E. jj = EE, B = p.ll, (1.71)
\vhere (J) E, J.1. are parameters of the media. In iron and some other
media the last equation is true only for weak fields. For strong fields
the equation is nonlinear. In nonisotropic (crystalline) media the
Cartesian conlponen ts of vectors on the left are linear functions of the
Cartesian conlponen ts of vectors on the right. In other words. we
replace the numerical multipliers in equations (1.71) by matrices.
In a strean1 of electric charge the density of convection current is
J('onv = pv,
(1.72)
\vherc p is the volume density of charge and. v is its velocity.
'l'hc force exerted by electric and magnetic fields on a charge q is
- - -
F = qE + qv X B. (1.73)
It should be stressed that this equation does not give all the forces
\vhich nlay act on a charged particle. For instance, it docs not include
forces of gravitation \vhich are generally relatively small. It docs not
include forces \vhich separate positive and negative charge in chemical
dry or liquid cells. It docs not include forces which eject electrons in
thern1al emission. All these other forces are importan t in electric
generators. In this book ,ve shall be concerned mostly \vith fields
outside electric generators.
In addition there are t\\yo basic equations \vhich connect electric
and magnetic quan tities and express the lau's of interaction bctu'cen
electric and 11lc1gnctic.fields:

Basic concepts and equations
51
1. Faraday-Maxwell equation
f E. ds = - :t f E.. dS
(1.74)
and
2. Ampere-Maxwell equation
f H. ds = f J.. dS + :t f D.. dS,
(1.75)
where
- - -
1 = loond + loony.
The integrations on the right are performed over any surface while
the integrations on the left are extended over the edge of this surface,
[Figure 1.30(a) J. The algebraic signs in these equations are deter-
mined by the right-hand rule, Figure 1.30(b, c), with respect to the
(a)
(b)
(c)
FIGURE 1.30 (a) The right-hand rule with respect to the positive directions of
"flux" a.nd round a closed curve encircling it: (b) if the observer is looking in the
arbitrarily chosen positive direction of the flux, then the positive direction round
a curve encircling it is clockwise; (c) if the positive flux is towa.rd the observer,
the positive direction round the curve is counterclock7J.Jise.

52
l/ec/ronulgl1cti( .ficlds
positive direction of the normal to the surface of in tegration and the
positive direction of integration round its edge. (If the handle of a
right-handed corkscrew is turned in the positive direction of integra-
tion, the corkscrew ,viII advance in the direction of the positive nor-
mal.)
Equations (1.74) and (1.75) apply to all media. In media for which
equations (1.71) arc valid, we have:
1. FARADAy-lVIAx\vELL EQUATION
f E, ds = -:t f 1l1I,. dS
(1.76)
and
2. AMPERE-1fAXWELL EQUATION
f H, ds = f (TE,. dS + :t fEE,. dS.
These are the equations for sourcejree regions, which are of primary
concern in this book. 'l'he sources or electric generators will be sur-
rounded by closed surfaces and their effect on the external fields will
be expressed by appropriate boundary conditions.
Equations (1.74) and (1.75) are i}tdepen,dent field equations. \Vc
can derive other equations which 1nust be true if equations (1.74)
and (1.75) are true. The surface of integration may be an almost
closed surface with just a small hole in it as shown in }"'igurc 1.31.
(1.77)
(Jft+ ':i: n ) dS
. .
. ...
. ... ...
. .
. . . -...
-: .:.: }.:(.: : .
. : <: :::}{;.:::i ;.::
'-/ .'. :,.....:;.
. . .. , ". '. '..e ,'3.:., .... -,
."...:....: . .. '. . . .-.:-." .,- -. .
.' :,....: ," -. ...- . iI' ... II  _. :...... ....s......
. ,,"'." .. . e. .". . I'  . ."... \. ,
'" ", ....;..."'. _,I ......,_.: '..I'.(:'.'..$.,. '.'....
'. .; ,'.....I;,.....::'.I.::.' 1t...:::..i.,'.:J.
. -:.. .... .. . . . t . . . ..' " . (, ': . '"
FH;URE 1.31 A pplicatioll of the A nlpere-J'/ ax'well equation to an
alnlost closed surface 7.iJitlz a shril1killl!. /zole.
Let us apply equation (1.75) to such a surface and assume that the
hole shrinks to a point. 1"he line integral must vanish in the limit if

Basic concepts and pquations
53
1/ is finite. l"herefore. the total current leaving any closed surface
.
1S hero
f I n dS + :t f Dn dS O.
(1.78)
Equation (1.78) is not really a consequence of equation (1.75).
It is a consequence of our definition of "displacement current," the
purpose of which was to make the tubes of "total current" closed so
that Ampere's equation (1.55) for steady currents would apply to
time-varying currents. [See equation (1.61) J. l"he definition involves
a hypothesis that displacement currents generate magnetic fields
just as true flow of electric charge does. 'rhis hypothesis has been
con1irmed by experience.
Rearranging the terms in equation (1.78), ,ve have
:t f Dn dS = - f J n dS,
where the right-hand side represents the time rate with which elec-
tric charge is entering the volume enclosed either by way of conductors
or as  streal1l, of charged particles. Hence if ,ve integrate this equation
from an instant when there was no charge in the interior of the
closed surface to the instant when the flow of charge stops, we have
f Dn dS = q,
(1.79)
where q is the charge in the interior. Again, this equation is not so
much a consequence of equation (1.75) as a prerequisite for it.
In the same manner, we obtain from equation (1.74)
 f Bn dS' = 0
at
and since there are no magnetic charges
( 1.80)
f BndS = O.
( 1.81)
There are occasions when it is desirable to postulate magnetic charges.
'fhen equation (1.81) will be similar to equation (1.79).
Equations (1.74) and (1.75) or (1.76) and 1.77) express the
interaction between electric and magnetic fields. l'!zey f orn1, the f oUllda-
tion of electromagnetic field theory and of all its applications. We shall
refer to them frequently throughout this book and they must be
"

54
Ji/l'ctrol1lClj!.llcl ie fields
thoroughly understood. It is strongly recommended that they be
rcmenl bcred in the follo\ving verbal forms.
1. 1\:IPERE-l\1AX\VELL LA \V. The total electric current (the sum
of CO}l'ectioll. conduction. and displacement currents) passing
through a given surface equals the magneto1110tive force (the
li,1te integral of the 'l1tag1zetic intensity) round the edge o.l the
surface.
2. F ARADAy-MAX\VELL LA \\T. The nzagnetic displacenzent current
(the tinze rate of change oflnag1tetic flux) passing through a
given surface equals the negative of the electromotive force or
the "voltage" (the line integral of the electric intensity) round
the edge of the surface.
'There are no free magnetic particles and there can be no magnetic
convection or conduction current. It is possible, however. to express
the condition of magnetized bodies in terms of equivalent magnetic
charge. If this is done, another term appears in the Faraday-l\laxwell
equation similar to that of true electric current in the Ampere-
l\Jax\vell equation. It is convenient sometimes to introduce such a
term deliberately to facilitate mathematical solutions of certain
problems.
Another obvious but importan t observation is that the voltage
round a closed cun'c is the SU1n of the voltages along the arcs into u'hich
E
D
B
FIGURE 1.32 Referril1?, to equatio11s (1.82) and (1.83).
the curve may be subdivided. See Figure 1.32. Then. the Faraday-
l\1avell equation may be staten as follows:

Basic concepts and equations
55
V AB + V BC + V CD + V DE + V EF + VPA
a<f>
at
( 1.82)
where cI> is the total magnetic flux linked with the curve. Similarly,
we can state the Ampere-l\1axwell equation as the sum of magneto-
motive forces round a closed curve
a 'I'
UAB+UBc+UcD+UDB+UEF+UPA=I+ ,
at
(1.83 )
where I is the total true (convection plus conduction) current and
a'li / at the total displacement current linked with the curve.
Two additional tautological statements come from the definition
of the "average" of a given quantity. Thus the average electric
intensity tangential to a given closed curve is
E:n = ; f E, ds,
where 1 is the length of the curve. Similarly, the average intensity of
the componen t normal to a surface is
Er =  f En dS.
Iaxwell's equations (1.76) and (1.77), without the convection
current, may then be written as follows:
arer
lE:n = - liS .
r at'
( aEr )
lln:n = S crEr + E at ·
Here it has been tacitly assumed that J.I., u, and E are constant through-
out the region. Otherwise the expressions on the right-hand side
should be in terms of B:cir, Jr, and Dr.
1.27 Boundary conditions
lVlaxwcll's equations (1.74) and (1.75) are assumed to be general
and applicable to all closed circuits, either small or large, and to all
media, either homogeneous or nonhomogeneous, either isotropic or
nonisotropic. l"'hey have been formulated on the basis of experimental
evidence, but no matter how great the number of experiments or how

56
1le(/r0I11{Jgllclic firlds
varied the experiments, it is impossible to claim that the equations
have been established in the most general form that \ve stated.
However, so far all the conclusions that have been made from these
equations over many years have been confirmed experimentally.
Mathematical restrictions on 11axwell's integral equations are:
The field quantities must be integrable and the time derivatives of the
integrals on the right of equations (1.74) and (1.75) must exist.
In situations which can be realized physically these mathematical
restrictions are not severe. The field quantities may be discontinuous
for instance, without invalidating the equations. The fIeld quan tities
may even be infinite provided their integrals exist.
If the field quantities are continuous and differentiable, the integral
equations can be converted into a set of partial differential equations
This conversion may be accomplished by applying the integral
equations to infinitesimal circuits. When the parameters of a medium,
J.L, (J, and E are continuous, there is no reason why these requirements
should not be satisfied. On the other hand, at an interface between
two media where one or more of these parameters change abruptly,
the field quantities cannot all be continuous. Equations (1.71) show
that if some are continuous, others have to be discontinuous. From
the integral equations we can find those quantities which must be
con tin uous.
(1)
'.: A
(2)
. .
.
.. . .'. 
c
. .
.'
FIGURE 1.33 The interface between two media.
Consider two media (Figure 1.33) and a narrow rectangle straddling
the interface. Let AB = DC = land BC = AD = s. From the
Faraday-Maxwell law \ve have
a
V AB + V BC + V CD + V DA = -ls - Br.
at
As s approaches zero the right-hand side as well as V BC and V DA
approaches zero. 'I'hus V AB  - V CD = V DC. This is true for any 1
no matter how small; but as 1 approaches zero, V AB/l approaches the
tangcntia] component of :£2 and VDc/l the tangential component of

Basic concepts and equations
57
El' Hence, the tangential COlnpOnCllt of electric intensity 'l1lust be C01l-
tinuous at the interface betu'cerl tU10 11ledia
E 2 . tan == EI,tan.
( 1.84)
Similarly, the tangential component of magnetic intensity "lust be
continuous at theillterface between two media
H2.tan = II I . tan .
(1.85 )
Applying equation (1.78) to a thin pillbox [Figure 1.34(a)] with
one broad face in one medium and the other face in the other medium
t Jl.nbl.n
(1)
. . . __.'t t '.):,.,' "';' :"",.:!.fI.l. .::,._.....=-!!  .:.. \.,.t._ :.$"'.'t' , .
.,.... '4,.; ... ""]' _e l > f ,
(2) .' ......  ?".-.,.'. ...' '''.f'.:t''''::'''''''':.. ".... '.'" ., · .
Ii' ..,,.... f ,. .- ,,.. 1-' $
-, @. I ."" '. · '..,' ... .... . . .. . .. -
. .. .. --: '-j. ._., . . . .-..._ . ," . :  ti
, .' · , ] D '.'
.,.,.,..t... +. 2 '.,.
 .' " .; r.... 2 , n , n '.,
. . .
. . , . . ..,
. I'
(a)
..' (b)
.. ,
. .l1li E .
FIGURE 1.34 (a) A thin "pill box" or a "wafer"; (b) a pill box straddling the
interface between two media.
[Figure 1.34(b)], we find that the normal component of the total elec-
tric current density must be continuous at the interface between two 1nedia
a a
J 2 ,ft + - D 2 ,n = ll.n + - Dl,n.
at at
(1.86 )
In thl same manner, we flnd from equation (1.81) that the normal
component of the -magnetic flux density must be continuous at the inter-
face between two media
B2.nor = BI.nor.
(1.87)
Similarly, from equation (1.79) we find that the nornlal COl1zponent
of the disPlacement density is discontinuous across the interface betwee1
two media and that the discontinuity equals the surface density of free
electric charge on the surface
Dl.nor - D 2 ,nor = qs.
( 1.88)
If there is no free surface charge, then the normal component of the
displacement density is continuous across the interface between
two media

58
1lectro1Ilag'!letic fields
Dl.nor = D2.nor.
( 1.89)
1.28 Discontinuities
1"hc vectors E and jj on two sides of an infinite uniform sheet of
charge of density qs per unit area, imbedded in a homogeneous
isotropic medium (Figure 1.35) are equal and oppositely directed.
From equation (108) we obtain their magnitudes
Dl = !qs, El = qs/2eo (1.90)
t Dl=  q..
1
D -D =--q
2 1 2 s
FIGURE 1.35 A unifornlly charged plane sheet.
We have assumed, of course, that there are no other charges present.
Otherwise we have only equation (1.88). _
Equation (1.77) implies that the tangential component of H is
HI" tan
D e .- C
A e  B
H 2 , tan
FIGURE 1.36 A uniform plane current sheet.
discontinuous across an electric current sheet. See Figure 1.36.

Basic concepts and LJl(luations
59
Let C be the current per unit width of the sheet perpendicular to the
flow lines. Then the magnctomotive force round a narrow rectangle
ABCDA is
UA/l + U nc + U eD + U DA := Cl
\vhere 1 is the length of AB, chosen to be intinitesimal \vhile the
length of BC is an infinitesimal of higher order. Thus the mmf U llC
and U DA are infinitesimals of higher order and the equation becomes
IH 2 ,tan - lH 1 . tan = Cl
or
H 2 . tan - HI,tan = C.
(1.91)
If the current sheet is plane, then from symmetry considerations
H 2 ,tan = -H I . tILn = C.
( 1.92)
A practical approximation to an ideal current sheet is current
in a thin conducting sheet. Here equations (1.91) and (1.92) apply to
II tan on the two sides of the sheet, Figure 1.37. In passing through
the sheet of finite thickness [[tan chanwges rapidly but continuously;
hl.
T
H
1, tan


H 2 , tan
FIGURE 1.37 A n infinite, thin conducting Plate carrying unifornl current.
If the thickness of the sheet is h and the conductivity u, the current
C: per uni t wid th is
C = uEh
and apI)roaches zero with h. Let us suppose that the conductivity
increases indefinitely while E and h approach zero in such a way
that the product remains constant. In this \vay, ,ve arrive at a mathe-
matical concept of a perfectly conducting sheet capable of supporting
electric current while tangential electric intensity is zero. It is a very
useful concept.

60
E/cc/r011Ulf!.11ctic fields
1.29 Step-by-step calculation of electromagnetic fields
We have seen that static electric fields are produced in diclectrics by
static distributions of charge and in conductors by steady electric
currents. The lattcr are accompanied by static magnetic fields. Let
us denote this combination of static electric and magnetic fields by
£(0) and fIcO). Assuming that there are no convection currcnts, Max-
well's equations (1.76) and (1.77) yield
f EO) ds = 0,
( 1.93)
f HO) ds = f O'EO) dS.
The electric field in dielectric regions is entirely independent of the
magnetic fIeld. The magnetic field depends only on the electric field
in conducting regions of space.
Let us now assume that these fields begin to vary slowly. We do
not expect that the spatial distribution of these fields will be changed
radically; but we do expect that the terms depending on time deriva-
tives will produce small changes in the spatial distribution, depending
on the time derivatives of the original static fields. Let us denote
these" corrcction fields" by E(1) and ii(l). Then from the same equa-
tions we have
f EO) ds = - f Jl.H(O) dS
I at n ,
( 1.94)
f 11(1) ds = f O'E(1) dS +  f EE(O) dS.
I n at n
Since these correction fields may in gencral also vary with time,
we will calcula te the second-order correction fields, E(2) and 1/(2), from
f E(2) ds = -  f JJ,H(l) dS
I at n ,
( 1.95)
f 11(2) ds = f 0'£(2) dS + , f EE(l) dS.
I n at n

Basic concepts and equations
61
In principle this step-by-step calculation of successive correction fields
may be continued indefinitely. l"'hus we express the solutions of
l\Iaxwell's equations as follows:
E = E(O) + E(l) + E(2) + ... + E(m) + ...,
(1.96)
f1 = jj(O) + H(l) + /1(2) + ... + Zl(m) +
. . .
where
f E(m+l) ds = - a J p.H(m) dS
, at n ,
(1.97)
.. f H(m+l) ds = J uE(m+1) dS +  J EE(m) dS.
· n at n
Adding term by term the infinite sequence of equations (1.93),
(1.94), (1.95), etc., we have
f [EO) + EP) + EF) + ... ] ds = - :t J Jl[HO) + HJll
+ If 2) + ... ] ds,
that is,
f E.ds = - :t J j.lHndS.
Similarly, we find that series (1.96) formally satisfy the Ampere-
Maxwell equation.
In the next few chapters we shall apply this step-by-step method
to specific problems. We shall find that although in certain situations
we can evaluate the successive terms indefinitely, under most con-
ditions we can evaluate only the first few terms-sometimes exactly
and sometimes only approximately. The method has serious limita-
tions which will be pointed out when a suitable occasion arises; but it
furnishes insight into the behavior of electromagnetic fields as they
start varying faster and faster. When usedjudiciously, the step-by-
step method yields good approximations when exact solutions are
hard, or impossible, to obtain.

2
Static and Almost Static Fields
2.0 Introduction
In this chapter we analyze the fields of several basic sources which
are shown in Figure 2.18 in connection ,,,ith a summary of the results
in Section 2.11. A glance at this figure, before studying the details,
,viII be helpful. Then we consider the properties of fields in the large
and introduce quantities which later ,viII be identified as "circuit
parameters," "lumped," and "distributed." Finally, we develop
approximate techniques for handling" ahnost static fields." Such
fields include the effects of the first-time derivatives of E and 8.
A field can actually be varying very fast and still be almost static if
it is confined to a "sufficiently small" region of space. The precise
meaning of "ahnost static" and "sufficiently small" will be con-
sidered in Section 3.13.
2.1 Potential
For static electric and magnetic fields Maxwell's equations (1.76)
and (1.77) become
f E, ds = 0,
(2.1)
f H, ds = f J n dS = I,
(2.2)
,vhere j is the conduction current density, J n its normal component,
and I is the total conduction curren t linked with the circuit of inte-
gration on the left. Equation (2.1) implies that the electromotive
force from any point A of the field to any other point B [Figure
2.1 (a) ] is independent of the path along which it is taken. Hence, we
can choose a fixed reference point B and define for any other point A
a unique quantity V A equal to the electromotive force from A to the
reference point. This quantity is called the potelltial V A of the field
at point A. The reference point is often chosen at infinity.
62

Static and almost static fields
63
Consider Figure 2.1(b) which shows a third point C. Since V AB =
V AC + V CB or VA = V AC + Vc, we conclude that V AC = VA - V e .
Thus the electromotive force from A to C equals the potential drop
from A to C.
B
B
A
c
(a)
(b)
FIGURE 2.1 A ssisting in the definition of electric potential.
The potential of the field of a point charge with respect to infinity
can be found by integrating equation (1.33) along a radius (since the
emf does not depend on the path of integration)
1 00 q dr q 00
V - ---
r 47rEr 2 47rEr r
q
(2.3)
47rEr'
This is also the potential outside a charged spherical conductor.
The potential inside is constant since in the interior of a conductor
no static field can exist.
The lines and surfaces of equal potential are called equipotential
lines and equipotential surfaces. Together with the lines of force they
form a good pictorial representation of the field, which is similar to
a topqgraphic map on which the contour lines are loci of points of
equal height. Just as in the case of topographic maps, it is customary
to draw equipotential lines corresponding to equal increments (or
decrements) in potential. Thus Figure 2.2 shows that where the
equipotentiallines are close the field is strong and where the lines
are far apart the field is weak. Figure 2.3 represents a map of the
cross section of the field of a charged conducting strip. The equi-
potential lines are confocal ellipses and the lines of force confocal
hyperbolas. The lines of force are seen to be the lines of steepest
descent.
In the case of a static charge distribution, the potential on the
surface of any conductor must be constant because if it were not
constant, there would be a tangential component of E and a flo,v of

64
Electro111agnetic fields
FIGURE 2.2 A cross section of a charged sPhere, equiPotentiallilles
(circles), and lines of force (radii).
charge. In electrostatic fields conducting surfaces are equipotential
surfaces. Any equipotential surface may be replaced by conducting
surface without disturbing the field. For example, a very thin un-
charged conducting elliptic cylinder confocal with the edges of the
conducting strip in Figure 2.3 will not disturb the field.
\ I
\ / /
\ I I
.......
"
..",," '.......
./ "
,/" II I \ \ "'-
/ / \ \
I \
FIGURE 2.3 A cross section of a thin charged conducting strip,
equipotentlalUnes (elliPses), and lines of force (h'yperbolas).
The potential of any given charge distribution is found by adding
the potentials of the individual charged particles,
v = 1:: q,. l
47l"Er n
(2.4)
where r n is the distance from the nth particle to a typical poin t in the
field. If the distribution is continuous, we subdivide it into infinite-
simal volume (or surface" or line) elements of charge and integrate.

Static and almost static fields
65
Thus
_ f p(-zt, v, w) dT
V (x, y, z) - ,
411"E r 12
,vhcre p is the volume density of charge and r12 is the distance between
the element of charge at point (u, v, w) and a point P(x, y, z) of the
field
(2.5)
112 = [(x _u,)2 + (y - V)2 + (z - W)2J1I2.
These results follow immediately from the definition of potential.
Since the electric intensity of a system of charged particles is the
vector sum of the electric intensities of the individual particles
E = £1 + E 2 + E3 +
the potential of the system is
. . .
v = f E · dS = f (E I + E 2 + E3 + ...). dS
AB AB
= f E).lIS + f E 2 .dS + f E3'dS + ...
AB AB AB
= VI + V 2 + V 3 + ...,
the sum of the individual potentials.
Reciprocally, we can obtain the electric intensity from the potential.
Consider a point A with the potential V and a neighboring point P
with the potential V + dV as shown in Figure 2.4. By definition, the
A
FIGURE 2.4 Two infinitely close points A and P, the electr£c in-
tensity E at point A, and its conlponent E. in the direction AP.
emf E, ds from A to P equals the potential drop, -dV, from A to P
E. ds = -dV.
Hence

66
1lectro111agl1('tic fields
dV
--
ds'
that is, the cOlnponent of E in a given direction equals the negative of
the derivative of the potential in that direction.
In particular, the Cartesian components of E are
E,=
(2.6)
aV aV aV
E;r; = - ax ' E = - ay , E;r; = - az . (2.7)
Equation (2.6) implies that E is in the direction of the maximum
derivative of the potential. The maximum derivative of a scalar
function V, taken with its direction, is a vector, called the gradient
of V and is denoted by grad V. Thus we can \vrite
E = -grad V.
(2.8)
It is important to remember that the concept of potential is based
on equation (2.1) and does not apply in all its generality to time-
variable fields. However, if strong magnetic fields are confined to
certain regions, then equation (2.1) is approximately true outside
these regions. Also in certain regions we may have strong time-
variable electric fields and relatively weak magnetic fields. In such
regions equation (2.1) is also approximately true and we can intro-
duce the concept of local potential.
2.2 Calculation of electric fields
If the distribution of electric charge in an infinite dielectric medium
(or the distribution of sources of current in an infinite conducting
medium) is knuwn, the calculation of electric intensity is straight-
forward. In Section 1.11, for instance, we obtained the field of a
double current source by adding the components of the vector current
densities of the individual point sources. Let us solve the same
problem \vith the aid of the potential function. To obtain the po-
tential for a point source we integrate Er given by equation (1.23).
The integration is similar to that in equation (2.3) and we have
I
V = -. (2.9)
41rur
Using the notation of Section 1.11, we write the poten tial for two
point sources, I at point Band -1 at point A (see Figure 2.5) which
are separated by distance l

Static and alnlost static fields
67
v=
I
I l(r2 - rl)
47ru r l
47rUr2
47r Ur lr2
As point P moves farther and farther away from the sources, the
lines AP, OP, and BP become more nearly parallel and
r2 - r --+ !l cos 0,
rl - r--+ -!l cosO,
r2 - rl  l cos 0,
'I r 2  r 2 - 112 cos 2 o.
B
To distant point P
l It
A
FIGURE 2.5 A current element in a conducting medium.
Hence at distances large compared with l, the potential equals ap-
proximately
Il cos 0
V=
47rur 2
(2.10)
'fhe error is of the order of (llr) 2. For an infinitesimall equation
(2.10) is exact at all distaces.
l-'he above method of obtaining equation 2.10 suggests that for
the infinitesimal dipole V = -laVol az, where V o = II47rur.
The clement of length along the radius OP is dS r = dr, and along
the meridian of radius r we have ds e = rdO. Using Equations (2.6)
and (2.10) ,we have
aV Il cos (l
Er=
--=
ar
,
27rur 3

68
1/ectrrnn(JK.l1et ic fields
Ee = -- =
aI' II sin 0
rao
41rur 3
'rhus the calculations are less laborious if ,ve use the poten tial func-
tion.
Potential of a uniformly charged line filament
As our second example we choose a unifonnly charged filament OA
of length I, no longer small. See Figure 2.6. The potential and the
field are independent of the angle tP between half-planes issuing from
z
p
P(p,lf>,z)
l
z
o
FI<,UkE 2.6 lssisting in the calculation of the potential of a unifornl line
(Izare a.A.
the filament. If q is the charge per unit length, the element of charge
at distance 'It from the origin is q du and
J l q du,
V = 0 41re vp2 + (u - Z)2 '

Static and almost static fields
69
ly introducing a nc\v variable of integration v = u - z, \VC have
v
j l-Z q dv
-z 47rEV p 2 + v 2 '
Since
d[v + V p 2 + v 2 ]
v+ V p2 + v 2
dv
V p2 + v 2 '
we obtain
q l - z + rl q (l - z + '1) (z + r)
V=-ln =-In
41rE -z + r 41rE p2
(2.11)
where
, V p2 + Z2, rl = V p2 + (l - Z)2.
If we multiply the numerator and the denominator in the first
expression by (r + z) (rl - l + z) we obtain a third form
q , + z
V = -In .
41rE '1 - l + z
Therefore
l - z + rl
-z + r
r + z
= cxp (41rEV /q) = k,
'1 - l + z
and
l - z + '1 = - kz + kr,
, + z = krl - kl + kz.
By adding and rearranging the terms, we obtain
(k - 1) (r + rl) - (k + l)l.
'rhus
k + 1 27rEV
r + rl = l = l coth
k - 1 q
lIenee the equipotential lines arc ellipses confocal with the ends of
the charged filamen t. equipoten tial surfaces are prolate spheroids,
and electric lines are confocal hyperbolas. Thus the map of the field
in a radial plane resembles the one in Figure 2.3.
Let us no\v examine the potential on a thin cylinder p = a,O < z < l,
around the filament. Not too near the ends \\'C may neglect a 2 in
comparison with Z2 and (l - Z)2. Hence. approximately
q 4z(1 - z)
'-ln .
47rE a 2
(2.12)

70
Electronzagnetic fields
At the ends z = 0, l we have
q 2l
V-ln-.
41rf a
Half way between the ends
q l
Y = -In-.
21rE a
Thus the potential at the ends is about half the potential in the
middle and approaches one half as l increases or a decreases since
In(l/a) becomes much greater than ln2. If ZI is the distance from the
middle, equation (2.12) becomes
q l q ( 4zi)
V  - In - + - In 1 - - ,
21rf a 41rf l2
(2.13)
,vhich shows that as l/a increases, the potential will be substantially
constant over an increasing central portion of the cylinder and then
drop to half of its value more rapidly near the ends.
Potential of t"'O equally and oppositely charged filaments
The potential of a pair of equal and oppositely charged filaments
[Figure 2.7(a)] may be obtained from equation (2.11). The lower
and upper limits in the integral for the potential Y- of the lower
filamen tare -l and zero to begin with and -l - z and - Z after the
change in the variable of integration. Hence
q -z + r
Y- = --In
41rE -l - z + r2
q l + z + '2
- --In
,
41rf z + ,.
where
T2 = V p2 + (l + Z)2.
Adding this to V in equation (2.11), we find the potential of both
filaments
q (l - z + 'I) (z + r)2
V = -In .
41rf p2(l + z + r2)
(2.14)
Axial cross sections of two equipotential surfaces are shown in
Figure 2.7 (b). Ncar point 0 the equipotential surfaces are conical.
Assuming that p and z are small compared with l, the above equation
becomes

Static and almost static fields
71
q z + " V P 2 + Z2
V=-!n .
27rE P
If z = kp, V is constant; but z kp is the equation of a cone.
z
l
z
l
p (p,cp, z)
(a)
(b)
FIGURE 2.7 (a) T'lVO uniformly but oppositely charged filaments; (b) an axial
cross section of t'lOO equiPotential surfaces.
T,vo perfectly conducting surfaces, if coinciding with the equi-
potential surfaces whose cross sections are shown in Figure 2.7 (b),
will not disturb the field. If they arc maintained at equal and opposite
potentials, the potential outside is given by equation (2.14).
Qn the surface of a thin cylinder p = a, -l < z < l, about the
charged filamen ts, we can neglect p2 in the expressions for " 1'1, and
'2, provided we are not near the ends or the middle of the cylinder.
Hence for z > 0, ,ve have
q 4Z2 (I - z) q 2z q l - z
V = - In = - in - + - In .
47rf a 2 (l +z) 27rE a 47rE l + z

72
J/C(tronla?l1ctic firld s
From synlmetry considerations
V(-z) = -1 1 (z).
l'hus the potential drop bct\vcen poin ts equidistant from the middle is
q 2z q l - z
V(z) - V( -z) = -In - + - In
7rf. a 27rf. I + z
q 21 q z q I - z
= - In - + In + - in .
7rf. a 7rE l 27rE l + z
As 2l/a increases, the potential differences becomes substantially
constan t over increasing portions of the cylinders, excluding the
decreasing central and end sections:
(2.15 )
2.3 Calculation of charge distributions
Frequently there are problems in which something is kno\vn about
fields and one has to find charge distributions. For instance, two
wires (an antenna) may be connected to a generator. See Figure
2.8(a). 1-'he impressed electromotive force Vi in the generator drives
z
z
+
+
Eo
z=Q
l
I
2 V o
M Vit +
0
.....,J

M
Q)
s::
Q)
t:)
- ! V o
2
2l t
(a)
(b)
FIGURE 2.8 (a) Tu'o conducting wires connected to a generator; (b) a conducting
u'ire in a Un'lfOr111 electric field.
electric charge from one wire to the other. In the static case, \ve kno\v
only that the potentials of the wires are constant and the potential
difference V o is equal and opposite to Vi. In a symmetric arrange-

Static and altllost static fields
73
men t the potcn tials of the \vires \vill be equal and opposite. In the
tin1c-variablc case. \VC kno\\O the electromotive force between the
input tcrn1inals of the antenna. l'UA = F'. and the component of
electric intensity tangential to the \vires is very small. From this
inforn1ation \VC have to find the charge (and current) distribution in
the \vires. Once we find it we can calculate the field.
Another example is shown in Figure 2.8(b). A wire is in a static
or time-variable field. We want to know what happens in the wire
and its effect on the field. The wire may be broken in the middle and
connected to a device absorbing energy (a receiving antenna). Exact
solution of such problems is possible only in a few special cases and
even then only \vith the aid of special mathematical methods. Usually
we have to be satisfied with approximate solutions.
\Vhile the calculation of fields from given charge distributions is
straightforward. the solution of inverse problems which we are con-
sidering in this and the following sections requires varying degrees of
imagination. l"here is no single method applicable to all problems.
One has to take advantage of special conditions. He who solved a
given problem or a class of problems for the first time had to be
inventive. Consider, for instance, the calculation of charge q(z) per
unit length on the wires diverging from A, B in opposite directions,
as sho\vn in Figure 2.8(a), from the given potentials. By swinging
the wires about A and B, we can make them parallel. Reciprocally,
parallel wires can be swung apart. Thus \ve have related problems
and the solution of one might help us solve the other. Let us see to
what extent it does.
If the length of parallel wires is considerably greater than the
distance s bet\veen their axes, we expect substantially uniform charge
distributions except near the ends. The field of each wire by itself is
radial. If q is the charge per unit length. the radial displacement
density is q/27rPI, where PI is the distance from the axis. Hence, the
radial electric intensityis q/27rEPl..Similarly, the radial electric intensity
of the field produced by the other charged wire is -Q/27rEP2 where P2
is the distance from its axis. Of course. the charge on one wire attracts
the opposite charge on the other \vire. and there is some nonuni-
formity in charge distribution round the axis of each wire. But if s
is fairly large in comparison \vith the diameters of the wires, the
nonuniformity is small. Hence the mean voltage bet\veen the \vires is
V o = r (q/27rEPddpl + fa (-Q/hEP2)dp2
a ,
= ( q / 7r f. ) In ( s I a) .

74
llectromagl1e/ic fields
From this equation we find the charge per unit length
q = eVa,
c = 1rE/ln (sla).
In the theory of "circuits \vith distributed parameters" the quantity
C is called the" capacitance per unit length."
The equation indicates that the capacitance between the opposite
elements of two wires depends only on the ratio, sla, of the mean
distance bctween the elements to the radius. If we use the same
formula for the capacitance between the corresponding oppositely
charged clemen ts, as we swing the wires apart to obtain the configura-
tion shown in Figure 2.8(a), \ve obtain
q(z) = C (z) V o ,
C(z) = 1rE/ln (2zla)
since 2z is the distance bctween the elements. Let us look at the
situation from another angle. 1'he potential of a charged particle
decreases as the distance from the particle increases. Hence, the
potential at any point on a thin wire will be determined primarily
by the charge density (per unit length) at that point. Since the
potential is constant along each wire, the charge density q(z), where
z is the distance from A, B, must be approximately constant. We can
use, therefore, prior knowledge of equation (2.15) for the potential
of two uniformly charged filamcnts and adjust it to makc V(z)
V( -z) = V o by assuming
q(z) -
V o
,
.11 (z)
(2.16)
where
1 2l 1 z 1 l - z
A (z) = - In - + -In - + - In
1rE a 1rE l 21rE l + z
q 2z 1 l - z
= -In - + -In .
7l"E a 27l"E l + z
Since A (z) is thc reciprocal of C (z), we observe that this res ul t
agrees with the previous one except near the ends of the wires.
1\S already noted, A (z) becomcs more nearly independent of z as
all decreases and may be further approximated by its average value,
1 jl 1 (l )
Ao = - A (z) dz = - In - - 1 .
l 0 7rE a
(2.17 )

Static and alOlost static fields
75
Let us try a third approach. Suppose that q(u) is the unknown
charge density per unit length on the upper wire, at distance u from
the midpoint of AB. The potential on an,d in the wire is V o /2. We
express therefore the potential of the surface charge, on the axis of
the \vire at rlistance z, and equate it to V o /2. 1"hus
vo = {+a 41rEVZ2( : z)2 + {. 47rEV;( : - z)2 '
where 2s is the distance i"lB. Here we take advantage of the fact
that the potential is constant throughout a conducting body to
avoid a double integral which would arise from surface integration.
The only approximation in this integral equation for the unknown
function is that we did not include the potentials of. the charges on
the flat ends of the wires. No method is known for solving this par-
ticular type of integral equation exactly. 1"0 solve it approximately,
we note that the integrand is large in the vicinity of u = z, where
q(u) = q(z). We replace, therefore, q(u) by q(z) and take the
latter outside the integral signs. rrhe integrals can then be evaluated
as in the preceding section. 'fhe effect of s is negligible and we obtain
equation 2.16). 'fhe charge density, q(z), thus found, varies slowly
\vith z. 1"his is another reason why our approximation of q(u) by
q(z) is justified.
It is possible to formulate an iterative procedure for obtaining a
series solution of the above integral equation; but this is beyond
the scope of this text and quite unnecessary for practical purposes.
To solve the problem shown in Figure 2.8(b), '\'C observe that if
the impressed field is uniform, its potential with reference to the
central plane is - EoZ. Since the total potential of the wire must be
constan t the paten tial of the field due to the charge displaced by
Eo must be
V(z) = Eoz.
'The integral equation for the unknown charge density becomes
f l q( u) du
V(z) = EoZ = .
-l 47rE v/ a 2 + (u - Z)2
As in the preceding problem \ve replace q(u) by q(z) and obtain
EoZ
A1(z) ,
q(z) -
(2.18)

76
llcd rOI1U/ }!.lIl'IIC .Ii c!tls
\vhcrc
A1(z)
1 l-z+va 2 +(I-z)2
-In
41rE -l - z + Va 2 + (I + Z)2
1 [/- z + va/!. + (I =Z-f2J[l + z + va 2 + (l + Z)2]
== -In
41rE a 2
1 4(l2 - Z2)
 -In
41rE a 2
1 2l 1 ( Z2)
 - In + - In 1 - - .
21rE a 41rE l'2
\Vhen II a is large, A 1 (z) is approximately equal to A (z). In fact, if
in the foregoing integral we had replaced q(u) by -q(z) in the
interval (-l. 0). where we know that the charge is negative, we
would have obtained .l1 (z) in the denominator of equation (2.18).
This would be a better approximation (based. of course, on a better
understanding of the physical situation).
\\' c see that under the influence of the impressed field the wire
has become similar to a dipole or has become" polarized."
2.4 Metal sphere in uniform electric field
If a neutral metal sphere is placed in a unifornl electric field of
intensity Eo [Figure 2.9(a)]. the charge on it \\ill be displaced and
the field affected correspondingly. 'rhc tangential component of the
total E must vanish on the surface of the sphere. For the impressed
field \ve have
E; = Eo cos 0,
E = -Eo sin O.
(2.19)
'rhc 0 component of the reflected field due to the displaced charge
must vary as sin 0 or else the total E fJ will not vanish for all values
of O. Referring to equation (1.36), we find that the lield of an electric
dipole has the proper dependence on () and ,vc assume the reflected
field to be of that form.
,
27rEor 3
E r -
..JfJ -
A sin ()
41rE o r3
(2.20)
A cos ()
E T
't -
T -
Hence at r = a
A sin 0
EJ + Es = - Eo sin 0 + = o.
41rEoa3

Static and alnlost static fit.lds
z
+ + Er
Eo
- -
(a)
77
(b)
FU;URE 2.9 (a) A tlle/al sphere introduced into an originally uniform electric
field; (b) the lines of force in the presence of the sphere.
Thus \VC tind the unkno\vn constant
A = 41rEf)(z,3 Eo.
(2.21 )
'rhe sphere in a uniform electric field acts as a dipole of moment A
for r > a.
For r < a the total field vanishes. l'herc the displaced charge
produces a field equal and opposite to the impressed fIeld. 1"'he
quan tity A / Eo is called the polarizability of the sphere.
()ncc .11 has been determined. \ve can obtain other characteristics
of the polarized sphere. 'fhe poten tial of an electric curren t dipole
is giycn by equation (2.10). By analogy, we obtain the potential of
an electrostatic dipolc,
v=
A cos () a 3 Eo cos ()
(2.22)
41rEor2 r 2
'J'hc surface density qs of the displaced charge equals the discon-
tinuity in the radial displacement density [see equation (1.88)J at
r = (1,. ']"hus

78
Jleclronlagllelic fields
qs = IJ r = Eo(E; + E) 3f.oEo cos {J.
The total displaced charge on the upper hemisphere is
2r r /2
q = f f qsa 2 sin (J d(J dcp = 311"EOa 2 Eo.
o 0
Figure 2.9 (b) illustrates the lines of force in the totallicld.
'rhere is another reason for assuming that the field of displaced
charge might be of the dipole type, equations 2.20. On account of
symmetry conditions, to every clement of displaced charge at (a, 0)
there corresponds an clement of opposite charge at (a, 1r - fJ).
'rhe two clements form a dipole of finite separation 2a cos 0. At a
large distance r from the cen tcr of the sphere, the field of such a
dipole becomes indistinguishable from the field of an infinitesimal
dipole of the same moment situated at the center of the sphere.
Hence, for r » a, the field of an en tire displaced charge is of dipole
type and, pending subsequent verification (or rejection), we could
assume tentatively that the field of the displaced charge is given by
equations 2.20 evcr)'\vhere outside the conducting sphere. The
foregoing calculations show that this assumption is indeed con-
sistent \\Tith the conditions which must prevail inside and on the
surface of the conducting sphere.
A question might be raised whether there exists a different charge
distribution or a differen t field of the same charge distribution
which also satisfies the requircd conditions in the presence of the
same impressed field Eo. Let this different reflected field be £'1
while the one we found is Er. The total fields in the two cases are
Eo + Er and Eo + El. The difference between them is Er - £1.
If this difference did not vanish, we would have to conclude either
that electric charges in a conductor could separate spontaneously
without any impressed forces or that a static electric field can exist
\vithout any charge. Furthermore, this would mean that if we were
to create a field in a certain region in a laboratory and then in tro-
duce a metal sphere. a charged particle at a given point would be
acted upon by two different forces or by one force on l\Iondays,
\\Tednesdays, and l.ridays and by another on 'Tuesdays and 'fhurs-
clays. l'his is absurd and we conclude that there is a unique re-
sponse of a metal sphere to an impressed field. Once \\e have de-
termined a solution which satisfies all the physical conditions of a
given situation. the solution must represent the reality.
If the impressed electric field is varying slovTly at the rate Eo. then
in the first approximation we may assume that it i still distributed
uniformly in space and that the reflected field is still given by equa-

Static and ahnost static fields
79
tions (2.20) ,vith the moment A varying at the tinle rate A. "rhe
total time-varying electric field generates a magnetic field. 'rhat part
of the latter which is produced by reflected field can be calculated
very easily on account of its symmetry about the z axis passing
through the center of the sphere. In this field, magnetic lines are
circles of radius p r sin (). "I'hc magnctomotive force round a typical
circle is 27rpll" = 27rr sin Olllf'. By the i\mpcre-l\IaxweIllaw, this mmf
should equal the total electric current linked ,vith the circle. "fhis
current can be obtained by integrating the radial displacement
current density over a portion of the spherical surface uf radius r,
concentric with the origin, which is enclosed by the circle. The inte-
gration is exactly the same ,vhich led to equations (1.57) and (1.64)
for magnetic intensities of fields produced by electric current cle-
ments in conducting and in dielectric media. Thus ,vc have
Ii sin {J
EOa 3 Eo sin 0
11" =
47rr 2
r 2
r > a.
l'hc intensity of the magnetic field produced by the time-varying
i1npresscd field can also be obtained if ,ve know that it is symmetric
and if we know its axis of symmetry. Uniform static and slowly
varying fields do not have to be symm.etric. l;or instance, consider
two parallel metal plates of arbitrary shape, large in comparison with
the distance bet\veen them and connected to an electric generator.
Between the plates. not too close to their edges, the electric field
is substantially uniform. But there is no axis of symmetry and the
preceding method of calculating Inagnetic intensity is inapplicable.
We can, of course, calculate the mmf round any circle; but we can
not assume that the magnetic intensity is tangential to the circle
and that its magnitude is the same at various points.
If, however. the plates are circular and have a common axis, then
the magnetic lines are circles coaxial \vith this axis. 'rhe mmf round
a typical magnetic line uf radius PI, is 27rP 1 lllf'p \vherc CPt is the angle
round the axis. "rhc electric displacemen t curren t linked ,vi th this
line is 7rpiEI,E o . "fhcrefore
Illf'l = EoPlEo.
Note that ,vhile the electric field is uniform. the magnetic field is not.
Of course, if Eo is not constan t, the nlagnctic field will vary \vith
time and ,vi]] generate an electric field El \vhich ,vill be superimposed
on Eo. 'rhis additional field ,vill be proportional to Eo and nonuniform.

80
}f;lectronlClglletic fields
'fhc metal sphere \vhich \ve have been discussing in this section
could be centered either on the axis of the plates or off the axis.
In the former case PI = P = r sin (J and the lines of the total magnetic
field are circles. In the latter case, the total magnetic fIeld is the sum
of two circularly symmetric fields with different axes of symmetry.
The lines of the total field are no longer circles.
2.5 Dielectric sphere in uniform electric field
In the case of a dielectric sphere in a uniform field, Figure 2.10,
boundary conditions (see Section 1.27) require that the tangential
component of E and the normal component of jj be continuous across
the surface of the sphere. As noted in the preceding problem, these
conditions can be satisfied only if the corresponding quantities vary
with () in conformity to the impressed field, equation (2.19). There
FIGURE 2.10 A dt"electric sPhere in an originally uniform electric field and a
sketch. of Hues of electric disPlacenlent.
are two types of fields which have the required dependence on (J: (1)

Static and alnl0st static fields
81
a uniform licld \;ith lines parallel to the impressed field, and (2)
a dipolc field, equations (2.20). 'fhe dipolc fIeld bCC()nlCS infinite
\vhcn r = 0 and therefore cannot exist in the in tcrior of the sphere
since \ve have no point charges at the center to cause such behavior.
The effect of the dielectric sphere on the external impressed field
must decrease with increasing distance from the sphere and, hence,
must be represcnted by a dipole t)'e of field. Thus, outside the sphere,
\ve add to the impressed field a reflectcd field given by equations
(2.20). Inside the sphere we assume a trans1niUed field
E t = B cos 8
r ,
E; == -B sin 8.
(2.23)
1"he boundary conditions are
E; (a, 0) + E; (a, 8) = E; (a, 8),
D;(a,O) + I)(a, 0) = D:(a, 0).
Substituting from equations (2.19), (2.20), and (2.23) and cancelling
sin 0 and cos 0, \""Ie have
A
-Eo +
41rEoa 3
- -B
,
A
EoEo + - =. fB.
21ra,3
Solving
A -
41ra3EO(E - fO) Eo
2fo + E
3EoEo
B -
- .
2EO + E
(2.24)
1'h us the larger the ratio (E/ Eo) is, the smaller the electric in tensi ty
B inside the sphere is. Extcrnally the dielectric sphere acts as a dipole
of momen t A. Note that the ratio of this momen t to the dipole
momen t of a conducting sphcre is (E - Eo) (E + 2Ea). 'fhe dipole
monlcnt l)er unit volume of the sphere is
I 3 EO (E - EO) Eo
]> = A (41ra 3 /3) = (2.25)
2Eu + f
l"he fact that the dielectric sphere acts as a dipole is not surprising.
...ll nlaterial media contain protons and electrons. In conductors there
are numerous free electrons and the impressed electrostatic field
displaces them to the surface. In dielectrics the electrons are bound
for the 'most part (all of them are bound in perfect dielectrics). l'he
impressed ficld, ho\vcvcr, displaces the bound electrons \vith reference

82
1lc(lr0111ClJ!.llel ic jieltis
to the protons. thus fornling tiny dipoles throughout the dielectric.
"Ve expect also a surface layer of displaced bound electrons on the
bottonl hemisphere and a layer of positive charge on the top. similar
to free surface charges on the conducting sphere. In both cases the
tic Id of the displaced charge opposes the impressed field in the in terior
of thc sphere. In the mctal sphere the opposing field is exactly equal
to the impressed field and the total in ternal field vanishes. 'Ihis is. in
fact. the condition from which we tind the amount of displaced charge.
In the dielectric sphere the opposing field weakens the total field
bu t does not destroy it altogether.
Displaccmen t of bound electrons in dielectrics or the polarization
of dielectrics by the impressed field is responsible for the values of
dielectric constants higher than that of vacuum. Thus inside the
sphere
E; = B -
3EO
Eo.
2EO + E
t 3foE
Dz = Eo.
2Eo + E
The difference
3 E o(E - fO)
D t Et - E'
z - Eo 1 Z - 0
2fo + E
equals the dipole moment per unit volume. called the polarization P,
equation (2.25). Thus in the interior of the sphere
fJ = EoE + P. (2.26)
It is possible to prove that this equation is general. Alternatively this
equation can be used to define P. Subsequently it can be sho\vn that
P is the dipole moment per unit volume.
2.6 Proximity effect
If charged conducting bodies are very far apart. the potential and the
ficld at any point is essentially the sum of the potentials and the
fields of the individual bodies. calculated on the assumption that the
other bodies are not prescnt. 'rhus the potential of two equally and
oppositely charged spheres, A and B, shown in Figure 2.11 is
V eo ) - veo) + V(O) - q
- 1 2-
41rEorl 41rfor2.
l'he superscript "zero" serves to remind us that \\'C have a "zero-
order" approximation, good only if the distance l between the ccn ters
of the spheres is large in comparison with the diameter 2a of each
sphere.
q
(2.27)

Static and ahnost static fields
83
Iecausc of the attraction between opposite charges, the charge
distributions \vill not be uniform. l-'he nonuniformity increases as
1/2a decreases. l"his proxi1nity effect can be calculated approximately
from the result obtained in Section 2.4 when 1/2a is still so large
that the field of one sphere is nearly uniform in the region occupied
by the other. The electric intensity Eo produced by the charge on
1
FHiURE 2.11 Illustrating the calculation of the proximity effect of two equally
but oppositel)' clzar£ed spheres.
sphere A at the center of sphere B is
q
Eo = .
41r E ol2
(2.28)
l"his is also the electric intensity produced by the charge on B at the
cen ter of A. This is a mean value of the electric in tensi ty impressed
by one charged sphere on the other. Under its influence, some positive
charge on each sphere is shifted to the right and equal negative
charge is shifted to the left. This" dipole type" displacement of
charge is superimposed on the original uniform distribution of charge,
thus producing greater charge densities on the sides of A and B
which face each other. The moment A of each dipole is obtained
from equations (2.21) and (2.28)
A = q(a 3 /1 2 ).
The paten tial of each dipole is
A cas ()
V d =
41rEOr2
qa 3 cos 8
-
41rEol2r 2

84
/f;/cc/ronUI1!.Jle/;c fir/cIs
\\hcrc r is the distance from its cen ter and () is the angle bet\vcen a
t)l>ical direction and the dipolc axis from the negative to the positive
charge. l'or the dipole superimposed on B this angle equals 0'1.; for the
dipole supcrimposed on A the angle equals 7r - o. Adding the dipole
potentials to the potential given by equation (2.27), we have the
next approximation to the potential produced by two chargcd sphcres
qa 3 cos (h qa 3 cos (}2
+ .
41rEol'l.ri 41rEol2ri
(2.29)
v o > =
q
q
41r E o'l
4 1rE or2
Let us now calculate the potential at a point Q on sphere B. l"here
'2 = a and
'11 = (l2 + 2 al cas O 2 + a 2 ) -1/2
= l-l[l + (2a,/l) cas O 2 + (a,ll) 2J-l/2
 l-1 - (all 2 ) cos O 2 .
Hence, the first and the fourth terms in equation (2.29) add to
q/41rEol. 1"hereforc on B
V o> -  - q
B -
47rEol 41rEoa
qa 3 cos 0 1
4 1') 2 .
1rE(}lIrl
(2.30)
In absolute value the second term is the largest. ]{clatively to it the
first is of the ordcr of ail; and the third of the order of (a 4 /l 2 ri). Evcn
when the distance between the centers of the spheres equals only two
diameters, the largest value of the last term is less than one per cent
of the principal term.
'fhe exact potential of B must be constant, of coursc. \Ve can take
only the first two terms in equation (2.30) and consider the third as
indicative of the magnitude of error we make. Rut cos ()l docs not
differ much from -1 and rl is comparable to t. So \ve shall have a
better approximation if \ve let cos (h = -1, r'}. = l instead of dropping
the term altogether. 1"hus the potential of n is approximately
q ( a. a 4 )
VB = - 411"Eoa 1 - I - "j; .
(2.31)
l"he poten tial of A is
V A = - VB = q (1 _ a. _ a l4 )
41rEoa 1 l.J.

Static and almost tatic fields
85
lhus the capacitance bct\vecn the spheres is
( (l, tlr4)-1
C = q/(V A - V n) = hEoa 1 - I - Z; ·
(2.32)
2.7 Magnetic scalar potential
The existence of electric potential is a direct consequence of equation
(2.1) \vhich states that the electromotive force round every closed
curve in the static field vanishes. 1-'he magnetomotive force, on the
other hand, vanishes only when the curve is not linked with electric
current. This means that magnetic potential can be defined only for
regions free from electric curren t. This restricts the usefulness of the
concept in the case of magnetic fields.
FI<iURE 2.12 ./1 closely fWOlOld solel1oid al1d nlanetic lil1es of force.
'fhc magnetic analog of an electric dipole is a short and very thin
closely \vound solenoid. See Figure 2.12. Inside the solenoid magnetic
lines are substantially straight except in the vicinity of the ends.
By analogy with the electric dipole we obtain [see equations (1.36) ]

86
Electr01nagnetic fields
the follo¥ling eq ua tions for the magnetic field outside the solenoid
4>l cos 8 4>1 sin ()
B,.= 27rr 3 , Be= 47rr3
(2.33)
4>1 cos () 4>1 sin ()
H,= , He= ,
27r1J.r3 47rr3
where cp is the magnetic flux emerging from one end of the solenoid
and converging to the other. 'The quantity cp1 is the monzent of the
solenoid or of the 1nagnetic dipole. Similarly, by analogy with equa-
tion (2.10) for a double current source (or a similar equation for an
electrostatic dipole) we have the potential of the magnetic dipole,
4>1 cos ()
u=
(2.34)
47r,ur 2
The magnetic dipole resembles more closely an electric current
element shown in Figure 1.9 than an electrostatic dipole shown in
Figure 1.16. l\Iagnetic lines and lines of current flow arc closed.
Inside the solenoid magnetic flux is from the south-seeking end to the
north-seeking. Equation (2.34) is valid only outside the solenoid.
If we imagine a semi-infinite, thin solenoid, we shall have a magnetic
point source with a radial magnetic field and a potential similar to
the electric potential of a point charge, equation (2.3), or the po-
tential of a point current source, equation (2.9) . But this potential
is of no help in the calculation of magnetic fields generated by given
curren t distributions. In such a calculation the curren t distributions
are subdivided into current elements of moment It = J dT (see
Section 1.23). l\Iagnetic lines of the fIeld of an element are circles
coaxial with the element and the magnetic intensity is given by
equation (1.64). '[he individual element is surrounded by displace-
ment current. In general the magnetomotive force around closed
curves does not vanish and the potential in the above sense does not
exist for a single clement even though it may exist for the entire
current distribution. at least in current-free regions (see Section 2.10).
2.8 Magnetic vector potential*
In vector analysis it is shown that any vector \vhose flux through
any closed surface vanishes every\vhere is the" curl" of another vector.
* This section is optional and may be omitted.

Static and almost static fields
87
Magnetic flux density i3 is such a vector (Section 1.26), and we have
- -
B = curl A.
(2.35)
The vector A is called the 1nagnetic vector potential. \Vhen so intro-
duced, this vector has no physical significance. Actually no matter
how \ve introduce it, it remains an auxiliary mathematical function,
a convenien t computational "gimmick." The follo\ving derivation
may make the concept seem less abstract.
1'he magnetic intensity of a current element in an infinite non-
conducting medium as shown in Figure 2.13(a) is [equation (1.64)J
It sin () lip
Ilq, = - -
47rr 2 47rr 3 '
where It is the moment of the element. The magnetic flux through the
z
from 00

r-------
I
B
z
B D
r--------,
L- ----jc
A
+q
I
-q
-----

to 00
q
I
-q
(a)
(b)
FIGURE 2.13 A ssisting the analysis of the field produced by an electric current
elenlellt.
rectangle A 00 BA shown in the figure where AB = Liz, is
jJ.llLiz foo
Li cJ> = P (p2 + Z2) -3/2 dp
47r p
jJ.llLiz jJ.I [Liz
= (p2 + Z2)-1/2 = .
47r 47rr

88
llectr(Jnl(lJ!.lIc/ic fields
Suppose that I starts varying \vith time. According to the Faraday-
l\tIax\veIlla\v, the time derivative of <I> will create an electric t1eld
\vhich \vill be superimposed on the field of the end charges of the
current element. The total counterclockwise emf round the circuit
A 00 BA equals the time derivative of .1<1>. The question is: How is
it distributed round the circuit? l"'here is no S)'11ll11ctry or anything
else to guide us. The student \vill now understand the statement made
in Section 1.29 that the step-by-step method of calculating time-
variable fields cannot, as a rule, be carried on indefinitely if ,ve use
1\11 ax\vell's laws in integral form.
Let us assume tentatively that the entire emf is along AB. In
other \vords we assume that the electric field created by magnetic
current is given by a vector P parallel to the current clement. Then
the clockwise emf is Fzz and from. the above result
Jljl
F =--
z 4'
'Trr
F p = F", = O.
(2.36)
To be on the safe side we denote our" error" by G so that the true
electric field is
- - -
E = F + G,
(2.37)
where G may have p and z components. From considerations of
symmetry all 4> components in our case should equal zero. The clock-
wise emf round AB 00 ,;1 is
f BodS = f FodS + f GodS = - :t (l1<1».
Vector F ,vas defined so that its line integral equals the right-hand
side term. 1"'herefore
f GodS = o.
(2.38)
In certain calculations of the work done in establishing magnetic
fields one has to in tegrate E.l round a closed circuit when I is the
same in all parts of the circuit. In such cases we can replace E. by F,
since the integral of G. cancels out. Note also that equation (2.38)
implies that G is the gradient of some scalar potential function.
'fhe integral of F round an infinitesimal closed circuit such as
ACDBA in "'igure 2.13(b) divided by the area enclosed, is the
component of curl ft in the direction perpendicular to the area. (see
Appendix I). In our example this is also the componen t of 13 in

Static and almost static fields
89
that direction. Thus
...: - JJ.jl
B = curl F = curl-.
47rr
Taking the time integral, we obtain
p.Il
B = curl-
47rT
for the current element of vector moment II. The magnetic vector
potential
A --
p.Il
(2.39)
47r"
of the current element has thus been obtained. In calculating mag-
netic fi.elds, it is easiest to add the vectors parallel to current elements
and then obtain B by differentiation. Nevertheless the calculations
are lengthy and one should take advantage of possible simplifications
in each specific case, as we shall in the next two sections.
In a homogeneous isotropic medium H is also the curl of a vector
A/,uo
z
p
p
A
11
o
FH,URE 2.14 Illlls/ratill? the calculation of the l1zagllelic field pro-
duced by a straight, 1tllifornl, electric cllrrelltfila1nellt.
2.9 Straight uniform current filaments
The field of a straight uniform current filament shown in Figure
2.14 can be calculated from the 1ield, equations (1.62), of a semi-
infmite current filamen t. 1"'his field is the sum of the field of current I '

90
1/f(tron1l1 J!.nrt ie fields
from point 0 to infinity and the field of current -I from A to in-
finity. 'rhus
I (1 + cos 8) 1(1 + cos 8 1 )
IIfP= - .
47rr sin () 47rrl sin 8 1
Since r sin () = 11 sin 8 1 = p, we have
I (cos 8 - cos 8 1 )
HfP= ·
47rp
In the vicinity of the fIlament not too near the ends, 0  0 and
8 1  7r; thus
(2.40)
I
lIfP -.
27rp
l'he longer the filamen t is, the greater is the range of p in which the
approximation is valid. The equation is exact for all values of p
if the filamen t is infinite in length.
2a
(a)
(2.41)
z
y
x
(b)
FIGURE 2.15 (a).1 circular turn of 'lL'1're carrying uniform current I. (b)
IllllstraHl1J!. 1110f!.l1et;c lines of jorre. lil1ked 'U'ilh the 'wire llnd lite (a/culaliol1 of
the 11uIJ!.lleti( iuteusit)' ou the axis of the rillJ!..

Static and alnlost static tlclds
91
2.10 Circulating current
'T'hc exact field of a circular turn of ,virc [see 14'igurc 2.15 (a) ] carrying
current J can be expressed as an elliptic integral. 1"'he solution is
much simpler for distances large compared with the diameter of the
loop.14"'irst we obtain the fled on the axis of the loop, Figure 2.15(b).
For each current element, II is perpendicular to the radius A P and
its magnitude is I ds/47rr 2 [see Equation (1.63) and note that point
P is in the equatorial plane of the element, {J = 1r/2J. The component
of this jj along OP is (I ds/47r1'2) sin {J. The radial components
cancel on account of symmetry. Since sin" = air, we have, for the
whole loop
a 2 J Sf
Hz =- = ,
2r3 21r(a 2 + Z2)3/2
(2.42)
where S is the area of the loop.,
At distances large compared with 2a, the field should be the same
as that of the magnetic dipole of proper moment. On the axis, Hz is
the radial component (in spherical coordinates) and we can identify
the moment 4?l in equations (2.33) by setting {J = 0 and comparing
with equation (2.42). 'rhus we find <l>l = III S and the complete field
I S cos {J I S sin (J
IIr = ,II, = , (2.43)
21rr 3 41rr 3
where r is now the distance from the center of the ring and {J is the
angle between the radius and the axis of the ring. The product I S
is called the area moment of the circulating current as contrasted with
the equivalent diPole moment p./ S.
A thin solenoid and a current ring have the same fields at points
not too close to them; but they are expressed differently. In a solenoid
we have a given magnetic flux but not the current. As the radius of
the solenoid approaches zero, the current in the winding must be
increased indefinitely in order to maintain the same field. On the
other hand, in a ring we have given current but not the flux. The
radius of the wire making the ring does not enter the expressions for
the field as long as this radius is a small fraction of the radius of the
ring. In fact, in deriving equations (2.43) we have tacitly assumed
that the radius of the wire is zero. In this case, the magnetic intensity
in the immediate vicinity of the wire equals I/21rp where p is the
distance from the axis of the wire. Hence the magnetic flux linked
\vith the wire ifi infinite.
If the electric current is varying with time the magnetic intensity
and the magnetic flux density will also vary. Hence a circle coaxial

92
Electromagnetic fields
with either the solenoid or the current ring is linked with magnetic
displaccmcn t curren t (that is, the time rate of change of magnetic
flux) and there will exist, in accordance with the Faraday-lvIaxwell
equation (1.76). an electromotive force round the circle. If the
coordinates of the circle are rand fJ, the radius is r sin fJ. ly symmetry
the electric intensity is uniform round the circle, and the emf is
27rY sin fJE tP . The radial magnetic flux linked with the circle may be
obtained by integrating Br over the spherical cap of radius r sur-
mounting the circle. Such an integration was performed in Section
1.22 where we obtained the magnetic field of a current dipole and
used it subsequently in Section 1.23 to obtain the magnetic field
produced by the electric displacement current of a time-variable
electric dipole in a dielectric medium. Thus noting the differences in
the algebraic signs in laxwell's equations, we have
cPl sin (J p.i S sin fJ
E", = - = - (2.44)
47rY 2 47rY 2
Comparing equations (2.43) with equations (2.33) and (2.34)
we have the scalar magnetic potential of the circulating current
IS cos fJ
U=
(2.45)
47rY 2
1"'he solid angle n of a cone is defined as the ratio of the area inter-
cepted by the cone on the surface of a sphere, centered at the apex
z
p
y
x
FIGURE 2.16 A cone, with its apex at point P, subtended by a circular ring in
the xy Plane and illustratin1. the definition of the solid angle.

Static and alnl0st static fields
93
of the cone, to the square of the radius. Assume that the loop is
infinitesimal and imagine a cone from some point P subtended by
the loop as in "'igure 2.16. The area in tercepted on the surface of the
sphere of radius r centered at P, is dS cos 0, ,vherc dS is the area of
the loop, and its ratio to the square of the radius is (dS/r 2 ) cos o.
By definition, this is the solid angle of the infinitesimal cone dn.
Hence the potential of an infinitesimal circulating current may be
written as
U = I dH/41r.
(2.46)
Consider a current loop of an arbitrary shape [see Figure 2.17J.
p
FHiURE 2.17 A cone subtended by a current loop of arbitrary shape.
On any surface bounded by this loop we imagine crisscross lines
subdividing the surface into infinitesimal elements of area. Imagine
a circulating current I round the boundary of each clement. These
currents cancel on the boundaries common to the elements. What is
left is the circulating curren t round the boundary of the en tire area.
For each element we have equation (2.46). Integrating, we have for
an arbitrary closed curren t
U = In/41r.
(2.47)
l\Iagnetic scalar potential, when it exists, is a many-valued function
of position. Imagine a circular ring, for instance. If we start with
n = 0 at infinity and approach the ring, n will increase. If we approach

94
L/l'ctr()111(lgllctic jie/cis
the plane of the ring outside the ring, U will approach zero again;
but if we approach it inside the ring, 12 will approach 27r. As \ve pass
through the plane of the ring, it will keep increasing. It is the exterior
solid angle that we use to preserve the continuity of potential. lhcn if
we approach the plane of the ring from below but outside the ring,
n will approach 47r, where originally it was zero. At all points, U
has an infinite number of values differing by 1'1-1, where 1t is an integer.
This is as it should be. The line integral of II round a closed curve
linked with current I once must equall or -I; but this integral
is also the magnetomotive force, and therefore the change in magnetic
poten tial, round the curvc. When obtaining ii by differen tiation w..e
have to change U continuously, and the constant nl docs not affect II.
2.11 Comparison of electric and magnetic fields
By now it should be clear that many similarities exist between
electric fields in conducting and nonconducting media and between
electric and magnetic fields. There are also importan t differences.
Analogies are very useful both in thinking and in calculations; but
serious errors can be made if the differences arc forgotten.
Figure 2.18 illustrates the following: (a) a point source of electric
conduction current I in a conducting medium; (b) a point charge in a
dielectric medium; (c) a point source of electric displacement current
I = q in a dielectric medium; (d) a point source of magnetic flux, <1>;
(c) a point source of magnetic current. 4>; (f) conduction current
between perfectly conducting concentric spheres; (g) electrostatic
field bet\veen conducting spheres, not nccessarily perfect; (h) electric
displacement current between perfectly conducting spheres; (i) a
double source of current Il in a conducting medium; (j) an electro-
static dipole of moment ql; (k) an electric current clement of moment
I = q! in a dielectric medium; (1) a magnetic dipole (solenoid) of
moment 4>t; (m) a magnetic current clement of moment <t>l; (n) a
ring of current of area moment 1 S; (0) a ring of slowly-varying cur-
rent; (p) two closely spaced parallel plates (double layer of charge)
with a voltage V between them so that the area moment is I'S;
and (q) a double layer with a slowly varying voltage.
IIost of these sources shown in Figure 2.18 and their fields are
idealiza tions of physical sources and their fields. l"hc field of a poin t
charge in an inflnite space is an idealization of the 1icld of a small
charge far removed from other bodies. l"he tield of a point source of
direct current in an infInite conducting medium is an idealization
of currcnt issuing from an open end of a thin insulated wire submerged

Static and almost static ficld
95
in a large lnctal tank tilled with some conducting fluid, for example.
The other end of the \vire and tank may he connected to the terminals
of a direct current generator. Case (c), ho\vcver, in \vhich \"'lC asume
direct displacement current emerging from the end of a semi-infinite
wire, can best be described as a mathematical model of a purely
hypothetical physical situation, rather than an idealization. 1"0
produce this situation in a laboratory we would need tremendous
voltages (increasing at tremendous rates) applied continuously along
the wire in a way that would neutralize the radial field voltage at the
wire itself. Otherwise, the uncompensated voltage distribution on the
wire would produce displacement current issuing all along the wire.
This \vould be superposed on the radial displacement current emerging
from the end of the wire. Such a con tinuous distribution of driving
generators is not necessary in case (a) in which the wire can be
insulated from the conducting medium. However, there is no sub-
stance with zero dielectric constant which could" insulate" the wire
from a dielectric medium. Nevertheless, the hypothetical case (c)
will help us obtain the field of the corresponding time-variable source
and many other fields which are physically realizable and of great
practical importance. Similarly, the current clement in free space,
case (k), is an abstraction, a mathematical model of a hypothetical
physical source. Using its field, we can calculate the field of any
physically realizable distribution.
'rhc electric fields of point sources (a), (b), and (c) are, column
by column
J r =
4 .,,'
'Trr"
IJ r =
q
4 .,,'
'Try"
. I
Dr=-
47rr 2
I
q
E --
r - ,
4'TrEr 2
r I dt
I
E -
r - .,,'
4'Trur
4'TrEr 2
(2.48)
Er=
I
v-
--,
4'Trur
v=
q
,
4'TrEr
If docs not exist
Here the analogy is almost complete. On the other hand, for magnetic
fields \ve have
II
'"
/(1 + cos 0)
4'Trr sin ()
II = 0
,
II", =
1(1 + cos 0)
(2.49)
4'Trr sin 8 .
'rhcre is no magnetic tield in the electrostatic case, and in (c) V
docs nut exist when I is time variable.

96
lle(/r()nla gllc! ic Jiclds
00
I
(a)
00
JJ.
Her
f
.
P(r,tJ,cpJ
Dr, Er
(b)
P(r,O,cPJ
.
Br, /1,.
 / 
(d)
(f)
26
(g)
00
Il'P
! .
f ]=q ·
P(r, O,)
. .
Dr, E,
(c)
00
. .
B" II,
 /! 
(c)
(h)
FIGURE 2.18 A nalogt"es betu)een electric and magnetic fields.

Static and ahnost static fields
97
H
f1 ,
I
E I) or Eline
q
I
H
E I D or Eline
lq
I
(i)
-q
(j)
I
(k)
p.
B or H line
E,
<I>
.
B
(I)
I
(m)
E
B or II line I D or Eline
p. Jl. I e
H
(n)
(0)
(p)
FIGURE 2.18 A nalogies between electr'£c and magnetic fields.

98
llec/rOl1l(lgllelic fields
l"'hc n1agnctic liel<Js of point sources (d) and (e) arc, column by
colunln,
ct>
B --
T - ·
471"r 2
. ri>
Br =-
47rr 2
cJ>
11, =
47rjJ.r 2
. 4>
Hr=
471"J.,Lr 2
(2.50)
<I>
u=-
471"jJ.r
U = does not exist.
1"he analogy with equations (2.48) is unmistakable. The field of a
magnetic point charge has been omitted because in the real world
magnetic charge docs not exist. On occasions it is convenient, how-
ever, to introduce fictitious magnetic charge in which case we have a
third set of equations corresponding to the middle column in equa-
tions (2.48).
'fherc is no electric field in the magnctostatic case; but in the
time-variable case we have
4>( 1 + cos 0)
EtIJ = .
47rr sin (J
(2.51)
l"'his equation is analogous to equation (2.49) except for the difference
in algebraic sign. 'fhe difference arises from a similar difference in
IVfaxwcll's equations.
In (a), (b), and (c) since there is no electric field tangcn tial to the
spheres to begin with, we can introduce perfectly conducting spheres
concentric with the point sources without disturbing the fields. 1"hus
the fIelds bctwecn the spheres, (f), (g), and (h), will be given by
equations (2.48) and (2.49). In (f) and (h) we can connect the" feed
wires" to the spheres and disconnect the remaining wire portions.
The fields inside the in tcrior sphere and outside the exterior sphere
will disappear and only those bet\veen them will remain. The reason
for requiring perfect conductivity of the spheres becomes clear when
we note that currents have to flow in them and an E8 field would
appear if the spheres were not perfectly conducting. If the con-
ductivity q bet\vccn the spheres is much smaller than the conductivity
of the spheres, E8 will be much smaller than Er and we have a good
approxima tion to the idal case.

Static and ahuost static fields
99
In (g) the spheres need not be perfect since there is no current.
Originally there is the charge q of the point source, -q on the inside
surface«>f the interior sphere, q on the outside surface, -q on the
inside surface of the exterior sphere, and q on its outside surface.
Connecting the latter to ground, \ve remove the exterior field. Con-
necting the point charge to the interior sphere, \Vc remove the field
inside it. Thus only the charge q on the outer surface of the interior
sphere and -q on the inner surface of the exterior sphere \vill remain.
There are no physical magnetic analogs of (f), (g), and (h).
'fhc fIelds of a double current source in an infinite conducting
medium (i), of an electrostatic dipole (j), and of a current element
in a dielectric medium (k) are, column by column,
It cos {}
V=
4 t)'
, 7rUr-
It cos {J
E T = ,
21rur 3
Il sin {J
E8 = .
411"ur 3 '
Il sin {J
Hip =
,
41rr 2
ql cos {J
V=
,
471"Er 2
V does not exist
ql cos 8
Er = ,
271"fr 3
ql cos {J
E T = (2.52)
27rfr3
ql sin 8
E 8 =
47rEr 3
ql sin 8
Ee = ,
47rEr 3
II = 0,
Il sin {J
lltp = ,
471"r 2
where in the third column I = q or q = f I dt.
Similarly the fields of a magnetic dipole (1) and magnetic currt,nt
element (m) are
u=
lIT =
4>l cos 8
47rJ,J.r 2
cJ>l cos 8
,
21rJ,J.r 3
cfll sin 0
H8 = ,
47rJ,J.r 3
E = 0,
U does not exist
<l>l cos (J
II, = 271"J,J.r 3
(2.53)
<Pl sin I)
lI8 = 471"J,J.r'!
<i>l sin {J
E -
tp- 41rr 2

100
Electrol1UlJ!.lletic .fields
lields of the circulating curren ts, constan t (J1) and time varia hIe
(0) arc
I S ens 0
,
27rr 3
lir
IS cos ()
27rr 3
lIr =
Hs
I S sin ()
47rr 3
I S sin ()
H s =
47rr 3
(2.54)
E=O
,
EfP =
J.Lj S sin ()
47rr 2
'fhc scalar magnetic potential in case (n) is many-valued. J 4 rom
equations (2.53) \Vc obtain its" principal" value by letting tf>l = J.LI S.
Finally, \ve have double layers' of electric charge, (p) and (q),
analogs of circulating currents, whose ficlds arc
v S cos 0 V S cos ()
E - E -
r - 27rr 3 r -
27rr 3
V S sin () V S sin ()
£s= 47rr 3 E s = 47rr 3 (2.55)
E V S sin ()
II = 0, II rp = 47rr 2
']"'he magnetic 11clds of a solenoid and a current ring are similar
but expressed in terms of t\\"o different quantities, dipole moment
<pI and area moment IS. We can express the magnetic flux 4> in terms
of the current I in the winding; but then ,ve have to in troduce the
nunlbcr of turns pcr unit length 11, and the area S of the cross section
of the solenoid. 1"hese three parameters /, Jl, and S are conveniently
cAl)ressed by one parameter <1>. As S approaches zero, I has to increase
indefinitely to maintain a constant <1>. Similarly VlC can exprcs-,; I in a
current ring in terms of 4>; but then ,ve have to introduce another
parameter, the radius of the \vire. In fact, equations (2.54) arc valid
\vhen the radius of the wire is infinitesimal so that the magnetic flux
<I> is infinite.
Analogously, the fields of an electric dipole and a double layer are
similar but one is expressed in terms of the dipole moment ql and the
other in terms of the area moment V S. The dipole moment is finite
for point charges when the voltage between them is infinite. lhc area
moment is finite ,vhen the thickness of the double layer is infinitesimal
and the charges are infmite. 'I'here is no "area" connected ,vith the
dipole.

Static and almost static ficld
101
(b)
FIGURE 2.19 A cross section of two coaxial c'ylinders: (a) equally but oppositely
charged; (b) carrying equal but opposite currents.
Two infinitely long. coaxial metal cylinders are shown in "\igure
2.19. In (a) they are equally and oppositely charged; in (b) they
carry equal and opposite currents, with current I in the inner cylinder
flo\ving out of the page. In case (a) the field exists only between the
cylinders. If q is the charge per unit length, the radial displacement
per unit length is q and the displacement density is q/27rp where p is
the distance from the axis. Thus
q
D =-
p 2 '
7rp
q
E p =-.
27rEP
(2.56)
The magnetic field (b) on the other hand, exists not only between
the cylinders but in the cylinders as well. The magnetomotive force
round a circle of radius p equals the enclosed current and hence \ve
have
lp
H =-
tp 2 2'
7ra
O < p < a
I
a < p < b
(2.57)
27rp
l(c2 - p2)
-
,
27rp (c 2 - b 2 )
b < p < c
=0
,
p > c.

102
1le(tro1nllKl1{'ti( Jields
If. in cases (a) and (b) b approaches infinity and a approaches zero.
\Vc have an infinitely long filament of either charge or current in free
space.
2.12 Images
In problems studied so far, we have dealt, for the most part, with
fIelds in infinite homogcneous media. In mixed media even the ficld
of a point source is considerably more complex and more powerful
methods arc needed for their calculation. Some cases, howevcr, can
be solved by elcmentary methods. One of these is the case of two
semi-infmite homogeneous media separated by a plane boundary.
z
z
E1 + E 2
A'
EI + E 2
q
A
+q
h
h
Air
Soil
or sea
water
h
Air
Conduct-
ing plate
h
-q
-q
(a)
(b)
FIGURE 2.20 A point charge q at A and its image - q at A': (a) above and
below the plane 'interface beLu'een air (or son1e dielectric) and a conducting medium,.
(b) above and below a conducting plate.
Consider a charge q at point A above ground or sea water at height
Jz as sho\\ln in Figure 2.20(a). Soil and sea water are conductors,
even though poor connuctors as compared to metals. Hence the
charge at A will pull a charge of opposite sign to the surface until
the clcctric field bclow the surface is red uced to zero and the tangcn tial

Static and alnl0st static fields
103
component on the surface also becomes equal to zero. Similarly if the
charge is above a thin conducting plane, J;igure 2.20(b). a charge of
opposite sign ,viII be pulled (from infinity) until the component
ta,ngen tial to the plane vanishes. This is the essen tial condition which
the field in the upper half space rou;t fulfill. To the original field of
the point charge at A, '\Fe must add another field which will make
the tota.1 ta,ngential component on the conducting surface equal to
zero. To find this field we imagine a charge -q at the mirror image
point A' below the surface. The resultant field of both charges
satisfies the above requirement. Thus we have the required potential
above the in terface
v=
q
q
47rEor2'
z > o.
(2.58)
47r E orl
The first term is the potential of the point charge q at A as it would
exist in an infinite medium. The second term is the potential of the
surface charge so distributed that it makes the total potential con-
stant on the surface. The differential of l' in a direction tangent to
the surface, and thus the ta,ngen tial componen t of the electric field, is
zero on the surface. As far as the oupper region is concerned, this added
potential equals the potential of the image charge -q below the
surface. This image charge is a virtual source, not a real source, since
actually there is no charge at A'.
The potential of the surface charge in the upper and lower half
spaces is symmetric about the surface. Therefore, in. tlte lower half
spa.ce the surface charge potential is equal to the potential of a
virtual charge -q at point A which thus cancels the potential of the
real charge.
The density of the" surface charge equals the normal component
of the total jj at the surface
aV
qs = D z = -EO - =
dz
q cos (}2
,
27r(h2 + p2)
(2.59)
,vhcre p is the distance from the axis AA' and (}2 is the angle indicated
in Figure 2.20.
In the case of two semi-infinite dielectric media, air and pure
water for instance (Figure 2.21) the field of a point charge in one
medium will penetrate the other. '"rhis field will displace the bound
electrons, and at the interface between two media a layer of bound
charge ,vill appear. Since the impressed field of the point charge at A
is the same as in the preceding problem, we conjecture that the
bound surface charge will be distributed in the same manner as the
free charge on a conducting plane and that above the interface the

104
llectronlaglletic fields
z
h
h
Air, fO
Pure \va ter, E
qr
FIGURE 2.21 A point charge q at A and its it11age qT at A' in the Plane inter-
face between two pure dielectrics, air and pure u'ater, for exan1ple.
field of this surface charge, the reflected field, might be the same as
that which would be produced by some image charge qT at A'. l"he
transmitted field below the in terface is the sum of the field of the sur-
face charge and the impressed field. \Ve conjecture that this field
might be the same as that \vhich would be produced by some charge
qt at A . Thus we assume the potential of the total field as follows:
v= q + qr ,
47rEor 1 411'" E or 2
z > 0
(2.60)
qt
,--
- ,
41rE r l
At the interface rl = r2 the tangential component of il must be
continuous. Hence the potential must be continuous and
z < o.
q + qr = q t ( EO/E) .
(2.61)

Static and almost static fields
105
The normal component of D must also be continuous. First we find
q cos (h qr cos (}2
D,= 2 + 2
47r1"1 47rr2
z > 0
qt cos (lI
-
4 2 '
'Trrl
z < O.
At the interface rl = r2 and (}2 = 7r - (}l, so that the continuity of D.
yields
q _ qr = qt.
From equations (2.61) and (2.62) we find
(2.62)
fO - E
qr = q,
Eo + E
2E
qt = q.
Eo + E
(2.63)
Note that as E  00, qr  -q and qt  2q.
Still another case is that of a direct current source in a conducting
medium, sea water for example, with a nonconducting medium
above as shown in Figure 2.22. Here the boundary condition is:
the normal component J z of the conduction current density must
vanish at the interface. For a source [ at point A this condition will
be satisfied by the addition of an image source I at point A' at the
same distance above the interface. Thus
I I
V = + ,
47rUrl 41r Ur 2
z < O.
(2.64)
On the in terface
I
V=
,
21rUrl
z = o.
This is twice the potential of current [ emerging from A. The po-
tential is continuous across the interface. Hence, for the electric
field above the interface we have
v=
I
I
z > O.
(2.65)
,
2'TrUrl
Erl -
2 2'
1r Ur l
At the interface the tangential component of E maintains the curren!
in the sea water just below. In sea water the normal component of E
vanishes since the normal componen t of the conduction current density
vanishes. 1"'he discontinuity in the normal component of E is due to

106
l/('c/rOI1Ul1?lleti( Jie/cis
a surface layer of charge of density
qs
00
IJ n
Eol COS 0 1
') .
27rCTri
I
I
I
J+ :
,
I
A'
It I!
Air
Sea \vater
p
h
FI<,t;I{E 2.22 i1 point source of direct current I ill a cOl1ductin;!. nlediul1l (such
as sea 'waler) at point A and its i1uQf!.e at A' in a lloncol1ductil1[!. 1Hedizl1H, such as
fllr.
'T'he nlagnetic intensity beIo\\' the interface is obtained by adding
t\VO magnetic intensities. One is due to the current I \vhich runs
through the \virc to point "fJ and then spreads radially fronl .:1.
'J'hc other is due to the reflection of this radial current fronl the
interface vlhich appears to cnlanate radially from the image source
at A '. 'rhus
1/" =
1(1 + cas 8 1 )
47rp
1(1 + cas ( 2 )
41rp
(2.66)
1(2 + cas 0 1 + cas 8 2 )
-
41rp
z < o.

Static and alnl0st static fields
107
...:\bovc the in terface the nlagnctolnotive force round a circle of
radiu p, coaxial \vith the wire, cqual - J since this is the only
current crossing the area of the circle. l'lhercfore
I
II = --
fJ 2'
1rp
z > o.
(2.67)
Note that II fJ is continuous at the interface.
2.13 Tubes of flow and equipotential surfaces
Examples in this section illustrate the manner in \vhich boundary
conditions affect fields in mixed media and explain why such fields
are usually complicated and hard to evaluate. In the first series of
relatcd examples we consider electric ficlds between two infinite
perfectly conducting coaxial cylinders, either equally and oppositely
charged or carrying equal and opposite currents. Two different media,
conducting in one case and dielectric in the other, are separated
either by radial planes, Figure 2.23 (a), or by a coaxial cylindrical
surface, Figure 2.23 (b).
Let us consider in detail the radial currcnt flow in case (a). When
the media are separated by radial planes, E is continuous across them;
E is at right angles to the cylinders because they are perfectly con-
ducting. These conditions are satisfied if we assume that E is radial
-[ or - q -[ or - q
\ E p \ J p or Dp
E p or i p or D p
1
)
(a)
(b)
FIGURE 2.23 1 cross section of two coaxial metal cylinders, either equally but
oppositely charf!.ed, or carryinf!. equal but opposite currents: (a) when two dif-
ferent, COllducti1lf!. or dielectric tl1edia are sepa.rated by radial planes; (b) when
the 1nedia are separated by a coaxial c'ylindrical boundary.

108
J/rclr01}11.1Hdi( Jil'lds
in both nlcdia
",4
E
p
o < <P < 21r.
p
llence
u)A
J p =-,
p
O<<p<a
u 2 A
a < <p < 27r.
p
Let K be the total radial Cllrrent per unit length of the coaxial pair
2"
K = 1 Jpp d.p
o
u1Aa + u2r1 (271" - ex).
Thus \ve can express i1 in terms of K and obtain
K
E " -
p -
[ula + U2 (21r - ex) Jp'
o < 'P < 271"
J p =
[ula + U2 (271" - a) Jp'
u1K
O<<p<a
(2.68)
u 2 K
[ula + u2(271" - a) Jp'
ex < <p < 21r.
If U2 = 0, so that the medium is a dielectric (air, for instance), the
radial current vanishes in the region but the electric intensity docs
not. Hence
J p = 0,
K
E p =-,
u)exp
a < <p < 21r.
Therefore
EoK
D p =-,
u)ap
a < 'P < 21r.
'rhus in this region the charge density on the inner cylinder is foK/ Ulexa
and on the outer cylinder - foK/ulab.
\Vhen the boundary between media is cylindrical as shown in
ligure 2.23(b), it is the radial current density or displacement
density, as the case may be, which is continuous. Thus, for dieletric

Static and alnlost tatic tields
109
media
D p
q
21rp'
E p
q
,
27rEIP
q
a < p < b
a < p<c
2.69)
c < p < b.
,
27rE2P
T,vo analogous situa tions exist in the case of rnagnetostatic fields
between coaxial cylinders carrying equal and opposite curren ts.
See l.'igurc 2.24. '['he fields are coniincd to the space between the
cylinders. l\fagnetic lines arc circles when the permeabilities JJ.l and
J.l.2 arc equal. It appears that they can still be circles when J.l.l  J.l.2,
provided B" is con tinuous across the radial boundaries and H" across
the cylindrical boundary. 'rhus in case (a) we set
A
Brp = -, 0 < 'P < 211"
P
A
II" = -,
JJ.IP
A
,
J.l.2P
-[
(a)
O<cp<a
(2.70)
a < cp < 211'".
)
(b)
FIGURE .1.4 A cross section of two coaxial cylinders carryin;!. equal but opposite
currents: (a) """hen two homof.eneous media 'with different permea.bilities are
separated by radial Planes; (b) when the media are sepa.rated by a coaxial cylin-
drical boundary.

110
l/ertronlaJ!.nrt ic Jir/tls
'rhc n1agnctomotive force round a magnetic line should equal the
enclosed current
'l.".
1" Il<pp dl{J = I
o
or
J.Ll 1 Aa + J.L21l1 (27r - a) - I.
(2.71)
Hence
.J'1 =
J.L2 a + J.Ll (21r - a)
and ,ve have the magnetic field, equation (2.70), in terms of the
curren t in the cylinders.
'rhe current is not distributed. uniformly round the cylinders.
'There is no magnetic field outside the coaxial pair or in the in tcrior of
the inner cylinder. By the Ampere-1\1ax,vclllaw the current is always
equal to the magnctomotive force round a closed path encircling
the current. Hcnce if we take a path along a magnetic line in region
(1) between the interfaces with region (2) and complete the path
with radiallincs along the interfaces. piercing thc inner cylindcr and
joined together in its interior, \ve find that current II in the enclosed
portion of the cylinder is
J.L2 a l
/1 = J.LT1Aa = .
J.L2a + J.Ll (27r - a)
l'he curren t in the remaining portion of the cylinder is
J.Ll (27r - a) I
/2 = J.L2 1 A (21r - a) = .
J.L2a + J.Ll (27r - a)
J.LIJ.L'l.I
(2.72)
We can take a section of the coaxial pair and form a toroid (see
Figure 2.25) with a perfectly conducting boundary. A coaxial pair
and two parallel planes form a link between the toroid and a generator.
l'he field inside the toroid is the same as in the preceding case and the
circulating currents in the two portions of the toroid are as given
above.
The tield inside a closely and uniformly wound solenoid of the same
shape and dimensions as the toroid is also given by equations (2.70)
and (2.72) in which I = nlw, where 11, is the number of turns and lw
the current in the winding. '[he major difference is that the dis-
con tinuity in H f/J across the winding is now the same round the
solenoid. Since H f/J in the interior is different in the regions with
different permeabilities, we must have a magnetic field outside the

Stat ic and altnost static ficld
111
.
(2) J.l2
.. I
! tl +-1
I
I
I
FlCiURE 2.25 A toroidal conductor carrying a circulating current
'with !'wo sectors of different pernleabilities.
solenoid. The sources of this field arc at the interfaces of the two
media. If one medium is iron and the other air, the iron becomes a
magnet. "fhe magnetic lines emerging from one pole of this magnet
and converging to the other leak out of the solenoid between the
adjacent turns of the winding.
rhe difference between a toroid with perfectly conducting walls and
a solenoid may be summarized as follows: In the toroid the field
cannot escape from the interior but the current can and docs re-
distribu te itself; in the solenoid the circulating curren t cannot re-
distribute itself but the field can and does escape the interior through
the gaps in the winding.
In the case of cylindrical layers with different pcrmeabilities as
illustrated in Figure 2.24(b)  we have
I
IIf/J = -,
27rp
o < cp < 27r
J.l1I
B =-
f/J 2 '
7rp
a < p<c
(2.73)
J..L 2 1
= -,
27rp
Let us examine the reasons why the solutions in the foregoing
c < p < b.

112
l/rctronlagl1cti( firlds
examples turned out to be simple. Suppose we start \\'ith a homog-
enous isotropic medium and ca]culat the l1eld. ,\r c can rcprcscn t it
graphically by drawing lines of flow tangential to E, and cquipo-
ten tial surfaces perpendicular to the lines of flow. The field may thus
E lines

(a)
(b)
FIGURE 2.26 (a) A tube of flow between two charged conductors; (b) an equi-
potentiallaj'er surrounding one of them.
be divided either in to tubes of flow, bounded by lines of flow, or in to
equiPotential layers, bounded by equipotential surfaces. In the pre-
ceding examples tubes of flow are radial sectors in the case of electric
fields and toroids in the case of magnetic fields. Equipotentiallayers
are cylindrical shells in the case of electric fields and radial sectors
in the case of magnetic fields.
By definition there is no flow of curren t, electric displacement, or
magnetic flux, as the case may be, across the lateral boundary of a
tube of flow. 1"he flow takes place from onc source to the other
through the ends of the tube. If the medium within a complete tube
of flow [see Figure 2.26 (a) ] is replaced with some other isotropic
medium, the boundary conditions along the lateral boundary are

Static and ahnost static fields
113
satisfied by the 5ame type of field configuration. The tangential
component is continuous and produces either more flow or less,
depending on the parameters of the new medium but does not produce
any flow across the boundary. l"hus the normal components of flow
remain equal to zero and their continuity is preserved.
Similarly the medium within a complete cquipotential layer sur-
rounding a source, Figure 2.26(b), may be replaced with some
other medium without upsetting the boundary conditions.
£1
D
FIGURE 2.27 A n intersection A B CD of a tube of flow and an equiPotentiallayer.
The boundary conditions will be upset if a portion of a tube of flow
(or of an equipotential layer) is replaced with some other medium.
See Figure 2.27. The voltage from AB to CD would produce more
flow between these surfaces if f2 > fl, and yet the amount of flow
should have been preserved. Alternatively the same flow would
lower the voltage between AB and CD and thus detach this volume
from the equipotentiallayer. The field configuration has to change
and the knowledge of the field for the homogeneous c case does not
help us solve the more general case.
2.14 Properties of fields in the large
Certain quantities may be associated with fields in the large. cFor a
given field these quantities depend on its geometry, on the physical
characteristics of the media and on the details of field distribution.
However, the same quantities may be associated with many quite
different fields. For example, current I in a tube of flow is propor-
tional to the voltage V across it. The ratio
G = [IV
is a property of this tube. The quantity G may be either calculated

114
ElectrOt71agnet£c fields
or measured. Usually the voltage which produces the current is
given. The resulting current can then be calculated if G, called the
conductan.ce of the tube, is known. The reciprocal of the conductance
R = I/G = VII
is called the resistance of the tube.
Tubes of flow are in parallel. Currents in the individual tubes are
added to obtain the total current. The conductances of these tubes
are also added to obtain the total conductance.
Equipoten tial layers are in series. The voltages across the indi-
vidual layers are added to obtain the total voltage across them all.
Hence the resistance of several equipotentiallayers is the sum of the
resistances of the individual layers.
Analogous quantities may be associated with electrostatic and
magnetostatic fields. In the next three sections we shall consider
such quantities for more general fields.
2.15 Resistance and conductance coefficients
Let II and 1 2 be electric currents emerging from two perfectly con-
ducting bodies K 1 and K 2 , imbedded in a conducting medium, Fig-
ure 2.28. Each body is connected, of course, with an insulated wire
to one terminal of a generator whose other terminal is connected to
a wire conveying current from infinity (or ground). 'The poten tials
of K 1 and K 2 with reference to infinity (or ground) are linear functions
of the curren ts
VI = rul1 + r12[2,
(2.74)
V 2 = r 21 / 1 + r22[2.
Analogous equations are true for any number of conductors. Co-
FIGURE 2.28 Two perfect conductors KI and K 2 serving as sources of direct
current in a dissiPative medium.

Static and almost static fields
115
efficicnts .r mn are called the 1nutual resistance coefficients when m  n,
and self-resistance coeificie'l'zts when In = n.
1"0 calculate these quantities we have to solve an appropriate field
problem. For example, if K 1 and K 2 are spheres of radii a and b,
small compared with the distance l between their centers, then ap-
proximately
II 1 2
VI = -+-,
411"0" a 411"O'l
(2.75)
II 1 2
V 2 = - +-.
411"0"l 411"ub
It is not just a coincidence thatr21 = '12. It is always true that
r mn = r nm ,
(2.76)
as we shall presently show.
1'0 obtain the resistance coefficients experimentally we should
disconnect one body, K 2 for instance, from the source of current
and measure the ratios
r 11 = V 1/ I 1, r 21 = V 2/ II. (2.77 )
A mutual resistance coefficient is seen to be equal to the potential
of one body due to a unit current emerging from another. The re-
ciprocity relation, equation (2.76), means therefore that the potential
of Km due to a unit current emerging from Kn equals the potential of Kn
due to a unit current ernerging fro111, Km.
If 1 2 = -II, the entire current emerging from K 1 flows into K 2 .
In accordance with the definition in Section 2.14, the resistance
between K 1 and K 2 is
R = (Vi - V 2 )/I 1 = T11 - 2r12 + 1'22.
Solving equation (2.74) for II and 1 2 , we have
II = g11 VI + g12 V 2 ,
(2.78)
(2.79)
/2 = g21 VI + g22 V 2 ,
where
gn = r22/D,
g22 = Tnl D,
g21 = g12 = -rI2/ D,
D = r11r22 - r12 0
'fhe gmn are called the conductance coefficients. \Vith a little thought
given to the method of solving a field problem, such as that leading

116
Elcr!rnnUlJ!.llc/ic ficlds
to cq ua tion (2.75), we conclude that the resistance coefficients are
essentially positi't'e. '"fhe self-conductances must also be positive: If
V 2 is equal to zero, then the current must flow out of Kl if VI is posi-
tive. Therefore D must be positive, and \;e conclude that the 111utuaJ
conductance coefficients arc essentially negative.
Some current emerging from Kl goes directly to K 2 ; the rcst goes
to infinity (or ground). Likewise, some curren t from K 2 goes to KI
and the rest goes to infinity (or ground). Hence, there is a net current
between KI and K 2 . This current, in the direction from Kl to K2' is
proportional to the potential drop VI - V 2 . The coefficient of pro-
portionality, G 12 , is called the direct conductance between Kl and K 2 .
Thus that part of II which goes directly to K 2 is G 12 ( VI - V 2 ). The
remainder, going to infinity, is G loo VI, where Glen is the direct con-
ductance to infinity. Similarly, that part of 1 2 which goes to Kl is
G 12 (V 2 - VI) = -G 12 (V 1 - V 2 ) . "Thus
II = G1ooV 1 + G 12 (V l - V 2 ),
(2.80)
1 2 = G 12 (V 2 - VI) + G 2oo V 2 .
co
C, \.d)
\\
KI
1/G 12
K 2
FIGURE 2.29 An equiva.lent 1!etwork representing properties in the large of the
field of two sources of current, such as Kl and K 2 in Figure 2.28.
Comparing with equations (2.79), we have
gll = G 100 + G 12 ,
g12 = -G 12 ,
(2.81)
g22 = G 200 + G 12 , g21 = - G 12 .
Thus we have proven the reciprocity thcorem and have shown once
more that the mutual conductance coefficients are negative (since
the direct conductance G 12 is essentially positive).
We have also shown that in, the large the current flow in the medium
may be represented by a network of resistors. See Figure 2.29. The

Static and ahnQst static fields
117
resistances in this nct\vork are the reciprocals of the following con-
ductances
Glee = gn + g12,
G 200 = g22 + g12.
G I2 = - g12,
2.16 Potential and capacitance coefficients
If the medium surrounding the conducting bodies in Figure 2.28 is
dielectric, and if ql and q2 are the charges on Kl and K2, then
VI = Pllql + P12q2,
(2.82)
V 2 = P21qI + P2'1fj2,
'v here the p are the potential coefficients. All equations are analogous
to those in the preceding section (except that the conductivity u,
,vherever it appears, should be replaced by the dielectric constan t E;
only the names of various quantities are different.
If charges are expressed in terms of poten tials,
ql = Cll VI + C12 V 2 ,
(2.83)
q2 = C21 VI + C22 V 2 .
The coefficients are called the capacita.nce coefficients.
If q2 - -ql and all other conductors, if any, are uncharged, the
ratio
C = ql/(V 1 - V 2 )
(2.84)
is called the capacitance between Kl and K 2 .
l"hc equivalent network of capacitors is similar to that in Figure
2.29. The direct capacitance C I2 will appear between the nodes Kl
and K 2 . The remaining capacitances will be C loo and C 200 .
2.17 Inductance coefficients
\\Tith some understandable differences the properties of magnetic
fields in the large are similar to those of electric fields. Let 1 1 and 1 2
be the currents in two conducting loops, Figure 2.30, 4>1 be the
magnetic flux linked with the first loop, and 4>2 be the flux linked
with the second loop. The magnetic field at each point of the field
is in part proportional to II and in part to 1 2 . The same will be true

118
Eleclronlagl1elic fields
of cJ>1 and cf>2,
cf>1 Lull + L 12 1 2 ,
(2.85)
<1>2 = 111 + 212.
II /2
... 
..
 /2
II
FIGURE 2.30 Two current loops.
Here Lu nd L 22 arc the self-inductances of the loops ano L 21 = L r.--
is the mu tual inductance.
If II and /2 arc varying \vith time and the loops are perfectly
conducting, the voltages bct\veen the terminals are
VI = 4>1 = L ll i 1 + L 12 i 2 ,
(2.86)
V 2 = 4>2 = L 21 i 1 + L 22 i 2 .
If we have two closely-wound coils with 1h and n2 turns, we should
replace in equations (2.85) II by 1111 1 and /2 by 1t212. Thus
cf>1 =n 1 L ll l l + n 2 L 12 ! 2,
<1>2 = n 1 L 21 / 1 + 1l 2 L 22 1 2 .
'fhen the voltages across the terminals of the coils arc
· 2. .
VI = n 1 cJ>1 = 1'[,IL 11 1 1 + 1ll'rl'2L I2 1 2 ,
(2.87)
· . 2'
V 2 = 11'2<1>2 = 1l1'1z, 2 L 21 I 1 + 1t2"L;.2 1 2.
1he quantities 111Lll and n2 are the inductances of the coils and
1l11t2L12 is the rou tual inductance. Actually the" transformer ra tios"
n1  n, and 1hn2 ,viII be reduced by the t1 ux l eakage betwecn the
tur hc o ils.  -
2.18 Transmission lines
Electric charge in a \virc ]YQ, as sho\vn in Figure 2.31 (a), moving
slowly back and forth in a uniform magnctic field is subject to a force

Static and alnlost static fields
119
x
Q
vet)
o

+
z
+
+
+
p
x
(a)
Q
/
('
I
I
I
I
I
I
I
I
I
1 (z, t) 1
I + + + A I  +i + I B
I -------------------
lp z
--- DI
I
I
I
I
I
I
I
I
.
I(z,t)

1 V(z.t)
s
vet)
Z
l
(b)
FIGURE 2.31 (a) A 'lLre moving in a nlagnetic field; (b) a wire sliding on
parallel wires in a magnetic field.
(Sections 1.17 and 1.18). This motional or induced force per unit
charge is
E; = -Bov(t), (2.88)
where Bo is the magnitude of the magnetic flux density and v(t) is

120
Electronll1g11r/ic jicltls
the speed of the wire. The density (per unit length) of the displaced
charge Q(x, t) may be obtained from equation (2.18) \vhere \VC use
.("1 (x) as the function defined in equation (2.16), since this is a better
approxima tion then A I (x). Thus
Bov(t)x
Q(x, t) = (2.89)
A (x)
The electric field around the wire can now be calculated as in Sec-
tion 2.2.
At distances large compared with the length s of the wire, the
electric field is essentially that of a dipole. Two elements of charge,
Q(x, t) dx and Q( -x, t) dx, form a dipole of moment 2xQ(x, t) dx
and the moment of the entire charge distribution is
f t'
P = 2 xQ(x, t) dx.
o
Replacing A (x) in equation (2.89) by its average value [equation
(2.17) ] we have approximately
p = - -f7jCoBoS3v (t) ,
(2.90)
where
7rE
Co =
in (sj2a) - l'
and a is the radius of the wire. The electric field is concentrated
around the wire since the field of a dipole decreases as the cube of
the reciprocal of the distance.
Suppose now that the wire is sliding back and forth on a pair of
parallel wires of length l, Figure 2.31 (b). The field voltage between
the ends of the wire is
v PQ = -sE; = BoS1J(t) ,
(2.91)
where s is the distance between the axes of the wires. The motional
voltage will displace electric charge from the upper wire to the lower.
Let q(z, t) be the charge per unit length on the lower wire. Except near
ends, the field around the wire due to this charge is given by equation
(2.56). Hence the transverse voltage from the lower wire to the
upper, due to this charge, is
f ' q(z, t) In (sja)
VI = Epdp = ,
G 211"E

Static and ahnost static fields
121
\vhere a is the radius of the \vire. 'fhe charge on the upper wire is
-q(z. t) and it produces an equal voltage. Thus the total transverse
voltage is
v (z, t) -
q(z, t)
C '
(2.92)
where C is the captuitance per unit length of the parallel pair
c=
In (s/a)
1rE
(2.93)
The current I(z, t) in the lower wire equals the time rate of increase
of the charge on AB, Figure 2.31 (b),
a jl I l a
fez, t) = - q(z, t) dz = C - V(z, t) dz.
at , , at
(2.94)
According to Faraday-Maxwell law the electromotive force round
the closed circuit ABCDA is
V AB + V BC + V CD + V DA =
atl>
--
at'
(2.95)
where cp is the magnetic flux (out of the paper) through the rectangle.
The magnetic intensity of the field produced by the current in the
lower wire is [eq ua tion (2.57) ]
I (z, t)
Hf(J = ,p > a.
21rp
The con tribu tion of this to <l> is
I ll' Il J.I. s
cI>1 = JJ,Hf(J dp dz = -In - I(z, t) dz.
a a , 21r a
There is an equal contribution from the field produced by the current
in the upper wire. Hence
cI> = I I LI ( z, t) dz, ( 2.96)
a
where L is the inducta1tce per unit length of the parallel pair
JJ, s
L = - In-.
1r a
(2.97)
If Rl and R 2 are the resistances of the lower and upper wires, per

122
1'/('(tr(Jl1U1J!.llrli( firlds
unit length, then
v.w = 1 1 RJ (z, t) dz,
z
l'DC
1
- f R 2 I(z, t) dz.
z
(2.98)
1"hcse equations are approximate. To understand the nature of ap-
proximations consider a wire (Figure 2.32) carrying current. "There is
a magnetic field inside the wire [see equation (2.57) J. As long as
the current is steady, the magnetic field is constant and does not affect
the distribution of current. But \vhen it is varying with time, the
time derivative of magnetic 'flux linked with a closed circuit (in a
Q p
t I I I 
__ _ ..!: 2!-':':1'.:-_ _ _
"'IGURE 2.32 A closed circuit At N PQ M in a radial plane of a conducting wire
carrying current I.
radial plane) will make V IN and V QP unequal. The current dis-
tribution will no longer be uniform. This will affect the resistances
and the inductance. If variations with time are" slow enough," the
effect will be small. If the wire is very thin, the flux linked with the
rectangle MN PQ is small, and the effect will be small even for more
rapidly varying fields. Later we shall demonstrate that the effect of
the time-variable magnetic field is to drive the curren t toward the
surface of the \vire (the "skin effect") and thus increase the re-
sistance.
The remaining voltages in equation (2.95) are
V/Jc = V(l,t), V DA = -V(z,t).
"raking all these results into consideration, we transform equation
(2.95) in to
V(z, t) = V(l, t) + / (R l + R 2 )I(z, t) dz + i l L  I(z, t) dz.
z  at
(2.99)
Equations (2.99) and (2.94) determine the relation between the
transverse voltage from onc wire to the other and the current in the

Static and alnl0st static fields
123
\vires. 'fhe transverse voltage at the sliding wire [see equation (2.91) ]
.
IS
v (0, t) = B(}5v(t).
(2.100)
Neglecting the time derivatives, we have initial approximations
VCO) (z, t) = V (l, t),
[(0) (z, t) = O.
Thus the voltage, generated in the moving wire may be "trans-
mitted" to large distances from it. Substituting the first of these
equations in equation (2.94), \ve obtain the next approximation to
the current
l a iJ
I(I)(z, t) =  C at V(l, t) dz = C(l - z) at V(l, t).
(2.101)
Even though C is small, for long wires the current in the sliding wire
and near it may be substantial. Substituting in equation (2.99), we
obtain the next approximation for the transverse voltage
a
VO) (z, t) = V (l, t) + ! (R 1 + R 2 ) C (l - z) 2 - V (l, t)
at
(J2
+ LC(l - Z)2  Vel, t).
at 2
These successive approximations can be continued indefinitely.
Simpler results are obtained if the dependence on time is sinusoidal,
V (l, t) = V (l) sin wt, since the time derivatives can be easily calcu-
lated and the end voltage V (l) can be related to the" generator"
voltage given by equation (2.100) in the present example. Of course,
equations connecting V (z, t) and I (z, t) do not depend on the kind
of generator we happen to connect to the wires.
If the parallel wires are shorted at the far end Be (see Figure
2.33) the equations connecting V (z, t) and [(z, t) are essentially
the same as in the preceding case except that V (l, t) is equal to zero
in equation (2.99), and current I (l, t) through the shorting rod should
be included on the right-hand side of equation (2.94) so that it
becomes
I(z, t) = I(l, t) + / C  V(z, t) dz.
z at
For the shorted pair the initial approximation is
[(0) (z, t) = I (l, t).
Substituting this in equation (2.99), we have
d
V(O) (z, t) = (R 1 + R 2 ) (l - z) I (l, t) + L(t - z) - [(l, t). (2.103)
at
(2.102)

124
l/{'c/rolllaglle/ic fields
'I'his nlay be substituted in equation (2.102) and the iterative process
may be continued indefinitely.
x
IQ fez, t)
..
1
1-- D, C
I I
I I t V(z,tJ
I vet) 1
 I
I I I s
, 1
1
I I l(z,t)
1++ A'

--- ---- ----- -- - -----
IP z z
I . l
FIGURE 2.33 A wire PQ sliding on parallel wires P Band QC, terminated by a
resistive rod B C, in the presence of a magnetic field.
More generally we may have some resistance R L (" load" re-
sistance) at the far end BG. In this case neither V (l, t) nor I (l, t)
vanishes. Instead we have
Vel, t) = RI(l, t).
Again, there is only one unknown quantity which can ultimately be
related to the generator voltage or the" input voltage" at z = O.
Thus a pair of parallel conductors may be used for transmitting
electric and magnetic effects to large distances from their source.
The name for such a pair is transl1zission line.
In equations (2.99) and (2.102) the quantity l does not have to
refer to the end of the line. If l is equal to z + z, where z is in-
finitesimal, we may express the increments  V and I in terms of I
and V and obtain aV jaz and aljaz in the limit. Alternatively we can
differentiate equations (2.99) and (2.102) with respect to z.
2.19 Coaxial transmission lines
A pair of coaxial cylindrical conductors as shown in Figure 2.34
constitutes a coaxial transmission line. Transmission equations are

--'-
Static and ahnost static fields
125
x
A
.
+ t V(z,tJ Di I( z, t) C
AI  B
I
.
--
y .
+
-
I

1
J
z
p
2a
2b
FIGURE 2.34 Axial and transverse cross sections of coaxial cylinders, shorted
with a conducting disk.
of the same form as in the case of parallel wires, with the only differ-
ence in the expressions for the inductance and capacitance per unit
length. For the coaxial line
J.I. b
L=-ln-
,
27r a
211"E
c=
In (bja) ,
(2.104)
where a is the outer radius of the inner cylinder and b is the inner
radius of the outer cylinder. The cylinders are assumed to be so
thin that the current is uniformly distributed throughout their
cross sections. In Chapter 4 we shall remove this restriction by
evaluating V AB and V CD in terms of I (z, t) for rapidly varying
fields.
The major difference between a parallel pair and a coaxial pair
is in the character of field distributions. In the former case the field
extends to fairly large distances in the radial direction. In the latter
it is confined almost entirely to the region occupied by the coaxial
line. In the time-invariable case there is no magnetic field outside
the outer cylinder. There is a weak electric field depending on the
resistance of the outer cylinder. In the time-variable case this field
will generate a magnetic field which will react back, etc. If the
coaxial pair is close to the earth, the outer conductor will make a
transmission line with ground return. As the frequency increases,
however, the external field will start decreasing very rapidly. This

126
1:/('c/r0l11l1J!.llr/i( fir/ds
\\ill be shown in Chapter 4 \\here \ve consider propagation of tlelds in
highly dissipa tive media.
2.20 Limitations of the step-by-step method of calculating
self-consistent fields
In Section 2.18 the step-by-step method of calculating self-consistent
fields (Section 1.29) \vas applied successfully to transmission lines.
Starting with the conditions imposed on the voltage and current at
the far end, we worked back to the generator. Equally well, we could
have started from the gencrator if the input voltage V(O, t) had been
given. The input current, however, depends on the conditions at the
far end. In step-by-step calculations we need 1(0, t) as well as V (0, t).
In the preceding examples it is possible to express the electric and
magnetic fields in terms of these two quantities, one known and
the other unknown, as accurately as we wish and then to determine
the unknown quantity from the conditions at z = t.
The situation is quite different in the case of an electric dipole.
Starting with the electric field, we obtained the first approximation
to the magnetic field. The success depended on the circular symmetry.
In Section 2.8 we tried to obtain from this field the second approxi-
mation to the electric field and found it impossible to allocate the
contributions to Ez: and E p (or to Er and Ee). This can be remedied
by applying l\Iaxwell's integral equations to differential circuits.
But another difficulty will remain. The field of an electric dipole is
not unique. If the dipole is surrounded by a concentric conducting
sphere of radius, = '0, or by a concentric dielectric shell, its field
will certainly be affected. It can be shown that the field of an
clectrostatic dipole is affcctcd less and less as the radius of the sphere
or the shell increases, so that in an infinite space it is determined
uniquely by local conditions. It is impossible to show that this is also
true for a time-variable dipole for the simple reason that it is not
true. To obtain a sufficien tly general field for a time-variable dipole
in order to satisfy the conditions which might exist at large distances
from it, we should start not only with a local electric field but also
with a local magnetic field, independen t of the electric field [in the
same sense that /(0, t), is independent of V(O, t) J. Unfortunately,
all we know about this local magnetic field is that it must be time-
variable to begin with, since no appropriate magnetostatic field can
exist besides the one associated with direct c*urrent element.
One way out of the difficulty is to start with assumed fields at
r = '0 and work backwards to the dipole. rrhere are simpler methods
of analysis, however, which will be explained in Chapter 5.

3
Energy Storage, Dissipation,
and Transfer
3.0 Introduction
All physical phenomena are accompanied by transformations of one
form of energy into another. Electromagnetic phenomena are no
exception. In Section 1.8 we calculated the work done by electric
intensity when maintaining electric current. Experience shows that
the spent energy appears as heat. The heat is distributed throughout
the volume occupied by the field and the energy is delivered somehow
from the generator to different regions of the field. The amount of
generated heat depends on the local conditions and only indirectly
on the generator.
When electric particles of opposite sign are separated against the
force of attraction, work is done. 'The corresponding amount of
energy, "electric energy," must then be associated with the sepa-
rated charges. Similarly, "magnetic energy" is associated with moving
charges (in addition to the kinetic energy). 'I'here is evidence that
electric and magnetic energies arc associated with electric and nlag-
netic fields rather than with electric particles themselves. If we
raise a body so that it acquires potential energy and then let it fall,
we can account for all its potential energy when it is transformed into
kinetic energy or in to heat. If we separate electric charges of opposite
signs and let them recombine, there is some residual energy for which
we cannot account unless we assume that the energy is in electro-
magnetic fields. After the process of separation and recombination
of charges has been fmished  the field does not disappear en tirely.
The residual field, the" radiation field," is a shell of E and jf traveling
and expanding outwards. Radar echoes bear witness.
In this chapter \ve shall develop the idea of storage of electric and
magnetic energy in various regions of a fIeld and the idea of flow of
energy from one region to another. 1'hese ideas are useful in studying
electric oscillations in which electric energy is tran5formed periodically
into magnetic energy and vice versa, and in calculating idealized
127

128
1lec/r0111ag71e/ ie fields
cquivalen t circuits for fields in actual physical structures. On the
basis of energy considerations. it is possible to assign definite nleanings
to the following expressions: a "primarily electric" field. a "primarily
magnetic" field, and a "slowly varying" field. Throughout this book
the student should pay special attention to energy storage, dissipa-
tion, and flow.
3.1 Energy conversion and flow
Consider the concrete situation shown in "igurc 2.33. A wire PQ can
slide along a pair of wires terminated \vith a rod Be whose resistance
is R,. Assume that the resistance of the wires is negligible in com-
parison with Rz. In the absence of a magnetic field, a force is needed
to set the wire PQ in motion. To maintain a constant speed Va a
force is needed to overcome friction. In the presence of a magnetic
field, an additional force F is needed to push the wire through the field.
rrhis comes about in the following way. The charge moving with the
wire is acted upon by a downward force. As the charge moves down-
ward, it becomes subject to another force acting to the left. Hence
an equal force F is required to push the wire to the right through the
field.
The work done by the latter force per second is F(dzjdt) = Fvo.
This work W must equal the work done, per second, by the motional or
induced voltage Vp \vhen driving the downward current 10 against
the field voltage V PQ; that is
W = Fvo = Vplo.
From equation (2.88) we find vtp = - Es = BovoS. Therefore
F = BoloS.
Since we have assumed that the resistance of the wires is negligible,
V Be = V PQ = VP, the \vork W is also the work done, per second,
in maintaining the current through the resistive rod BC. An equivalent
amount of heat appears in the rod.
rfhus mechanical energy is converted into electrical energy, trans-
ferred to a distant resistor BC, and there converted into heat. To trace
the transfer of energy in greater detail we shall consider a simple field
configuration as illustrated in Figure 3.1. The figure shows a longi-
tudinal section of width w of a pair of coaxial cylinders of nearly
equal radii, which locally are ahnost parallel planes. The field is
substantially uniform in every transverse cross section. When the

lncrgy storage, dissipation, and transfer
129
w
I
FIGURE 3.1 A strip of width 'Z£} of a pair of coaxial cylinders of large and nearly
equal radii.
current 10 is steady the Faraday-l\Jaxwelllaw yields
V PQ = V PB + V BC + V CQ .
IVf ultiplying by 1 0 , we have
VpQ/ o = VPBl o + VBcl o + VcQlo.
(3.1 )
( 3.2)
l"he first term on the right is work done per second by V PR driving
current 1o, that is., the power dissipated in the lower strip. 'fhe second
term is the power dissipated in the terminating strip, and the third
tcrm is the power dissipated in the upper strip. The left side equals
the work done per second by the impressed voltage, driving current
1 o against the ficld voltage V PQ; that is, the power leaving the gen-
erator.
In the present case the magnetic intensity llJJ is constant through-
out the field. '"fhe longitudinal electric intensity Ez:. 1 at and in the
lower strip is also constant. Also Ez:. 2 is constant at and in the upper
strip. If the strips are identical Ez:. 2 is equal to - Ez:. 1 . rrhe transverse
voltage V(z) and transvcrse electric intensity Ez(z) vary with z.
Since
v PQ = hEz(O),
10 = u,/l JJ,
V PH = lEr,1
V QC = lE'.2
V BC = hE;&(l) J

130
lleclronlalletic fields
equation (3.2) becomes
hwE.(O)Il ll = IwEz.1H" - lwEZ:.2H" + lzwE;r:(l)H y .
l"'he left side suggests that power enters the flcld uniformly through-
out its cross section, and that p(Td.Jer jlow per unit a.rea in the z direction
.
IS
Ex(O)lly.
l"'hc flow is in the direction of the advance of a right-handed cork-
scrc,v ,\?hose handle is turned from E z to II Y' rrhe right sidc suggcsts
tha. t ptnver is lea'ing the space between the strips in accordance \vi th
the same rule. l"'hus the first term represents the down\vard flow intu
the lower strip; the second, the upward flow in to the upper strip;
and the third, the for\vard flo\v into the resistor Be.. 1"'hus, the po\vcr
fio\v per unit area may be represented by a vector, called the Po)'nting
'vector,
j5 = E X II.
( 3.3)
3.2 Distribution of magnetic energy
Uniform speed can not be attained instantaneously. An infinite force
,vould be required. f\S the speed v(t) of the sliding \vire in Figure 2.33
increases from zero to Va, the curren t I (t) increases from zero to 10.
Sin1ilarly the currcnt in Figure 3.1 must increase gradually from zero
to 10. Equation (3.1) does not describe this transition period. 1"'he
equation that does describe this period contains the term acf.>/at
representing the magnetic current linked with the closed path
J)BCQP. In accordance with the Faraday-lYfaxwelllaw, we add the
vol tage a <I> I at to the right side of cq ua tion (3.2). The corresponding
increase in the input voltage is
, a<l> ally (t)
V PQ = =}.I.hl. (3.4)
at at
In equation (3.2) direct current 10 should be replaced by I (t) and
there \\"ill be additional po\ver en tering the field,
, aII,,(t)
V J)Q/ (t) = }.I.hl u,H y(t).
at
Additional energy entering the field during thc transition interval
(0, T) is
f T V'QI(t) dt = JJ./rlwfT llu(t) allu(t) dt = JJ.hlwI1;(T) (3.5)
o 0 at

Energy storage, dissipation, and transfer
131
since 11 11 (0) == O. Thereafter energy enters at the rate given by
equation (3.2) and we already know \vhat happens to it. The pre-
ceding equation suggests that the extra energy that entered the
field remains there and is associ a ted with the magnetic field. If the
current were permitted to decrease back to zero, the magnetic field
\vould disappear. During this period, the decreasing magnetic flux
would give risc to an input voltage in a direction opposite to that of
V;>Q and energy \vould flow back to the generator.
Equation (3.5) indicates that energy is distributed throughout
the magnetic field and that the amount stored per unit volume is
.J.l.I12.
( 3.6)
In terms of electric current, equation (3.4) becomes
I J.l.hl al(t) aI(t)
V PQ = = Ll
w at at '
where L is the inductance per unit length of the strip transmission
line and Ll is the total inductance of the loop P BCQ. Magnetic
energy of the entire field. equation (3.5) can then be expressed as
em = .LlI.
(3.7)
3.3 Distribution of electric energy
In the preceding section we assumed that during the transition period
the current is independent of z as it actually is thereafter. But during
this period the transverse voltage is rising and there is a transverse
displacement current bet,veen the strips anel the corresponding
charging currents in the strips [see equation (2.102) J.'fhc input
current equals I (t), as assumed in the preceding section, plus the
total charging current. l"'he energy entering the field ,viII be greater
than that calculated in the preceding section. 'rhe extra amount will
be stored in the electric field between the strips.
'fo remove unnecessary complications, let us assume that at the
far endBC in Figure 3.1 there is no connection bet\veen the strips,
so that only the charging current exists. 'fhe density of charge on
the lo,ver strip is [) J; - EE z . and the total charge EwlExo Hence the
charging curren t is
I (t)
aE;z
- Eu'l-
at '

132
Electro"nlagnetic fields
and the energy en tering the space bet\veen the strips
i T V I (t) dt = iT f;wlhE z aE z dt = !f;wllzE;( T) (3.8)
o 0 at
since Ez = 0 at t = O. Thus, the energy stored per unit volume of the
electric field is
EE2.
(3.9)
In terms of the voltage V between the strips, the stored electric
energy after completion of the transition period is
1 wl
e = - - V 2 = JClV 2
e 2 Jz 2,
(3.10)
where C is the capacitance per unit length of the strip transmission
line and Cl is the total capacitance.
\Ve have considered simple fields in order to focus attention on
essential factors. Expressions (3.3), (3.6), and (3.9) are general
and can be deduced from the analysis of fields which are either be-
tween coaxial cylinders of arbitrary radii or outside parallel wires.
In most situations there arc complicating factors which are best
ignored until the essen tials are understood.
3.4 Oscillations in a cavity
Figure 3.2 (a) represents an axial section of a metal cylinder \vith a
coaxial plunger. Suppose that ,ve somehow displace a charge q
from the lower face of the cavity to the bottom of the plunger and
then let the charge go. Alternatively suppose that the cavity consists
of the two sections, one of which is a circular plate capacitor as shown
in Figure 3.2 (b). The capacitor can be charged and then the rest of
the cavity can be slipped over it. 'rhe opposite charges will tend to
unite and current I = -tj will flow in the walls of the cavity. At the
moment the positive charge q has left completely the bottom face
of the plunger and neutralized the negative charge on the bottom
of the cavity, the current continues to flow and the positive charge
starts accumulating on the bottom of the cavity. The negative charge
will be accumulating on the bottom face of the plunger. This accumu-
la tion stops \vhen the electric field of the displaced charge stops the
current and then reverses its direction. Oscillations continue until the
energy of the electromagnetic field inside the cavity is dissipated as
heat in the walls. The equation for such free or natural oscillations
can be obtained as follows.

Energy storage, dissipation, and transfer
133
z
(b)
FIGURE 3.2 A cylindrical cavity: (a) with a coaxial cylindrical plunger: (b)
the Sal'11e with the bottnnz of the plunger and the opposite section of the cavity
shown separatel-y to de1110nstrate that they constitute a circular plate capacitor.
To begin with there is an electric field between the bottom of the
cavity and the bottom face of the plunger. If we assume that the
charge is distributed uniformly, displacement density is D z = -q/rra 2
and the electric intensity is Ez: = -q/E7ra 2 . The stored electric energy
.
IS
Ij'j2rja q2
e 4J = - fE;p dp dcp dz = -,
2 0 0 0 2C,
(3.11 )

134
1lcr/r()111aJ!.lleti( jirlds
\vhere
E7ra 2
C t =-
S
(3.12 )
is the total capacitance of the field.
In reality the field is distributed nonuniforn1ly. l;rom the nature of
the above calculations, ho\vcver, it is clear that there is stored electrlc
encrKY proportlonal to the square of the charge. Only the value of the
coefficient C't may be somc\vhat different.
}\5 the charge starts moving, a magnetic field is created. If we
assume that I is constant in the torus, that is, if the displacement
currents from the lateral surface of the plunger to the bottom of the
cavity (indicated by dotted lines in :Figure 3.2) are neglected,
I -q
II", = - =-.
21rp 27rp
(3.13)
Then the stored n1agnetic energy \\Till be
1 l h 1 2r b
em = - f J.l.ll;p dp d<{) dz = Lll,
2 0 0 a
( 3.14)
where
p.h
Lt = In (b/a)
27r
is the total inductance of the torus. Again it should be noted that
equation (3.14) is exact for some value of the coefficient Lt even
though we may not be able to compute Lt exactly.
As the curren t flows through resistive ,valls, some energy will be
dissipated in heat. rrhe dissipated power is proportional to the square
of the curren t
(3.15)
w = Rtq2,
(3.16)
where Rt is the total resistance.
Equation (3.16) represents the time rate of decrease in the total
energy of the cavity
d (1L ej2 + q2 ) = _Rtq2.
dt 2 t 2C t
After differentiation and cancellation of rj, we obtain
(3.17)
Ltij + RlJ + .J.... = O.
C t
(3.18)

Energy storage, dissipation, and transfer
135
'fhis is a second-order linear differential equation with constant
coefficients which possesses exponential solutions of the form
q = Ae pt .
( 3.19)
Substituting in equation (3.18) and cancelling A exp (pt) , we have
L t C t p2 + RtCtp + 1 = O.
Solving, we obtain
R, .J?i 1
P1.2 = - 2L,  4Li - LtC t '
Hence the general solution is
q = A exp (PIt) + B exp (P 2 t).
The square root is real if
R, > 2 y1 LtjC,
in which case the discharge is nonoscillatory. In metal cavities, how-
ever, we have
(3.20)
(3.21)
R, « 2 y1 Lt/C,
and equation (3.20) becomes
Pl.2 = - -X jw,
(3.22)
(3.23)
vihere
Rt
 = 2L/
w=
1
yI LtC;
(3.24)
Thus
q = e- ft (Aejc.rt + Be-;wt).
Since q is essentially real, B must be the complex conjugate of A.
Let A = !(M + jN); then B = !(M - jN) and
q = e-ft(M cos wt - N sin wt). (3.25)
'This equation represents exponentially decaying oscillations of fre-
quency f = w/27r. From equations (3.12), (3.15), and (3.24) we find
1
w = . (3.26)
a yI (h/2s) In (b/a)
In free space
1
 r- = 2.998 X 1()8  3 X 10 m/sec.
v P.oEO
(3.27)

136
Electronlagllctic fields
Assuming
a = 2 em,
b = 4 cm,
h = 4 em,
s=lmm
,
we have approximately
w = 4 X 10 9 ,
f = 640 me/sec.
3.5 Damping constant
The quantity  defined in equations (3.24) and (3.25) is called the
damping constant and the ratio /f the logarithmic decrement. Both
quantities are measures of the rapidity with which the fields in the
cavity decrease and the rapidity with which their energy is dissipated.
When the cavity walls are good conductors, the natural frequency
of oscillations, w/27r, depends little on the dissipation of energy and
can thus be determined on the assumption that the walls are perfectly
cOI?ducting. This often makes it easier to calculate the frequency
and the field distribution within the cavity. From the field distribu-
tion it is possible to obtain the power dissipated in the walls of the
cavity since the tangential component of H gives the current per unit
length at right angles to itself. The damping constant is then de-
termined from the ratio of energy dissipated per second to the total
energy content.
Thus we reV-Trite equation (3.17) as follows:
de
- = -Way = -ke
dt '
(3.28)
where Way is the average dissipated power and
k = Way/S
is 'the fraction of total energy it rcpresents. Solving equation (3.28)
we have
S = Soe- kt ,
where So is the energy content at t = o. At some instant e is entirely
clectric and is proportional to the square of the amplitude of charge,
q. At other instants it is entirely magnetic and is proportional to q.
Therefore, the damping constan t for the field in tensities is
Way
t - lk - -
" - 2 - .
28
Let us apply this equation to the problem in Section 3.4. Neg-
( 3.29)

Energy storage, dissipation, and transfer
137
lecting dissipation in equation (3.18), we solve the equation and
obtain
q = ria sin wt,
where w is given in equations (3.24). The cosine term, appearing in
equation (3.25), need not be included since it would merely shift
the origin of time (irrelevant in the present case). The average
dissipa ted power is
W av = t-lRtl'/ [ sin 2 wt dt
o
= !t-lRlq [ (1 - cos 2wt) dt
c
(3.30)
lR .2 t
= '2 ,qa as  00.
At some ,instants the entire energy of the field is magnetic (when
sin wt = 1 and cos wt = 0). Hence
 lL '2
o = 2 ,qa.
(3.31)
On the average, half of the energy is electric and half is magnetic.
From equations (3.29), (3.30), and (3.31) we obtain
 = Rt/ 2 Lt
which agrees with equation (3.24).
3.6 Equivalent circuit for parallel wires shorted at the far end
Consider Figure 3.3(a) which shows a pair of parallel wires connected
to a genrator at one end and shorted at the other. Suppose that the
internal generator voltage is varying slowly. As a first approximation,
the current in the loop will be the same at all points and the input
voltage may be obtained from equation (2.103) if we let z = 0
al o
V(O, t) = Rllo + Ll-.
at
In this equation Rand L are the resistance and the inductance per
unit length of the pair and Rl and Ll are the equivalent" lumped"
resistance and inductance of the loop.
As the voltage varies faster, the transverse displacement currents
become significant and the input current will be larger than 10 by an
amount equal to the total displacement current (or the charging

138
Electro111agl1etic fields
current), equation (2.102). The input voltage will also be affected.
In the equivalent lumped circuit in Figure 3.3(b) we include a
capacitor in parallel \vith the equivalent resistor and inductor. The
value of the equivalent capacitance may be obtained from improved
Rt = Rl
t  (O) -..c II C t =V3Cl
I 0 ....
It l l
(a) (b)
FIGURE 3.3 (a) A pair of parallel wires shorted at the right end; (b) its equiva-
lent circuit when the voltage impressed at the left end is varying slowly.
voltage and current distributions; but the quickest and most effective
method is to calculate the energy of the electric field. Referring again
to equation (2.103), we observe that in the present notation,
V(z, t) = (RIo + L a:e O ) (l - z) = V(O, t) (1 - ).
The voltage is distributed linearly with z. From equation (3.10) we
can obtain the stored electric energy per unit length. Therefore the
total energy in the presen t case is
8. = C[V(O, t) J2 { (1 - y dz.
Hence
Be = iCl[V(O, t) J2 = Ct[V(O, t) J2,
where C t is the equivalent lumped capacitance (total capacitance).
This capacitance is smaller than Cl because the voltage is not dis-
tributed uniformly along the wires.
If the loop is suddenly disconnected from the generator, the
current away from the open ends will continue to flow until the
charges accumulated near the ends stop its flow and then reverse its
direction of flow. Oscillations will ensue. The natural frequency of

Incrgy storage, dissipation, and transfer
139
oscillations may be obtained from equations (3.24), (2.93), (2.97),
and (3.27)
Y3 ,,'3 v3 X 3 X 10 8
w- - -
- lV JJG - l - l
(3.32)
1'his frequency is so high that the application of the equivalent
lumped circuit idea might be seriously questioned. In the next chapter
the exact value
7r
w=
2l
will be obtained. On comparison, we find that the approximate
value is only 10 per cent higher than the exact value. This i5 very
encouraging since exact solutions are available only for simple geo-
metric configurations. No exact solution is available for the cylindrical
cavity \vith a coaxial plunger (see Figure 3.2) ; but \vith some rela-
tively simple improvement in the method described in Section 3.4
the natural frequency can be calculated \vith an error of only a frac-
tion of one per cen t.
3.7 Equivalent circuit for parallel wires open at the far end
Another configuration is sho\\'n in Figure 3.4. First there is a ca-
pacitance Cl bet\\'een the two wires. rrhe charging current is a linear
R t =  Rl, L =.!.LI
t 3
0 V\I\,  1
t  (O)  C =Cl
'0
t T
.
I I
, + I ;-1
0
(a)
(b)
FIGURE 3.4 (a) A pair of parallel 'wires open at the riJ!,ht el1d; (b) its equivalent
circuit 'It'hen the t'oltaKe irnpressed a.t the left end is t1ar'yin£ slou'ly.

140
Electro111aglletic fields
function of the distance z from the generator (assuming that the
capacitance per unit length is constant, that is, the wires are of con-
stant radius)
l(z) 10 (1 - i).
Both the magnetic energy and dissipated po\ver are proportional
to the square of the current and the results indicated in Figure 3.4(b)
follow.
3.8 Equivalent circuit for a parallel plate capacitor
Suppose that a voltage Va is applied uniformly bct\veen the edges of
two parallel circular plates of radius a, Figure 3.5. If q is the total
z
t
I
J(p) J(p) ++ t V o
 ....
a I
r-
I
FIGURE 3.5 One half of a circular Plate capac£tor with a voltage V o impressed
uniformly round the edges of the capacitor.
charge on the 10Ylcr plate, the charge density is q/'Tra 2 . l"his is equal
to the displacement density Dz:. Hence Va = hEz: = qh/E'Tra 2 and the
equivalent lumped capacitance is
q E7ra 2
C, = = -. (3.33)
V o It
1'hc charge on the Io\ver plate \\'i thin the circle of radius p is qp2 / a 2
and the charging current at distance p from the axis is
I (p) = qp2ja2.

Inergy storage, dissipation, and transfer
141
This is also the upward displacement current enclosed by the cylinder
of radius p. Hence
I(p) qp
Hip = - --
27rp 27ra 2 .
(3.34)
The stored magnetic energy is
1 fh f 2rfa p.1z
em = - JJ,Il;p dp dcp dz = - q2 = ! Ltq2.
2 0 0 0 167r'
(3.35)
Thus the equivalent lumped inductance in series with the circular
pIa tc ca paci tor is
Lt = J.Lh/87r.
(3.36)
3.9 Equivalent circuit for a slotted toroidal conductor
Let a toroidal conductor be open at p = a and so connected to a
generator that current 10 enters uniformly round the lower edge
z
hI
4
fo
I
r
b

FIGURE 3.6 One half of a toroidal conductor.
p = a and leaves round the upper edge as shown in Figure 3.6. Here
10
Hip = --, a < p < b (3.37)
27rp
and
8m = !LtI5,

142
1leclroI11aglle/ ic fields
\\hcre
p-oh
Lt == - In (b/a)
27r
(3.38)
is the total inductance of the toroid.
-\pplying the Faraday-l\fax\\?cll la\v to a rectangle .A1BCD in a
t)})ical radial plane and neglecting the resistance, 'vc have
f hjb aBf{) J.Lojoh b
V(p) = l"AD = - - dp dz = - In -. (3.39)
o p at 27r p
Hence
J.LJo b
E == -In 
z 2 '
7r P
and the stored electric energy is
f h f 2W' f b Eoll ·
Be =  EO E; p d p d'P dz = (J.LoI 0) 2 p ,
o 0 a 47r
(3.40)
,vhere the integral P may be evaluated by parts
h (b)'l 1 b ( b)2
P f p In - - dp = j In - d (p2)
a p 2 a P
( b)2 b b b
= p2 In - + f p In dp.
p a a P
After another integration by parts, we obtain
b (b)2
P = l(b 2 - a,2) - a2In- - a2 In- .
a (l
(3.41)
Substituting this in equation (3.40) and equating to
Be = !C t [V(a)J2
\vhich defines the equivalent lumped capacitance, ,ve find
27rE O P
C -
t - h [In (b/a) J2'
(3.42)
(3.43)
3.10 Use of equivalent circuits
In Section 3.6 \\1e found the frequency of free oscillations on a pair
of \vires shorted at one end fronl energy considerations. 'fhe magnetic
energy ,vas calculated on the assumption that the magnetic field was

Energy storage, dissipation, and transfer
143
uniformly distributed as it is when the current is time-invariant.
From this field distribution we then obtained the distribution of the
electric field and the corresponding electric energy.
Charge density on the lower wire, being proportional to transverse
voltage, varies as (l - z). Hence the charging current varies as
(l - Z)2. The magnetic field produced by this current was neglected.
The approximate frequency was found to differ from the exact by
10 per cent.
L . : II
I
.
I.. 112 II.. 1/2 .1
(a)
Ll !Ll
6

......
!.Cl lCi ...J
C 1 C 2 ..-41C""1
2 2 II
N
..J
(b)
FIGURE 3.7 (a) A transmission line of length l, open at one end and shorted
at the other; (b) the calculation its equivalent circuit for obtaining an improved
value of the line's lowest natural frequency.
We now obtain an improved value for the natural frequency.
Note Figure 3.7(a) which shows a pair of wires of length l divided
into two sections, each of length l12. In Figure 3.7 (b) each section
is represented by an appropriate equivalent circuit. The natural
frequencies of the combined network are obtained by equating the
sum of the admittan.ces (or the impedances) in each direction to
zero. Thus
jwC 1 1 - W2C2
+ =0
1 - w 2 L1C 1 jw
(3.44)
and finally
(w 2 LCl2)2 - 60(w 2 LCl2) + 144 = O.

144
1le(lr()nl(ll1rli( fields
Solving. we obtain
1.58
WI = . ,
l
7.58
w,) =
- l.y;;;;'
(3.45)
In equation (3.32) the factor v3 = 1.732 \vas compared \vith the
exact value 1r/2  1.57. 'fhe error in the new value, 1.58, is smaller
than one per cen t.
'[he new equivalent net\vork also has a higher natural frequency,
the significance of which will be discussed in the next section.
3.11 Higher modes of oscillation
In Section 3.6 it was found that slowly varying current in parallel
wires connected to a generator at one end and shorted at the other
is distributed nearly uniformly along the \\Tires. Hence, the magnetic
energy is also distributed nearly uniformly along the entire length.
On the other hand, the transverse voltage vanishes at the shorted
end and increases linearly with the distance from the shorted end.
'I'hcreforc, the electric energy increases with the square of the distance
from the shorted end. Similarly in Section 3.7 it was shown that when
wires are open at the far end, electric energy is distributed nearly
uniformly along the \vires while magnetic energy increases as the
square of the distance from the open end. In the analysis of electric
oscillations on a pair of parallel \vires shorted at one end, it ,vas
natural to su bdivide the distributed field in to t\\TO regions; one, near
the open end, where the electric field is relatively strong and the
magnetic field is \\Teak, and the other near the shorted end where the
magnetic field is strong and the electric field relatively weak. During
the oscillations, energy fluctuates bet\\Tecn these regions and is con-
verted from one type to another.
If the fields are varying rapidly, all that can be said is: in the
inl1nediate vicinity of the open end the magnetic field is weak and in
the int1nediate vicinity of the shorted end the electric field is \veak.
Nothing can be said about the intermediate region. In fact, we know
that electric and magnetic energies are continuously distributed
along the wires and a possibility exists that oscillations may take place
bet\veen adjacent sections of the parallel pair. Starting with oscilla-
tions on a shorted pair as sho\vn in Figure 3.8(a) and assuming that
t\VO such pairs are placed back to back, one arrives at a mode of
oscillation in a pair open at both ends, as shown in Figure 3.8 (b).
Since for the assumed directions of curren ts in the \vire there is no

Energy storage. dissipation. and transfl"r
145
---
++

+ +
----""""'-
---...
..
(a)
(c)
7,
-- ..... ' -...... ++
+ +  --------=-
 -
, /
 - - , +" -<"-
+  '--=..'" +
.
(b)
(d)
FIGURE 3.8 lIiglzer 1nodes of osct'llatioll in sections of transnzissio1l lines.
current in the shorting bar, the bar can be removed \vithout dis-
turbing the fields. l"or the san1e length of \vires the frequency of
oscilla tions in this mode is t\\rice as high as in the shorted pair.
'rhis argument leads to a sequence of other possible modes of
oscillation \vith increasingly higher frequencies such as those illu-
strated in Figures 3.8(c <1). It is nu\v clear that the n1cthods de-
scribed in the preceding sections for calculating natural frequencies
of electric oscillations arc restricted, in their present fornz, to the 1110des
'Lilith thc I07i)est frequcncies. j/ronz thc start, 7.oe assu1Jzcd that the en tire
field could be divided into only two regions. one \vith a strong electric
ticld and the other \vith a strong magnetic field. Although the equiva-
lent circuit in I"igurc 3.7 has a. mode of oscillation \vith a higher
frequency. \\'e cannot expect this frequency to be a good approxin1a-
tion to the next mode of oscillation in the shorted pair sho\vn in
J'igurc 3.g (c), since the actual field distribution for this nl0dc docs
not at all rcsenlblc the assumed distribution. In fact, the frequency
of this mode is three times as large as that of the lo\vest nl0dc \vhile in
equation 3.45) \ve have W2  4.Rwl.
"fhis docs not mean that our Dlcthod cannot be modified so that it
\vill beconlc applicable to higher modes. \\' c \vould only need to find
a \\'ay of calculating proper values of equivalent lumped inductances
and capacitances.

146
JI{'(lron1(lf!.l1{'/i( fields
3.12 Comparison of strengths of electric and magnetic fields
}\/lax\vcll'g equations in1ply that tinlc-variablc fields are ncver purely
electric or purely magnetic. \Vhere they arc "varying slo\vly," such
fields may be only "primarily" electric or "primarily" magnetic.
\\'c have already considered a number of examples. \Vc no\v give a
more precise meaning to the words "prinlarily electric," "primarily
nlagnctic," and" slowly varying."
'rhe physical dimensions of the E and jj arc different. Hence we
cannot compare thc magnitudes of 11 and II directly in order to decide
\vhich is the larger. Iut we can compare stored energy densities and
total stored energies. If 8e is the electric energy stored in a given
region of an electromagnetic field and 8m is the stored magnetic
energy. then in this reKion the filed is pril1za.rily electric or prinza.rily
l1zaKnetic depending on whether
8e »e m or 8m» 8 e .
Similarly at a given point the 1JZagnitlule E of electriciJltcnsity is
greater than, equal to, or smaller than the 1nagnitude II of1naKnetic
intensity depending upon whether
f:El > J.l.II2,
.f:E}. = J.l.II'!,
EE'! <J.l.II2.
1."hus one may consider clectric and magnetic fields equally strong if
E = 1]11,
1]= .
(3.46)
Since the magnetic flux dcnsity B equals J.l.11. this equation 111ay also
be \vri t ten as
E = cB
,
c = 1/.
(3.47)
'rhe quantity 1] has the physical dimensions of resistance and is
called the intrinsic impedance of the 11tediul1Z. [n frce space
710 = V J.l.o/EO = 376.7  1207r ohms.
The quantity c has the dimensions of velocity and is called thc
intrillslc velocity of the l1zediunz. In free space
Co = 1/ = 2.998 X 10 8  3 X 10 8 m/sec.
'rhese two quan tities playa very important role in the propagation of
electromagnctic waves.
In free space a magnetic field of intensity equal to one ampere per
nlcter is comparable in strength to an electric field of 377 volts per

Energy storage, dissipation, and transfer
147
meter. Similarly a magnetic field in which the flux density is one
gauss (10- 4 webcr/m2) is comparable in strength to an electric field
of 30000 volts/meter.
3.13 The meaning of "slowly varying field"
It is now possible to define more precisely a "slowly varying field."
Consider the example of two parallel wires of length l shorted at the
far end (Section 3.6). The stored magnetic energy is
em = !LlI2.
In terms of the voltage at the near end, the stored electric energy is
ee = iClV2.
If I = Ia sin wt, then V = Lli = wLlla cos wt. Therefore
em = ! LlI sin 2 wt = tLlI (1 - cos 2wt),
ee = iw2CL2l3I cos 2 wt = T\zw2CL2l3I(1 + cos 2wt).
The ratio of the average electric energy to thc average magnetic
.
energy IS
av (8 )
e = lw 2 LCl2. (3.48)
av (em)
As long as this ratio is considerably smaller than unity, the field is
primarily magnetic and is" varying slowly." In terms of the equivalen t
lumped parametcrs this condition is
w 2 L t C t « 1
or
w « l/ V LtC t .
Similarly from equations ( 3.33) and (3.36) we find that if
w « l/ V C t Lt = 2Y2/a,
(3.49)
the field between parallel mctal plates of radius a is primarily electric
and is varying slowly.
The smaller is the region occupied by a field, the greater is the
range of frequencies in which the field can be either primarily electric
or primarily magnetic. As the dimensions of physical capacitors and
coils become smaller, the structures become more nearly ideal circuit
elements.

148
llectronul1!.1Zelic firlds
3.14 Electric networks
f\ system of physical resistors. inductors. and capacitors connected
together constitute a physical electric net\vork. If the net\vork clc-
n1cnts arc \vell designed. they may be approxinlatcd over a large
frequency range hy ideal resistors. inductors. and capacitors. The
p 0 lV AI L K
Rt l.Jt
(7:l {i:\,
C t
0 
A C D E F G
FIGURE 3.9 A 11. electric llelu'ork with tu'o 'independent nleslzes.
voltages bet\vcen the terminals of ideal clements are rela ted to the
currents through tht. clements as follows (see Figure 3.9) :
dI.\! L dV BC
V () N = l t l 0 N . V J\J L = L t , I Be = Ct. ( 3 .50)
dt dt
where 1<t. Lt. and e't are the IUlnped resistance, inductance, and
capacitance of the respective elcnlen ts. It is understood that at
sufficiently high frequencies it \vill be necessary to replace a physical
resistor by an equivalcnt network consisting of an ideal inductor in
series \vith an ideal resistor. shunted by an ideal capacitor as 5ho\\'n
in Figure 3.10(a). 'fhe equivalent circuits for physical inductors and
capacitors arc sho\vn in Figure 3.10(b, c).
.i\pplying the Faraday-l\1ax\vcll la\v to a closed circuit in 'Figure
3.9. \\/e have
l/ AB + l'/JC + 1'(,]) + V DK + V KL + llLM + VllN
+ l'.vo + llop + V PA =
dcl>l
--
dt
lhc connecting leads arc usually good conductors and their resistances
may he neglected. '['hen
v uc + V DK + V LM + V NO + V PA -
d <1>1
dt
(3.51 )

Energy storage, dissipation, and transfer
149
Rt
Lt
C t
.
(a)
(b)
(c)
FI<;URE 3.10 (a) 7'he equivalent network for a physical resistor cOlltains an
ideal huluctor in series 'with an £deal resistor and an £deal capacitor ill shunt
'with both. (b) 1'he equivalent netu)ork for Q, physical inductor is allalaf!.ous /0
the resistor network but has dz:fferent orders of nUIJ!.lIitude of the constituent re-
sistances and inductances. (c) r'he equivalent llet,oork Jor II physical capacitor
C01lsists of an ideal capacitor in series 'with an ideal inductor and an ideal resistor.
'l'he magnetic flux is a linear function of thenzcslz currents II and 1 2
[see equation (2.85) J. rrhe coefficient IJll is the self inductance of the
first circuit and L J2 is the mutual inti uctancc of the t\VO circuits.
In practical circuits Lll and 112 arc usually negligible in comparison
with other inductances in the network. Equating the right-hand side
of equation (3.51) to zero, we obtain the first Kirchhoff equation.
The nlcsh currents in the nct\vork have a meaning only to the
extent to \vhich the displacement currents bct\veen the various leads
may be neglected. At sufficiently high frequencies it nlay be neces-
sary to include the cflcct of these displacement currents by intro-
ducing ideal capacitors bet\veen every pair of leads. Even these
approximations nlay he inadequate \vhen the frequency is high
enough.

4
Waves
4.0 Introduction
In this chapter \Vc shall develop basic concepts associated with
simple types of time-harmonic electromagnetic \va ves: characteristic
impedance (or, more generally, wave impedance), propagation
constant, phase constant, phase velocity, wavelength, attenuation
constant, and reflection coefficient for impedance discontinuities.
'fhe chapter is concluded with sections on nonuniform transmission
lines, image parameters, and a qualitative analysis of \\Taves in hollow
tubes.
4.1 Maxwell's laws of interaction between time-harmonic electric
and magnetic fields
Quantities varying sinusoidally \vith time, called time-harmonic
quantities, may be represented by complex quantities in the sense
that the actual quantities are real parts of the complex quantities.
If the real quantities are replaced by their complex representations in
any linear set of equations, then the real parts of the solutions of the
resIting equations are solutions of the original equations. Let
E(u v w)e iwt
" ,
II (u, v, w) e iwt
(4.1 )
be complex vectors representing time-harmonic electric and magnetic
tields of frequency f = w/27r. '[he coordinates (u, v, w) may be any
set of coordinates, usually orthogonal, such as the Cartesian set
(x, )', z), the cylindrical set (p, cp, z), or the spherical set (r, 0, cp).
Each component of E and II is a complex quantity whose amplitude
equals the amplitude of the corresponding sinusoidal quantity and
whose phase is the initial phase (the phase at t = 0) of that quantity.
Let us substitute expressions (4.1) into Iax\vell's equations (1.76)
and (1.77) for source-free regions. In the presen t case the differ-
150

\,r aves
151
entiation \vith respect to t is equivalent to multiplication by jw and
ei",t f E.(u, v, w) ds = -jWJ.l ei",t f II n(U, v, w) dS,
e;..t f II.(u, v, w) ds = (u + jWE) ei"'l f En(u, v, w) dS.
( 4.2)
If two complex quantities are equal, their real and imaginary parts
are separately equal. Thus
B('ll, v, w; t) = re E(1/" v, w) e iwt ,
( 4.3)
ll(u, v, w; t) = re II(u, v, w) e iCIJt ,
are solutions of l\1axwell's equations.
In equations (4.2) the exponential time factor may be canceled.
Dropping specific reference to coordinates, we have
f E. ds = -jWJ.l f II n dS,
( 4.4)
f II. ds = (u + jWE) f En dS.
4.2 Equations for time-harmonic fields in transmission lines
In Sections 2.18 and 2.19 we derived equations for transmission lines
consisting of two thin parallel wires and two thin coaxial cylindrical
conductors. The "thinness" was assumed to ensure uniform dis-
tribution of current in the conductors. The word" thin" was left
vague except in a qualitative sense: if the fields were varying more
rapidly, the conductors \vould have to be thinner for the equations
to remain valid. We shall now remove the restriction of "thinness"
and in deriving the equations study more carefully all the assump-
tions \ve make.
Figure 4.1 shows radial and transverse cross-sections of two
coaxial metal cylinders. On the left they are connected to a generator
which creates a radial voltage distributed uniformly around the
cylinders. Let V (z) be the transverse voltage at distance z from the
left end, and I (z) the current in the inner cylinder, which can be

152
l:lc(tronl{lKllet ic fields
p
z
p
l
26
FIGURE 4.1 Axial and tranSt'erse cross sections of two coaxial cylindrical
conductors.
either a solid or a hollow shell. At z = l the cylinders are terminated
by a plug which can be either a good conductor, so that the trans-
mission line is effectively shorted, or a thin resistive film deposited
uniformly on glass or some other nonconducting material. Let us
assume therefore that neither V (l) nor I (l) vanishes. l"heir ratio
will be determined by the impedance of the plug. Applying the first
of IVfax\vell's equations (4.4), to the closed path ABCDA, we have
l b
V AB + V BC + V CD + V DA = -jWJ1. f f Hop dp dz,
z a
\vhere
l
V AB = f E.(a, z) dz,
z
V Be = V(l),
I
V CD = - V DC = - f E.(b, z) dz,
z
V DA = -V(z).
In these equations a is the outer radius of the inner cylinder and
Ez(a, z) is the longitudinal electric intensity on its outer surface.
Similarly Ez(b, z) is the electric intensity on the inner surface p = b

\\.ave
153
of the outer conductor. }.{earranging the terms, \\'c obtain
I' (z)
I
l' (I) + J [E.(a., z)
z
I h
E.(b, z) ] dz + jw/J. f  ll., dp dz.
4.5)
If the dielectric between the cylinders is not perfect, there is a
radial conduction current as well as displacement current. If (J and E
are the cond uctivity and dielectric constan t, the radial densities of
these curren ts will be U E p and jWEE p . Hence
J lJ2r
I (z) = I (l) + ((J + jWf) Ep(p, z) pdq; dz,
z 0
(4.6)
where the integral represents the total radial current through a
cylindrical surface of radius p, coaxial '\vith the cylinders. This equa-
tion is always correct if p is equal to a. As written, it is correct only
\vhen the function
pEp(p, z) = A (z)
(4.7)
is independent of p. 1"his condition implies that radial current flows
straight frum one cylinder to the other, and that none of it is diverted
in the longitudinal direction bet\veen the cylinders.
If the cylinders are not perfectly cond ucting, there is always
a longitudinal electric field of intensity E z and longitudinal currents
of density (u + jWE) Ez exist in the medium bct\veen the cylinders.
Currents in the two cylinders are in opposite directions; therefore
Ez(a, z) and Ez(b, z) are in opposite directions. Hence. Ez(p, z) is
smaller bet\veen the cylinders than on their surfaces. How large are
these longitudinal currents in the dielectric in comparison with the
currents in the cylinders? In metals the conductivity U m is very large.
In copper, for example, U m = 5.8 X 10 7 . In dielectrics the con-
ductivity is very small. In pyrex glass, for instance, U = 10- 12 . Even
in such poor dielectrics as sand in \vhich (J = 0.002 the longitudinal
cond uction currents arc negligibly small. 'fhe radial curren ts cannot
be neglected as readily because they increase with the length of the
line. Similarly the ratio of the longitudinal displacement current
density bet\vecn the cylinders to the conduction density in the
cylinders isjwf./u m . For glass, E = 4 X 10- 11 . The frequency must be
very high indeed for this ratio to be comparable to unity. True, the
cross-section of the dielectric bet\vcen the cylinders is larger than the
cross-section of the conductors; but not large enough to permit

154
j.:I('ctro11ulj!.llrtir firlds
:-;ignitican t total longitudinal displacemen t curren ts a t frequencies
\vhich arc not exceedingly high. ']"hus \ve arc justitied in neglecting
these currents for the present. \\That happens \vhcn the longitudinal
currents become substantial \vill be considered later.
\V care no\v in a position to express the various terms on the right
in equations (4.5) and (4.6) in terms of V(z) and fez). 1'he quantity
Ez(a, z) is duubled if I (z) is doubled; that is, whatever its value is,
Ez(a, z) is proportional to I (z). Similarly Ez(b, z) is proportional to
- I (z). rrhus we set
Ez(a, z) = ZII (z),
Ez(b, z) = -Z 2 1 (z),
(4.8 )
where Zl and Z'I. are called the internal impedances, or the surface
impedances, per unit length of the corresponding conductors. At low
frequencies they are simply the resistances per unit length. In the
next section we shall derive the equations from which these quantities
may be determined.
Since we decided to neglect the longitudinal displacemen t curren ts,
the magnetomotive force around a magnetic line of radius p, 27rpH rp,
equals the current in the inner cylinder and
I(z)
llrp = -.
27r'P
Referring to equation (4.7), we have
( 4.9)
b
V(z) = f E p dp = A (z) In (b/a),
a
that is,
V(z)
pEp = In (b/a)'
(4.10)
Taking these results into consideration, vve transform equations
(4.5) and (4.6) into
l
V(z) = V(l) + f Zl(z) dz,
z
(4.11 )
l
l(z) - 1(1) + f YV(z) dz,
z

\\'aves
155
where
Z Zl + Z2 + jwL,
J1. b
IJ = In -
,
27r a
( 4.12)
27r(u + jWf.)
Y = G + jwC = .
In (b/a)
In these equations Z and Yare called, respectively, the series im-
pedance per unit length of the transmission line and the shunt ad-
mittance per unit length. In the present case they are independent of
z and may be taken outside the integral signs. We deliberately left
them under the integral signs since the derivation of equations
(4.11) remains valid to the exten,t that longitudinal displace1nent cur-
rents are negligible, even when a, b, and the thicknesses of the cylinders
are functions of z.
One should note that equations (4.11) are equations in the large
for fields bet\veen the cylinders, from which E and II at various
points may be obtained with the aid of equations (4.9) and (4.10).
Equations for parallel wires and other transmission lines are also
of the form of equations (4.11). Only the parameters Z and Yare
different. The real part of Z is called the resistance per unit length
and the imaginary part the reacta'1tce per unit length. The real and
imaginary parts of Yare, respectively, the conductance and the
susceptance per unit length.
4.3 Field in the interior of a metal cylinder
When w = 0, the longitudinal current in a metal cylinder, such as the
inner cylinder of the coaxial line in Figure 4.1, is distributed uni-
formly. If w is not equal to zero, the magnetic current linked with a
closed path, such as A'M'MAA', will generate a circulatory electro-
motive force which "rill make V A' /.1' unequal to V A Itl and will upset
the uniform distribution. 1"he radial current density must be con-
tinuous across the outer surface of the conductor. Just outside, this
density is (0" + jWf.) Ep(a + 0, z), where" a + 0" indicates" just
outside." Just inside it is umEp(a - 0, z). Therefore, Ep(a - 0, z)
is an exceedingly small fraction of Ep(a + 0, z). Thus in the 11xetal
cylinder \ve may neglect the radial electric field.
Let Ez(p, z) be the longitudinal electric intensity at distance p
from the axis, and I (p, z) be the longitudinal current enclosed by the
cylindrical surface of radius p. Applying the Faraday-l\Iaxwell law

156
llcc/r()nlll?llcl£c ficlds
to rectangular path ./1' [' M A A' in a radial plane as ShO\\TI1 in };igure
4.1, \Vc have
J Z+' JZ+' jw JZ+' Ja
Ez(p, z) dz - Ez(a. z) dz = -- p-l/(p, z) dp dz.
z , 211" z p
The current I(p, z) is a slo\\rly varying function of z since its varia-
tion depends on the weak radial current in the dielectric medium
between the cylinders. Thus, even \\rhen S is large compared with 2a,
the above equation yields
sE,(p, z) - sEz(a, z)
jWJJ Ja
- -- s p-l/(p, z) dp
21r p
or
Ez(p, z)
jWJJ Ja
- Ez(a, z) - - p-l/ (p, z) dp.
21r p
(4.13)
Also
[(p, z) = fr [ umE.(p, z)p dp dl()
o 0
= 21ru m [ pE.(p, z) dp. (4.14)
o
Note the similarity between these equations and the transmission
line equations (4.11). In fact, equations (4.13) and (4.14) are field
transmission cquations in the radial direction. In the next scction we
shall obtain the solution of thesc equations for low frequencies, and
latcr in this chapter for high frequencies.
4.4 Step-by-step solution of transmission equations
Let us now apply the step-by-step method of dealing with l\1ax\\rell's
equations for interaction betwecn elcctric and magnetic fields to the
special case represented by equation (4.11). We let
V(z) = Vo(z) + V 1 (z) + V 2 (z) + ...,
( 4.15)
/(z) = /o(z) + /1(Z) + /2(Z) +
\\rhere
Vo(z) = V (l), /o(z) = I (l),
l
Vn(z) = J ZI n - 1 (z) dz,
z
l
In(z) = J YVn-l(z)dz.
z

\Vavcs
157
If Z and Yare independent of z
V1(z) = Z(l - z) I(l), I1(z) == Y(l - z) Vel)
V 2 (z) = ZY(l - z)2V(l), 1 2 (z) = !YZ(l - z)2l(l)
1
V 3 (z) == - Z2Y{l - z)3l(l)
2.3
1
1 3 (z) = - ZY2(l - z)3V(l).
2.3
(4.16)
Successive terms appear more symmetrical if we let
VZY = r,
V Z/Y = K
Z == Kr
,
Y = K-1r.
( 4.17)
(4.18)
so that
In equations (4.16) we substitute r 2 for each product ZY, and use
equations (4.18) for the remaining Z or Y. Then
V (z) = V (l) [1 + -.!.. f2(l - z)2 + -.!.. f4(l - Z)4 + ...J
2! 4!
+ K I (l) [f (l - z) + -.!.. fa (l - z) a + -.!.. f5 (l - z) 5 + ... J
3! 5!
or
V(z) =V(l) cosh r(l - z) + KI(l) sinh r(l - z).
Similarly
fez) = l(l) cosh r(Z - z) + K-IV(l) sinh r(Z - z).
V(z) and l(z) may be expressed in terms of the input voltage
V (0) and the terminal impedance
Zt = V(l)/l(l)
(4.19)
and in many other ways suitable for each particular occasion.
In the interior of a metal cylinder the parameters Z and Y depend
on the variable of integration. Nevertheless the successive integrations
can be carried out and the solutions may be expressed as infinite
series which can be recognized as those for certain Bessel functions.
We now evaluate a few terms in order to obtain some idea of what
happens to the resistance of a wire at low frequencies. From equation
(4.13) we have the initial approximation
Ez(p, z) = Ez(a, z)

158
Elcc/ronla?llelic fields
\vhich gives the uniform current distribution. Substituting in equa-
tion (4.14). we obtain
I (p, z) = 7rU m P2 Ez(a, z).
Returning to equation (4.13), we find
Ez(p, z) = [1 - tjwJJ.u m (a 2 - p2) ]Ez(a, z).
Continuing, one obtains
I (p, z) = [ 7rUmp 2 - 17rjwJJ.up2(a2 - !p2) ]Ez(a, z).
If we stop with these approximations, the internal impedance per
unit length, equation (4.8) becomes
Ez(a, z) 1
Zl= =
lea, z) 1ru m a 2 (1 - ljwJ1.u m a 2 )
If the frequency is such that wJ1.u m a 2 « 8, ,ve have
1 . 1 jwJJ.
Zl = (1 + ijwJJ.u m a 2 ) = + -.
7rU ma2 7rlT ma2 81r
( 4.20)
rrhe first term is the dc resistance per unit length of the cylinder.
rrhe second term indicates that the conductor has internal inductance,
J,L/81r. This can be found from energy considerations as readily as the
internal inductance of a capacitor, equation (3.36), \vas found.
At low frequencies they arc the same, since it does not matter whether
the magnetic field is produced by displacement current or conduc-
tion current. In both instances we start with uniform current dis-
tribu tions.
To this order of approximation in equation (4.20) the resistance
is not affected by the frequency. From the next approximation to the
curren t \ve have
s
ZI 1 = 7ru m a 2 (1 - ijwp,u m a 2 - 7\-w2J1.2ua4).
Inverting and calculating ,the resistive component to the vorder of
this approximation, \ve obtain
1
Rl = (1 + rl"2"w2p,2ua4).
7ru m a 2
For copper
Rl  Ro(1 + 1100f 2 a 4 ).
If a = 1 em = 10- 2 m and f = 100 cps, the increase in resistance is
11 per cent. For higher frequencies, one could hardly use the formula

\\'avcs
159
since the correction term would be too large and higher order cor-
rections would be needed. If a = 1 mm, the same relative increase
in resistance would occur when! = 10 000.
4.5 Equivalent circuits
To focus our attention on a section (Zl, Z2) of the transmission line,
let Z = Zl and 1 = Z2 in equations (4.11). In the in tegrals we replace
fez) and V(z) by their mean values in the interval (Zl, Z2). We then
have
v (Zl) - V (Z2) = Z (Z2 - Zl)[ mean,
( 4.21)
I (Zl) - I (Z2) = Y (Z2 - Zl) V mean.
Let us assume that Z2 - Zl is so small that the voltage difference on
the right side of the first equation is small compared to V mean, and
the current difference on the right side of the second equation is small
compared to I mean. This automatically excludes sections which are
Z(Z2 - Zl)
111(

t 2 t
Y (Z2 - Zl)
V (Zl) V (Z2)
 
I (Zl) I (Z2)
(a)

 Z (Z2- Zl)
t
Y (Z2 1 _ Zl) 1
V (Z2)
V (Zl)


I (Zl)
I (Z2)
(b)
FIGURE 4.2 Equivalent networks for short sections of transmission
lines: (a) a n network; (b) a T network.

160
llecironlClglleiic fields
either shorted or open at z = Z2. With this restriction the difference in
transverse voltages across the ends of the section equals the voltage
drop across a lumped impedance Z(Z2 - Zl). The difference in currents
equals the current through a lumped shunt impedance I/Y(z2 - Zl).
These lumped impedances can be divided symmetrically as in Figure
4.2 (a, b). In case (a) we have a "IT network" and in case (b) a" T
network." Thus, transmission lines may be represented by a sequence
of either IT or T networks.
Equations (4.21) are not good enough for obtaining equivalen t
circuits of shorted or open sections, unless, of course, we subdivide
them into still shorter sections and therefore have many II or T
networks in cascade. For a shorted section we use equations (4.16)
and, setting, V (l) = 0, we take
V(z) = V1(z) + V 3 (z) = Z(l - z)[1 + tZY(l - z)2J!(l)
I (z) = Io(z) + 1 2 (z) = [1 + !Z Y (l - z) 2J1 (l).
That is, we take the first two effective approximations for V(z) and
I (z). The input admittance of a shorted section of length l - z is
I(z) 1 + !ZY(l - z)
-
V(z) Z(l - z) [1 + lZY(l - Z)2J
1
= [1 + !ZY(l - z)2][1 - ZY(l - Z)2]
Z(l - z)
1
= [1 + lZY(l - Z)2]
Z(l - z)
1 1
= + 3Y(l - z)
Z(l - z)
as far as the quantities of the order of (l - z) are concerned. We have
already obtained this result by the energy method (,vhich is' simpler)
when Z is equal to jwL and Y is equal to jwC. The factor one-third
enters because the voltage vanishes at z = 1 and hence there is little
electric energy stored in the vicinity of z = l.
The case of an open section is similar.
4.6 Differential equations
Differentiating equations (4.11) with respect to z and noting that
under the integral signs z is a "dummy variable," which can be re-

".aves
161
placed by u, let us say, we obtain
dV
= -Zl
dz '
df
dz
-YV
,
(4.22)
where Z and Y may, in general, be functions of z.
The conditions at the terminal end of the line have disappeared
and must be included in the calculation of the arbitrary constants
which appear in the general solutions of these differential equations.
4.7 Characteristic impedance
If Z and Yare independent of z, equations (4.22) may be solved
by the usual methods. There is a certain important property, however,
which can best be highlighted by posing the following question: Is
there a terminal impedance such that the ratio of V(z) to I(z) is
independent of z, thus causing the input impedance to be equal to
the terminal inlpedance independen tly of the length of the line?
One way to anSVler this question is to assume
V(z) = KI(z)
( 4.23)
and see what happens. Since the K we seek does not depend on z
dV(z) = KdI(z).
Equations (4.22) become
dl(z)
K dz - -Z f(z),
(4.24)
dI (z)
dz
- -KY f(z).
Dividing
K = Z/KY
or
K = y Z/Y,
( 4.25)
we have our answer: It is "yes" and we have the value of such a
terminal impedance. This impedance is called the characteristic
impedance of the transmission line. Later this concept will be gen-
eralized to apply to nonuniform transmission lines. In order that.K

162
Electro111aj!.l1ctic fields
be physically realizable the algebraic sign of the square root should be
chosen in accordance \vith the condition
re K > O.
(4.26)
For nondissipative transmission lines
K = v' L/C.
(4.27 )
4.8 Propagation constant
Substituting from equation (4.25) in (4.24) we have
dI
dz
-rI
,
( 4.28)
where
r = Z/K = vzy.
I = Ae-r.
( 4.29)
( 4.30)
Therefore
The normally complex quantity
r = a + j{3
(4.31 )
is called the propagation constant. Its real part a is the attenuation
constant and the imaginary part the phase constant. The names are
self-explanatory if equation (4.30) is written in an expanded form,
I = Ae-aze-j{Jz.
( 4.32)
Thus a is the relative rate of change in the amplitude per unit length,
and fj is the rate of change of the phase, also per unit length. For
nondissipative line, a = O.
From equation (4.23) we find
v = KA e-r.
( 4.33)
4.9 Phase velocity, wavelength
Assuming a non dissipative line and introducing the time factor in
the expression (4.32) for the current, we have
I e iCIJt = A ej{ClJt-fJ) .
1"he actual current, the real part of this expression, is a time-harmonic

\\iavc
163
function for a fixed z. Also it is a sinusoidal function of z at a given
instan t t. ']'1he phase
<I> == wt - {3z
will appear constant to an observer moving in the z direction with a
velocity
dz w
dt 13
( 4.34)
v = --
This is the phase velocity.
The distance from crest to crest is called the wavelength A. When z
increases by A, the phase decreases by 27r. Hence
{3A = 27r,
A = 27r/{3,
{3 = 27r/A.
( 4.35)
From equations in the preceding section we obtain
{j = wVLC,
1
v =
VLC'
27r 1
A - -
- wVLC - fVLC
( 4.36)
and
fA = v.
4.10 Transfer of power
The average power dissipated in the terminal resistance K (we are
still assuming a nondissipative line terminated by its characteristic
impedance) is
p = KI,
( 4.37)
where Ia, is the amplitude of the current through K. Since Ia, is inde-
pendent of z, this is also the power entering the line at z = 0, and
presumably carried by the wave to the terminal resistance.
4.11 Attenuation constant
When dealing with slightly dissipative transmission lines, considerable
simplification may often be achieved by neglecting dissipation to
begin with, solving the idealized problem, and obtaining the effect of
dissipation from the idealized solution. The methop is based on the
assumption that slight dissipation does not affect the fields locally
and is significA.nt only when we compare the strengths of fields at
two widely separated places in the line.

164
llr(/r()n1(lJ!.l1r/ if .ficlds
I.or instance, if R is the series resistance and C; the shunt con-
ductance. both per unit length of the line. the average po\ver dissipated
per unit length is
JfV RI + CV =  (R + K2G)I,
\vhcre fa and Va are the amplitudes of the current and voltage.
'fhe power }) carried by the wave is given by equation (4.37). Hence
the relative decrease in po\\rer, per unit length, is
lV R
= - + KG.
P K
Since power is proportional to the square of the amplitude of current
(or voltage), the relative rate of change of the amplitude of either is
half of this ratio
a=
W R
= - + JKG.
2P 2K 2
( 4.38)
4.12 Reflection
So far \\rc have considered waves in a line terminated into its char-
acteristic impedance K so that the ratio V (z) /1 (z) is independent of
z. If the terminal impedance
V(l)/I(l) = Zt  K,
the difference Zt - K gives rise to a backward or reflected wave.
By analogy with equation (4.30) for the current wave traveling in
the direction of in creasing z (the distance from the origin of the wave) ,
\\re obtain the following expression for the current wave traveling
back from z = l in the direction of increasing l - z (or decreasing z)
Ir = Be-r(l-z). (4.39)
The voltage of the reflected wave is obtained from equation (4.22)
by differentiation. l'hus
Vr = - KBe- fO - z ). (4.40)
The sum of the inciden t wave given by equations (4.30) and
(4.33) and the reflected \\rave is the total wave
I (z) = Ae- fz + Be- fO - z ),
V(z) = KAe- rz - KBe-f(L-z).

\Va\'cs
165
'fhe ratio 13/ i1 nlay be determined from the terminal condition
\l (I)
I (l)
Zt
.It erl - B
K .
Ae-rl + B
l"'hus
K - Zt
B == Ae- rl .
K + Zt
Note that A exp( - rl) is the complex amplitude of the incident
current wave at the lrnpedance discontinuity at point z == l.
Summarizing
I (z) = A [e- rz + ke-r1e-r(l-z)],
( 4.41)
v (z) = KA [e- rz - ke-rle-rCl-z)],
where
K - Zt
k =
K + Zt
(4.42)
is the reflection coefficient for the current \vave, and -k is the reflection
coefficien t for the voltage wave.
The constant A may be expressed either in tcrm of the input volt-
age or in terms of the input current.
4.13 Input impedance
From equation (4.41) we derive the input impedance
V (0) 1 - ke-- 2rl
Zi = /(0) = K 1 + ke- 2rl "
( 4.43)
4.14 Standing waves
Suppose that the line is nondissipative and shorted at z - t. Then
k = 1 and equations (4.41) may be transformed into
I ( z ) = A' cos {j (l - z),
( 4.44)
v (z ) = j K A' sin {j (l - z),
where A' - 2A cxp( -j{3l) is a new constant. At all points the

166
Eleclronzaglle/ic fields
current is either in phase or 180 0 out of phase, the amplitude is
sinusoidally distributed, and we have a standing wave pattern.
Similarly there are standing waves when the line is open at z = t.
4.15 Modes of oscillation
Suppose that the line is shorted at z
1(0) = 0 and A' = 0 unless
l and open at z - O. Then
cos I3l = 0,
{3nl = (n+ !) 'n',
( 4.45)
where n = 0,1,2, e e e. Since 13 = w VLC , this can happen only for
the following frequencies and corresponding wavelengths
W n =
(2n + 1)'n'
2lVTC '
4l
An = .
2n + 1
( 4.46)
The corresponding current and voltage distributions are
. . (211. + 1) 7rZ
I (z) ex: SIn {3nz = SIn ,
2l
( 4.47)
(2n + 1) 7rZ
V (z) cx: jK cos I3nZ = jK cos .
2l
"These equations express possible free oscillations, that is, oscilla-
tions which do not require continuous excitation. "rhey will arise if
we momentarily connect the line to a generator and then disconnect
it, or if ,ve touch the line some place with a charged body, or if we
merely move a charged body or a magnet past the line. In such a
case all of the 1nodes of oscillation corresponding to different integral
values of 11, will usually be excited. 1'0 excite a pure mode it is neces-
sary to obtain at some instant a charge or current distribution
conforming to equations (4.47) for the particular n of the desired
mode.
From equations (4.46) we find that the length of the section must
equal an odd number of quarter-wavelengths,
An
1 = (2n + 1) -.
4
( 4.48)

\\"aves
167
Current and voltage distributions for 11- = 0, 1 art shown in Figure
3.8(a, c) with obvious forms for higher values of 1t.
4.16 Propagation in highly dissipative media and skin effect
Equations (4.13) and (4.14) express radial propagation of fields in
the interior of a metal cylinder at a typical place along the line. E
and H", arc the only field components present and their relative direc-
tions are such that power is flowing in to the cylinder from the di-
electric medium between the cylinders. Thus their relative values are
independent of z and if one is concerned only with these, the z co-
ordinate may be ignored and a simpler notation, Ez(p), I(p), and
H",(p) may be used to express the dependence on p. Differentiating
these equations, we obtain
dE z (p) jwJJ.
= - I(p),
dp 27rp
dI(p)
= 27ru m pE z (p).
dp
( 4.49)
Suppose that our conductor is a thin cylindrical shell whose outer
and inner radii are a and a - Iz, respectively. Let us introduce a new
variable,
u = a - p,
which represents the distance from the outer surface of the shell,
in to our equa tions. thus obtaining
dEz(u)
du
JWJ.I.
= - [(u),
27r(a - u)
dl(u)
= - 21rO'm (a -tt) Ez(u).
dtt
Neglecting u in comparison with a, we have
dEz(lt) WJl
=j- [(u),
du 27ra
( 4.50)
dI (11,)
= - 27rau m E z ('lt).
du
Equations (4.49) and (4.50) are of the same form as equations
(4.22). In equations (4.50), however, Z and Yare constant and we

168
J/rrlr(}nlllJ!.nc/ic fields
can use the results already obtained
I (u) == ,:1 e rl + Be-I'(h-u),
Ez(u)
K,tc- rll - KIJe-r(h-u)
,
\vherc
1
K=-'i  '
27ra U m
r == yj WJlU m .
(4.51 )
Since /(11) = 0,
B = -..4e- rh
and
leu) = A (e- ru - e-2rh+ru)
( 4.52)
Ez(u) = K./l (e- ru + e-2rh+ru).
In particular, the internal impedance of the conductor is
Ez(O) 1 + e- 2rh
Zl = = K .
1(0) 1 - e- 2rh
For copper Jl = 47r X 10- 7 and Urn = 5.8 X 10 7 so that
r = yj w47r X 5.8 = (1 +j)27rvs:Bj.
(4.53 )
( 4.54)
'fhe attenuation constant is seen to be large even at moderate fre-
quencies so that the field decays rapidly with increasing distance
from the surface, as the frequency increases. The internal impedance
approaches its ultimate value
1 jWJl
Zl = K = - -.
271"a Urn
( 4.55)
Of course, the propagation constant is expressed in nepers per
meter. It is more instructive to express it in nepers per millimeter
r = 0.01510(1 + j).
If f = 10 000, then at the depth of one millimeter from the surface
the magnitude of the exponential terms in equation (4.53) is
C- 2ah = e- 3 . 02 = 0.048.
l"hus at high frequencies the field and the current in a conductor
are confined to a thin layer exposed to the field in the dielectric
medium. l'his phenomenon is called the skin effect and is a direct
consequence of the high attenuation of waves in conducting media.
.

\\'ave
169
4.17 Nonuniform transmission lines
If the transmission line is nondissipative bu t the parameters IJ and C
are functions of the distance along the line. the transmission equa-
tions (4.22) become
dl/ (z)
dz
- -jwL(z)I(z),
( 4.56)
dI (z)
= -jwC(z)V(z).
dz
Let us introduce a new independent variable, the Phase integral,
tJ = w jZ VL(z)C(z) dz
o
( 4.57)
so that
dfJ = w y L(z) C(z) dz.
If V, I, L, and C are expressed in terms of fJ, equations (4.56) be-
come
dV(fJ)
dfJ
- -jK(fJ)I({}),
( 4.58)
dI(fJ)
d{}
jV ({})
= -
K (fJ)'
where
K({}) = V L({})/C(fJ).
(4.59)
Introducing new dependent variables V(f}) and i({}) such that
V({}) = [K({})]lV({}) ,
I (fJ) = [K (fJ) J-j! (fJ) (4.60)
and substituting in equations (4.58), we find
dV ... K'...
- -J.[ - - V
d{} 2K '
di
- -
d{}
K' ...
-J.Y +-[
2K '
(4.61 )
where
K' = K'(fJ) -
dK({})
d{}

170
llc(tr(}n1(l?lletic fields
If K' « 2K, approximate solutions are
1 == A e-}" + Be i ",
v = A e- i " - Be i ".
At z = 0, f} o. It {J = () at z = l. We may express A and B in
terms of 1(0),11(8), and 8. ']'hus
A -[1 (8) + 11 (t l ) Je iO ,
B - [1 ()) - 17 (8) Je- i6 .
Substituting in the above equations and letting f} = 0, we have
expressions connecting V and 1 at one end of the line with V and 1
at the other end. l"hus
1(0) = 1(8) cos 8 + jJ!(8) sin 8"
V(O) = 11«(j) cos 8 + ji(8) sin 8.
From these and equations (4.60) we obtain the relations bet\veen the
voltages and curren ts at t \VO ends
 K(e) V(a)
/(0) = [/(e) cos e + j sin 8] 
K(O) K(e)
( 4.62)
V(O) = I K(O) [V(O) cos 0 + jK(O)I(O) sin 0].
'JK(O)
4.18 Image parameters
Equations (4.62) arc identical with equations for a two-pair trans-
ducer in terms of its inzage parameters
VI = KI (V 2 cos 0 + jK 2 I 2 sin 0).
K 2
( 4.63)
II = : (1 2 cos 0 + jK"2 I V 2 sin 0),
where Kl and K 2 arc il1lage i1npedallces and e is the image transfer
constant. If the transducer or the transmission line is terminated at
the second pair of terminals into its image impedance K 2 , the im-
pedance seen at the first pair of terminals is Kl. Also, if the. transducer
or the line is terminated at the first pair of terminals into its image

\\"a\"cs
171
impedance l\). the inlPcdance ccn at the second pair of lernlinals
is A..
If the transducer or the line is either shorted (l'2 0) or open
(I'!. = 0) at the second pair of terminals. the corresponding im-
pedances seen at the first pair arc
Similarly
Hence
Zl.,h = jK 1 tan fl,
Zl,oP = -jK 1 cot 8.
(4.64)
Z'}..,h = jK 2 tan A,
Z2.0p = -jK 2 cot 8.
( 4.65)
Kl = vlZ l"hZl.OP ' K 2 = vlZ 2. ah Z2,OP
tan e _ -jZI..h _ -jZ2..h = '_ ZI..h = ,= Z2..h .
KI K 2 "Z\,op V Z2.op (4.66)
4.19 Waves in hollow tubes
At high frequencies t\VO conductors are not essential for guiding
electric wavc. ()nr may suffice. Figure 4.3 depicts a hollow metal
I
/ II I
I /
I
+
1 D
2:lsh

a I
I
I
I
I
I
I
/
I
/
I
/
/
FIG URE 4.3 A rectangular waveguide with a wide ridge.
tube consisting of two parallel strips of width a, distance b apart,
and t\\'O cylinders, each of rectangular cross-section 5, connecting
the edges of the strips. Let V = bE, be the transverse voltage between
the strips and 1 = all t the longitudinal curren t in the lower strip.
In addition there will be lateral curren ts in the cylindrical walls and

172
llcctro'l1agnetic fields
longitudinal magnetic flux produced by these currents. Applying
the ];'araday-fax\vell la\v to a rectangular path A BC'DA . \vhcre
AB dz, \ve have
dll jwp,b
- - --I.
dz a.
(4.67)
The longi tudinal curren t I \viII decrease by an amoun t dI in dis-
tance dz due to a leakage in the form of the vertical displacement
current jWfaE t dz = (jwfajb) V dz and hunt leakage in the form of
conduction currren t. 1'10 obtain the latter \ve note that the electro-
motive force round the closed path .AQPD.t1 equals - V (if we assume
tha t the \valls of the tube arc perfectly conducting). On the other
hand, this must equal the magnetic current -jwCf> linked with the
path. l"'herefore. <P = V jjw, B = V jjwS, II = 1-' jJw}J.S and the
shunt conduction current II dz = V dzjjw}J.S. rrhere is an equal
leakage current via the left cylinder. Hence
dI (iWfa 2)
= - - + V.
dz b jw}J.S
( 4.68)
']'hese equations are approximate since \ve have assumed that the
transverse magnetic field Et, II t is uniform and is confined to the
region between the strips, that the longitudinal magnetic field is also
I /
/ /
/ /
/
/
/
I
/
I
/
I
bI
I
I..
a
2a
/
/
/
/
/
I
/
/
/
,I /
/ /
, /
/
,
.1
I
FIGURE 4.4 ,,1 'waveguide of rectanf,ular cross section.

\\'aves
173
uniform and confined to the cylindrical portions of the tube, and
that there is no electric field there. These approximations are ana-
logous to those we made in analyzing electric oscillations in cavities
and can be improved. The approximations are particularly rough .In
the case of a tube of rectangular cross-section as shown in Figure 4.4
where the middle portions of the top and bottom may be thought of
as ,: two parallel strips." In this case the exact solution is k.qown and
our intuitive thinking can be checked. Since, in this case, S =!a.b,
we have
dV
dz
jW}J.b
- --1
,
a
dI
( jwEa 4)
- - + V.
b jW}J.ab
( 4.69)
dz
The propagation constant is
r = I - W2JlE.
V'a 2
( 4.70)
When wVJ;e < 2/a, r is real so that V, I, and the field decay ex-
ponentially with increasing z (assuming that they are excited at
z = 0). When wVJ;e > 2/a, r is pure imaginary and waves may
travel in the tube. The critical frequency and the corresponding
wavelength are
2
We = aVJ;e'
Ac = 7ra.
( 4.71)
The exact value is Xc = 4a. For the tube in Figure 4.3 where b is
small compared with the total height, the results should be more
accurate.
..

I I
'.
1=0
/=0


1=/ max
FIGURE 4.5 The relative distribution of transverse displacement currents asso-
ciated with waves along parallel wires.
The case just considered is analogous to a two-conductor trans-
mission line shown in Figure 4.5 in that two portions of the tube act

174
JI('(lronl(Jf!.neli( firlds
as the forward and return conductors. In both instances \\.c have
transverse displacenlcnt currents. But at low frequencies in a hollow
tube thcse displacement currents are negligible in comparison \\'ith
the lateral conduction currents. and the "strip line" or center portion
is effectively short-circuited at its edges. As the frequency increases,
the shunt inductive reactance increases, the transverse conduction
currents decrease, and parallel strips become in effect separate con-
ductors.
Thcre is another possibility of wave propagation in which the
entire current in a hollow conductor returns as a longitudinal dis-
I
 D c
vt
-
v
..
I
FIGURE 4.6 Illustrlltinf!. the nature of possible waves in the interior of a circular
cylindrical conduc/inf!. tube.
placement current. See Figure 4.6. Since at low frequencies displace-
ment currents are feeble, this possibility can occur only at high
frequencies. In Chapter 6 we shall develop a method for solving such
problcms exactly, at least for simple shapes of tubes. It is instructive,
however, to obtain at least qualitative results from direct physical
considerations. Assuming a circularly symmetric field, we have
radial and longitudinal components of electric field E p and Ez.
:Iagnetic lines of force arc circles and we have only flip' If the tube is
perfectly conducting E z must vanish at its surface and so we assunl,e
E. = Eo (1 - :: ).
(4.72)
l"he actual distribution of E z is undoubtedly different. It would be a
remarkable accident if Yc were to guess it exactly; but here we are
concerned with qualitative aspects of a possible ficld. We now obtain

\\'aves
175
. .
In successIon
( _ p2 )
J z == j wEE o 1 l) ,
Q."
2r P (p4 )
/,(p) = j j J,pdpdcp ==jwE7rE o p2 - _,
o 0 2a 2
I = /,(a) = ljwE7ra 2 E o ,
21
Eo=
jWE7ra 2 '
I,(p) 1 . ( p3 )
IItp(p) = 27rp = 'i.)WEEo p - 2a 2 '
j a 3Jl
Btp dp = -hjWJla 2 EEo = - I.
o 87r
Hence, applying the Faraday-Maxwell law to a rectangle ABCDA
in a radial plane, where AB = dz, we have
dV _ -jw ja Btp dp _ Eo = _( 3 jw /J. + . 2 ) I. (4.73)
dz 0 87r }WE7ra 2
Here V = V AD is the transverse voltage from the axis to the peri-
phery of the tube.
The rate of change in the longitudinal current depends on the
radial displacement currents and must be proportional to the trans-
verse voltage
dI
dz
-jwCV.
(4.74)
There seems to be no simple way of estimating C without a better
understanding of what happens inside the tube.
The propagation constant is
r = /2C _ 3W2/J.C . (4.75)
'E7ra2 81r
At low frequencies r is real and there are no traveling waves. The
critical frequency, for which r = 0 and above which traveling waves
arc possible, is given by
4
wcV;;a = - = 2.31. (4.76)


176
1lcclro111aJ!.llclic fields
'fhe exact value (to three figures) is 2.40. "fhe critical wavelength is
comparable to the diameter of the tube.
Eviden tly, similar types of fields can exist bctween coaxial cylin-
ders. Indeed they arc required if the vol tage distribu tion impressed
at the input end of the coaxial linc docs not conform to thc radial
field we have been considering. Below a certain critical frequency
these fields are attenuatcd with increasing distance from the end
and represent just an "end-effect." '[he critical wavelength is of the
order of the distance betwecn the cylinders. l"hus, regardlcss of the
distribution of applied voltage, the propagation in coaxial pairs is
governed by equations (4.11) until very high frequencies are rcached
when t.hese equations apply only if we maintain a proper distribution
of the input voltage.

5
Spherical Waves
5.0 Introduction
In this chapter \VC shall examine some sjrnple types of spherical
waves, that is, waves traveling in all directions from the center of
their excitation. "fhe simplest problem mathematically is to find
the field associated with an electric current wave in a wire starting
from some point and going to infinity. l"he solution of this problem
may then be used for solving other problems, namely: problems
concerning waves between coaxial cones, cylinders, or parallel planes;
problems concerning waves guided by thin diverging cones or parallel
wires; and finally the problem of waves excited in free space by an
electric charge oscillating back and forth bct\veen t\VO nearby poin ts.
We have seen that electric generators do work \vhen exciting waves
guided by parallel \vires and that therefore \va ves carry energy
away from the generators to distant points. Similarly, spherical waves
in free space carry a\vay energy from their sources. From the la\\ of
conservation of energy we conclude that the density of this radiant
energy varies inversely as the square of the distance from the source
of these \vaves (the same energy must pass through every sphere
con cen tric with the source) .
1\5 in the case of \vaves along wires the phase of spherical waves is
retarded \vith increasing distance from the source. Hence, the waves
arriving at a given point in space from t\VO or more sources \vill
interfere either conf;tructivcly or destructively. "[his in terference
produces directile radiation.
5.1 Maxwell's equations for circularly symmetric fields
In Chapter 1 \ve derived expressions for electric and magnetic fields
generated by a semi-infinite direct current filament and a direct
current clement. "fhose expressions should be approximately correct
in the case of slowly varying curren ts since there is no reason to
expect any abrupt change in the nature of the fields. In Section 2.8
177

178
}f;/rc/ronlll£1lelic fir/lis
,\.c attenlpted to obtain the effect of the time-variable magnetic
field on the electric field of the clemen t and failed because \VC \\Jere
unable to apportion this efTect betwecn E, and E p . Integral equations
( 1.76) and (1.77) are too gross for analysis of the details of fields
unless we kno,v more about their general character from other
physical considerations. In order to calculate the fine-grained effects,
we need equations expressing local conditions in the vicinity of a
typical point. Such conditions are obtained by applying equations
(1.76) and (1.77) to infinitesimal mesh circuits formed by coordinate
z
jWEE r
(a)
z
H", t d,fJ",
y
y
c
jWEE,
(b)
FHiURE 5.1 Illustrating the derit'atlon of AI llxwell' s differential equations for
circularly s)'nln1elric fields fron1 Al Clxwell' s 'integral equations.
lines. General partial differential equations may be obtained in this
manner for any system of coordinates. We shall confine ourselves to
fields in which magnetic lines are circles with cen ters on the z axis
and clectric lines arc in planes passing through the z axis. 'rhus there
are only three componen ts to consider: II rp, Er, Es, each independen t
of 'P.
Consider a typical magnetic line surmoun ted by a spherical cap
of radius T, concentric with the origin, as shown in r"igure 5.1 (a). 'rhe
magnetomotive force round this line should equal the total radial
"

Spherical \\'a vcs
179
displacement current through the cap
j 21r e
27rr sin () II <p = j jWEE,.r2 sin 0 dO dl{!
o 0
= 27rjwEr 2 j9 Er sin 0 dO.
o
Canceling 27rr and differentiating with respect to 0, we have
a
- (sin () H tp) = jwer sin OE T . (5.1)
ao
Next we view Figure 5.1(b) which shows two Inagnetic lines on a
typical cone coaxial with the z axis and separated by distance dr.
The differential increment in the nlagnetomotive force should equal
the displacemen t curren t in the direction of decreasing (J through the
s trip bounded by these lines
a
- (27rr sin 0 II.,,) dr = -jwEEe(2rrr sin 0) dr,
ar
that is,
a
(r11 ",) = -jwe(r E(J).
ar
Finally, consider Figure 5.2 which illustrates an infinitesimal
(5.2)
z
D
y
:r
FIGURE 5.2 A differential circuit in a radial plane.

180
1lectro11lagllr/ic firlds
circuit ABC]) in a radial plane bounded by arcs of concentric circles
whose radii are rand r + dr. and by radii 0 and 0 + dO. 'rhe countcr-
clock\\'ise electromotive force round the curvilinear rectangle should
equal the magnetic current through the rectangle in the direction of
increasing cp
V AR + V IlC + V CD + V DA = jw jJ. II tpr dO dr.
Since
a
V DC - VAil = - (V.\fN) dr,
ar
a
V lJC - V AD = (V pQ ) dO,
ao
V}'fN = EeT dO,
V PQ = Er dr,
we have
a aE
-- (rEB)drdO + ---.!.drdO =jwjJ.H.prd(Jdr.
ar ao
Hence
a
- (rEe)
ar
aE r
- -jwJl(rll,,) + -.
ao
( 5.3)
Thus we have three partial differential equations connecting
Er, E9, and II tp for any electromagnetic field with circular symmetry
in \vhich electric lines are in planes passing through the z axis. There
is a similar set of equations for any field in which electric lines are
circles coaxial with the z axis. 'This set includes fields generated by
uniform circulating curren ts.
5.2 Waves on semi-infinite wire
In Chapter 1 we obtained the following expression for a semi-infinite
direct current filament on the z axis [Figure 5.3 (a) ] issuing from the
cen ler 0 of a spherical frame of reference
1(1 + cos 8)
II tp = .
411'"T sin 8
Let us no\v inquire whether there can exist a time-harmonic field
with the same dependence on o. To answer this question \VC assume
1 + cos 8
Iltp(r,O) = R(r)
sin 0
(5.4)

Spherical \vaveS
181
z
I
I
I
z
I
I
,
t l
P(r,O)

EfJ
t 1
o.
(a) (b)
FIGURE 5.3 Current filaments along the positive z axis: (a) with a point charge
at the origin 0; (b) with a small charged sphere concentric with the origin.
where R(r) is an unknown function of r only. If no such field can
exist, then equation (5.4) will be inconsistent with Maxwell's equa-
tions (5.1), (5.2), and (5.3). If it can exist we should be able to find
proper expre&')ions for II f{J' E" and E8.
Substituting from equation (5.4) into equation (5.1), we find
R(r)
E,=- . .
} WE*'
(5.5 )
Substituting from equation (5.4) into equation (5.2) \ve have
1 1 + cos 8 d(rR)
rE8 = -- .
jWE sin () dr
(5.6)
Substituting from equations (5.4), (5.5), and (5.6) into the re-
maining equation (5.3), we obtain
d 2 (rR)
dr 2 = -(j2(r R),
13 = wV;;.
(5.7)
Solving, we have
rR = Ae- iJ3 , + Be iBr .
(5.8)
The first term of this solution represents waves traveling radially
away from the center O. 'fhe second term represents waves converging

182
.lle(/r0111a£lleti( jie/lis
to the cen tcr. for example, \\Taves reflectcd from the inside of a con-
ducting sphere concentric \vith O. Both types are kno\\rn as spherical
'Z.£)a'es .
j;'or \vaves traveling out\vard the current I (r) in the \virc is
I (r) = 27rr sin OIl tp(r, 0) as 8  0
47rr R
47r ..1 c- jfjr ,
and
1(0) = 10 = 47rA,
A - lo/4rr.
Hence,
I oc-jr (1 + cos 8)
II =
f(i 4' ,
7r7 SIn 0
E8 = 11 11 tp
( 5.9)
l o c- if3r
47rjwEr 2 '
'fhc radial electric in tensity decrcases faster than the meridian
intensity. 'fhe ratio of their magnitudes is
E,=
11= .
t E,/ E81 = l/pr = X/27rr.
(5.10)
'rhis ratio is less than one-sixth at a distance of one wavelength from
the origin. 'rhus at larger distances E is nearly perpendicular to the
\vire as it should be if the \vire is perfectly conducting and if there
arc no generators in series \vith it. Near the origin the component of
E tangential to the wire is large. \Ve need an equal and opposite
electric in tcnsi ty to main tain the curren t in the wire and to generate
the fIeld described by equations (5.9). 1\t r = 0 this intensity is
infinite. 1\ more realistic physical situation is obtaincd by assuming
a sn1aIl metal sphere concentric with O. as sho\vn in Figure 5.3(b).
It is possible to modify the solution so that it \vould represent a \vavc
generated by an impressed voltage, concentrated in the in1mediate
vicinity of such a sphere. 'I'his modifIcation is signiticant only in a
restricted region around O.
5.3 Waves between coaxial cones
In1ugine another semi-infinite \virc along the negative z axis, carrying
a progressive current "a've I oc- jfjr . For this \vavc positive charge is

Sphl'ri(al \va \.C
183
accunlulatin at O \vhile an equal but negative charge is accumu-
lating there due to the current \vhich i tlo\ving out of 0 along the
positive:: axis. ']"herc is no net charge accunlulation at point 0 and
fer the combined field of t,vo tilaments Er O. 'T'hc contribution
of this second current to the other field components may be obtained
from equations (5.9) if ,ve replace 0 by 7r - O. Hence, for two 'loires
loe- itJr
II I(J = . '
27rr SIn 8'
E8 = T}II fP'
(5.11)
Since there is no radial electric fIeld. we may insert two per-
fectly conducting cones, coaxial with the z axis without disturbing
the tleld. as illustrated in l;igurcs 5.4(a, b). Hence, equations (5.11)
are proper expressions for the flcld between such coaxial cones. Elec-
tric lines are strictly along meridians. 'fhc voltagc along a typical line
.
IS
f 92 7]J 0  f82 dO
V (r) = Eer dO = - e-]T -=--
Rl 21l" 81 sin 0
7]1 0 . f82 d(OI2)
= 211" e-J{Jr 8. sin (Oj2) cos (0/2)
= 7)T o e-j{Jr f82 d tan (0/2) ,
27r 61 tan (0/2)
where 0 1 and O 2 are the half-cone angles. Thus
V(r) = Kl o e- if3r ,
7] [tan (02/2)]
K = -In
21r tan (lh/2) ,
(5.12)
where K is the characteristic impedance of a biconical transmission
line (see Section 4.7).
In the case of two equal cones, l"igure 5.4(a), of internal half-
angle"', 8 2 = 7r - '" and
K = ; In cot ().
(5.13)
\Vhcn "'/2 « 1
7] 2
K = In-.
1r '"
(5.14)

184
J/C(tro111Cl?llctic fields
z
z
y
y
() 2 7T-t/J
x
(a)
(b)
FIGURE 5.4 Coaxial cones.
5.4 Waves between coaxial cylinders
If \\.C assume that in Figure s.4(b) the cone angles approach zero
in such a way that rOI and r8 2 remain constant, then the cones ap-
proach coaxial cylinders. Since r sin (J = p, equations (5.11) yield in
the limit
l o c- ifJz
11 tp = ,
27rp
E p = 1111 tp.
(5.15)
Equations (5.12) become
1:" (z) = Kloe- ifJz
11 O'J
K = In - -
211" 8 1
11 rf)2
In
27r rOI
(5.16)
11 b
In -
,
27r a
\\.here (J, and b arc the radii of the cylinders.
,.

Spherical \vaves
185
5.5 Waves between parallel planes
As (b - a) (b + a) decreases, the coaxial cylinders in Figure 5.5
approach parallel planes. If
R = (a + b)
(5.17)
is the mean radius, and if x is the radial distance from the mean
x z
,
,
,
FIGURE 5.5 A section of coaxial cylinders of larJ{e and nearly equal radii and
bent C'ylindrical coordinates.
cylinder, equations (5.15) may be expressed as
Ioe- jf3z
Il=H=
I(J JI 27r(R + x)'
E p = E = 1//1J1
loe- jfJz ( X x 2 )
= 1--+-....
27rR J R2
'fhc maximum value of x is ,(b - a). Hence to the extent to which
(b - a.) (b + a) is negligible in comparison \vith unity, we have a
uniform field
10
11J1 = - e-j{Jz
27rR '
Ex = 11/ 1 Y'
(5.18)
Since b = R + ,Jz and a = R - Iz where h = b - a, the char-

186
J';lc(/r0111a1l{'/it Jirlds
actcristic inlpedance nlay be expressed as
71 1 + ( 11 / 21 )
In
27r 1 - (11,2R)
K
1] ( Ii /z2 /13
211" 2R - 81? + 24R3 -
. . . )
1] (h /12 /z3
- - -----
27r 2R 8R2 24R3
.. .).
lIenee
1]1z ( /z2 )
K=- 1+-+---.
27r R 121<2
(5.19)
5.6 Waves guided by thin diverging cones
The field of two diverging progressive current filaments, OA and OB
as sho\vn in Figure 5.6, may be obtained from equations (5.9). If the
/A
()
p
-00-

z
1"1< . CHI.:: 5.6 Dh,'crging thin conical conductors with a C01unlon apex.

Splu\rical ,va ,.cs
tR7
currents arc equal and opposite at points equidistant from 0, there is
no accunlula tion of charge at 0 and the radial field vanishes. "fhc
transverse field may best be expressed in terms of mixed coordinates
(r. 8'. 'P') and (r, 0", If") ",-here 8' and 8" are the angles made by a
typical radius OP \\.ith O1 and OB, \vhile 'P' and q;" are the angles
fixing the position of P around OA and OB, respectively. ]"he total
field is the sum of two fields, one associated \vith the Jilament OA
and the other \vith OB,
Ioe- i13r ( 1 + cas (}')
HfP' =
47rr sin 8'
E 9 , = 1]11 fP'
( 5.20)
Htpfl -
Ioe- ifjr ( 1 + cas 0")
47rr sin 8"
E 9 " = 17 11 tp'"
Electric lines run from one filament to the other on spherical surfaces
concentric with o. Without disturbing the field we can introduce
perfectly conducting conical surfaces around each filament provided
the electric lines cut these surfaces at right angles. '[his condition is
fulftlled exactly by certain circular cones around OA and OB, whose
axes, however, do not coincide with either OA or GB. But as the cone
angles become smaller, their axes approach OA and OB.
The characteristic impedance of the transmission line formed by
thin diverging cones is
r 1] j" 1 + cas 8 TJ j" cas (8/2)
K = dO = - do
27r sin 8 27r sin ((} /2)
 
17 sin ({) /2)
= - In
7r sin (1/;/2)'
(5.21 )
where {} is the angle bet",.ecn the axes of the cones each of angle 21/;.
5. 7 Waves guided by parallel wires
'J'wo diverging cones. as shown in Figure 5.6, approach a pair of
parallel wires when {} and l/I approach zero in such a way that r{} and
r1/; remain constant. Point 0 recedes to infinity. Angles 8' and 0"
approach zero for all points in an increasingly large region around
the wires: p' = r sin 8' and p" = rO" are the distances from the axes
of the wires. and a = r sin1/; is the radius of a wire.

188
h:/cclronIU K.llc/ ic ficlds
5.8 Waves generated by an electric current element
l"hc field of an electric current element AB of moment It (an oscil-
la ting electric dipole), };igurc 5.7 , may be obtained by superposing
the fields of t\VO semi-infinite progressive current filaments. One of
00
It !I
Per,e)
r'
B
I
FH;URE 5.7 Illustrating the calculation of the field of a current ele-
nlellt 11 B fronl the fields of two semi-infinite current filanlents.
the filaments starts at A and the other starts at B. They are 180 0
out of phase so that beyond B there is no current. Using equation
(5.9). we have
I e- ifJr ( 1 + cos 8) I e-jfJle-ifJr' (1 + cos 8')
H= - .
47rr sin 8 47rT' sin 8'
'fhe factor cxp (- j{3l) in the second term arises from the fact that
the phase of the progressive current filament which starts at A is
retarded by an amount (3l on arrival at B. Noting that " sin 8' = r
sin O. we obtain
I e- jfJr
ll = F,
411'"T sin 8
(5.22)
\\hcre
1/ = 1 + cos 0 - e-jpl-j{J(r'-r) (1 + cos 0').

Spherical \\'a yes
189
As l approaches zero, A P and BP become more nearly parallel and
T - r'  I cos 0
e-j{Jl-j{J(r'-r)  1 - jl3l - jl3(r' - r)
 1 -j{3l(l - cosO).
Hence
F = 1 + cos 0 - (1 + cos 0')[1 - j{3l(l - cos 0)]
= 1 + cos 0 - 1 - cos (}' + j {31 (1 + cos 0') (1 - cos 0).
Since the difference between' (J' and (J is infinitesimal, the substitu-
tion of cos (J for cos (J' in the last term will alter F by an infinitesimal
of the second order (note that (31 is an infinitesimal). Thus
F = cos (J - cos (J' + j{31 sin 2 (J.
As 0' approaches 0,
cos 0 - cos 0' = 2 sin! (0' + 0) sin! (0' - 0)
 (0' - 0) sin o.
Referring to Figure 5.7, we observe that 0' - 0 = Be/, = (II') sin 0
except for infini tesimals of higher order. l"herefore
F = (llr) sin 2 0 + j{31 sin 2 (J
and equation (5.22) becomes
j{3/1 ( 1 )
11 rp = - 1 +  e-i{Jr sin o.
47rr J{3r
The remaining field components may be obtained by differentia-
tion from equations (5.1) and (5.2)
1]fl ( 1 )
Er = --:; 1 + -::-- e-j{Jr cos 0,
27rr" J{3r
(5.23)
1]= 
j w}J.I I ( 1 1 )
E(J = 1 + - - e-j{Jr sin (J.
47rr j {3r /32r 2
At distances large compared with the wavelength, (/3r)-1 « 1 and
.{3II 'Il
II J 'R. J 'R.
. f/J = - e- JJJr SIn 0 = - e- JJJr SIn 0,
47rr 2Xr
(5.24)
(5.25)
E(J = 1]11 rp,
1]Il
Er = - e- j {3r cos o.
27rr 2

190
Elec/ronlagl1etir fields
As the distance from the curren t elemen t increases, the radial electric
intensity decreases much faster than the meridian intensity. l'hus in a
distant field E and jj are perpendicular to the radial direction of wave
propagation. The field is relatively strong in the equatorial plane of
the clemen t and van ishes on its axis.
Any linear distribution of current may be subdivided into current
elements of moment !(s) dS, where lIS is an element of length and
I(s) is the current at a typical point of linear distribution. Any
volume dstribution 01 current may be subdivided into elements of
moment J dv, where J is the current density and dv is an clement of
volume. Thus the field of any current distribution may be obtained
by integrating the field given by equations (5.23) and (5.24) over
the region occupied by current.
5.9 Waves above perfectly conducting planes
In the equatorial plane of a current element, Er vanishes and E is
perpendicular to the plane. Hence the plane may be made a perfect
conductor without perturbing the field. Thus equations (5.24) and
(5.25) apply to a vertical curren t elemen t of momen t Il/2 just above a
perfectly conducting plane.
E(J

 l
1
FIGURE 5.8 l-Vaves between an infinite conducting cone and a conducting plane
(a cone of 180 0 angle).
An instructive comparison can be made between waves in free
space and waves in the half-space above a perfectly conducting
plane, on the one hand, and waves beween coaxial cones or between a
cone above a perfectly conducting plane, on the other hand. See
Figure 5.8. In the case of cones, including the 180 0 cone which is a
plane, the field varies inversely as the distance from the apex at which
a voltage is applied [see equations (5.11)]. 1"he variation with () is

Spherical \\'aves
191
rela tively small if the cone angle 21/1 is equal to or larger than 120 0
since sin 60° = 0.866. The same can he said of the field of the current
clement in the region 7r/3 < () < 27r/3 when r » X. In fact, the
difference becomes particularly significant only at distances smaller
than X. If r = X, {3r = 211'" = 6.28 and the contributions of 1/{3r and
1/ ({3r) 2 to the field in tensi ties are already rcla tively small.
Both the differences and the similarities become understandable,
if \ve note that conducting cones provide an easy path for radial
electric current. On the other hand, in free space, for a given voltage
drop the radial displacement current is small unless the area available
for the flow is large. In other words, in the presence of cones, for a
given radial current there is no radial voltage drop and the en tire
impressed voltage is transmitted, in a wave-like manner, to large
distances. In the case of a current element in free space, on the
other hand, most of the impressed voltage is consumed in driving
radial displacement current in the vicinity of the element. Whatever
is left at larger distances is then freely transmitted beyond.
5.10 Radiation
Let us examine the field in the vicinity of a current element. For this
purpose we shall expand expressions (5.23) and (5.24) in power
series in r. Thus
e- iBr = 1 - j {3r + ! ( j (3r ) 2 - . ( j (3r ) 3 + -rh ( j (3r ) 4 + ...,
e- i /3r 1
- - - 1 + ! ( j (3r ) - i ( j (3r) 2 + 7I ( j (3r) 3 + ...
j{3r j{3r
e- jfJr
1 1.
. -. +! - 11 {3r + -iT ( j (3r) 2 + ...
(J{3r) 2 J(3r
(j{3r) 2
Hence
( It (32It j{33 r Il ) .
[lip = - + - - + ... SIn 8,
411'"1 2 87r 127r
( It jW}J.Il 1J{32Il j1J{33rIl)
Er = - - - + cos 0,
2-rrjwfr 3 4-n-r 67r 167r
( Il jW}J.Il 1J{32Il 3j1J{33rIl).
E8 = + + - - sin 8.
47r}wEr 3 87rr 67r 321l"
For slowly varying electric fields the first terms of these expressions
were already obtained directly from Coulomb's law and from the
Ampere-Maxwell law.

192
1./('(tr(JnUll!t1lc/ ic fields
'fhc componcnts of E parallel to and perpendicular to thl clements
arc
E z = Er cos 0 - E8 sin 8
It (2 C05 2 8 - sin 2 0)
47rjwEr 3
jwp,It jwp,ll sin 2 (J 1J{32Il
+ --+...
47rr 81rr 67r
and
E p = Er sin 8 + E8 cos (J
3Il sin 0 cos (J jwp,/l sin 0 cos (J
-
4-rrj WEr 3 87rT
j1J13 3 r It sin (J cos (J
161r +
. . .
l\Jost of the terms are in time quadrature with current 1 (as signified
by "jw") and rcprcsen t the reactive field around the clemen t. 'I'hc last
term in E z is 180 0 ou t of phase with I. A voltage equal to - E zt is
needcd to drive current in the clement against its field. rfhe average
work done by this voltage per second is
p =  re [-(Ezt)I*],
(5.26)
where the asterisk denotes the complex conjugate of I. Substituting
the last term of E z \ve have
p = 1;71" «(3l}2II*,
(5.27)
where 11* is the square of the amplitude of the current.
In free space, TJ  1201r and
P = 407r 2 (ljX) 21 1*.
(5.28)
"fhis is the average pO\'vcr contributed to the field by the electric
generator driving current I. 'I'he medium is nondissipative. Since
energy is consumed, \ve have to conclude that the wave carries it
a\vay radially at rate P. This is the radiant energy.
l"'igure 5.9 illustrates a thin resistive sheet of large radius r, con-
centric with the current clement. Let thcre be a perfectly conducting
sphere of radius r + (Xj4). The \vave which penetrates the tirst
sphere will be totally reflected from the second and a standing 'Nave
\\'ill be formed between the two spheres. At the perfectly conducting
sphere, E8 vanishes and II fP is maximum. Just outside the resistive
sphere, E8 is maximum and II tp vanishes. Current will flow along the

Spherical \\'aves
193
Perfect conductor
FIGURE 5.9 l/lustrat-ing the concept of pourer carried by a spherical wat'e in free
space.
meridians of the resistive sphere in response to E8. Let J(J be the
current per unit length perpendicular to the lines of flow. Assume
that the resistance per unit length is 'TJ so that E(J = 'TJJ(J. The Ampere-
1axwelllaw requires that H",(r - 0) - lIrp(r + 0) = I(J. Since
IIrp(r + 0) = 0, we have H",(r - 0) = J 8 = 'TJ-1E(J. Hence both
E(J and II", of the dipole wave [see equation (5.25) ] will be continuous
at the resistive sphere and no reflected wave will be generated inside
this sphere. 'J'he average power dissipated per unit area of the spherical
sheet is
llr =  E(JJ: =  E(JIl: = .'TJH I:
5.29)
which is in agreement with equation (3.3) for power flow. 1"he total
po\vcr dissipated in the resistive sphere may be obtained by sub-
stituting for II", from equations (5.25) and integrating over the
sphere
2r r
p = ?71jj HJl:r 2 sin 0 do d<{)
o 0
'TJ(32l2 I 1* j2rj r
= sin 3 8 do dip
3271"2 0 0
= ...!!..- ((3l) 2J 1* .
121r
'Ihis is equal to the power, equation (5.27), contributed to the field
by the curren t clemen t.

194 Electronzagnetic fields
5.11 Interference and directive radiation
Consider Figure 5.10 which illustrates the distant field of two parallel
current elements, each of moment Is, a distance l apart. The com-
bined field may be obtained from equations (5.25). If P is a distant
point in the direction at an angle", to the line AB joining the elements,
/
/'
/'
./
/
/
/'
/:
/'
/'
o'"='/
\C /"
.f..../
/"
/
/'
Is A
l
B Is
FIGURE 5.10 Two current elements.
then the difference between A P and BP is substan tially the projection
of AB on AP
A P - B P = l cos "'.
The effect of this difference on the amplitudes of the individual
fields of the clemen ts is of the second order; but the effect on the
phases is important unless l is a very small fraction of "A. Thus for the
total field, we obtain
.J
H <p = !....! I exp (- j fJr) + exp [ - j fJ (r - 1 cos ) ] I sin /1
2"Ar
jI s exp (-j{3r) sin 0
= [1 + exp (j (3l cos "') ]
2"Ar
j J se- j{Jr sin (J
= exp (j{3l cos "') 2 cos (!{3l cos 1/1).
2"Ar
Waves arriving at a distant point from different sources may be in
phase so that they will reinforce each other or they may partially, or

Spherical \va ves
195
totally, destroy each other. If l = X/2. they will destroy each other
along the line joining the sources, 1/1 = 0 or 7f'. The maximum rein-
forcemen t will take place in the plane perpendicular to the line
joining them.
"fhis phenomenon is called the interference of waves and is respon-
sible for directive radiation. l\10rc and more of the total radiant
energy can be thrown in certain directions by properly arranging
more and more sources of radia tion.
\Ve can also control directivity by proper relative phasing of the
individual sources. If the moment of the source at B is Is exp ( -j{3l)
so that its phase lags by (3l = 27f'l/X, the phase of the source at A,
we have
j I se- ifJr sin f)
/1 rp = {1 + exp [j{3l (cos 1/1 - 1) J}.
2Xr
The waves reiniorce each other in the direction 1/1 = O. In particular
if l = X/4, the bracketed term becomes
1 + exp [j7r (cos '" - 1) 12J.
Its value is 2 in the direction AB, and zero in the opposite direction.
5.12 Current distribution in thin wires
"Electric current element" is a mathematical abstraction, a mathe-
matical model of a hypothetical physical situation. It is useful for
computing fields of known current distributions since such dis-
tributions can be subdivided into elements. Currents in conductors
are determined by voltages impressed on them by other fields such
as fields of electric generators. These calculations are rarely exact.
In Section 5.3 we considered fields between infinite coaxial cones and
found that the currents in the cones and the transverse voltage
(along the meridians) are sinusoidal. Such waves can be excited
by connecting the apexes of the cones to an electric generator. If the
impressed voltage is V o , the input current will be /0 = Vol K and at
any distance r the current will be 10 exp (-j{3r). If we introduce a
metal sphere of radius r = l, the traveling wave will be reflected from
the sphere and will converge to the center. The reflected current will
be
/oexp (-j{3l) exp [-j{3(l- r)J = Ioexp (-2j{3l) exp (j{3r).
The situation is more complex, however, if the cones are just termi-
nated at distance r = t. The current at r = l must vanish and there

196
Electromagnetic fields
will be a reflected wave converging toward the center. The electric
field of the forward and backward waves at r = 1 will excite a field
in free space beyond r = 1. This field cannot conform to the field
between the cones since the latter requires the presence of conducting
cones. The field will be distorted both outside and inside the spherical
surface r = t. All that we can say at this stage is that the total field
will retain circular symmetry and that it will satisfy Maxwell's
partial differential equations obtained in Section 5.1. The solution
is very complicated since Ee and HfIJ have to be matched on the
sphere r = t. The simple waves in Section 5.3 are called the pril1ipal
waves.
Approximate solutions can be determined when the conical (or
cylindrical) wires are thin. In Section 5.1 we obtained the exact
field associated with an infinitely thin current filament along the
positive z axis with the source at the origin O. This field is given by
equations (5.9). Suppose that starting from z = 1 we superimpose
another semi-infinite current filament which cancels the original
filament from there on. Its fIeld may be obtained from equations
(5.9) by making the following substitutions:
10  -/0 exp (-j{3l),
r -+- rl,
o -+- Ol.
Ee -+- Eel'
II rp -+- H .pI'
Er  Erl'
where (rlo ( 1 ) are spherical coordinates with reference to another
origin at z = t. Thus we shall obtain the exact field of an infinitely
thin progressive current filament of finite length, extending from
z = 0 to z = t. At z = 1 there is a point charge (Io/jw) exp (-j{3l).
'fhis charge can be removed by superimposing an equal and opposite
charge with a progressive current wave traveling from z = 1 in the
negative z direction. This current wave can be canceled from z = 0
to z = - 00 by another current wave along the negative z axis.
In this manner we obtain the exact field of a standing current wave
\vith a source at z = o. At z = l the current vanishes as it should at
a free end of a wire. The field is the resultant of four fields of the type
in equations (5.9).
The current distribution is sinusoidal just as it is in a pair of
parallel wires. This we would expect from physical considerations.
Figure 5.11 shows the transition from parallel wires to wires going in
opposite directions. In case (a) when the separation between the
wires is small in comparison with their length, the end effects are
negligible at both ends A, Band C, D. As the wires are spread apar t,

Spherical waves
197
+ +
B  D
A _ _ + + C
D
(a)
+
A
+
[) B
A
B
+
(b)
+
C
(c)
FIGURE 5.11 Evolution of a linear antenna from a sectlon of two parallel wires.
the end effects remain negligible at A, B and become more pro-
nounced at C, D. The circular arcs from one wire to the other show
the electric lines along the meridians for the principal waves. For the
complete field the electric lines become increasingly distorted as the
distance from A, B increases. When the wires are thin, most of the
energy is near the wires. We recall that E p = q/27rf.oP and H f{J = I/27rp,
where p is the distance from tne axis of the wire. Energy densities are
proportional to the squares of the absolute values of E p and H 'P. If
the total energy is kept constant as the radius of the wire approaches
zero, the field will approach zero except in the vicinity of tlte wires.
It is this local field that determines primarily the nature of the
current distribution. Thus the spreading of the wires affects sig-
nificantly the distribution of the more distant field and relatively
little the distribution of curreLt and charge on the wires.
Let us now determine the series inductance and shunt capacitance
per unit length for principal waves on thin diverging wires of constant
radius a. We subdivide them into infinitesimal sections and consider

198
r:/cd r0111111!.llC/ ir .ficlds
each section as a section of a thin conc. 'l'hat is ,ve think of a cylin-
drical \\'ire as a 'cone" \\'ith a gradually decreasing half-cone angle
'" <If 'z, \vhere a is the radius of the \vire. 'fhe characteristic inl-
pcdance, equation (5.14), \\'ill be
TJ 2z
K = In .
1r a
(5.30)
'The velocity of propagation and the phase constant should remain
essentially unaltered. If Land C are the inductance and capacitance
per unit length, we have
{3 = wVLE = wV;;,
Hence
LC = J.l.E,
L=K,
Using equation (5.30)  we find
J.l. 2z
L = - In ,
7r a
K = Vi /c .
K2 = Lje
C = V;;/K.
(5.31 )
7rE
c=
In (2zja.)
(5.32)
Logarithmic functions are slowly varying functions and can be
approximated by their average values
J.l. (2l )
Lav = 11" In;; - 1 ,
1rE
C av =
In (2lja)
(5.33)
1
The value of C av is essentially equal to the one we can obtain from
the electrostatic field considered in Section 2.3. There is an additional
term -In 2 in the denominator which becomes less significant as
2lla increases. Different methods of approximation are not expected
to yield identical results unless higher order terms are all included.
Equations (5.32) were obtained for'infmitely long wires when electric
lines arc essentially the meridians. When the wires arc terminated
some electric field will exist outside the spherical surface of radius l,
passing through the ends of the wires. 1"his "end effect" is not in-
cluded in equations (5.33). If it is included, the t\VO approximations
agree.
5.13 Short antenna
An antenna in Figure 5.12 is called" short" if {3l«l, where 1 is the
length of each antenna arm. On such an antenna the charge dis-

Spherical wa \'CS
]C)9
z
",
"
./
Generator
FIGURE 5.12 A short antenna.
tribution is substantially constant and the current distribution is
linear.
I(z) - 10 (1 - ),
O<z<l
(5.34)
= 10 (1 + D,
-l < z < -0,
where To is the in put curren t. The genera tor" sees" the pair of ,vires
as lumped circuit consisting of a total capacitance
C't = Cl
(5.35)
in series with a total inductance
Lt = Ll. (5.36)
The distant field may be calculated from equations (5.25) by sub-
dividing the current into elements of moment I (z) dz and integrating

200
Elrc/r0l11l1J!.llC/ jcfirlds
their ilclds from z ::: -I to z = t. In any direction 0 the distance from
a typical clement is r - z cos (J. Since it was assumed that (:Jl « 1, the
c ff eet of the expression
{:3z cos (} < (:Jl cos (} « 1
on the phase of waves arriving from different elements to a distant
point is negligible. Hence, the pair of wires acts as a current clement
of moment
p = t fez) dz = 210 { (1 - ) dz = lol,
-l 0 I
that is, the distant field of the current in two short wires of total
length 2l is the same as that of a current clement of length l. Hence
the radiated power is given by equation (5.28). 1"hc generator sees a
resistance. called the radiation resistance Rrnd which can be deter-
mined from
(5.37)
p = RradJOJt = 407r 2 I ll'A I 210ft .
(5.38)
Thus
Rrad = 807r 2 IliA 1 2 .
When (l/A) is very smalL the resistance of the wires may be
considcrably larger than Jrlld and it should be included in the equiva-
tRw l
Rrad
Cl
1 Ll
FJ<jURE 5.13 The eguit'alellt circuit for a short antenna.
len t circuit for the short an ten na (see Figure 5 .13 where Rw is the
resistance of the \vire pcr unit length).
5.14 Half-wave antenna
'Ibe input impedance of a short antenna is capacitive and large.
Additional inductance can be added in series to tunc this capacitance

Spherical \\'a \'1"5
201
out to make the delivery of power from the generator easier. Another
way is to increase the length of the antenna until it becomes sclf-
resonant. In Section 5.12 we found that the current in thin \vires is
essentially sinusoidal. At the ends it must vanish. Hence, at distance
z from the cen ter
I (z ) = 10 sin (3 (l - z),
O<z<l
(5.39)
== 10 sin (3(l + z),
-l < z < o.
The voltage along the meridians from the upper to the lower wire is
1 dI
V(z) = - . = -jKavlo cos (3(l - z).
JWC av dz
(5.40)
These expressions are only for the principal wave and take no account
of radiation. Thus when l = Xj4, the voltage and current distribution
become
V(z) = -jKavlo sin {jz,
(5.41 )
I(z) = 10 cos {3z.
The distant field is calculated again from equations (5.25), this
time by including proper phase factor, Figure 5.14,
j{3Ioe-ifJr j";../4
H '" = sin 8 cos {3z exp (jl3z cos 8) dz
41rr -";../4
(5.42)
loe- ifJr COS (7r cos 8)
-
21rr sin 0
To obtain the radiated power we substitute in equation (5.29) and
in tegrate over a sphere of radius r
f 21"f1"
P = 7] H ,/1;r 2 sin (J dO dr;
o 0
7] * f1t/2 CO:;2 (1r COS (J)
= 1010 do.
21r 0 sin (J
By changing variables of integration and assuming that the
an tcnna is in vacu urn (7]  1201r), we can reduce this in tcgral to
f 21t 1
P = ISI o ft
o
- cos u * *
du = 15 Cin27r 1010 = 36.541 0 1 0 . (5.43)
1(,

202 l/ectromagllc/i{" .fields
z
t "",
z:-A/4 "",
./'
./'
/'
\
\
\ ,/
t "",
, "",
'"", ,,/
r
Generator p
,
t I
I
J
/
I
/
/
/
z=-,\/4
FIGURE 5.14 A half-wave antenna.
Hence. the input resistance of the half-wave antenna in free space is
R = 73.1 ohms.
(5.44)
Incidentally, this result sho\vs that in the vicinity of z - O the
voltage distribution given by equation (5.41) is not good enough.
5.15 Retarded potentials
In Chapter 2 \ve introduced a potential function V such that
E = -grad V.
(5.45)
'fhis function is useful in calculations involving electrostatic fields.
1'hc deiini tion ,vas based on the assumption that there \vas no time-
variable magnetic field. rrhe above equation implies that the line
integral of ii round a closed curve vanishes. Hence, in a time-variable
eletromagnetic iicld E cannot be expressed as the gradient of any
scalar function, no matter hov. it is defined. l"his docs not mean,

Spherical waves
203
ho\vever. that a part of E could not be so expressed. Such a generalized
potential function might still be useful in field calculations.
Since l and ij for an oscillating dipole contain a phase retardation
factor exp ( -j{3r), this factor should be expected in any function
from \vhich they might be obtained. 'l'hus \Vc assume a retarded
potential of a poin t charge q
V o
qe- ifjr
47rEr
(5.46)
As (3r approaches zero, either because A approaches infinity or r
approaches zero, this potential approaches the electrostatic potential.
l"or t\VO point charge situated on the z axis, charge q at z = 1/2 and
-q at z = -1/2 \ve have
qe- if3Tl
1 " -
, -
47r Er l
qe- ifjr2
47rEr2
Assuming that l is in1initesimal, \VC retain only infinitesimals of the
first order in
rl = r - l cos 0,
r2
, + l cos 0
and in V itself. Hence
qc-j(jr
V = ,) [r2 Cxp (j{3l cos 0) - '1 cxp (- .i{31 cos 0) J.
47rr"
l{cplacing the cxponentials by the first t\VO terms in their po\ver
series, \ve find that the bracketed expression becon1es
72 - rl + . (rl + r2)j/31 cos 0
\vhich is equal to
I cos 0 + j{3rl cos o.
lIenee. for the dipole
qle-i{jr
V = (1 + j{3r) cos O.
47rEr
Since q = l/jw.
II
T = . <) (1 + j{3r) e-i(jr cos (J.
411'J wEr -

204
Electrornagllctic fields
The negative of thc gradient of V has the following spherical
components
aV 'T]ll ( 1 ) jwp,Il
- = - 1 + - e-i{Jr cos 0 + e-j{Jr cos 0
ar 21rf 2 j (:Jr 411"r '
_ aV = jwp,Il ( _ ) e-i{Jr sin O.
rao 411'"r j{3r {32r2
Subtracting these expressions from Er and E8 for the current element,
we have
aV jwp,Il
E,. + - - e-j{Jr cos 0,
ar 411'"r
aV
E8+
rao
jwp,Il . .
e-{(jr SIn O.
411'"r
Expressions on the right are the rand () components of a vector whose
Cartesian components are
jwp./l
0, 0, - e-i{Jr.
47rr
1"'he vector A whose components are
p,I I e- j{J r
Az=
47rr
is called the retarded vector potential of thc current element (situated
at the origin along the z axis). rrhus we have
Az = 0,
A II = 0,
(5.47)
E = - jwA - grad V
(5.48)
for a typical curren t elemen t, and therefore, by the principle of
superposition" for any number of discrete clements or for any con-
tinuous distribution of elements. Thus if p dv is an element of charge
and j dv an element of current, \\PC have from equations (5.46) and
(5.47)
f pc-ifJr dv
V= ,
47rEr
- f p,J e- ifJr dv
A= ,
411'"r
(5.49)
where the integrations extend over the entire volume occupied by
charge and current.

6
Normal Modes
6.0 Introduction
Little imagination is nceded in order to calculate the field of any
given source distribution in an infinite homogeneous isotropic medium.
Thus, Coulomb's la\v gives the field of a single stationary electric
particle. The field of any number of such particles is obtained by
superposition of the fields of separate particles. If the number of
particles in an element of volume is large, the summation may be
approximated by integration (the charge is "smoothed" over a
volume). To obtain the magnetic field of a given direct current dis-
tribution, the latter is subdivided into current elements. The field
of a single clement is obtained directly from physical laws ; then the
superposition principle is applied. 'fhe field of a time-varying current
distribution is obtained in the same manner; only the field of a typical
current clement is somewhat more complicated.
In the case of two or more homogeneous media, boundary con-
ditions must be satisfied at the interfaces bctween the media. Some
imagination is needed for solving such problems. In Sections 2.4 and
2.5 solutions were synthcsized for conducting and dielectric spheres
imbedded in a uniform electric field in an infinite homogeneous di-
electric. In Section 2.12 the field of a point charge in a semi-infinite
medium above a conducting plane was constructed by superposing
an "image field" on the field of the point charge in an infinite medium.
There are no simple rules for solving field problems under all
conditions. Someone \vith imagination discovers a method for solving
a specific problem or a class of problems and passes it on to posterity.
One might think that the larger is the class of problems, the more
po\verful is the method. So it is, in a sense. Not infrequently it happens,
however, that the" more power-ful" method yields a form of solution
more difficult to in tcrpret than another form obtained by a "less
povlerful" method. Various methods often complement each other.
In this chapter \ve shall illustrate by examples a method of calcu-
lating fields which "works" for homogeneous media bounded by
coordinate surfaces in Cartesian, cylindrical. and spherical systems.
205

206
El('c/ronUl/?l1elic fields
It consists of solving problems piecemeal. lirst, one looks for certain
special types of solutions of Laplace's or l\Iax\veIl's partial differ-
en tial equations. 'l'hcn one selects those solutions \vhich satisfy the
required conditions on S01Jle boundaries. Finally. one combines these
solutions to satisfy the ren1aining boundary conditions as \vell as the
conditions at the source of the field.
l"he examples in this chapter have been selected to illustrate the
various phases in the application of the method. '1'0 present a broad
view of the method, all phases are considered in the lirst example.
Subsequent examples stress either one particular phase of the method
or some feature of a particular physical situation. fost problems
involve only plane boundaries and Cartesian coordinates. I)roblcms
involving cylindrical, spherical, and conical boundaries present no
new features except for a greater complexity of some mathematical
functions.
6.1 Direct current in conducting plates
Figure 6.1 shows a direct current I in a conducting plate of \vidth w.
Assume that the thickness of the plate is suddenly increased at


I
z
a
FIGURE 6.1 COllduct£llg plates of equal width w, but of different thickness,
joined together and carrying direct current I.

Normal morles
207
z = 0 and is decreased to its former thickness at z = l. Assume also
that the flo\v is uniform in the y direction.
In Section 2.1 it ,vas sho,vn that if the magnetic field is time-
invariable, the electric intensity can be expressed as the negative of
the gradient of a potential function V. Thus
E = -grad 1 ' .
(6.1)
'Vc also know that the total current passing outward (or inward)
through a closed surfcc is zero. This implic5 that the divergence of
the current density] vanishes (see l\ppcndix I) ; thus
divJ = O.
(6.2)
If the conductivity u of the plate is constant, we also have
div E O.
(6.3)
Substituting from equation (6.1) into (6.3), ,ve obtain Laplace's
equation
div grad V = LlV = o.
In Cartesian coordinates this becomes (see Appendix III)
(6.4)
a 2 v a 2 V aq(
-+-+-=0.
a ') a ') a ')
x" yo. zoo
(6.5)
In our problem we have assumed that the flow is uniform in the y
direction. Hence
a 2 v iJ2V
-+-=0.
ax 2 az 2
(6.6)
'rhere arc several methods for solving such equations, one of which
is the 1nethod of separation of variables. One assumes solu tions in the
form of a product of two functions, each depending on one variable
only
V(x, z) = X(x)Z(z).
(6.7)
Substituting in equation (6.6) and dividing by the product XZ, \ve
obtain
1 d 2 X 1 d 2 Z
--+ -- = O.
X dx 2 Z dz 2
The first term is a function of x only and the second of z only. 'fheir

208
Electromagnett'c fields
sum can vanish only if both terms are independent of both variables.
Thus we set
1 d 2 d1:Y
-- =k
X dx 2 '
1 d 2 Z
-- - -k,
Z dz 2
(6.8)
where k is a separation constant which may be either real or complex.
The values of this constant are restricted, however, by the boundary
conditions. The current density in the region 0 < z < l, for example,
must be parallel to the z axis at the poundaries x = 0 and x = a.
That is, the normal component J:r: of J and therefore E:r: must vanish
at x = +0 and x = a - O. From equations (6.1) and (6.7) we have
aV
E:r: = - = -X'(x)Z(z)
ax '
( 6.9)
aV
E z =
-X(x)Z'(z) .
az
The boundary conditions require that
X'( +0) = 0,
X'(a - 0) = O.
(6.10)
Since the general solution for X is
X(x) = Ae-Vk:r: + BeVk:r:,
(6.11 )
we have
X'(x) = yIk(-Ae-Vk:r: + Be Wcz )
so that
Vk(-A + B) = 0,
Yf,(-Ae- Vka + Be Vka ) = O.
rrhe first equation will be satisfied if
k = 0' or B = A.
(6.12)
In the first eventuality the second equation is satisfied automatically.
In the second case we must have
e-V'ka = eV'ka or e 2Vka = 1.
Let
Vk = p +jq
where p and q are real quantities. Then
(6.13)
e2pa+2jqa = e2pae2jqa = 1.
Since the absolute value of the quantity on the left must equal unity,

Normal modes
209
we must have p = O. The remaining factor equals unity when
qa = n1r, q = 1t1rja, It = 1,2,3, .... (6.14)
Substituting in equation (6.13), we have
k = - (11,7r / a) 2. ( 6 .15 )
These values of the separation constant k are called the proper values
or eigenvalues or characteristic values of the boundary value problem
and the corresponding functions
X(x) = 2A cos (n7rx/a)
(6.16)
are the proper functions or eigenfunctions or characteristic functions.
Note that the first possibility in equation (6.12) is included ifn = 0
is included in the sequence of integers in equation (6.14).
Substituting from equation (6.15) into the right-hand equation of
the set (6.8) and solving, we obtain
Z (z) = Cne-nrzla + Dnenrr./a, if n,  0,
(6.17 )
= Co + DoZ, if 11, = o.
Thus there is an infinite set of solutions for equation (6.6), all of
which satisfy the boundary conditions at x = O. a. The most general
solution may be expressed as an infinite series,
00
Vex, z) = Co + DoZ + L: (Cnc nr * + Dnenr.,a) cos (n1rXja),
n-l
( 6.18)
where the arbitrary constants associated with X(x) have been
absorbed into C n and Dn. Substituting in cquation (6.9), we have
00
Ex = L: (u1rja) (Cncnrzla + Dne"r.la) sin (n1rXla)
n-1
( 6.19)
1 '"
-4 -
.JZ -
00
_ Do + L: (n'K / a) (C "C"rz/a - D"e"r'ia) cos (1t1rx/ a)
n-l
for 0 < z < t.
The still unknown constants C n , Dn can be dctermined from the
boundary conditions at z = 0 and at z = l which can be imposed
either on V or on one or the other componen ts of E or j depending
on a particular problem. In the present case, current J flows in the z
direction, on the average, and the coefficients may be expressed in
terms of J.(x, 0) and J.(x, l). Multiplying E. in equation (6.19) by
rT to obtain J., and integrating over the area 0 < x < a, 0 < Y < w,

210
Electronzag1Zetic fields
,,,hen z = 0 or z = l, we have
- Douaw = ED l a J. (x, 0) dx dy
o 0
l W l a J.(x, l) dx dy I,
o 0
that is,
Do = -[ /uaw.
( 6.20)
The remaining coefficients are found if we multiply both sides
of the expression for E z by (j cos (nz7rx/a) to obtain Jz cos (l1l7rxja),
integrate over the same areas, and use the fact that
l a cos (l17rxla) cos (m7rxla) dx = 0,
o
'11,  tn,
= Ia
2 ,
n = 11t.
Thus, \ve find
1n7r 1 10 10. m7rX
-2 O1tJ(C m - Dm) = Jz(x, 0) cos - dx dy,
o 0 a
(6.21 )
1n7r l tD 1 4 11l7rX
-2 uw( Cme-mrl/a - Dmemrl/a) = ] z(x, l) cos - dx dy.
o 0 a
The current enters the middle plate and leaves it only when 0 < x < h
(see Figure 6.1). Therefore
C m - Dm = FI.m
Cme-mrl/a - Dmemrl/a = F 2 ,m,
where
2 10. m7rX
FI,m = - Jz(x,O) cos - dx
m7rU 0 a
2 1 4 1n7rX
F2,m = - ]z(X, l) cos - dx.
m7rU 0 a
Solving equations (6.22), we have
Fl - F e-mrl/ a
C m = ,m 2,m
1 - e- 2mr l/ a
- F2.me-mrl/a + FI,me-2mrl/a
Dm =
1 - e- 2m 7:l/a
(6.22)
(6.23)
(6.24)

Normal modes
211
If l/a is large, we have approximately
C m = FI,m,
D = - F 2 e-m-r lJa
m .m ·
(6.25 )
In fact even if l = a, these equations are good approximations since
exp (-71") = 0.043. For larger values of m the approximations are
good even when l is smaller than a. In other words the coefficients C m
are determined largely by the conditions at the face of the middle
plate where the current enters and the coefficients Dm are determined
by the conditions at the face where the current leaves.
Changing the subscript m back to 1t, we find J  from equations
(6.19) (6.20), and (6.25)
I 00 1"l7r 1t7rX
f = - + L (] F1,ne-n-r,/a cos-
aw n-l a a
00 1t7r n7rX
+ L (J F 2 ,n e -nr(l-z)/a cos -. (6.26)
n-I a a
Except in the vicinities of z = 0 and z = l, J, is substantially con-
stan t over the cross section of the plate. The 'end effects" are con-
.fined to distances front the junctions smaller than the thickness of the
plate.
Integrals given in equation (6.23) can be calculated exactly if J
is given at each face. In the problem formulated at the beginning of
the section only the total curren t I is specified. The boundary con-
ditions at z = 0 and z = 1 are as follows:
] (x. 0) = 0, J ,(x. l) = O h < x < a,
fz(x, -0) - ],(x, +0), 0 < x < It, (6.27)
] (x, l - 0) - ] ,(x. 1 + 0), 0 < x < Iz,
Ez(x, -0) = Ex(x, +0), 0 < x < Iz,
E;r,(x, 1 - 0) = E;r,(x. l + 0). 0 < x < lz.
1'he last four equations represent the continuity of the normal com-
ponent of currcnt dcnsity and tangential component of electric
intensity. For this particular problem there exists a method for
satisfying all of these conditions exactly.
In most problems, however, one has to rely on methods of suc-
cessive approximations. In the present case, for example, \ve can
start by satisfying the conditions which seem to be the most im-
portant from the physical point of view. The current enters the thick
plate from the thin plate through the junctions between them.

212
llectrollulllctic fields
Hence the first equation in the set (6.27) must aI\vays be satisticd.
From the results already obtained we know that the current dis-
tribution in the thin plate is uniform except within a distance h
from the junction. 'fhis nonuniformity is a secondary effect. Uniform
current enters the junction and spreads. As it spreads, the flow lines
in the thin plate will become curved but the perturbation ,viII not be
as great as in the thick plate \vhcre the current spreads to the upper
boundary. 'T'hus \ve start \vith the assumption
I
Jz(x,O) -- o < x < h,
,
wJz
(6.28)
=0 It < x < a.
,
Hence from equations (6.23) and (6.25) in which \ve let 1n cqua11',
\VC find
2! fh n7rX 21 a n7rh
en = Fl n = COS - dx = sin -.
, 1l7rulzw 0 a 'n 2 1f' 2 uhw a
(6.29)
Confining ourselves to a single junction, we let l = cx). 14"rom
equations (6.19) and (6.29), we then obtain
co 21 1t7r1z 1'Z7rX
Ex = L sin - sin - e- n 7l' z la,
n-I lZ,7rU/zW a a
z > O. (6.30)
'rhus although \ve started with an assumption, implied in equation
(6.28), that Er, = 0, ,vhen z = O. \ve flnd that once the current has
entered the middle plate the lines of flow become curved. Even at
z = 0 when x > It, Ex :;t. 0 and there is current spreading to\vard
the upper boundary of the middle plate as it should.
Since Ex must be continuous when z = O. Ex is different from zero
when z < o. There \viII be an end-effect in the thin plate \vhich \vill be
represented by exponential terms, decreasing in the negative z
direction
co 01l"X
Ez(x. z) = L fl a e a 7l' z lh sin -,
a-I h
z < O.
(6.31 )
Coefficients Aa arc the coefficients of the sine series for Ex(x. 0) In
the interval 0 < x < h
2 jh 01l"X
.I1a = - Ez(x. 0) sin - dx.
It 0 Jz
(6.32)

Normal modes
213
Substituting from equation (6.30) and integrating, we have
Aa = (_ )a+1 41 f. a sin 2 (nll"lz/a) .
7r 2 (J/zw n-I n[a 2 - (nh/a)2J
When Iz/a approaches zero, the numerator under the summation
sign approaches zero as (h/a)2 and Aa approaches zero as h/a. If
h/a approaches unity, Aa also approaches zero. In any case the end-
effect is wiped out within a distance from the junction comparable
to h.
Comparing equation (6.31) with equation (6.19), we find Ez(x, z)
and the longitudinal current density in the thin plate
( 6.33)
I 00 aX
]z(x, z) = - -  uAaearzlh cos-,
wh a-I h
z < O. (6.34)
The first term is the one with which we started [see equation (6.28) J.
This expression, at z = 0, instead of equation (6.28), can now be used
for calc'ulating coefficien ts C n in the thick plate. From the first term
the expressions given by equation (6.29) are obtained. From the
sununation we find the correction terms
1 2 ( - ) ah 2 sin (n7rh/ a) 00 A a
C n = . (6.35)
7r 3 a a-I a 2 - (nll/ a) 2
If h/a is much less than one, these terms are of order (h/a)4 in com-
parison with the coefficients C n .
In principle, the sequence of successive approximations can be
continued indefinitely. Calculations become more laborious but at
each stage they are straightforward.
6.2 Direct current in stratified plates
'rhe problem in the preceding section is representative of a wide
class of problems involving direct-current fields, electrostatic fields,
magnetostatic fields, and time-variable fields. The method of solution
is quite general; but its implementation varies with different classes
of problems. 'rhc problems in this and the following sections are
deliberately simplified to illustrate rlifferent features of the method.
Figure 6.2 shows a plate made of two parallel plates with different
conductivities (JI and (J2. The plates are in contact so that current
can pass from one to the other. We have seen that the first important
step in our method of solving a concrete problem is to find solutions
of Laplace's equation which satisfy the boundary conditions along the

214
Electro'Ynagnetic fields

z
x
a[ UJ
b 0'2
;:
I
FIGURE 6.2 A laminated conducting Plate.
plate. The next step was to satisfy the conditions at the ends of the
plate. In the present problem we are concerned primarily with the
first step.
Assume that the field is independent of the y coordinate so that one
has to solve equation (6.6). The boundary conditions are
Jz(O, z) = 0, Jz(a + b, z) = 0, (6.36)
]z(b - 0, z) = ]z(b + 0, z), Ez(b - 0, z) = Ez(b + 0, z).
(6.37)
The first two conditions are the same as in the preceding problem.
The last two are needed because we have a discontinuity in the
conductivity of the medium. The solutions that went into the series
given by equations (6.18) and (6.19) are continuous functions of x
and z. In the present problem Jz is continuous and therefore Ez
discontinuous across the interface between two media. Hence, some
modification of solution is needed.
\Ve already know that our equations possess solutions varying
exponentially in the general direction of current and sinusoidally at
right angles. Thus let
Ez(x, z) = A sin rx e- rz , 0 < x < b
(6.38)
= B sin r(a + b - x)e- rz ,
b < x < a + b,
where A, B, and r are unknown constants. The dependence on x

Normal modes
215
has been deliberately chosen to satisfy equations (6.36). The 1irst
boundary condition in equations (6.37) "rill be satisfied if
0'2A sin rb = O'lB sin rat
(6.39)
Since
E -
z -
aV
ax
we have
v (x, z)
- J E",dx
= A r- 1 cos rx e- rz ,
o < x < b,
(6.40)
b<x < a+b
= -Br- 1 cos rea + b - x) e- rz ,
except for a function of z only, which we take as a separate solution,
Vo(z) = Ao + A 1 z.
(6.41 )
Therefore
E,(x, z) =
aV
= A cas rx e- rz
a '
z
o < x < b,
= -B cosr(a + b - x)e- rz ,
(6.42)
b < x < a + b.
The second boundary condition in equation (6.37) will be satisfied
if
A cos rb = -B cos rat
(6.43)
Dividing equation (6.39) by equation (6.43), ,vc have the character-
istic equa.tion for r
0'2 tan rb = - <11 tan rat
(6.44)
I;or each root r n of this equation we have a set of solutions, given
by equations (6.38), (6.40} and (6.42). rhe constants .11 and B
arc not independent but arc related, as in equation (6.43). To satisfy
the cq ua tion, let
An = ln cos l'n a .
Bn = -Pn cos rnb,
(6.45)
where P n is, so far, arbitrary.
Equation (6.44) may be solved numerically or graphically. Let
b/a = k,
ra = It.
( 6.46)
(6.47)
l"hen
tan ku = - (0'1/0'2) tan ti.

216
1lcctro1JuIgllcl ic fields
The left side can be plottcd versus ku and the right side versus 11.
\\7 e pick the values of ku andzt which correspond to equal ordinates.
I;rom these values \ve plot u, versus k (ku) ju. Since the tangent
function is periodic and is conlposed of an infinite number of branches,
\ve can obtain it f (k). If a and b arc given  we find k, the corrc-
sponding values of u and tinally the corresponding values of r = ula.
l'he general solution, consistcnt \vith thc boundary conditions given
by equations (6.36) and (6.37). is the sum of solutions for different
roots of the characteristic equation (6.44). 'I'hus from equations
(6.41), (6.42), and (6.45)  \ve have
co
]z(x. z) = Pofo{x) + L Pnfn(x)e- rnz ,
( 6.48)
n-l
where the P n are arbitrary constants, and the characteristic functions
or eigenfunctions are
fo(x) = 0"2,
O < x<b
= 0'1, b < x < a + b,
( 6.49)
fn (x) = 0'2 cos r na cos r nX,
O < x<b
= O'} cos r nb cos r n (a + b - x),
b < x < a + b.
Similar expressions can be obtained for the remaining field quan tities.
lrom what we learned in the preceding section, we kno\v that
equation (6.48) can represent only the current injected into a plate
\vhich extends indefinitely in the positive z direction. If there is another
junction at z = l, for example, we should include a series in which
r n is replaced by - r n.
End conditions may be satisfied in much the same manner as in the
preceding section. 1"hu5 if J z(x, 0) is given, the coefficients in the
series given by equation (6.48) can be determined if both sides are
multiplied by appropriate functions of x and integrated over the
interval (0, a + b). In the preceding problem the ease \vith which
such coefficients \vere obtained was due to the orthognality property,
equation (6.21). of the cosines. In the present case the functions
defined by equation (6.49) arc also orthogonal. l'hus
j fJ+h
[O'(x) J-1n(X)fm(X) dx = 0,
o
n  1n,
(6.50)
\\Therc O'(x) = 0"'2 if 0 < x < band O"(x) = 0'1. if b < x < a + b.
'[0 prove, substitute from equation (6.49), integrate, and use the

!\ ormal modes
217
characteristic equation (6.44). l"hen if both sides of equation (6.48)
for z = 0 arc multiplied by [u(x) J-m(X) and integrated, only one
unkno\vn coefficient }) m \vill be left on the right side.
If there are more layers \vith different conductivities, the char-
acteristic equation becomes more complex; but the method remains
essentially the samc. If the conductivity is a continuous function of
x. there arc still solutions varying exponen tially \vith z; but the
cigcnfunctions fn(X) arc no longer sinusoidal. Instead they are
solutions of a certain ordinary differential equation. l"'his equation
can be obtained fronl equations (6.1) and (6.2). It should be noted
that equations (6.3) and (6.4) arc no longer valid since they are based
on the assumption that (J is at least piccc\vise independen t of x.
\Vhcn (J is a function of both variables x and z, a further modifica-
tion of the mcthod becomes necessary. Various field quantities can
still be expressed either as sine or cosine serics in x for any particular
value of z. 'fhis is known from the theory of Iourier sieries. 'fhe
coefficients, however. arc more general functions of z and the separate
terms of the series do not satisfy the licld equations and thus do not
represen t possible fields. Some of these questions \vill be considered
in Chapter 9.
6.3 Direct current in expanding plates
Figure 6.3 sho\vs the top vie\v of a thin expanding plate. \Ve assume
that electric current en ters the plate along a circular arc liB of radius
a and leaves it along another arc CD of radius b. On the average the
flow is in the radial direction. The coordinate system that fIts this
p
b
FIGURE 6.3 The top view of a thin Plate u'hose width increases at a ulliforl1l,
rale.

218
Electronzagnetic fields
geometry best is the cylindrical system. If the field is uniform in the
z direction, Laplace's equation (see Appendix III) becomes
p  (p av ) + a 2 V = o.
ap ap iJcp2
The componen ts of E are
(6.51 )
aV
E - ---
P ap'
Ef{J =
aV
pacp
(6.52)
The boundary conditions are
Ef{J(p, 0) = EI(J(p, t1) = 0,
(6.53)
where t1 is the angle of the expanding plate. To make the problem
definite we should also have some boundary conditions on the cir-
cular arcs AB and CD such as a description of the manner in which
the current enters and leaves the plate.
The method of solution is essentially the same as in the preceding
two problems. We assume a product solution
V(p, cp) = R(p)<J>(cp).
(6.54)
To satisfy the boundary conditions given by equation (6.53), we
should have
cp' (0) = cp' (t1) = O.
(6.55)
Equations for Rand cf> will be obtained if we substitute the product
solution in equation (6.51), divide by Rif>, and decide that each
term must be a constant. From our experience with a similar problem
in Section 6.1 we can avoid a repetition of details by anticipating that
cP should be some sinusoidal function and by choosing the form of the
separation constant more directly suitable for obtaining this result.
Thus
p !!... (p dR ) = v 2 R,
dp dp
Eq ua tion (6.55) \viII be satisfied if
d 2 cJ>
dcp2
(6.56 )
- - -v 2 <1>.
cI> = A cos vcp,
sin v{} = 0,
(6.57)
v = l1/rr/tJ,
n = 0, 1, 2, ....

Normal modes
219
The equation for R has solutions of the form
R = Bpa.
Substituting in the equation, we find
(6.58)
a 2 = v 2
,
a = ::I::v
(6.59)
so that the general solution is
R = Bnpnr/" + CnP-nr/".
(6.60)
If n, is equal to zero, equation (6.60) reduces to only one inde-
pendent solution. The general solution is obtained from equation
(6.56) as follows:
!!: (p dR ) = 0,
dp dp
dR
p = Co
dp
(6.61)
R = Co In p + Bo.
Combining these results, we have
a:> [(p)nr/" (a)nr / "] n7rcp
V(p, cp) = Bo + Co In p + :.; B b + C; cos.
(6.62)
Here new arbitrary constants have been introduced: B corresponds
largely to the potential distribution on the arc CD. where p = b,
and C corresponds largely to the potential distribution on the arc
AB, where p = a. As bja increases, the potential distribution on one
arc affects less and less the potential distribution on the other.
This expansion should be compared with the series given by equation
(6.18) for the potential in a plate of constant width and the series
given by equation (6.26) for the current density in the general direc-
'tion of flow. Power functions and exponential functions are related.
'}'he former approach the latter when ?J approaches zero in such a way
that p{} remains constant.
The logarithmic term in equation (6.61) is particularly important.
The radial electric intensity and current density derived from it are
inversely proportional to p and are independent of cpo This term con-
trols the total radial curren t in the expanding plate while the re-
maining terms represent "end-effects."
1'he problem of expressing the unkno\vn constants in equation
(6.62) in terms of radial current distributions on AB and CD is
exactly the same as in the case of plates of uniform width.

220
llectronlagl1etic fields
6.4 Direct current in bent plates
Figure 6.4 shows the upper view of a thin plate bent into a circular
strip. The principal difference bet\vecn this problem and that of an
expanding plate is in the direction of flo\v. In the present case the
D
B
p
FIGURE 6.4 A top view of a thin bent plate.
fIo\\" is, on the average, in the 'P direction. 'fhe boundary conditions are
Ep(a, tp) = Ep(b, tp) = 0,
that is,
R'(a) = J<'(b) = O.
(6.63)
Since 11 is no longer given by equation (6.57), \VC \vrite the general
solution, equation (6.60), in the form
1< = Bp' + Gp-'.
(6.64)
Differentiating and using the boundary conditions given by equations
(6.63), we have
v (Ba"-l - ('a-.- 1 ) = 0,
( 6.65)
V (Bb,-l - Cb-,-l) = O.
Both equations arc satisfied if
v = O.
In this case the proper solution for <I> in equations (6.56) is
cf) = ,ilICP + Ao
(6.66)
(6.67 )
while R = const.

Normal modes
221
If II is not equal to zero, equation (6.65) yields
C/B = a 2 , b 2 ,
(b/a)2' = 1
cxp [2v In (b/a) ] = 1 = e i2nr , Il = 1, 2, 3, ...,
(6.68)
(6.69)
that is,
II = jk n ,
1t7r
k =
n In (b/a) ,
n = 1, 2, 3, ....
(6.70)
For each characteritic value of 11, equation (6.68) determines
the ratio of arbitrary constants
en Bna2ikn.
lIence equation (6.64) becomes
R = Bnaikn[(p/a)ikn + (p/a)-ikn]
= Bnaikn {exp [jk n In (pia)] + exp [-jk n In (pia)]}
(6.71 )
= P n cos [k n In (pia)],
where P n is a new arbitrary constant.
Substituting the characteristic values from equation (6.70) into
equation (6.56) for <1>, we have
cf> = }.f neknf{J + N ne-kn'P.
(6.72)
Collecting the results, \ve have
00
V (p, cp) = A tCP + A 0 +  (Jf neknrp + N ne-knrp) cos [k n In (pi a) ]
n-)
(6.73)
where the constants k n are given by equation (6.70). Various field
components can be obtained by differentiation.
As in previous problems the remaining unknown constants can be
expressed in tcrms of V (p" cp) or J f{J(p, cp) at the ends of the plate
where cp = 0 and cP = {J. Conveniently enough the characteristic
functions given by equation (6.71) and thcir derivatives turn out
to be orthogonal. It is simpler to prove this directly from the differ-
ential cquations 6.56), and the boundary conditions given by
equation (6.63) than by direct calculation of the integrals. 1hus

222
l/ectronlaglletic fields
for any two characteristic values, we have
 (p dRn ) -kp-lRn,
dp dp
 (p dRm ) = -kplRm'
dp dp
Iy multiplying the first equation by Rm, the second by Rn, and sub-
tracting, we obtain
d (dRn dRm)
dp pR m dp - pR n dp = (k;. - k;.)p-IRnR ml
where the left side is a complete derivative. 1\J ultiplying by dp and
integrating from p = a, to p = b, \ve obtain
( pRm dR" _ pR n dRm ) b = (k;. - k;.) t p-IR"Rm dp.
dp dp a a
In view of the boundary conditions given by equations (6.63) the left
side vanishes. Hence,
(6.74)
b
f p-IRnRm dp = 0,
a
k  k.
(6.75)
6.5 Electric current filament between perfectly conducting parallel
planes
To illustrate the solution of problems involving static magnetic
fields, consider an infinitely long strip of direct current I between a
pair of perfectly cond ucting parallel planes at x = 0 and x = a.
See Figure 6.5. \\T C assume that the strip is parallel to the )' axis
and that the planes are extending to infinity in the positive and
negative z directions. After solution has been obtained, we let the
width s of the strip approach zero and thus determine the field of an
infinitely thin current filament. In Chapter 2 it \\ras concluded that in
current-free regions the magnetic intensity of a static field may be
expressed as the negative of the gradicn t of magnetic paten tial. 'rhus
the problem reduces to solving Laplace's equation (6.6), and satisfying
suitable boundary conditions.
The component of II normal to a perfectly conducting surface must
vanish. Hence
H:z:(O, z) = H:z:(a, z) = O.

Normal modes
223
x
a
1
.. z
y
FIGURE 6.5 A cross section of two parallel conducting p!anes and an infinite
current striP of width s.
In Section 6.1 we already obtained solutions for E=equation (6.19)J
\vhich satisfy these conditions and we can adapt them to the present
problem.
At z = :f: 00 the field must vanish. Hence
00
II:z; = L A me- m l'Z l a sin (1n7rxja), z > 0
m"l
00
- L Bmemrzla sin (m7rxja), Z < 0
m"l
(6.76)
00
llz = L A m e- m l'z/acos (l1t7rxja), => 0
m-l
00
= - L Bmemrz/a cos (m7rxja) , = < o.
m-l
'1\J agnetic lines of force go around the strip so that by symmetry \ve
kno\v that
II z ( x, - z) = II z (x, z),
(6.77)
11:z; (x, - z) = -II % (x, z).
Therefore
Bm = -Am.
(6.78)
By the Ampere-l\ilaxwcll law the discontinuity in [Ix across the

224
1lectro111agnetic fields
curren t strip equals the curren t per unit length nornlal to the lines
of flo\v. Assuming that the current is distributed uniformly, \ve have
llx(x, +0) - ll:r;(x, -0)
J Is,
Xo - s < x < Xo + s.
Elsc\v here across the plane z
tion (6.77)
Ilx(, + 0) = 112s,
= O.
0, II x is continuous. In vic\v of equa-
Xo - s < x < Xo + s
ou tside the in terval.
(6.79)
1'hc coefficien ts A n of the sine series for II x (x. +0) are
2 a
An = -1 llz(x, +0) sin (nrrx/a) dx.
a 0
Substituting from equation (6.79) and integrating. \ve find
2/ . 1l1rS . ntrXo
An = - SIn - SIn -.
1Z1rS 2a a
(6.80)
As S approaches zero
I 1n1rXo
Am-sin-.
a a
(6.81)
These equations together with equation (6.78) determine the mag-
netic field.
If the current I is alternating slowly, there will exist an elcctric
field parallel to the curren t,
E(x, z) = -jWp.l z II, dx
o
(6.82)
co a
= jWJi. L - llm e- mrz / a sin (11Z1rxla),
n-} m1r
z > o.
6.6 Point charge inside a hollow metal tube of rectangular cross
section
Solution of three-dimensional static problems involves more ticld
components. more terms in series expansions. but no ne\v features.
Consider, for instance, a point charge q inside a hollo\v nlctal tube of
rectangular cross section. In the preceding example a line lilamen t
was rcplaced by a strip of finite \vidth in order to express the condi-

Normal modes
225
y
"
",
.,,/
"
I · (x .YO)
L.-s
 x
a
/'
z
FIGURE 6.6 A conducting cylinder of rectangular cross section and a rectangle
of electric charge.
tions at the source of the field in a form [see equation (6.79) ] con-
venient for calculating the coefficients A n. In the presen t case the
point charge at (xo, yo) is replaced by a charge distributed uniformly
over a rectangular area around (xo, )'0). See Figure 6.6. 'rhus if s
and It arc the sides of the rectangle and Q is the surface density of the
charge, then q = Qhs. After the solution ha5 been obtained, \Vc allo\v
Iz and s to approach zero and Q to incrase correspondingly.
\Ve start with Laplace's equation
a 2 l r a 2 V atV
-+-+-=0.
ax 2 a y 2 az 2
(6.83)
Since the potential on the conducting boundary must be constant,
let it be equal to zero. From our experience \vith previous problems
\ve know that linear differential equations with constant coefficients
possess cxponcn tial and sinusoidal solutions. 1'0 satisfy the boundary
condition on the surface of the tube we select the following set of
functions:
V mn = sin (1n7rx/a.) sin (n7ry/b)Zmn(z ;
111., n = 1, 2, 3, · · .. ( 6.84 )
'"fhe general solution is the sum of these solutions taken over all
in tegral values of 11t and n. From the theory of Fouries series it is
known that an almost arbitrary function of x and y can be expressed
as a double sine series in x and y. In our case we expect the coefficients

226
Ji.lfc/r0111agnc/ic fields
to depend on z. The coefficients are obtained by substituting from
equation (6.84) into (6.83)
d 2 Z mn 2
= rmnZ mn .
dz 2
2 (l1l7r)2 (1t7r)2
r mn = - + - .
a b
(6.85)
Hence
Zmn = .I1mne-rmnz + lmnermnz,
(6.86)
\vhere Amn and Bmn are arbitrary constants of integration. 'fhc
potential must vanish at infinity. 'rhus the terms containing negative
exponents must be proper for z > 0 and those with positive exponents
for z < O. The potential must be continuous across the plane z = 0
containing the charge. Therefore Bmn = Amn and
. 11l7rX . 1t7rY
Vex, y, z) = L: Amn Sin - sin - cxp (-rmnZ),
m,n a b
z > 0
. m7rX . 1t7rY
- L: Amn SIn - sin - exp (rmnZ),
m,n a b
z < O.
(6.87)
The component of displacement density normal to the surface
charge
aV
Dz = EEz = -E-
az
is discontinuous. The discontinuity equals the density Q of the
surface charge. Hence
4 lY .1n7rX. 1'l7rY
2tr mn Amn = - Q sin - sin - dx dy,
ab a b
\vhere the integration is extended over the area of the rectangle
wi th sides Iz and s
-  s < x - Xo <  s , - It < Y - Yo <  h.
If one is interested in obtaining the field of a point charge, there
is no need to carry out the integration explicitly. One should merely
observe that as hand s approach zero, the coordinates x and y in
the integrand approach constant values Xo and )'0. The sine terms
can then be taken outside the in tegration sign. l"he remaining in tegral
\vill represent the total charge q. Thus
2q . 1J'l7rXO . n7r)'o
Amn = sin SIn -.
Er mnab a, b
f\
\.:-

Normal modes
227
6.7 Normal modes of field distribution and wave propagation
All the solutions we have obtained in this chapter have one feature
in common: they arc represented by infinite series of functions
satisfying field equations and certain boundary conditions. Each
term of an infinite series, such as the series given by equations (6.87),
represcn ts a possible .tield distribution and is called a '11lode of field
distribution. l\lorc specifically these tield5 are called nor1Jzal or ortho-
gonal 11lodes because each can exist without the others. To generate
the field corresponding to one particular pair of values of m andn
in equation (6.87), one needs only establish in the plane z = 0 a
distribution of potential or of surface charge, proportional to the
prod uct of the corresponding sines.
Throughout the rest of this chapter we consider time-harmonic
fields \vhich involvc wave propagation, and thc field generated by a
particular source expressed as the result of superposition of normal
1nodes of propagation.
6.8 Transverse magnetic (TM) waves between perfectly conducting
parallel planes
In this and the following sections we consider waves between per-
fectly conducting planes. Suppose that the boundaries shown in
Figure 6.1 are such planes and not the boundaries of a conducting
plate. The medium betwecn the plancs is now assumed to be a perfect
dielectric. Furthermore let the planes extend indefinitely in the y
direction, and assume that theiield is uniform in this direction.
l\Iaxwell's equations in Cartesian coordinates (see l\ppendix II)
become two independent sets, one connecting II II' Ex. and Ez and the
other" connecting Ell, II x, and II z. 1"he first one is
aE aE z ally
az - -jwJ.lIl + ax ' jWEE z = ax
(6.88)
all
y
az
-jwEE z .
J>aralIcl planes can be considcrcd as coaxial cylinders of infinitcly
large radii. 'rhus our problem is related to the one studied in Chapter
4. 'fhere it ,vas assumed that the longitudinal displacement currcnts
were negligible. I.,Ct us see if such an assumption is consistcn t ,vith
equations (6.88) the C( fin,'\-graincd" field cquations. One immcdiate

228
llec/ronlagnetic fields
consequence is that [Ill is independent of x (see the second equation).
l;'.rom the other t\VO equations \ve find that li:& and iJEzj ax arc also
independent of . Hence Ez is proportional tu .11 + Bx. Since E;:
vanishes at both perfectly conducting boundaries, A = B 0
and Ii;: vanishes identically. Iquation (6.88) then reduces to
dEx
liz
-jWJ.LIIlI'
dII"
- = jWEEx.
dz
(6.89)
l"hcir solutions are
lIt = IIte- ifjz ,
II; = Iloc ifJz ,
(3 = w.
Et = rJ11te-i/Jz
E; = - rJI1ocj(jz
rJ = .
(6.90)
'fhc first pair repre5cnts \vaves traveling in the positive z direction,
and the second represents \vaves in the negative z direction. Both
lield intensities E and II are at right angles to the direction of propa-
gation. Such \vaves are called tranS7.'crse clectronzagnetlc waves CfEl\1
\vaves) .
Other \vaves consistent \vith the field equations (6.88) are called
tralls'crse 1nagnclic wa'es ('l"l\f \vaves) because I} is perpendicular
to the direction of propagation. 1"hesc "raves arc not only possible,
they are necessary because 1"El\I \\1aves alone do not lit the physical
conditions at junctions \vhere the distance bet\veen parallel planes
changes. See )4"igurc 6.1. l\t z = 0, for instance, Ex is not independent
of x; it differs from zero in the interval (0, lz) and vanishes in (h, a).
Fron1 our experience \vith static problems we conclude that Ez
can be proportional to sin (11l7rxja). If this is the case, E;r, and II 11
arc proportional to cos (l1t7rxja). 1"hus
Ex = E(z) cos (11t7rxja),
.""
(6.91 )
II 11 = fj (z) cos ( nZ7rX j a) .
If \ve eliminate E;: from the first equation of the set (6.88) by sub-
stituting from the second and then using equation (6.91), we obtain
d It ( '11Z 2 7r 2 ) ..
d = - jWJ.L + -:---;; II,
Z }WEa"
( 6.92)
dH
dz
-jwEE.
A

N ornlal nlodes
229
'fhe gcneral solution of these equations is
n = Ae-r'mz + Bermz
E = KmAermz - l\mBerm,
( 6.93)
where r m is the propag ation constan t and Km the UHl,' ,,'e {nlPcdance
r m =  _tJ2 + 11:: 2 =  11::2 _ 2 ,
Z" !' /.
fi m = m/}WE.
(6.94)
In these equations A is the ,vavelength of plane ,vaves in the dielectric
medium of permeability J.L and dielectric constant f, so that fX =
1/. Observe that at lo,v frequencies all propagation constants
arc real and the ,vaves are attenuated. 1"hcy appear only as end
effccts at junctions (Figure 6.1). .l\s the frequency increases, more
and more \va ves beconlc tra ve ling ,va vcs. 'fhc cut-o ff wavelengths Ac
and frcquencies fr, arc given by r m = o. 'rhus
Ar ,m = 20,/ 'In.
fr,m = nt/2ay JlE .
(6.95)
'I'hc existence of these higher order nzodes of propagatiort was an-
ticipated fruln physical considerations in Section 4.19.
6.9 Transverse electric (TE) waves between perfectly conducting
parallel planes
l"he set of field equations, complementary to thc set given by equa-
tions (6.88), is
aE y
az
jWJ.L/I x,
jWJ.LII  =
a Ell
--
ax
(6.96)
a/Ix alI
- = jWEE lI + -.
a  
 u
Waves described bv these equations are called transverse electric
wa'es ('rE waves) because E is perpendicular to the direction of
propagation.

230
l/ec/ronlaf!.lle/i( fields
Again anticipating that }!;y and [Ix are proportional to in (11l7rx/a),
let
E y - £ (z) sin (1Jl7rx/a).
(6.97)
11;& Ii(z) in (nz7rx/a).
1'he negative sign is introduced to make the final form of equations
siInilar to the transn1ission line equations in Chapter 4. Eliminating
II z from equations (6.96) and substituting from equations (6.97),
\\'e have
d fI ( 1n 2 7r 2 )
- - - jWE +  fl.
d':J Jwp,a
Here the propagation c onstant and the wave impedance are
r m = '_ (32 + 1;t 2 7r 2 .
'\j a 2
dJ
dz
-jwp,H,
(6.98)
K m = jWjl/ r m .
(6.99)
1'hc cut-off frequencies are the same as for rfl\f waves. There are no
\va ves corresponding to 11Z, = O.
'I\vo perfectly conducting planes y = 0 and :)' = b may be intro-
ducl \vithout perturbing the fIeld since they will be perpendicular
to I. Hence. 'T'E \vaves of the type considered here may cxit in
hollo\v metal tubes of rectangular cross section, that is, in rccla11Kular
'U)(/'i.'CKuidcs.
6.10 Waves in perfectly conducting rectangular waveguides
In this section \ve shall consider wa,res in perfectly conducting \vavc-
guides of rectangular cross section. Let the boundaries be
:r=O, X=l1. y=O, ),=b. (6.100)
\\"e start by inquiring \vhcthcr such a waveguide can support '[:\1
".aves for \\'hich II z 0 and 'rI vw.avcs for \vhich Ez O. Xo risk
i involved. If there arc no \vaves in \vhich either onc or the other
conlponent vanishes. \ve shall discover it aftcr thc appropriate
ubstitutions in the field equations.
For 1'1\1 \vaves E z lTIUst vanish on the boundaries dcfined in equa-
tions (6.100). Let u aSSUlllC \vaves traveling in the positive z direc-
tion. 'rhus
l1lrrX llrry
Ez = il sin - sin - cxp (- fmnz) ,
a b
111, 11 - 1, 2, 3, ...
(6.101)
For the remaining components we write similar expressions in which

Normal modes
231
either one or the other sine function is replaced by the corresponding
cosine function as required by field equa tions. 14'or example, the first
equation given in Appendix II is
aE z aE II
a,y
iJz
-jW}.LIl z .
The derivative of Ez with respect to y is proportional to C05 (n7ryjb).
1"his equation cannot possibly be satisfied unless Ell and //;r, arc also
proportional to cos (1t7ryj b). I-Ia ving writ ten the expressions for the
various field components. we substitute them into field equations
and obtain linear algebraic equations for the complex amplitudes of
the components. These equations are then solved in terms of A.
In this manner. the follov'ing results arc obtained:
A jWEll7r nt7rX n7r)'
//x= 2 sin-cos-exp(-rmnz),
xmnb a b
AjWE11Z7r 11t7rX. 1t7rY
1/ 71 = 2 cos-sln-exp(-fmnz),
Xmna a b
(6.102)
Ex = KmJI lI ,
Ell = -Kmnllx,
r mn = ( : r + (,1 2 - #2,
X;n = ( : y + (' y,
Kmn = r mn/jWE.
Similarly for TE waves, start with
m7rX n7rY
Hz = Bcos-cos-exp (-fmnz),
a b
nt, n, = 0, 1 2,
(6.103)
In the present case either 1n or n may equal zero, but not both,
without yielding a field identically equal to zero. As in the preceding
case one obtains
BjwJ1.m7r m7rx. n7rY
Ex = cos - sIn - exp (- rmnZ),
xnb a b
BjwJ1.m7r 11l7rX n'7rY
E7I = - ? sin - cos - exp (- r mnZ),
Xna a b
(6.104)
Hz = - K;EII' H7I = K;Ez, Kmn = jW}.L/r mn .
The expressions for Xmn and r mn arc the same as those for 'fM waves.

232
1lectro1ua?tlleli( fields
1"1he double index '111" 1l is used to designate a particular \vave or
mode of propagation. l"'hus one may refer to a TErnn \vavc and
1"'E mn D10de or a 1"l1\f mn \'lave and 'l'l\f"w Inode. If a. is greater than b,
then the 'rllO nludc has the lo,vest cut-off frequency \vhich is given by
WcVE 'Tria.. fc 1!2aVE. Xc = 2(1..
']"'his \va ve is called the donziJlant wa.'i.'e. At lo\vcr frequencies no \va ves
can travel inside the metal tube.
6.11 Natural oscillations in metal cavities
If the rectangular 'va veguidc considered in the preceding section is
closed \vith t\VO perfectly conducting planes z 0 and z = l a metal
cavity is forlncd. (icneral expressions for possible fields in such a
cavity should include positive exponential functions of z as \vell as
the negative. l"'he conlponcnts Ex and E y nlust vanish \vhcn z == 0, t.
l"'his condition cannot be satis1ied \vhen r mn is real. 1"'herc must be
\\avcs traveling back and forth and adding in phase. Le t
I'mn j#mn. #mn = # - ( lI:7r Y - c7r r
(6.105)
In order that Ex and Ell could vanish at z = 0, I they must be pro-
portional to sin I3mnz and it is necessary that
sin 13m,} = 0,
{3mrJ = p7f",
p = 1, 2,
(6.106)
Substituting in the preceding equation \VC find
( "11r)2 (n1r)2 (P1r)2
{32 = W 2 P.E = --;; + b + I ·
'fhis equation determines the natural frequencies of cavity oscilla-
tions.
(6.107)
6.12 Attenuation
The ,valls of any actual waveguide are not perfectly conducting.
Po\ver \vill be dissipated and \vavcs \vill be attenuated. Since the
power dissipated per unit length of the guide is a small fraction of
the po\ver carried by the wave, the attenuation constant can be
clculated by using equation (4.38). In order to apply this equation
one has to assume that the field distribution over a typical cross
section of the guide is not affected much by the conductivity of the

Normal nlodes
233
\valls. \V c eXIlCct this to be the case when the conductivity is large.
It is the tangential component of II that determines the current in
the \valls. If the conductivity is infinite. the tangential component
of Ii is zero. "Then the conductivity is large, this component is small.
In Section 4.16 we obtained the ratio of Etan to the total current in a
conductor; that is. to the tangential magnetomotivc force. From it
\\re fInd the ratio of Etnn to /1 tan. ']"hus at high frequencies [see equa-
tions (4.53) and (4.55) and note that for a cylinder 1(0) 27ralltanJ
we have
Etan = 1]('/1 tan)
77c = vi j;;; /qc = 1(,( 1 + j). (6.108)
Using equation (3.3), we obtain the formula for the average power
absorbed by the walls of the guide per uni t length,
P = !Rc I EtanIltn ds = Rc I HtanI1tan ds,
(6.109)
where the integration is round the periphery.
1"'hc average power carried by the \\rave is also obtained by applying
equation (3.3) to the transverse components of the field. Thus
T'V = . II Et X In dS,
(6.110)
where the in tegration extends over the cross section of the guide.
We now apply these equations to a special case of a TrvI \\?ave
between parallel planes. The attenuation which \\re are studying is
that of traveling waves. when in absence of power dissipation, the
propagation constant given by equation ( 6.94) is a pure imaginary,
 ';n27r2
r m = j{3m, 13m = - - {32.
a 2
From equations (6.91) and (6.93) we have
II y = A cos (m7rx/ a) e- jfjm %
Ex = K,Jl y ,
Km = 13m/ WE .
At the conducting planes x = 0 and x = a the tangential!l is
11 II = ::l:: A e- i"m % .
Hence the average power absorbed per unit length and unit width
by both planes is
P = RcAA*.

234
l/eclromaglletic fields
The average power carried by the wave, per unit width, is
f a fll nz7I"x
lV =  E:rH: dx = Km,,4A * cos 2 - dx = iKmaAA *
o 0 a
if n1,  O. Other\vise lV = !K o a..1A *. 1"hcrefore
a
P /2W :::::: 2/cl Kma,
m rf; O.
6.13 Damping constant
'I'he same principle may be applie(l to the calculation of damping
constants of natural oscillations in metal cavities. There the average
absorbed po\ver is given by
p =Vc ff HtanHt.n dS,
(6.111)
where the integration is extended over the surface of the cavity.
'The total stored energy is
e = t fff Il.j'1* dv + lEfff E.E* dv,
where the first term represents the average magnetic energy and the
second the average electric energy. Since the two are equal,
e = ! fff Il.Il* dv.
(6.112)
Consider, for example a mode of oscillation for \vhich 1n = It = 1,
P = 0, and E z is given by equation (6.101); that is,
. 1I'"X . 7I"y
Ez = A sin sin -.
a b
Note that according to equation (6.106), P = 0 implies that r ll = O.
From equation (6.102) we find
AjWf7r . 7rX trY
II % = SIn - cos -
xi1b a b '
1r 2 1r 2
xii = - +-
a 2 b 2
II" =
AjWf7r 7rx. TrY
2 cos - sin -.
Xlla a b

Normal modes
235
At t,vo ,valls of the cavity, x = 0 and x a, \\'C have
AjwE7r . TrY
II y = =F 2 sIn
Xua b
Hence the absorbed power is
j ljb 1r21('w2E2bl
PI == l?c J1Jl H ; dy dz = 4 ') AA *.
o 0 2Xlla"
For the ,valls)' = 0 and y = b we find
j lja 1r 2 lcW2E2al
P2 = lc Ilz(x, O)lI:(x, 0) dxdz = AA*.
o 0 2Xlb2
For the walls z = 0 and z = l
j bja 1r21?cw2E2ab (1 1 )
P3 = lc (IIzII: + llyll;) dx dy = 4 - + -:; AA *
o 0 4Xll b 2 a"
l w 2 E 2 ab
C AA*.
4 . 2
Xu
The stored energy is
j tjbja ,uw2E2abl
8 = !,u (Hz/I: + 111/11;) dx dy dz = 2 AA *.
o (J 0 8Xll
From these expressions the damping constant [see equation
(3.29) where Way is the present P]
 = P/28 = (Pi + P2 + Pa)/28.
is obtained.
6.14 Waveguides and cavities of general shapes
In this chapter we have considercd fields, waves, and oscillations
bounded mostly by parallel planes. A similar analysi:) can be carried
out \\,hen some of the boundaries are cylindrical. The only difference
is that we use cylindrical coordinates and some of the functions are
Bcssel functions instead of circular functions. Except for small values
of the argument. 13cssel functions resemble sines and cosines or ex-
ponential functions. Spherical coordinates are suitable for fields
bounded by spheres, cones, and planes. Spherical coordinates can
also be separated in the field equations and general solutions can be
obtained as infinite series of suitable product solutions.

236
Elec/ronuzgl1ctic fields
Only a few coordinate systems permit separation of variables.
Even in those that do, this method is not necessarily the best. In
Chapters 8 and 9 \VC shall study a method \vhich is far more general
than the method of separation of variables.
\\'hile mathematics depend on shapes of boundaries, the physical
aspects of fields can be understood by studying what happens \vhen
the boundaries are relatively simple. Other boundaries can be thought
of as deformations of simple boundaries to which the fields have to
adjust.
6.15 Excitation of guided waves
Just as in free space, the basic source of waves in hollow tubes is the
curren t clemen t. Once its field is determined, the field of any given
current distribution can be calculated by integration. The current
element usually excites all modes. Depending on its position, how-
y
",
"
,/
",
./
tI"'"
.. x
.s-
FIGURE 6.7 ",1 conducting cylinder of rectangular cross section and a transverse
current fila 111en t.
ever, some may be missing. In some instances it is simpler to compute
the amplitudes of the excited modes directly from the given dis-
tribution. For example, in I'igure 6.7 when a uniform o::icillating
curren t is main tained in a thin wire parallel to the short side of a
holIo\v rectangular tube. the field is indcpenden t of the )' coordinate
and the field intensities are Ey. II z, II z. 1"he excited modes are the
1"E mo modes considered in Section 6.10. l"he amplitudes of these

Normal mode
237
modes can be calculated as in the static case considered in Section 6.5.
In fact one can use some of the results obtained in that section.
From equation (6.98) we have for a typical mode
Em(z) - Eme- rmz ,
z > 0
 E e rm '
m ,
z < 0
since EJI is continuous at z = O. Hence
f1 m (z) = K;lEme- rmz , z > 0
=K;"l Emcrm z , z < o.
Referring to equation (6.97), we have the general form for the field
00
E7I = - L: Eme- fmz sin (m7rxla), z > 0
m-l
00
 - L: Eme rmz sin (nz7rxla), z < 0
m-l
(6.113)
00
fix = L:K;lEme- rmz sin (1n7rxla), z > 0
m-l
00
- - L: K;lEmcr".z sin (m7rxla), z < O.
m=:ll
As {3 approaches zero, 11;r, approaches the static expression in equa-
tion (6.76) \vith Am = K;lE m . The discontinuity in l1x is related to
the current in the same manner, regardless of the frequency. Hence,
for an intinitely thin current filament we use equation (6.81) so that
Kmi m7rXo
Em = - sin
a a
(6.114)
rrhus \ve have the amplitudes of the transverse field components.
l'he longitudinal component \vhich can be obtained from the second
equation in the set given by equation (6.96), approaches zero as w
increases. 1\t high frequencies the waves tend to become trans'erse
electronzagnctic 'i.()(J,'i)CS ( 'fEN! waves) .
rrhcre is another method of calculating the amplitudes of the
various modes \vhich is equivalent to the above but has a physical
meaning. ]"hc current I is driven against its field, work is done, and

238
Electromagnetic fields
energy will flow into the field. The average work per second against
the field of the 1nth mode i
EmbI* sin (1117rXo/a.).
The average power carried in the 1nth mode in both directions from
the plane z 0 is
jbjO (-E,Jl:) dx dy = ab(l/K:')EmE:'.
o 0
Equating the two and canceling Emb, we have
(a/K:)E: = 1* sin (m7rxo/a),
E: = (K:/* fa) sin (nz7rxo/a).
Taking the conjugates of both sides, we obtain equation (6.114).
This nlcthod is particularly easy to apply to a curren t clemen t.
We start with a typical mode \vhich may exist in a hollo\v tube as
determined in Section 6.10, on both sides of the clement. 'The com-
ponent of E parallel to the element must be continuous. l"his con-
dition connects the amplitudes of the fields on both sides of the
clement. Then \Vc calculate the po\ver expended by the emf in1presscd
on the current element and the po\\rcr carried by the field. and thus
determine the amplitude of the mode.

7
Reflection and Scattering
7.0 Introduction
In this chapter we are concerned with the effect of irregularities or
"discontinuities" in media on waves. Depending on the point of
view, the effect can be described as either" reflection" from dis-
continuities or "scattering" by discontinuities.
7.1 Reflection at a junction of transmission lines
In Chapter 4 we considered wave propagation in a uniform trans-
mission line. The basic integral equations (4.11) and differential equa-
tions (4.22) are expressed in terms of two variables, the transverse
voltage V and the longitudinal current I, which represent, in a gross
sense, the intensities of the electric and magnetic fields associated
with the transmission line. These equations also contain the primary
para1neters, the series impedance per unit length Z and the shunt
admittance per unit length Y, which, in addition, are properties of
fields in the large and from which the secondary parameters are
obtained: the characteristic impedance K, equation (4.25), and the
propagation constant r, equation (4.29). If we take a section of a
transmission line and connect one pair of its terminals to a generator
which impresses a voltage of some frequency on the line and the other
pair of terminals to a device having impedance K, we shall find that
the ratio V /1 equals K at all points along the line and thus is inde-
pendent of the length. Furthermore the waves are traveling from the
generator to the other end. Usually there is some dissipation of energy
in the line so that the waves are attenuated with the increasing dis-
tance from the generator. In studying wave phenomena it is con-
venient to assume nondissipative lines and media so that r = j{3,
where the phase constant {3 equals the rate of change in the phase of
V and I per unit length.
If the line is terminated into some impedance other than K, a
239

240
Electromagnet1:c fields
\\"ave traveling from the impedance to the generator is originated,
cqua tion (4.41). This is the reflected wave. Its amplitude and phase
in relation to the inciden t wave are expressed by the reflection co-
efficient, equation (4.42), depending on the difference bctv."ccn the
terminal impedance and the characteristic impedance.
o
0
Kl
vt ...
I
0 )J
z=o

... K 2
o
z=
FIGURE 7.1 Reflection of 'wat'es at a junction z =  bet'lL'een lra.l1snl1'Ssio1l lines
with different characteristic 'i11Zpedallces.
In Iigure 7.1 t\VO uniform nondissipative transmission lines are
joined together. Let
Kl = V LI/C l ,
K 2 = yL 2 /C 2 ,
(7.1)
{31 = w V LIC l ,
{32 = w V L 2 C 2 .
Assume that the second line is infinite or terminated in the char-
acteristic impedance K 2 and that the generator is to the left of the
junction z = . As far as the line to the left of z =  is concerned,
the other line is just a device with impedance K 2 . lIenee from equa-
tions in Section 4.12 we obtain
I (z) = A {cxp (-j{31Z) + k exp (-jf31) exp [-j,Bl ( - z) J},
(7.2 )
v (z) = KIA {cxp (-jf3lZ) - k exp (-j,Bl) exp [ -jf31 ( - z) J},
z < t
_ l;,
v."here the reflection coefficien t for the curren t \va vc (ratio of the
amplitudes of the reflected and incident \vaves) is
Kl - K 2
k = .
Kl + K 2
(7.3 )
Note that there is no reflection if K 2 = Kl even though 132  {3l
so that the phase velocities in the two lines are different.

Reflection and scattering
241
Both the voltage and the current must be continuous across the
junction. Hence
I (z) = I () exp [ -jI32(Z - )]
(7.4)
= A exp (-j{31) P exp [ -j{32(Z - )],
Z > ,
where the transmission coeffi-eient for the curren t wave is
2KI
p=l+k= .
Kl + K 2
Thus at the junction the amplitude of the incident current wave
is multiplied by the transmission coefficient and then the wave
continues to travel with a different velocity.
For the voltage wave we have
(7.5)
v (z ) = K 21 (z) ,
z > .
(7.6)
Hence
V(z) = KIA exp (-j{31)qexp [-j{32(Z - )],
z > 
(7.7)
where the transmission coefficient for the voltage wave is
2K 2
Kl + K 2 .
q=l-k-
(7.8)
7.2 Reflection from a discontinuity in a transmission line
Any discontinuity in impedance gives rise to reflection. Consider a
uniform transmission line and introduce a lumped impedance Zz in
series with the line at the point z = , as illustrated in Figure 7.2.
To solve this problem one could take advantage of short cuts \vhich
are developed in comprehensive treatments of networks and trans-
mission lines. Here \ve use the basic principles from \\rhich the short
c
,0 
K K
vt I \
I \ ...
... I \
I \
 I \
I I \
0 Jr A bv\A B
z 0
z=e
FHIURE 7.2 Reflection of 'wat'es fronz aninlpedance discontilluity at
Poil1t z = .

242
Electronzagnetic fields
cuts are developed. The differential equations (4.22) assume con-
tinuous parameters and are valid on either side of the lumped im-
pedance but not in the infinitesimal region including it. rrhU5 \ve
\vrite separate solutions, with different arbitrary constants of inte-
gration, for z <  and z >  and join them together \vith appropriate
boundary conditions. For z <  \ve take equations (7.2) \vith the
subscripts dropped and leave the reflection coefficient k arbitrary.
1"0 the right of the discontinuity \ve have
I (z) = Be-jfJ(ze),
(7.9)
V(z) = KBe-j{j(z-).
l"he current passing through the lumped impedance is a defInite
quantity so that it is continuous across the impedance discontinuity
I(+O) =I(-O).
(7.10)
Applying the Faraday-Maxwell law to the circuit ABCA we have
V AB + V JJC + V CA = 0,
(7.11)
assuming that the physical dimensions of the lumped impedance
are so small that the magnetic displacement currrent linked \\rith the
closed circuit is negligible. In this equation
V A /J = Z II () ,
V Be = V ( + 0) ,
VCA=-V(-O). (7.12)
By substituting in equation (7.11) and rearranging the terms, we find
V( + 0) - V( - 0) = -Z,l().
(7.13)
Equations (7.10) and (7.13) are the boundary conditions from
which k and B can be determined. Thus
B = A (1 + k)e-j(jE,
KB - KA (1 - k)e-j{JE = -Z,B.
1"he second equation may be written as
(K + Zl)B = KA (1 - k)e-j{JE.

Reflection and scattering
243
Dividing by the first equation, we obtain
1 - k
K+Z =K .
I 1 + k
so that,
k -
Zl
2K + Zz
(7.14)
Therefore
2K
B = A e- jIJE .
2K + Zl
(7.15 )
Naturally the reflection coefficien t is independen t of the amplitude
A of the incident wave. The amplitude of the transmitted wave
equals the amplitude of the incident wave multiplied by a certain
transmission coefficient.
7.3 Reflection of plane waves at normal incidence
Waves generated by a current clement are spherical (see Section
5.8). Waves generated by any current distribution in a finite region
are obtained by superposition of such spherical waves. Thus they
are also waves expanding in all directions; at large distances their
amplitudes are inversely proportional to the distance from some
point in the region occupied by the sources. If we consider only a
limited region of space at large distances from the origin of waves,
the waves appear essentially plane and uniform. 'I'heir amplitudes
appear constant over a wavefront, that is, a surface of constant
phase. Their amplitudes also appear constant in the direction of
propagation. In such a region equations (5.25) can be written as
follows:
II JI = II oe- jp z Or
Ex = TJH lI ,
7.16)
11 = v;;:;;,
{3 = wy JJ.f. .
Here the Cartesian coordinates have been substituted for the spher-
ical: x for 8, y for 'P, and z for r.
It is much easier to solve certain important problems for plane
waves than to solve them for spherical \vaves. For this reason the
concept of uniform plane \vaves is very important even though
such waves cannot be realized physically except in a restricted

244
Electronlagllelic fields
region-and even there only approximately. One such problem is
that of reflection of waves from a plane interface between two semi-
infinite media. First, let us consider the case of normal incidence.
See Figure 7.3. Equations (7.16) are so similar to the equations
for the current and voltage in a transmission line [equations (4.30)

H"
/-LI,lI
.. -.....
FIGURE 7.3 Reflection of un1form plane 'wat'es incident normally on
the interface bel7.veen t'wo 1zo»zo£eneolls senli-infinite n1edia.
and (4.33) ] that we can write the answer to the problem from the
results obtained in Section 7.1. The magnetic intensity corresponds to
the current (per unit length), the electric intensity to the voltage
(per unit length), and the intrinsic impedance.,., to the characteristic
impedance K. Thus from equation (7.3) we obtain the reflection
coefficien t for II,
k = 7]1 - 712
7]1 + 7]2
and from equation (7.5) the corresponding transmission coefficient,
(7.17)
2711
p= .
7]1 + 7]2
If the interface is in the plane z = 0, the equations for Hand E are
(7.18)
1171 = [f o exp (-j{31Z) + kIlo exp (j{31Z),
= plIo cxp (-j{32Z) , Z > 0,
z < 0
(7.19)
Ex = 711/I o cxp (-jf31 Z ) - k711I1oexp (jf31 Z ),
= 7]'2pllo exp (-j{32Z), Z > o.
z < 0

H.eflection and scattering
245
These equations satisfy the boundary conditions at z = 0 which re-
quire the continuity of E:r. and B".
7.4 Reflection of plane waves at oblique incidence
In the case of oblique incidence, as shown in Figure 7 .4, let the
magnetic vector be parallel to the interface. Equations (7.16) for
PI, El
FIGURE 7.4 Reflection and refraction at oblique incidence.
the incident wave may be written as follows:
H; = 110 exp (-j{31S), E = 7JlII, (7.20)
where s is taken in the direction of propagation and u is at right
angles to it and to the y axis (\vhich comes out of the paper). If {}
is the angle of incidence, as sho\vn in Figure 7.4, then
s = x sin fl + z cos fl. (7.21)
"rhus
H; = 110 exp [ -jf31 (x sin {} + z cos {}) J. (7.22)
1hc componen ts of the incidcn t electric vector in the x and z direc-
tions are
E; = E cos f} = 'TJl1/o cos f} exp [ -jf31 (x sin {} + z cos {})],
(7.23)
E: = -E sin {} = -7JIHo sin f} exp [-j{31(X sin iJ + z cos {})J.

246
llectronlag71etic fields
'"Ihis wave impinges on the interface and excites waves below it
,vhich are traveling away from the interface z = O. The tangential
components of E and jj must be continuous at the interface at all
points. This condition cannot be satisfied unless the field in the
lower medium is proportional to
exp (- j131X sin {}).
(7.24)
From the situations considered in the previous sections we conclude
that, in general, ,vc cannot satisfy both con tinuity conditions without
a ,vavc originating at the interface and traveling away from it in the
upper medium. Physically this means that the waveS excited in the
lower medium will generate, in their turn, reflected waves in the
upper medium. The field intensities of the latter should also be
proportional to the quantity (7.24).
For the reflected waves we can write immediately
H = kIlo exp [ -j{31(X sin {} - z cos {})],
(7.25)
where the reflection coefficient k is yet to be determined. This ex-
pression has the required dependence on x and it represents a wave
moving away from the interface. The reflected wave is just another
uniform plane wave and the angle of reflection equals the angle of
incidence.
The E vector must be normal to jj and to the direction of propa-
gation and, if II is coming out of the paper, the E vector's direction
should be as indicated in Figure 7.4. 1"hus
E = -T/lkIlo cos {} exp [-jI31(X sin {} - z cos {})],
(7.26 )
E = -T/lkHo sin {} exp [-j{31(X sin {} - z cos t1)J.
We know that the field in the lo\ver medium is proportional to the
quantity (7.24) and that the wave is traveling away from the inter-
face. "rhus for the transmitted "avc we assume
II = pHo exp (-j{31X sin {}) exp (-jl3z z ) ,
(7.27 )
where p is the transmission coefficient and {3z is the phase constant
in the z direction. To determine this constan t we turn to lVIax\vell's
equations (Appendix II) and set a/a)' = O. Thus
aE z aE z
az
- - jWJJ. 2 H 11
ax
(7.28)
az
-jwE2Ez,
ally .
- = )WE 2 E z .
ax
all JI

Reflection and scattering
247
Substituting from the last t\VO equations into the first, we have
a2IIu a 2 II II
- + - = -/3lIl/'
ax 2 az 2
/32 = w.
(7.29)
Substituting from equation (7.27), \ve flnd
-{3i sin 2 fl - /3; = -/3
and
I3z = vi /3  - /31 sin 2 fl.
(7.30)
For /3z we take the positive sign to assure that the wave described in
equation (7.27) is traveling away from the interface. The component
of E parallel to the interface may be obtained from equations (7.27)
and (7.28)
E:J; = ({3z/wE2)pHo exp [-jl3lx sin {} - j/3zzJ.
(7.31 )
The coefficients k and p can now be determined from the continuity
of the tangen tial field componen ts
H(x, -0) + II;(x, -0) = H(x, +0)
E;(x, -0) + E(x, -0) = E;(x, +0)
or
110 + kIlo = plIo
1]111 0 cos fl - TJlklI o cos {} = (/3:e/ WE2) pH o.
Hence
TJI COS {} - ({3 z/ Wf2)
k =
TJI COS {} + (/3 z/ WE2)
p = 1 + k,
(7.32)
where /3z is given by equation (7.30).
Depending on the parameters of the media and the angle of inci-
dence, the phase constant /3z can be either real or imaginary. When it is
real, the transmitted wave described in equation (7.27) is a uniform
plane \vavc since its amplitude I plIo' is constant. rrhus we assume
that it is a \vave traveling at some angle", to the normal, as shown in
Figure 7 .4. l"his is the angle of refraction. In optics the transmitted
wave is called the refracted wa'i)c. l"'hus we rewrite equation (7.27) as
II; = plIo exp [ -j/32(X sin'" + z cos "') J.
(7.33)

248
Electronzagnetic fields
Of course, this means that
132 sin ,y = 131 sin f}
and
132 cos,y = 13z: = V l3i - l3i sin 2 .
Both {31 and 132 are proportional to w. Canceling w, we have
(7.34)
sin ,y = n sin fl,
'11, = V JJIEI/ JJ2 E '},
and
cos,y = V I - n 2 sin 2 fJ.
(7.35)
\Vhen vacuum is the lo\ver medium, the quantity n is called the
refracth,e index of the upper medium. Otherwise it may be called the
relative refractive index.
The cocfficien t of reflection can now be expressed as
711 cos fl - 712 cos '"
k = .
711 cos fl + 712 cos ,y
(7.36 )
We have seen that the coefficient of reflection given by equation
(7.17) at normal incidence is analogous to that in transmission lines,
equation (7.3), \vhere it equals the ratio of the difference bet\vecn two
characteristic impedances to their sum. 1'he present equation is also
of this form. In fact, it could have been derived immedia.tely if one had
recognized that a unif ornt plan.e wa've tra.leling at an, angle to the inter-
face can be considered as a phase pattern, exp (- j131X sin iJ), traleling
nornza11y to the interface. The wa'e impedance normal to the interface
\vould then be
E;
---: = 7]1 cos fl.
II;
Similarly for the transmitted ,va ve
(7.37 )
E;
= 7]') COS .".
II t - 'Y
"
1'he relation between {} and '" \vould still have to be determined
from the field equations; but the entire derivation would have been
much shorter. Furthermore the concept of \vavc impedance enables
one to write the expression for the reflection coefflcicn t \vhen the
lo\ver medium is stratified. 'rhus a great degree of generality can be
attained. Ho\vever, the value of this concept would not have been
appreciated without the straightfor\vard analysis based on tirst
principles.

Reflection and scattering
249
There is an important special case in which k = 0 and there is no
reflection. This happens when
7]1 COS f} = TJ2 COS 1/1.
From this and equation (7.35) we find the angle of incidence
1 - (2El/ lE2)
1 - (El/Ez)2
sin f) =
When J.L2 = 1, this simplifies to
1
sin {) =
v I + (El/E2)
1
-
v i + 11,2
(7.38)
and
sin t/; =
v I + 1Z 2 '
n
The angle of incidence for which there is no reflection is called the
BrC"ciJster angle.
Equation (7.35) indicates that there may exist a certain critical
angle f},. for which the angle of refraction 1/1 is 90° so the transmitted
wave is traveling parallel to the in terface. This angle is given by
sin {)c = l/n.
(7.39)
l"'his can happen only when n > 1. In such a case. cos 1/1 and (j, are
imaginary for any angle of incidence greater than {Je. Iquation (7.33)
indicates that the transmitted wave will then be attenuated ex-
ponen tially as the distance from the interface increases. In the
expression ( 7.36) for the reflection coefficien t the n umcra tor an d
denominator become conjugate complex. 'l"'hus the absolute value of
k is unity and the amplitude of the reflected \vave equals the ampli-
tude of the incident \vave. 'rhis is total reflection.
So far our discussion has applied to incident uniform plane \vaves
\vith thc magnetic vector parallel to the in terface. as had been as-
sumed in the beginning of this section. A similar analysis can be
made for the case in \vhich the electric vector is parallel to the in ter-
face. In the formula for the reflection coefiicient the cosines \vill be
replaced by the secants as the concept of normal ¥lave impedance
immediately indicates. "Then neither E nor II is parallel to the
interface the ¥lavc can be resolved in t\VO components, onc with
E })arallel to the interface and the other with ji so disposed.

250
Electronlagnetic fields
7.5 Waves at grazing incidence over imperfect ground
A perfectly conducting plane introduced at a right angle to the
electric vector in a uniforn1 plane \vavc docs not perturb the field.
Thus a uniform plane \vave n1ay skim the surface of such a plane, as
sho\,'n in Figure 7.5 (a) , and in vacuum
11 11 = 110 cxp (- j (3oZ) ,
Ex = Tlolio cxp (-j{3oZ) ,
(7.40)
/10 = w V J.LoEo ,
170 = V J.Lo/ Eo.
1\n imperfectly conducting plane, on the other hand, does perturb
the field since 111/ in1plics that an electric current is in the direction
of propagation and therefore has a driving electric intensity Ez.
"rhus if the medium belo\v the plane x = 0 is a homogeneous dissi-
pative n1edium. there must exist a component of E in the direction of
propagation. Propagation in such a medium was discussed in Sec-
tion 4.16 and the relation bet\\'cen E and II at the interface was
obtained in Section 6.12, equation (6.108). In the present case this
equation becomes
E z = TlcI! 1/'
TIc = vjwp./(u c +jWE).
(7.41 )
The dielectric constan t, \,'hose effect is negligible in metals at fre-
quencies belo\v 10 12 cycles per second, should not be neglected in the
case of soils. Hence
Ez/ Ez = Tlc/170'
(7.42 )
In obtaining this ratio it \vas assumed that the field givcn by
equations (7.40) is not affccted exccpt for the appearance of E z at and
near the surface. rrhe assumption is reasonable as long as the absolute
value of the ratio given in equation (7.42) is small compared with
unity \vhich is thc case of sea \vater and various soils.
I:quation (7.42) indicates that the electric vector is inclined toward
the interface. Since lz and E z arc not in phase, their resultant will
rotate during the cycle. It may be shown that this resultant \viU
describe a narro\v ellipse. Figure 7.5 (a).
7.6 Wave antenna
l\n interesting application of the \vavc tilt ovcr ground is a wa,'i)e
antennlL consisting of a long \vire parallel to the ground. Figure
7 .5 (b). rrhis \virc and the ground form a transmission line. Let us
suppose that this line is termina tcd by its characteristic impedance K

J{eHection and scattering
251
x
A B
K K,l' K
z= z=l
(b)
FIGURE 7.5 (a) Wave tilt at grazing incidence; (b) a wave antenna.
at both ends. The longitudinal field of a wave skimming the ground
will be impressed on the \vire. The voltage impressed on a typical
element AB of the wire at z =  may be obtained from equations
(7.40) and (7.41).1"'hu5
V AD = Ez() d = 17,11 0 exp (-j{3o) d.
This voltage sees impedance K in both directions. that is, the total
impedance 2K. Hence the current through AB due to this voltage
will be
dIet) = (7]c/2K) 110 cxp (-j{jot) d.
1"'his current is propagated in both directions. 1"'hus at point z
dI(, z)  (7]c/2K)H o cxp (-j{3ot) exp [-r(z - )Jd, z > 
- (7],/2K) 110 exp (-j{3o) exp [- r (e - z) J d, z < t;
that is
,
dI (, z) - (17c/2K) 110 exp [( r - j/3o) J e- rz d, z > 
- (7]c/2K)Ilo cxp [ - (r+ j/1o) tJ e rz d, z < .
The propagation constant r is complex since the \virc and ground
have resistance. Its imaginary part, however, is nearly equal to j{3o,
particularly at high frequencies. 'rhus in the forward direction the
induced currents add in phase. In the backward direction they tend

252
Elec/ronzagllctic fields
to cancel. At z = t the total current is
I(l) = (7'Jc/ 2K )Hoe- rl { cxp [(r - jt10HJ d
o
= ( c/ 2K ) Hrrrl exp [(r - j/3o)lJ - 1 .
7J !' '(3
-Jo
If I' is equal to a + j{3o,
1 - e- a1
l(t) = (TJc/2K)Hoe-j(jol
a
When a is small, al may be small even if l is large. l'hcn
l(l) = (7Jc/2K)IIo Cxp (-jf3ol) t.
The current through the impedance K at z = l will be proportional
to the length of the wire.
Thus power may be abstracted from a passing wave. Note that
morc powcr will be received when TIc is relatively large and none
when 71c = O. The wave antcnna \vorks better over poor ground than
over good ground. 1'hc antenna \vould be inoperative over perfect
ground.
7.7 Scattering by a discontinuity in a transmission line
'There is another method for obtaining the effect of a discontinuity
on \\ave propagation in a transmission line, See Figure 7.2. If instead
of the lumped impedance ./1/3 there \vcrc a generator, the generator
,vould start \\aves in both directions. Similarly the voltage across
the lumped impedance also generates \\aves in both directions. 'rhese
\\raves arc called scattered ,OQ,,'es. They arc superimposed on the
o
vt
".
I

.--

o
z=o
Q-J\/\/\,-o
A z=e B
FIGURE 7.6 Scattering of u)at'es by an inzpedance discontinuit,y.

Reflection and scattering
253
original inciden t or primary wave, Figure 7.6. From this poin t of
view the equations for the primary wave are
[p(z) = Ae- JP .,
(7.43)
Vp(z) = KAe- 1fJz
on either side of the discontinuity. For the scattered wave we have
I. (z) = Be-1IHz-c) ,
z > 
(7.44)
= Be- jfja - z ) , z < ,
where B is the scattered current through the discontinuity. The
corresponding voltages are
V, (z) = KB exp [-jl3(z - )],
= -KB exp [ -jl3( - z)],
z > 
(7.45)
z < .
In this approach the continuity of current, equation (7.10), is
satisfied automatically since the primary current is continuous.
The total current through the lumped impedance Zl is
I P ( ) + I. ( ) = A e- ifJE + B.
The voltage discontinuity in the boundary condition (7.13) equals
2KB since the primary voltage is continuous. Thus
2KB = -Zl(Ae-jfJt + B)
and
Zl
B = - Ae- ifJE .
2K + Zl
To the left of the discontinuity the scattered wave and reflected
wave are identical. Hence the coefficient of A in equation (7.46) is
the reflection coefilcient. This agrees with equation (7.14). To the
right of the discontinuity the transmitted wave equals the sum of
the primary and scattered \vaves. 1"'his agrees with equation (7.15)
(note that the B in this equation equals A cxp (-jl3) + B in this
section) .
1"his approach is particularly good when the discontinuity is small
and one is interested only in the first-order effect. The argument
would run as follows: In the first rough approximation the primary
wave is not affected by the discontinuity and the current through it is
(7.46)
Ae- i8E .

254
Electromagnetic fields
Therefore the voltage drop across it is
V An = Z z Ae- iJ3f .
This is equivalent to a gcnerator \\.ith a voltage
-ZzAe-jJ3
driving curren t in the positive z direction. The impedance seen by
this generator is 2K. Hence the current through the discontinuity
due to this voltage is
- (Zl/2K) Ae- jj3f .
This is the scattered current B, ,vhich agrees with equation (7.46)
to the extcnt that Zl is negligible in comparison with 2K.
7.8 Scattering by a small perfectly conducting sphere in free space
Consider a small perfectly conducting sphere of radius a in the path
of a uniform plane \\rave, as shown in Figure 7.7. By" small" we mean
that the difference in the phase of the wave at any given instant is
+
+
E z
+
..
FIGURE 7.7 Scattering of u)at'es by a srnall conducting sPhere is due
to a fluctuating charge as in an electric diPole and to circulating
currents as in a ring.
negligible over the volume occupied by the sphere. Thus it may be
assumed that the sphere is in a uniform electric field of intensity Eo
and a uniform magnetic field of intensity 11 0 , ".here Eo and 110 are
the fIeld intensitics of the incident wave at the center of the sphere.
Under the influence of Eo electric charge will be displaced up and
down \\.ith the frequency of the wave and the sphere will beconle an
oscillating dipole. Its electric field is given by equation (5.24).
If {3a « 1, the field in the vicinity of the sphere varies with the

Reflection and scattering
255
distance r from the center as if it \vere an electrostatic field. In Sec-
tion 2.4 we determined the electrostatic moment of this dipole,
equation (2.21). l"'he moment of the corresponding current clement
equals the time derivative of the electrostatic moment
jw47rEOa3 Eo.
Substituting this for It in equations (5.23) and (5.24)  we obtain
the field of the electrically polarized sphere which is a part of the
total scattered field.
1"'hc other part of the scattered fleld is due to Ho. At the surface
of the sphere the radial component II, of liD must vanish. Hence
there must be circulating currents on the sphere \\?hich produce an
equal and opposite radial magnetic field. The sphere becomes an
oscillating magnetic dipole. Its moment may be obtained by a
method similar to that in Section 2.4. The moment is
- jw27rJ.Loa 3 H 0
in the y direction.
In the case of a sphere more exact expressions for the dipole mo-
ments can be obtained. Indeed, the exact scattered field for a sphere
of any radius can be calculated. The present approximate method,
however, applies to small obstacles of shapes for \vhich the exact
field is impossible to find while the corresponding electrostatic and
magnetostatic problems can be solved.
7.9 Scattering by a small perfectly conducting sphere above a
perfectly conducting plane
1"'he method in the preceding section applies to more complex situa-
tions such as the one shown in Figure 7.8 where a uniform plane
wave strikes a perfectly conducting plane with a small sphere above
it. Here the primary wave, impinging on the sphere, is the resultant
of the wave incident on the plane and that reflected from it. At the
centcr of the sphere there are two components of E, one vertical
and the other horizon tal. Thus the sphere will act as two electric
oscillating dipoles and a magnetic dipole. The waves generated by
these dipoles are reflected from the plane. The reflected wa veswill
appear to be coming from dipoles on an image sphere. The inlage of
the vertical dipole is in the same direction as the dipole. The same is
true of thc image of the horizontal magnetic dipole. The image of the
horizontal electric dipole is in the opposite direction. '1"'hesc conclu-
sions are drawn from the fact that at the reflecting plane the tan-

256
Electro1nagnetic fields
-,
,. \
{ J
'_/
FIGURE 7.8 Scattering of waves by a small sphere above a conducting plane.
gential electric intensity must vanish. Since the dipoles have end
charges we can use the results of Section 2.12. The currents have to
follow suit.
The reflected dipole fields will act on the sphere just as the primary
wave did. Ho\vever, their fIelds are relatively weak.
7.10 Scattering by a long rectangular loop
In the problem of scattering by a sphere we calculated the electric
dipole moment from the electric intensity of the incident wave and
suggested that the magnetic dipole moment due to the circulating
currents could be obtained from the magnetic intensity. This is the
quickest method for small obstacles. Both moments can be obtained
from the electric in tensity: the electric dipole momen t from the
incident E at the center of the sphere and the magnetic dipole moment
from the change in E from the near end of the sphere to the far end.
Thus in the vicinity of the sphere \\rith the center at z = O. \ve have
E o e- it3z = Eo - j{3EoZ + .... (7.47)
The first term drives the charge up and do\vn. The second term has
opposite signs at z = -a and z = a so that it drives the charge round

Reflection and scattering
257
the sphere. This term is small in comparison with the first but it is
easicr to drive the current in a closed conducting path, as in a coil,
than in an open pa th, as in a ca paci tor.
To illustra te these poin ts we shall consider a narro\v rectangular
loop of length l large in comparison with the separation s between
the axes of the long wires, Figure 7.9 (a). Let the field at the center
E I
....... . 0
z=_ll z=Q
2
+
l Is f
z=  I
+ +
:0 t
(a) (b)
FIGURE 7.9 Scattering by (a) a two-wire line shorted at both ends and (b) by a
snlallioop.
of the loop be Eo and 110. The voltages i1npressed by the incident \vave
on the short sides of the loop at z = - [ and z = +!l are, respec-
tively,
VI = EoS exp (j{3l/2),
V 2 = EoS exp( -j{3l/2).
(7.48)
No voltage is imprcssed on either of the long sides since there is no
component of E tangential to them. As long as l » s, the loop is a
.parallel \vire transmission line. Let us assume that the dissipation in
the line is negligible and that its charactcristic impedance is K =
(L/C)t. 'rhe phase constant {3 = w(LC)i = w(E)i equals that of a
plane ,vavc. Hence
I (z) = Ae- jf3z + Be J13z ,
V (z) = KAe- iPz - KBej{Jz.
The total voltage across the short sides must vanish. Thercfore
K"te i13l !2 - KBe- jfjl / 2 = - VI,
KAe- iPl / 2 - KBi13 I / 2 = - V 2 .
Solving for .4 and B and substituting from equations (7.48), \ve have
/1 = -EoS/K,
B = 0,
and
I (z) = - (EoSl K) e-i{Jz,
It (z) = - EoSe- ifjz .
(7.49)
Suppose no\v that (3l/2 « 1 or [/A « 1/7l'; then
I(z)  -EoS/K,
V(z)  -EoS.
(7.50)

258
]lec/ronlagl1etic fields
That is, the incident \'lave \vill induce a uniform current round the
U small" loop, Figure 7.9 (b), and equal and opposite uniform charge
distribu tions on the longer sides of the loop. If C is the capacitance
per unit length, the total charge on the upper \vire is ClsEo and the
elcctrosta tic momen t of the electric dipolc is Cls2 Eo. ,!'he area mo-
ment of the circulating current is -Is! = -ls2EoIK. and the equiva-
len t magnetic dipole momen t is - J.LIs2 Eo/ K.
Let us no,v calculate the curren t in the loop directly from the
magnetic intensity 110 at the center of the loop. By the Ampcre-
l\Iax\vell la\v the induced electromotive force equals -jwJ.Llsllo. To
obtain the curren t \Vc divide this by the inductive reactance of the
loopjwI). lienee
I = -J.l.IloS/L = -J.l.TJ-lEoS/L = - (J.L€)+IEoS/L = -EoS/K
.
SInce
LC = J.L€ and K2 = LIC,
and, therefore,
L = K (J.L€) I,
c = (J.L€) ilK.
(7.51 )
The result agrees \vith equation (7.50).
If we calculate the second term in the power series for the ex-
ponential function in the expression (7.49) for I (z), we shall obtain
the charging current associated with the electric dipole. This current
is small in comparison with the circulating curren t.
Once we have found the current and charge distributions we can
obtain the local and distant fields either with the aid of retarded
potentials (5 48) or from equations (5.9) and (5.23). The distant
field can be calculated from equations (5 25).
7.11 Scattering by a short wire
In Section 2.3 \ve obtained the distribution of charge displaced on a
thin \vire by an electric field parallel to the \vire. If q(z) is the charge
per unit length at distance z from the center, the moment of t\VO
elementary charges q(z) dz and q( -z) dz = -q(z) dz is 2zq(z) dz.
The total moment is the integral of the elementary moment from
z = 0 to z = I. rrhe time rate of change of this moment is the moment
of the equivalent current element \\rhich can be substituted in equa-
tions (5.23) and (5.24) to obtain the distant field scattered by a
short thin \vire in the path of a uniform plane wave. See :Figure 7.10.

Reflection and sea tteri ng
259
+
+
+
..
2/
FIGURE 7.10 A wave imPinging on a short thin wire makes it an
electric dipole.
7.12 Scattering by a half-wave wire
In Section 5.12 we concluded that the current and transverse voltage
on thin diverging wires (see Figure 5.11) are given essentially by the
principal waves. The equations of propagation of principal waves
are the same for diverging and parallel wires. For parallel wires
the inductance and capacitance per unit length are constant while
for diverging \vires they vary with the distance from the origin
A, B [see equation (5.32) J. The functions involved are logarithmic
(j,nd hence slowly varying. Thus, they may be approximated by their
average values [see equation (5.33) J. Therefore the approximate
voltage and current in diverging wires satisfy uniform transmission
line equations (Chapter 4).
In the problem of scattering of waves by a wire of arbitrary length
2l, as shown in Figure 7.11 (a), in the path of a plane wave, the current
distribution has to be determined first. If the electric intensity at the
wire is Eo the voltage impressed on a typical element of the wire
is Eo dz, that is, 2Eo dz on t\VO elements equidistant from the center
of the wire A, B. This represents an increase in the transverse voltage
\vhen z is increased to z + dz. At A, B the wire may be continuous,
or broken, or have some resistance or impedance as in a receiving
antenna. Let us assume that it is continuous. Figure 7.11(b) illu-
strates the analogous problem for parallel wires. If z is the distance
from A, B the equations for the voltage and current are
dV
= -jwLI + 2Eo,
dz
dI
dz
-jwCV.
(7.52)

260
Electronzagnet£c fields
l Eod
B
E 
B C ,. : j :

A ./1 
Eod
(a) (b)
FIGURE 7.11 Illustrating an appro:xin1ate calculation of the induced charge
alld current dtOs/ribut£o1Z hz (a) a thin unOre of any length by cornparing the u'ire
u'ith (b) a tu'o-'wire transnzission line u,itlz t'oltaJ!.es impressed in series.
Eliminating V, we have
d 2 [
dz 2 = -{j2J - 2jwCE o , {j = wVLC = w. (7.53)
This is a nonhomogeneous differential equation and its general
solution is the sum of its particular solution and the general solution
of the homogeneous equation (Eo = 0). Equation (7.53) is satisfied
by a constant I = /0 which may be found from
_(32/0 - 2jwCE o = O.
Thus
[0 = 2 Eo/jwL.
Hence the general solu tion is
2E
A · 0
I(z) - cos {3z + B SIn {3z + :--,
JwL
I3A. (3B
V(z) = - SIn {3z - - cos (3z.
jwC jwC
Since V(O) = 0 and I(l) = 0, we find
B = 0,
2Eo
A =
- jwL cos (3(
and
2Eo 2Eo
I(z) = - cas {3z + -. (7.54)
jwL cas {3l jwL
From this current distribution we can obtain the scattered field.

H.eflection and sea tteri ng
261
This field will be approximate since the equations for V and I
are based on the assumptions that the wire is so thin that only the
principal waves need be considered and that the residual field and
hence the radiation may be neglected as far as their effect on the cur-
rent distribution is concerned. The effect of radiation becomes par-
ticularly important \vhen the length of the \vire equals >"/2. In this
case the first term in equation (7.54) becomes infinite. If the resist-
ance of the \vire had been included, the amplitude ,vould have been
large but finite. However, for good conductors the resistance of the
wire is not an importan t factor in limiting the amplitude of current.
It is the radiation resistance that is important. We may take it into
considera tion as follows.
As I3l approaches 1r/2, the first term in equation (7.54) becomes
more and more dominant. Thus we expect that when (jl = 1f'/2
and the length of the wire is equal to one half of a wavelength, the
curren t is given essen tially by the sinusoidal term
I (z) = C cos (3z
(7.55)
of some amplitude C. The dotted line in Figure 7.12 shows the shape
,
\
\
\
\
E \
, X
... I -
I 2
I
,
I
,
I
I
I
FIGURE 7.12 Scattering by a half-wave wire.
of this distribution. If Eo is the incident electric intensity at the
wire, the average work done per second by it, when driving the
current, is
/4 >..
p = . f EoC* cos [Jz dz = - EoC*.
- 21f'
(7.56)
rrhis must equal the radiated power which for this current distribu-
tion is given by equation (5.43),
P = 36.55 CC*.
(7.57)

262
ElccLronlagne/ic fields
By equating the two expressions, we find
XEo
c=
73.17r
and
XEo
I (z) = cos I3z.
73.17r
(7.58)
The distant scattered field may be obtained from equation (5.42)
by setting 10 = XE o /73.17r.
1"'hus at resonance the induced current and the incident electric
intensity are in phase and, sufficiently far off resonance, they are in
quadrature [see equation (7.54) J.
7.13 Scattering by a half-wave receiving antenna
Power may be absorbed from a plane wave if a wire is broken in the
middle and the ends are connected to a resistance or, in general, to
some impedance. See Figure 7.13. The maximum power is absorbed
\
\
\
B \
\
,
)IL I A
I -
A , 2
I
I
I
I
I
I
FIGURE 7.13 Scattering by a Italf-wave receiving antenna.
\vhen this impedance is the conjugate of the impedance which \vould
be seen by a generator across these ends ,vhen the antenna is used for
radiating po\ver. In such a case the reactance is "tuned out" and the
resistances are "matched." The half-wave antenna already has zero
reactance so that the load resistance should equal 73.1 ohms [see
equation (5.44) J. The current in the half-wave antenna is still given
by equation (7.55); the average work done by the incident field
per second is given by equation (7.56). Part of this \vork is absorbed

Reflection and scattering
263
by the load and part is reradiated since the current in the antenna
generates a spherical wave. Thus instead of equation (7.57) \ve have
P = 73.1 CC*
and
I (z) -
XEo
cos I3z.
146.2
(7.59)
The intensity of the scattered field, equation (5.42), is only half of
that for a wire without a load.
7.14 Scattering in waveguides
Any obstacle in a waveguide, such as the wire in Figure 7.14, will
perturb the field of a passing wave. The perturbation can be expressed
as a series of modes and their amplitudes can be obtained from the
boundary conditions at the surface of the obstacle.
y
..",.
"
/'
t:t:
o
.,
","
'"
,/
"
,,"
a
Z
FIGURE 7.14 Scattering by a transverse wire in a rectan.gular 'luat1eguide.
Suppose that a dominant wave (TE lO ) impinges on a perfectly
conducting transverse \vire of radius c. Let the, maximum amplitude
of this \vave be A so that
E = A cxp (-j/31Z) sin (7rx/a),
131 = V ffl - (71" / a) 2. ( 7 .60)
Assuming that the axis of the wire is at z = 0, we conclude that the

264
Electromagnet£c fields
sea t tered fIe ld \vill be of the form (6.113). 1'1h us
CD
E: = - L: Emexp (-rmz) sin (m7rx/a),
m-l
z > c,
(7.61)
ex>
- - L: Em exp (fmZ) sin (nx7rx/a),
m-l
z < -c,
,vhere
r m = vi (m7r/a.) 2 - (32.
In the region -c < z < c the fIeld will consist of modes propagating
in both directions because the current in the \vire is distributed on
its surface and it may be subdivided into infinitely thin current fIla-
ments, each of \vhich excites modes propagating in opposite directions.
On the surface of the wire the total Eu must vanish. This is the
condition from which the coefficients Em will be determined. If the
wire is thin, E and therefore E: \vill be substan tially uniform on its
surface. One \vould expect that the scattered field due to current I
in the wire is nearly equal to the field which would be generated if I
\vere on the axis of the wire. Therefore, the coefficients in equation
(7.61) are given approximately by equation (6.114), and outside
the wire for z > c,
CD m7r mrX
E: = - 'L,a-1Kmlexp (-rmz) sin sin-. (7.62)
m-l a a
For z < -c, z should be replaced by -z. '"fo the extent to which E
is the same round the wire, it can be calculated at some point of our
choosing, at point (xo, c), for example. The intensity of the incident
fie ld \\rill be taken equal to that on the axis of the \vire
Eo = A sin (7rxo/a). (7.63)
Thus from the boundary condition we have
co
Eo = I 'L, a-1Km exp (- r me) sin 2 ( nz7rX o/a).
(7.64)
m-l
rrhU5 I may be expressed in terms of either Eo or A.
The ratio
co
Z = Bob/! = 'L, {b/a)Km cxp (- rmc) sin 2 C l1l7rX o/a)
m-l
is the impedance seen from the \\rire. If a < A < 2a., only the 1'1ElO
mode ,viII be traveling. rrhc remaining modes will be attenuated and

Reflection and scattering
265
will represent a local field associated with the current in the wire.
Only K 1 \vill be real
Kl = WJ.L/111,
1"'herefore
K m = j WJ.L j r m for In = 2, 3, ....
Z = R + jX,
where
R = ;: sin 2 ( 7rx o/a) = :b [1 - ( ;a yr i sin 2 ( 7rx o/a),
(7.65)
(X)
jX = (jwJ.Lbja) L: r;lexp (-rmc) sin 2 (m7rxo/a).
m-2
To the \vire each half of the guide will appear as a transmission line
of characteristic impedance
2"1 b [ ( X )2]-j
K = 2R = ---;; 1 - 2a sin 2 ( 7rx o/a).
(7.66)
The wire itself will have an inductive reactance X. See Figure 7.15.

jX
K
..
. K
FIGURE 7.15 Equit'alent transmission line and shunt reactance
representi1ZK the physical situation in Fillre 7.14 under certain
conditions.
The series for X converges slowly. However, an approximate value
may be obtained from the nearly static case and the correction term
may be expressed as a more rapidly converging series. From equa-
tions (6.81) and (6.82), \ve have
jWJ.LI (X) 1 . l1Z,7rXo . tn7rX
E II (x, z) = - - L: - e- m7rz / a SIn - SIn -.
7r m-11n a a
(7.67)
In this case, the reactance of the wire is
Eob/I = -E1Jb/I = jwLo,
J.Lb 00 1 m7rXO
Lo = - L: e- m7rc / a sin2.
1r m-lm a
(7.68)

266
lle(lronl(lg'let icfi cld s
It is possible to sunl the series (7.67). l"he product of the sines may
be expressed in terms of exponen tial functions
. nZ7rXo . l1Z7rX 1117r(X - xo) l}l7r(X + Xo)
SIn - SIn - =  cos - . cos
a a a a
= } cxp [j nZ7r (x - Xo) / a J + ....
l"hese exponentials we substitute in equa tion (7.67). Setting
7r
W = - (z + jx),
a
.
J 7rX o
Wo=-.
a
(7.69)
we obtain
jwJJ./ 00 1
E y = - - L - {exp [-1n(w - wo)J + exp [-1n(w* - wti)J
47r m-11n
- exp [ -nt(w + wo) J - exp [-nx(w* + wri) Jl.
Since
ex> 1
L pm = - In (1 - p),
m-1 1 11-
we have
jwJJ./ [1 - cxp (-w - wo)J[l - exp (-w* - wti)]
Ell = - -In
47r [1 - exp (-w + wo) J[1 - exp (-w* + wti) ]
jwJJ./ sinh .! (w + wo) sinh  (w* + wri)
- - - In
47r sinh .(w - wo) sinh (w* - wri)
jJJ./ sinh  w + Wo I
- - -...:- In .
27r sinh  I w - wo I
(7.70)
From equations (7.69) we flnd
7r 7r r l
I w - Wo I = - y z2 + (x - XO)2 = ,
a a
7r 7r
I w + Wo I = - Y Z2 + (x + XO)2 =
a a
where r1 and 72 are, respectively, the distances from a typical poin t
(x, z) to the point (xo, 0) through which the current fIlament is

Reflection and scattering 267
passing and from its mirror image in the plane x - 0 to the same
point. Hence
jwJJ.! sinh (-rrr2/2a)
E" = - - In (7.71)
27r sinh (7rrl/2a)
Setting 11 = c and '2 = 2xo, a closed form for the inductance of the
wire is obtained at frequencies for which X » 2a,
JJ.b sinh ( 7rX o/ a)
Lo = In (7.72)
27r sinh (7rc/2a)
In this formula Xo should be greater than 2c. Otherwise the current
distribution round the wire will depart too much from uniformity on
account of the proximity to the wall of the guide. If 7rxo/a « 1, the
hyperbolic functions may be approximated by their arguments and
IJ.b 2xo
Lo = In -. (7.73)
211" c
'fhis is the inductance of the wire near a single conducting plane.
In our scattering problem a < X < 2a. The major difference be-
tween the impedance (7.65) seen by the wire in this case and the
low-frequency impedance (7.68) is in the first term of the series. In
the low-frequency case the first term is reactive while for the fre-
quencies we are considering it is resistive. The series for the reactance
in equations (7.65) begins with 111- = 2. The first approximation to
this reactance may be obtained by setting r m = m7r / a for m > 2.
Thus we obtain the series (7.68) except for the first term so that
. . ( JJ.b . 7rxo)
JX = JW Lo - ;- e- rc1a 5m 2 -;; .
(7.74)

8
Coupled Oscillations
8.0 Introduction
Starting with oscillations in coupled circuits, we develop the idea of
concentrated coupling between modes of oscillation in distributed
circuits and then the idea of distributed coupling between such
modes. This idea of coupling between modes of oscillation leads to
the representation of structures with distributed circuit parameters
by equivalent networks with lumped circuit parameters.
8.1 Oscillations in two coupled circuits
In Chapter 3, we considered several examples of a periodic transforma-
tion of electric energy into magnetic, and vice versa when the elec-
tric energy is concentrated mostly in one region and the magnetic
energy in another. Next in complexity is the case of two separate
concentrations of electric energy, Figure 8.1 (a, b). Case (a) suggests
three separate coils with three corresponding concentrations of mag-
netic energy, one of which is common to both circuits. Case (b) sug-
gests two coils with overlapping magnetic fields. 1"he circuit equations
are the same for both cases
( jWLll + . 1 ) II + jwL1212 = 0,
JwC n
jWL12Il + (j wL 'l2 + . 1 ) /2 = 0,
JWC 22
where in the first case L 11 = L 1 + L.v, 2 =  + L.w, and in the
second, La = L 1 , L 22 = L 2 . In both cases, L 12 = LM, en = C 1 , and
C 22 = C 2 .
From these equations we obtain the ratio of the currents
(8.1 )
II
- -
1 2
w 2 L 12 C ll
1 - w 2 L 22 C 22
w2L12C22
(8.2)
1 - w 2 L u C ll
268

Coupled oscillations
269
£1 L 2
Cl  C 2 Cl  G' C 2
£1 £2
(a) (b)
FIGURE 8.1 CouPled circuits.
I Hence,
(w 2 L ll C ll - 1) (w2L22C22 - 1) = w4Ll2CllC22,
and
LllC ll + L 22 C 22 :f: V(LUC ll - L 22 C 22 )2 + 4Li 2 C ll C 22
. (8.3)
2(L ll L 22 - L12)C ll C 22
There are two natural frequencies. For each frequency we may
obtain the ratio of the currents in the two meshes from equation
(8.2) .
2 _
Wl,2 -
8.2 Beats
An important special case arises when the natural frequencies of
two coupled circuits are equal in the absence of coupling
LUC ll = 2C22 = w(i'"2.
(8.4)
In this case, equation (8.3) becomes
w5 (1 ::I:: k)
2 _
WI,2 -
1 - k 2 '
where
L 12
k=
v' LI1L22
(8.S)
is the coupling coefficient. Hence,
Wo Wo
WI = VI _ k ' W2 = V i + k'
One of these natural frequencies is higher than the frequency of the
uncoupled circuits, and the other is lower.
(8.6)

270
llectronlaglle/ic fields
For the higher frequency WI the curren t ratio is
1 1 /1 2 = -VI J 22/L u = -p
(8.7)
and for the lo\vcr frequency
J 1 /1 2 = VL22/Lll = p.
(8.8)
tor the lo\ver frequency the relative directions of currents in the
two meshes are as shown in Figure 8.1. For the higher frequency
the direction of current in one mesh is opposite. l"hus, if A is the
complex amplitude of the current in the second mesh for the higher
"lode of oscillation then.
II (t) = - pA exp (jw1t),
1 2 (t) = A exp (jwlt). (8.9)
Similarly, if B is the complex amplitude of the current in the second
mesh for the lower mode of oscillation, then
II (t) = pB exp (jW2t),
1 2 (t) = B exp (jW2t). (8.10)
The two modes of oscillation may exist simultaneously, and in
general
II (t) = - pA cxp (jwlt) + pB exp (jw 2 t) ,
and
(8.11)
1 2 (t) = A exp (jwIt) + B exp (jW2t).
If B = - A, then there is no current in the second mesh at the
instant t = O. At any other time
II (t) - - pA [exp (jw1t) + exp (jW2t)]
- -2pA exp [j(WI + w2)t/2] cos  (WI - W2)t,
1 2 (t) = 2j A exp [j(WI + w2)t/2] sin (WI - W2)t.
(8.12)
Thus, the oscillations appear to be taking place with the angular
frequency !(Wl + W2), while the amplitudes of the currents are
varying periodically \vith the frequency  (WI - W2). \Vhen the
amplitude of the current in one mesh is zero, the amplitude of the
currcnt in the other mesh is maximum. In addition to a pcriodic
transfer of clectric energy into magnetic, we have also a periodic
complete transfer of energy from one mesh to the other.
In each mode of oscillation, however, the energy is evenly divided

Coupled oscillations
271
bct\veen the two nleshes. To show this, multiply equation (8.7) or
equation (8.8) by its conjugate
lIlt /1 2 It = L 22 / Ln-
Hence,
 LuJllt =  L 22 / 2 1f.
.
(8.13)
Thus, the maximum energies stored in the coils are equal. l"'he
amplitudes of II and 1 2 are constant and hence, in each mode of
oscilla tion there is no transfer of energy between the meshes. Thus,
equa tion (8.13) reprcscn ts the total energy in each circuit.
If LI = L 2 = L, P = 1, and we have a symmetric circuit, Figure
8.2, which can be easily analyzed from symmetry considerations. In
L L L L
c
'0
c
c
f?;
c
(a) (b)
FIGURE 8.2 Two modes of oscillation in s)'nzmetric coupled circuits.
case (a) the inductance LM may be considered as two inductances in
parallel, each equal to 2LM' 'I'he two halves of the complete circuit
may then be separated without disturbing the oscillations. The total
inductance in each half is L + 2I.Jv, and the natural frequency is
1
w = v' (L + 2 /.JM
(8.14)
In case (b) there is no current in the mutual inductance L. u , and the
same current flo\vs through both inductances. 'fhe inductance of the
big mesh is 2L and the capacita:nccC. Hence, the natural frequency
-
IS
w=
1
vrc'
(8.15)
8.3 Concentrated coupling between sections of transmission lines
'fhe exchange of energy bet\vecn two identical coupled circuits takes
place no matter how small the coupling coefficient is [see equation
(8.5) J. The smaller k is, the lower is the frequency of exchange.

272
Electrol1zaglletic fields
Consider now a long section of a two-wire transmission line, "shorted"
in the middle. The shorting bar has a small inductance, quite neg-
ligible in comparison with the total inductance of each scction; but it
is common to both sections and provides coupling between them.
Thus, the originally equal natural frequencies will be slightly altered
-- -
.......
---
-
."",
:/
'..........
'Ill
III  --=- -=- .:- ; -:: 
.........-.
--
---  .-.. -
(a)
.,..- -- -------- -- .-.
..,." -.....
."", .........
: .. ........
I I :
...... .... .",
.......--- ".
-- ---------- ---- --
- --
(b)
FIGURE 8.3 Two modes of oscillation in a two-wire transmission line U shorted"
in the middle.
and there \vill be two modes of oscillation as indicated in Figure
8.3 (a, b). If oscillations are excited only in one section, there will be
a gradual transfer of energy to the other section. After all energy
has been transferrcd, the reverse flow of energy will begin.
8.4 An equivalent network for a shorted section of a uniform non-
dissipative transmission line and its admittance in terms of
resonant frequencies
Let us consider a uniform nondissipative transmission line of length l
\vhich is connected to a generator of zero internal impedance at z = 0,
and shorted at z = t. Figure 8.4(a). Let the impressed voltage be
Vi exp (jwt). '"rhus, if Vi = 0, \VC have a section shorted at both
ends, Figure 8.4(b). 1"hc complex amplitudes of the transverse volt-
age and longitudinal current V (z) and I (z), satisfy equations (4.22)

Coupled oscillations
273
I i Vi  V(Z) t
G enerator
z=o
I(z)
II Ilv(z) t
I(z)
II

.
z I z=Q
z l
(a)
(b)
FIGURE 8.4 (a) A transmission line driven by a generator with zero internal
impedance (so that the impressed voltage is independent of the line impedance)
and shorted at the far end z = 1; (b) a line shorted at both ends.
in the interval 0 < z < l. These equations are
dV
- = -J.wLI
dz '
dI
- = -jwCV.
dz
(8.16)
In Chapter 4 the solutions were obtained in a form particularly
suitable to the analysis of wave propagation in transmission lines.
In this section we shall express the solutions in a different form and
obtain a network with lumped circuit parameters which is equivalent
to the line section with distributed circuit parameters.
Whatever the current I(z) may be, it can be expressed as a cosine
.
serIes
CJ)
I (z) - L In cos (n7rz/l),
n-o
o < z < l,
(8.17)
where
10 = 1- 1 [ l(z) dz,
o
(8.18)
l
1" = (211) 1 I (z) cos (n7rzll) dz.
o
Let the voltage be expressed as a sine series so that the boundary
condition V (l) = 0 is satisfied automatically. Thus,
CJ)
V (z) = L V n sin (n7rz/l) ,
n-l
o < z < l,
(8.19)
where
V" = (211) t V (z) sin (n1rZ/1) dz.
o
(8.20)

274
Eleclro111ag11etic fields
Since V(O)  0, the sine scries (8.19) does not converge uniformly
in the closed interval (0, l). At z = 0, there is the condition
V(O) = Vi
(8.21 )
to supplcment the series for other values of z.
Taking I(z) from the first equation in the set (8.16), and sub-
stituting in equations (8.18). \ve have
1 i l dV 1
10 = - - dz = --:-- [Vel) - V(O) ]
jwLl 0 dz JwLl
Vi
jwLt'
(8.22)
2 jl dV n7rZ
In = -- cos dz.
jwLl 0 dz l
Integrating by parts, we find
1ft = -  [vel) cos n1r - V(O) + 11-1r jl V(z) sin ll7rZ dZ].
JwLl l 0 l
Using equations (8.20) and (8.21), we obtain
2
I =_(Vi__l n7r V). (8.23)
n . It 2 n .
}w oJ
Similarly, the second equation in the set (8.16), and equation
(8.18) may be used to express the V n in terms of the In
V.. = - jl dl sin 1t7rZ dz
jwCl 0 dz l
- - rI(l) sin n;' - /(0) sin 0 - 1l7r jl I(z) COS 1Z7rZ dZ].
}wCl l 0 l
The lattcr integral is given by equations (8.18). Thus,
n7rI n
V" = . (8.24)
jwel
Substituting this in equation (8.23), \ve find
Vi
I - (8.25)
" - (jwLl/2) + Cn 2 7r 2 /2jwCl).
Thus, all the coefficients in the series for I (z) and V (z) have been
determined.

Coupled oscillations
275
In particular the input currcn t is
CX)
I(O)=2:/n
n-o
Vi CX) 1
= - + Vi 2:
jwLl n-I (jwLl/2) + (n 2 7f2/2jwCl)'
and the input admittance
y. _ /(0) _  +  1
In - Vi - jwLl  (jwLlj2) + (n 2 7r 2 j2jwCl). (8.26)
The first term is the admittance of an inductance Lt. The typical
term in the summation is the admittance of an inductance Ll in
La
LI
Cl
L 2
C 2
L3
C 3 eft
------L
Ln
FIGURE 8.5 An equit'alent network for the Physical transnzission line shorted
at the far end.
series with a capacitance 2Cl/n 2 7f2. Hence, Y in is the admittance of
the network shown in Figure 8.5, whcre
Lo = Ll,
Ln = Ll,
2Cl
C --
n - ·
'n 2 7f2
(8.27)
The input admittance becomes infinite, and the impedance zero,
at the following frequencies
wo = 0,
Wn =
1
V LnC n
n7r
(8.28)
lva.
1'lhese are thc resonant frequencies of the transmission line section in
Figure 8.4(a), and also the natural frequencies of the section shorted
at both cnds, Figure 8.4(b). In terms of thesc frequencies ,ve have
1 co 2jw
y. --+ L
In - jwLl "-1 Ll(w - w 2 ).
(8.29)

276
Ilectro111agl1etic fields
As W  0, \ve have,
1 2jw L:a) 1 1 jw2Cl L:a) 1
y. -+- --+ -
In · Ll Ll 2 · Ll 2 2 ·
JW . 71-1 W n JW 1(' ,,-1 n
Since
a) 1 7l"2
"""-=-
L..J 2 6'
n-l11-
\ve have
1
Y in  - +jwCl.
jwLl
This low-frequency equivalent circuit for a shorted section of the
line was obtained directly from energy considerations in Section 3.6.
\\.7ith the aid of this result, equation (8.29) may be expressed as
(8.30)
1 t 2jw [ 2 1  ]
Y in =  + .jwCI + w 2
JwLl ,,-1 Ll Wn-
(8.31)
1 CD 2jw 3
= - + 'UwCl + E Llw(w -
jwLl · ( 2 )
In this expression, the series is more rapidly convergent than the
series in equation (8.29).
8.5 Another equivalent network for a shorted section of a uniform
nondissipative transmission line and its impedance in terms of
antiresonant frequencies
A different equivalent network may be obtained if a different set
of functions is used to represent fez) and V(z). Suppose the voltage
and the curren tare represen ted by the following series:
 (2n + l)7rz
V (z ) = £...J V n cas ,
n-O 2l
o < z < l,
(8.32)
I(z)
a) . (2n + 1) 7rZ
-L:I"sln ,
2l
o < Z < l,
n-O
= Ii
Z = 0
,

Coupled oscillations
277
where Ii is the input current. The cosine functions have been
selected to satisfy the boundary condition V (l) = O. This time it is
the current that is represented by a nonuniformly convergent series.
The coefficien ts in these series are
2 l' (2n + l)1rZ
V n = - V(z) cos dz,
l lJ 2l
(8.33)
2 j' (2n + l)7rz
I n = - I (z) sin dz.
l 0 2l
In this representation there is a fixed input current which ideally
can be obtained when the internal impedance of the generator is
infinite so that the impedance of the line has no effect on the input
current, Figure 8.6(a). When Ii - 0, the line is open at z - 0,
1+ Ii  V(z) t
z=o
l(z)

II V(z) t
z 1 z 0
l(z)
I I
II-
z=l
(a)
(b)
FIGURE 8.6 (a) A transnzission line driven by a generator of infinite internal
impedance (so that the input current is independent of the line adnzittance) and
shorted at the far end z = 1; (b) a line open at one end and shorted at the other.
Figure 8.6(b). This is dual of the case in the preceding section where
the internal impedance of the generator was assumed to be equal to
zero so that the impedance of the line had no effect on the input
voltage All real generators have a finite internal impedance. If this
impedance is explicitly inserted in series with the external circuit,
\ve have in effect a generator of zero impedance. Similarly, if the
admittance of the generator is explicitly inserted in parallel \vith the
external circuit, we have in effect a generator of infInite impedance.
From equation (8.32), we obtain the input voltage
(X)
V(O) = L V n ,
n-o
and the input impedance
co
Zin = V(O)/Ii = L Vn/Ii.
n-Q
(8.34)
The method of expressing the V n and the In in terms of Ii is the

278
Elec/ronlagnc/ic fields
same as in the preceding section. "Tc substitute into equations (8.33)
the expressions for 1" (z) and T (z) in terms of the derivatives dI (z) jdz
and tll' (z) jllz from equations (8.16). 1"'hc integrals arc differentiated
by parts and equations (8.33) are used once marc. "rhus. we obtain
Ii
V =
n jw(Glj2) + [(2n + 1) 2 7r 2j8jwl.J]'
(8.35)
(2n + l)7rV n
In = .
2jwLl
l"'hc input impedance can nO\\T be expressed as
00 1
Zin = L
1l=OjWC n + (ljjwL n ) '
(8.36)
C'n
Cl.
Ln = 8Llj (211, + 1) 27r 2 .
"rhis is the impedance of parallel resonan t circuits connected in series,
Figure 8.7.
Lo L 1 L 2 L 'J
-----
00
o
Co
Cl
C 2
en
FIGURE 8.7 A n alternate equivalent net1.L'ork for a transrnission line shorted
at the far end.
"rhe input impedance becomes infinite at the resonant frequency
of each parallel circuit [which is also the natural frequency of a cc-
tion open at one end and shorted at the other, Figure 8.6(b) J. "fhesc
frequcncics arc
W n =
1
v LnGn
(2n + 1)7r
2lVLE ·
(8.37)
l'hcrefore
,
ex) 2jw
Zin = L 2 .
n...Q Cl (W n - w 2 )
(8.38)

Coupled oscillations
279
As w -+ 0, this impedance approaches
2jw  1 8jwLl  1 .
Z in = - L..J - = £...J = JwLl
C l ",-0 W 7r 2 n-O (2n + 1) 2
( 8.39)
since the sum equals 7r 2 /8. Here Ll is, of course the direct current
inductance of the loop. "fhe input impedance can now be expressed
as follo\vs:
00 2jw (1 1 )
Z. = J'wI.J + '"' - - -
m £...J Cl 2 ? 2
n-O Wn - W" W n
(8.40)
'"' j8I.Jw 3
= jwLl + L..J . 2
n-O (2n + 1)21f2(w n - w 2 )
This is the impedance of the nct\\'ork sho\\'n in !1"'igure 8.8. '1"'he
primary circuit is the transmission line section, regarded as a simple
C' C' C' C'
1 2 3 n
.1 LI
2
1.. LI
2
1. Ll
2
!LI
2
LM,l
L M ,2
L M ,3
LM,n ----
-TTT
A B
<Xl
o
FIGURE 8.8 A third type of equivalent network for a /ralls11z£ssion line shorted
at the far end.
loop of inductance Ll \vhich is coupled to parallel resonant circuits
for \vhich
C ' -
n -
8el
.
(2n + 1)21J"2
2/.J
LM,n = (2n + 1)'/1"'
(8.41)
l"'he impedance between .4 and B is
('X) J.w 3 L1f
Z . - . Ll '"" .' · n
In - JW ,+ L..., (1['2) ( 2 _ 2).
n-O .J I W", w
(8.42)

280
Electromagnetic fields
8.6 Equivalent networks for nonuniform transmission lines
In this section network equations will be derived for a section of
length 1 of a nonuniform transmission line, Figure 8.9. Let us assume
V(z, t) t
I (z, t)

z 0
It
E i(z, t)
z 1
FIGURE 8.9 A nonuniform transmission line, shorted at both ends, with a
distributed 'impressed series voltage varying arbitrarily with time.
a general time-variable case, and a distributed impressed series
voltage. Thus, let
Ei(Z, t) = impressed voltage per unit length,
I (z, t) = longitudinal current,
V (z, t) = transverse voltage,
g(z, t) = electric charge per unit length.
In this case the transmission equations are
av(z, t)
dZ
aI(z t)
- -L(z) a; + Ei(Z, t)
av(z, t)
- -C(z) .
at
(8.43)
al (z. t)
az
(8.44)
Since
q(z, t) = C(z) V(z, t),
equation (8.44) may be written as
aI (z, t) aq (z, t)
- -
az at
If the line is shorted at both ends
(8.45)
(8.46)
therefore,
V(O, t) = V(l, t) = 0;
q(O, t) = gel, t) = o.
(8.47)
At any instant, I (z, t) may be expressed as a cosine series, and

Coupled oscillations
281
at all instants by such a series with coefficients depending on time
ex)
I (z, t) = L In (t) cos (n1rzjl).
(8.48)
n-O
In this representation, lo(t) is the current circulating round the loop,
and I n(t) the current passing at the antinodes of the 'nth space
harmonic, taken by itself. The current I net) equals qn(t), the time
rate of charge qn (t) passing through these antinodes.
Substituting from equation (8.48) into (8.46), we have
aq(z, t) 00
= L (n7rjl)qn(t) sin (n7rzjl).
at n-l
Integrating from t = 0 to t = t we find,
ex)
q(z, t) - q(z, 0) = L (1t1r/l)qn(t) sin (n7rz/l)
n-l
ex)
L (n7r/l)qn(O) sin (n7rz/l).
n-l
Hence,
CX)
q(z, t) = L (n7r/l)qn(t) sin (n7rz/l) + Q(z),
(8.49)
n-l
,vhere Q(z) is a static charge distribution maintained by a static
impressed field. This distribution may be found from equations
(8.43) and (8.45). 1'1hus,
d Q(z) i
-- - E (t'7)
- et I.J .
dz C(z)
Suppose that such distribution, if any, has been found and deleted
from the time-variable distribution (8.49). Then,
00
q(z, t) = L (n7r/l)qn(t) sin (n7rz/l).
(8.50)
tI.-l
Equations (8.44), (8.46), and the boundary conditions are satis-
fied on account of our choice of time variables qn(t) and the forms
of l;'ourier series. It remains to satisfy equation (8.43). First we
substitute from equation (8.45)
a rq(Z t)] al(z t)
- ' + L(z) , = Ei(Z, t).
az _ C(z) at
This equation must be satisfied for all z. To ensure this we expand

282
Electrontaglletic fields
both sides in a cosine series in z, and equate the coefficien ts. Thus,
1 1 (a rq(Z, t)] aI (z, t») nZ7rZ
- + L(z) cos - dz
o az C (z) at I
1 1 nZ7rZ
= 0 Ei(Z, t) CDS l dz,
11t = 0, 1, 2, ....
(8.51 )
The first part of the integrand on the left may be integrated by parts
l ' l1l7rZ a [q(Z, t)] nZ7rZ q(z, t) I
cos-- dz = cos-
o I az C (z ) l (' (z ) 0
l ' (m7r/l) sin (m'7rz/l) q(z, t)
+ .
o C(z)
The first tcrnl on the right vanishes, and equation (8.51) becomes
j 1 (m7r/I) sin (l1t7rz/l)q(z, t) l ' m7rZ a/(z, t)
dz + L(z) cos - dz
(; C (z) 0 l at
l ' tl17rZ
= Ei(Z. t) cos - dz.
o l
Substituting from equations (8.48) and (8.50), we obtain
Looi 0 (t) + LOli I (t) + 1 4J2 i 2 (t) + ... = 11  ( t)
. [. ql(t)]
LmJo(t) + LmII1(t) + -
C ml
(8.52)
+ [L m2 i 2 (t) + q2(t) ] + ... _ V (t)
C m2
for 11t = 1, 2, 3, ..., w here
1
LOrn = 1 L(z) CDS (m7rz/l) dz,
o
Lmn = 1 1 L(z) CDS (rIl7rz/l) CDS (1t7rZ/l) dz, rIl, 11 = 0, 1, 2, ...,
o
(8.53)
. l' sin (nz7rz/l) sin (n7rzjl)
l/C mn = dz,
o (12/1111t7r 2 )C(z)
1'1t, 11, = 1, 2, 3, ...,
I
V(t) = f. Ei(Z, t) CDS (m7rz/l) dz,
o
m = 0,1,2, ....

Coupled oscillations
283
Equations (8.52) are the mesh equations for a network. The
induct3.ncc and capacitance in the1nth mesh are given by Lmm, and
e mm . l"'he mutual inductances and capacitances are Lmn, and em"
when 1n  ft.
8.7 Lagrange's equations in circuit theory
Equations (8.52) can be derived from Lagrange's equations if the
kinetic energy is equated to magnetic energy em, and the potential
energy to electric energy 8 e . Both 8m and 8e should be expressed in
terms of the" generalized coordinates" of the system, qn(t), and the
"generalized velocities," I n(t). Thus,
1 l
8m = - f L(z)[I(z)]2 dz.
2 0
(8.54)
By substituting from equation (8.48), we have
1 fi - mz 1tZ
em =- L L(z)/m(t)/n(t) cos - cos - dz
2 0m ,n l l
= ! L Lmnlm(t)In(t),
m.n
(8.55)
where the inductance coeffIcients are given by equations (8.53).
Similarly,
1 fl [q(z) J2
ee = - dz
2 0 C(z)
= 1 L q",(t)q,,(t)
2 71l, n Cm n '
(8.56)
where the capacitance coefficients are also given by equations (8.53).
1"'he "generalized forces" arc obtained from their definition
oW = L V(t) oqm
( 8.57)
m
where oW is the work done by the impressed electric intensity during

284
llect r0111ag1lcl ic fields
a virtual displacemcn t of the system (Oql Oq2, ...). Since
l
oW = f. Ei(Z. t) [I (z. t) ot] dz
o
and
I (z, t) at = L qm (t) at cos (nl7rz/l)
m
- L cos (1117rZ/l) aqm,
m
we have
l
v (t) = f. E; (z, t) cos (11I1l"z/l) dz. (8.58)
o
Lagrange's equations are
 [o(Sm - Se)] _ a(Sm - Se) _ .
V (t).
dt aIm aqm
By substituting from equations (8.55) and (8.56), we obtain the
network equations (8.52).
(8.59)

9
Generalized Telegraphist's Equations
9.0 Introduction
Developments in this chapter run parallel to those in the preceding
chapter. Instead of coupling bctween circuits we consider coupling
between transmission lines. Instead of coupling between modes of
oscillation we have coupling between modes of wave propagation.
Mathematically this idea of coupling between modes of propagation
leads to the conversion of l\1axwell's partial differential equations
with given boundary conditions into a set of ordinary differential
equations analogous to the equations for coupled transmission lines
developed by Lord Kelvin, the so-called" telegraphist's equations."
Kelvin's equations are approximate while the generalized telegra-
phist's equations obtained in this chapter are exact.
9.1 Coupled transmission lines
The transverse voltage and longitudinal current in two-conductor
transmission lines satisfy a simple pair of linear differential equa-
tions (4.22). From the results obtained in Sections 2.15, 2.16, and
2.17 it may be concluded that for three conductors, Figure 9.1, we
should have the following equations:
dV I
dz = -ZuIl - Z121 2 ,
dI I
- - - Y n VI - Y 12 V 2 ,
dz
(9.1)
dz
-Z12I 1 - Z2212'
dl 2
--
dz
- Yl2 VI - Y22 V 2 .
dV 2
The cocfficients in these equations depend on the geometry and
physical properties of the system. The equations are obtained from
Maxwell's equations in precisely the same manner as for two con-
ductors. Thus the Faraday-lVIaxwell equation is applied to rectangular
285

286
Electronlaglletic fields
I
V 2 t ..- /2
/2 ...
I
Vi t  II
/1 .,
....
E :
I
I
c: 1
:
F
D
A
z
FIGURE 9.1 Three parallel wires, constituting a s')'stenl of two couPled trans-
mission lines.
circuits ABCD and DCEP to obtain the rates of change of VI and
V 2 with z. The Ampere-l\1axwell equation is needed to obtain the
magnetic intensity of the field and the magnetic current linked with
these circuits. The rates of change of II and /2 with z are obtained
by calculating the transverse leakage currents.
For It + 1 conductors there are n pairs of equations in which the
derivatives dV m/ dz are linear functions of II, 1 2 , ... and dI m/ dz
linear functions of VI, v 2 , .... The coefficients Zmm are called the
series self-impedances per unit length and Zmk, m  k, the mutual
impedances, also per unit length. Similarly, Y mk is the self or mutual
shunt admittance per unit length, depending on whetherm = k or
111,  k.
If the coefficients are independent of z, equations (9.1) possess
exponential solutions
VI = Ae- rz ,
II = Be- rz ,
(9.2)
V 2 = Ce- rz ,
1 2 = De- rz .
The equations for the unknown constants are obtained by substitu-
tion in equations (9.1)
r A = ZllB + Z12D,
rB = YllA + YI2C,
(9.3)
rC = ZI2 B + Z 22 D,
rD = Y 12 A + Y 22 C.
These are homogeneous equations in A, B, C, and D. They possess
nonvanishing solutions only when the determinant of the system
of equations vanishes. Thus we shall have a fourth-order equation
for the propagation constant r. In fact this equation will be quadratic
in r 2 so that if r is a solution, then - r is also a solution. Physically,
this was to be expected since the propagation constants in both direc-

Generalized telegraphist's equations
287
tions should be the same. Thus there will be two distinct propagration
constants, r l and r 2 . For each value, the ratios B/ A, Cj A, and D/ A
may be determined from equations (9.3). The waves corresponding
to the t\VO values of r are two l1lodes of propagation. In the case of
n + 1 conductors therc will be 11, modes of propagation.
9.2 Weak coupling
Even for two coupled transmission lines the solution is rather
complicated. 1"he complcxity increases rapidly with the number of
transmission lines. Approximate solutions can be obtained in the
important special case of weak coupling when Z12 and Y12 are small in
comparison with Zll, Z22 and Y ll , Y22, respcctively. First the coupling
is neglected and each pair of equations is solved separately. Then
the solutions are substituted in the neglected terms and the resulting
nonhomogeneous equations are solved. This technique can be simpli-
fied still further as shown below.
Consider two coupled transmission lines, Figure 9.2. Suppose that
Kill
Kl
+
K 2
K 1 r 2
K 2
FIGURE 9.2 Two couPled transmission lines.
a voltage is impressed on one end of one transmission line while its
other end is terminated into its characteristic impedance KI. Let the
other line be terminated by its characteristic impedance K 2 at both
ends. In the first approximation
I1(z) = A e- r1 z,
VI (z) = KI Ae- r1Z ,
1 2 (z) = 0,
V 2 (z) = O.
(9.4)
The quantity -Z12I 1 in equations (9.1) is the series voltage per

288
Electromagnetic fields
unit length impressed on the second line. At z -  the elementary
impressed voltage is
-Z12Il() d = -ZI2Ae-rl d.
This voltage sees an impedance 2K 2 so that the current at z =  is
- (Z12/ 2K 2) Ae- r1t d,
and at other points
- (Z12/2K 2 ) A exp [- rl - r 2 (z - t) J d,
z > 
(9.5)
- (ZI2/2K2)A exp [-fit - r2( - z)J d,
Thus the total induced current is
- (Z12/ 2K 2) Ae- fz .!' eCf2-fllE d+
o
z < .
l
- (Z12/2K 2 ) A efz. J e-Cfl+fzlE d. (9.6)
z
Integrating as indicated, we have the total current produced by the
coupling voltages impressed in series,
Z I2 A [e- rlZ - e- r2Z e- ru - exp [ - fll - f 2 (l - z) J]
-- + .
2K 2 f 2 - r l r l + f 2
(9.7)
From equations (9.5) we find that the transverse voltages induced
in the second line are
- Z12Ae-rlEe-r2(Z-E) d,
z > 
(9.8)
+ Z12A e-rlte-r2(E-z) d,
z < .
Integrating with respect to  f.rom  = 0 to  = l, we obtain the
total transverse voltage due to the series coupling
[ e- rlZ - e- r2Z _ e- rlZ - exp [ - fIl - r 2 (l - z) J].
- !Z I2 A (9.9)
r 2 - r 1 r 1 + r 2
At z =  in the second transmission line we have an induced shunt
current - Y12V 1 d, one half of which goes to the left and the other
half to the right. From there on the induced current is propagated
along the line. Thus the elementary induced current is
-! Yl2 V l e- r2 (z-E> d = -! Y12KIAe-rlte-r2(Z-) d,
z > ,
(9.10)
! Y12 V 1 e- f2 (r z ) d =  Yl2K 1 A e-rle-r2(rZ) d,
z < .

Generalized telegraphist's equations
289
Integrating this with respect to  from 0 to l, we obtain the total
current in the line produced by the impressed elementary shunt
currents
[ e- r1o _ e- rz .
-!Y12K 1 A
r 2 - r 1
e- r1Z - exp [ - fit - r 2 (t - z) J ].
r 1 + r 2
(9.11 )
The sum of this and expression (9.7) is the entire current induced
in the second line by the wave traveling in the first
e-ru - e- rtZ
1 2 (z) = -!A(Z12K'2 1 + Y12K 1 )
r 2 - r 1
e- ru - exp [-r1l - r 2 (l - z)J
+ !A (Y12K 1 - Z12K"21) . (9.12)
r 1 + r 2
From equations (9.10) we obtain the elementary transverse
voltage induced in the second line if we multiply the expression for
z >  by K 2 and that for z <  by - K 2 . Then we integrate over the
interval (0, 1) and add to equation (9.9) to obtain V 2 (z).
In the same manner one can calculate the transverse voltage and
longitudinal current induced back into the first line by 1 2 (z) and
V 2 (z). This should be added to I1(z) and V 1 (z) in equations (9.4).
The correction would be of the second order in small quantities Z12
and Y 12 and in many practical applications is negligible.
There is one very important exception, however. If r 2 = r 1 , then
the first integral in the expression (9.6) equals z. In this case when
the length of the line is large, the induced V 2 (z) and 1 2 (z) will also
become large (for large z) and higher order approximations are
necessary. This" degenerate" case is best treated by the exact method
of Section 9 .1.
9.3 Directional coupling
In the preceding section it was concluded that the case in which the
propagation constants of the coupled lines are equal requires special
a t ten tion. In this case
Zll Y n = Z22 Y22,
or if the lines are nondissipa tive
LUC ll = L 22 C 22 . (9.13)
Consider first a subcase in which Lu = 2 = L and therefore ell =

290
Elcctronlagnc/ic fields
C 22 = C. Equations (9.1) becon1c
dV l . dI I
- = -jwLII - }wL I2 1 2 ) - -jwCV 1 - jWC 12 V 2 ,
dz dz
(9.14)
dz
-jWL1211 - jwLI2,
dI 2
dz
- -jwC 12 V 1 - jwCV 2 .
dV 2
-=
Assume exponential solutions of the form (9.2), substitute them in
equation (9.14) and obtain a set of linear algebraic equations of the
form (9.3). From there on proceed as outlined in Section 9.1 and
obtain the expressions for the propagation constants of two possible
modes of propagation and the corresponding amplitudes.
One may also take advantage of the symmetry of equations
(9.14). Adding the equations in the left column, and also in the right
column, we have
d(V I + V 2 )
dz
- -jw(L + L 12 ) (II + 1 2 ),
(9.15)
dell + ]2) .
== -Jw(C + C 12 ) (VI + V 2 ).
dz
Thus the sums of the transverse voltages and longitudinal currents
in two transmission lines are propagated \vith the phase constant
131 = wy(L + L 12 )(C + C 12 )
(9.16)
and velocity
VI =
1
v (L + L I2 )«(' + C 12 )
(9.17)
l'hc characteristic impedance is
K, = L + L l2 ,
C + C 12
Subtracting equations in the left and right columns, we have
d(V l - V 2 )
= -jw(L - L 12 ) (II - 1 2 ),
(9.18)
dz
(9.19)
dell - 1 2 )
= jw(C  C 12 ) (1'1 - V 2 ).
dz

Generalized telegraphist's equations
291
Thus the differences between the transverse voltages and longi-
tudinal currents are propagated with the phase constant and velocity
given by
V2 =
1
v eL - LIz) (C - CI2).
(9.20)
/32 = w y (L - L I2 )(C - C 12 ),
1"'he characteristic impedance is
K 2 = L - L12 ,
C - C 12
(9.21)
'l'he symmetry in equations (9.14) reflects the physical symmetry.
Consider for instance three equidistant parallel wires, Figure 9.3.
I . .
+ I ... -<41 I
-1 ...
21 0 .
-""'lIIII 
+ I ...... + I ....
I .
(a) (b)
FIGURE 9.3 Two n10des of propagat£on in symmetric coupled transmission
lines.
It is evident that these wires may support two independent modes of
propagation. In the mode (a) equal currents in the sante direction
in the outer wires return via the middle wire. In the mode (b) there
is no curren t in the middle \vire and the outer wires act as a trans-
mission line.
l{eturning to equations (9.15) and (9.19), \ve have
II (z) + 12(z) = Ae-jlZ,
TI(z) - I 2 (z) = Be- i {J2 Z ,
for waves traveling in the positive z direction. Hence
II(z) - 1A e- it3u + 1 Be- j (j2 Z
2 ,
(9.22)
/2(Z) - !Ae- it3u - Be-itJt.z.
Suppose that at z = 0 a voltage is applied across the first line but
not across the second line. For instance, in the case of three wires,
Figure 9.3, a voltage may be applied between the two lower wires

292
Electroma gn eit'e fields
and the terminal of the upper wire left floating. 1"hen /2(0) - 0
and B = A. Equations (9.22) become
II (z) = !A (e- if3u + e- if32Z ) = A exp [ -j ({31 + (32) z/2] cos  ({31 - (32) 
(9.23)
1 2 (z) = !A (e- if3u - e- if32Z ) = jA cxp [ -j ({31 + (32) z/2] sin  ({32 - (31) Z.
At z = 0, / 1 (0) = A and 1 2 (0) = O. At distance z = l such that
 I {32 - {31 l = !7r,
(9.24)
the current in the fIrst line equals zero and the amplitude of the
current in the second line is A. l"hus if the two lines are uncoupled
beyond z = l, the energy delivered to the first line at z = 0 ,viII have
passed entirely into the second line and there will be no wave in the
first beyond z = l. If the lines remain coupled, thc energy will start
flowing back into the first line. 1"hc phenomenon is similar to beats in
coupled circuits (see Section 8.3). In the present case it is called
directional COlt piing. As the coupling bet\veen the lines decreases,
{32 - {31 also decreases and l increases.
9.4 Waves in stratified media between perfectly conducting parallel
planes
In Section 6.8 \ve considered wave propagation in a homogeneous
dielectric bet,vcen two perfectly conducting parallel planes. Suppose
nov\" that the medium is nonhomogeneous. Let us keep the assump-
tions that the fIcld docs not depend on the y coordinate, Figure 6.1,
that the magnetic intensity is parallel to the planes, and that 1J. is
constant. Eliminating E and E" from equations (6.88), \ve obtain
 (C 1 all u ) +  (Cl all u ) = - w 2 JJ.111/'
ax ax az az
If E depends on both coordinates, it is easy to establish that this
equation has no solution in the product form
(9.25 )
II II = X (x) Z (z) .
(9.26)
The method of separation of variables no longer \vorks. Ho\vcver. if E
is a function of only one coordinate, then the variables can still be
separated.
Suppose, for example, that E = feZ) varies only in the direction of

Generalized telegraphist's equations
293
propagation. By substituting from equation (9.26) into (9.25) and
dividing by XZ, we obtain
feZ)  [ dZ ] = -w2E(Z) _  d 2 X .
Z dz feZ) dz X dx 2
This equation cannot be true unlcss the last term is constant just as
in the homogeneous casc. Referring to equations (6.91) we have
X (x) = cos (m7rx / a) , m = 0  1, 2, ....
The particular form and thc values of em are dictated by the fact that
Er, and hence X'(x) must vanish at the boundaries x = 0, a. The
equa tion for Z becomes
E(Z) :!: [ dZ ] = _ [W2E(Z) _ m 2 7r 2 ] Z.
dz E(Z) dz a 2
When E = E(X), equation (9.25) becomes
E(X) d [ dX] = -w2E(X) _  d 2 Z
X dx f(X) dx Z dz 2
This equation possesses solutions, cxponen tial in z,
Z = e- rz .
(9.27)
Hence
d [ 1 dX]
E(X) - - = - [W 2 ,uE(X) + r 2 Jx.
dx E(X) dx
This is essentially the same differential equation as equation
(9.27). The important difference is that r is not known and must be
determined from the boundary conditions
(9.28)
X'(O) = X'(a) = O.
(9.29)
There will be an infinite set of characteristic values r m and corre-
sponding characteristic functions X m which will satisfy equations
(9.29). These functions are orthogonal in the interval (0, a). To
prove this we take two solutions, X m and Y n, and write equation
(9.28) as
d [ 1 dXm]
---
dx f(X) dx
d [1 dXn]
dx E(X) dx
[ r2 J
- - W2 + E(;) X m ,
[ r2 J
- - w 2 p. +  Xn.
E(X)

294
Elcctronzu£nctic fields
1 ultiplying the first equation by -"Y n dx, the second by X m dx and
subtracting, we find that the left-hand side is a complete differential
[ 1 (dXm dYfl)] r; - r
d - X n - - "'(m - = XmX" dx.
E(X) dx dx E(X)
Integrating over the interval (0, a), we have
1 a
- (XnX:n - xmX) = (r - r)
E(X) 0
f a 1
- YmXndx.
o E(X)
In view of the boundary conditions (9.29) the left side vanishes and
f a 1
- Xm(X)Xn(X) dx = 0, if r n  r m .
o E(X)
This is as far as we can go \vithout specifying E(X). The complete
solution of the problem of stratified media depends essentially on our
ability to handle the second-order differential equation (9.27).
In general, onc has to resort to approximations or to numerical
methods.
9.5 Waves in completely nonhomogeneous media between perfectly
conducting planes
If E is a function of both coordinates the method of separation of
variables fails. In this case we shall not attempt to solve the partial
differential equation (9.25). Instead wc shall deal directly with the
first-order l\Iaxwcll's equations (6.88). Two are propagation equa-
tions, in the z direction,
aE x
--
az
aE z
-J.wIIII +-
r" JJ ax '
aH y
iJz
-jwEEx,
(9.30)
and one is a coupling equation between the transverse and longi-
tudinal fields,
1 all 11
Ez,= -.
jWE ax
For a given value of z, Ez and II v can be expressed as cosine series
in x. This follows from the theory of Fourier series. Hence this is true
(9.31)

Generalized telegraphist's equations
295
for all values of z when the coefficients are permitted to depend on z.
Thus we assume
Vo(z) ex)
E:r, = + L NmVm(z) cos (m7rx/a),
a m-l
lo(z) ex)
11 y = + L Nmlm(z) cos (m7rx/a).
b m-l
(9.32)
In these equations Vo(z) is the transverse voltage between the
planes and Io(z) is the current which would flow in a strip of \vidth
b (in the y direction) if the wave were purely transverse electro-
magnetic. The constan ts i.V m are chosen so that the average complex
power flow through the area ab is given by the following expression
ex)
p = ! L Vm(z)I:(z),
m-o
(9.33)
\vhich is always valid for any number of uncoupled transmission
lines. Since
bfa
p = ! J Ex H : dx dy
o 0
ex)
= ! Vo(z)I:(z) +  L !abl\l V m(z)I:(z),
m-l
we have
N m = (2/ab) = Iv.
(9.34)
l"'hus in the present case J.V m is independent of n.
Integrating equations (9.32) over the rectangle x = 0, a and y = 0,
b, we obtain
Vo(z) = b- I fbfa Ex dx dy = fa Ex dx,
o 0 0
(9.35 )
lo(z) = a-I fbfa H y dx dy = (bja) fa lly dx.
o 0 0
l\1ultiplying equations (9.32) by cos (1t7rx/a) and integrating over

296
Electronlaglletic fields
the same rectangle, we find that the terms for which m  n vanish.
Thus
f bfa 17rX fa n7rX
Vn(z) = N Ez, cos - dx dy = bN E,; cos - dx,
o 0 a 0 a
(9.36)
f bfa 1t7rX fa 1t7rX
In(z) = N H1I COS - dx dy = blV H1I cos - dx.
o 0 a 0 a
In the present case the field is independent of y and we could have
dispensed with integration in this direction. In waveguides, however,
the field is usually a function of both coordinates and the integration
has to be performed over the cross-sectional area. Furthermore, in
anticipation of the final form of the results we have chosen the
coefficients V n and In deliberately in such a way that their physical
dimensions are those of voltage and current. The preceding equations
give these coefficien ts in terms of the transverse field componen ts
just as equations (9.32) express the transverse field in terms of the
coefficien ts.
Differentiating equations (9.35) and (9.36) with respect to z,
we have
dVo(z) fa aE:r,
= -dx,
dz 0 az
dlo(z)
dz
_ (bja) fa all" dx,
o az
(9.37)
dVn(z) faaEr, n7rX
= blV - cos - dx,
dz 0 az a
dI n (z) fa al/ 1I Il7rX
= blV - cos-dx.
dz 0 az a
To obtain the equations connecting the V n and the J n, it is neces-
sary only to substitute from IVlaxwell's equations (9.30) into (9.37)
and to perform the integrations. l"'hus
dl 1 0 (z) fa fa aE z
= -jwJJ. 11 11 dx + - dx.
dz 0 0 ax
The last term equals Ez(a) - Ez(O) = O. Substituting 111J from

Generalized telegraphist's equations
297
equations (9.32) and integrating, \ve have
dVo(z)
dz
WJla
- -j b Io(z).
(9.38)
Similarly
dlo(z)
dz
- -jw(bja) { E(X, z)E", dx
o
(9.39)
(X)
- - L YOmVm(Z),
m-O
where
Y oo = jw(bja 2 ) l a E(X, z) dx,
o
l a m7rX
YOm = jw(bja)N E(X, z) cos -- dx,
o a
m  o.
In the same manner
dI n(Z)
dz
1 4 n7rX
- -jwbN E(X, z) E:r, cos - dx
o a
(9.40)
(X)
- - L Y nm V m (z) ,
m-O
where
l a 1t7rX
Y nO = jw(bja)N E(X, z) cos - dx,
o a
l a m7rX 1t7rX
Y nm = jwbN2 E(X, z) cos - cos - dx,
o a a
'm  0
l a 1n7rX n7rX
= (2jwja) E(X, z) cos - cos - dx.
o a a
For the middle equation in the set (9.37) we have
dV n l a n7rX l a aE z 1t7rX
- = -jwJ.l,bN H'JI cos - dx + bN - cos - dx. (9.41)
dz 0 a 0 ax a

298
Electromagnetic fields
In view of equations (9.32) and the orthogonality of the cosines, the
first in tegral on the right is
f a n7rX
II y cos - dx = .alVln(Z).
o a
The second integral we integrate by parts
(9.42)
] a a E 1t7rX n7rX a n7r 1" 117rX
---...:.. cos - dx = Ez cos - + - Ez sin - dx.
o ax a a 0 a 0 a
The first term vanishes at both limits since E z vanishes there. In the
second term we substitute from equation (9.31), integrate by parts
once more, and substitute from equations (9.32). rrhus we have
n7r 1 a 1 all 11 . n7rX 1t7r . '1l7rX a
- - - SIn - dx = - II E- 1 sIn - +
. . 1/
Jwa 0 E ax a Jwa a 0
1l7r 1 a d [ ll7rX]
- :-- Il y - E- 1 sin - dx.
Jwa 0 dx a
rrhe first tefm on the right vanishcs and the second becomes, aftcr
the substitution from equations (9.32),
n7r1o(z) 1 a d [ . 117rX]
E- 1 sIn - dx
jwab 0 dx a
 1l7rN J m (z) fa 1n7rX d [ . 1t7rX]
L...J . cos - E- 1 sIn - dx.
m-l jwa 0 a dx a
1"hc first term may be integrated immediately, and it vanishes.
The integrals in thc second term may bc integrated by parts. 1"hus
the en tire expression becomes
co 1l11'l7r 2 1V 1 ll .n7rX 1n7rX
- L:lm(z) . .) f-Isin-sin-dx.
m-l Jwa" 0 Q, (L
1"'his is OUf final expression for the second integral in equation (9.41)
rrhe first integral is given by equation (9.42). Therefore
dV n(Z) 00
= - L: Znmlm(Z) ,
dz
m-l
21111l7r 2 fa 1Z7rX 11l7rX
Znm = f- 1 sin - sin - dx,
jwa 3 0 a a
111 . 11
(9.43)
2n27r21U Jl7rX
Znn = jWJJ + . E- 1 sin 2 - dx.
Jwa 3 0 a

Generalized telegraphist's equations
299
Thus an infinite set of ordinary differen tial equations (9.38),
(9.39), (9.40), and (9.43) of the type connecting the voltages and
curren ts in an infinite system of coupled transmission lines has been
obtained. Here there are no "transmission lines" in a physical sense;
instead one could speak of coupled ',nodes of propagation. Summarizing
the results and isolating the terms involving the self-impedances
and self-admittances from the mutual terms. we have
dVo(z) Wp'a
- -j b Io(z),
dz
dlo(z) 00
-YooVo(z) - 2: YOmVm(z),
-
dz m-l
(9.44)
dV n(z) 00
-ZnnI n(Z) 2:' ZnnI m (z),
-
dz m=l
dI n(z) CX)
-YnnTn(Z) - 2:' YnmVm(z),
-
dz m"""O
where the primes denote the omission from the summation of the
terms for ,vhich ln = n. For the principal mode the admittance Y 00
depends solely on the average dielectric constant over the interval
o < x < a.For the higher modes the self-admittance also depends
largely on the average dielectric con stan t,
f a fa 2n-rrx
Y nn = (j w j a ) E (x, z) dx + (j w / a) E ( x, z) cos dx
o 0 a
except \vhen E(X, z) is a periodic function of x, proportional to
cos (2n-rrxja), so that the last term becomes significant. 'rhe mutual
admittances tend to be small in general. 1-'hey all vanish if E (x, z)
is independent of x. If E(X. z) = Eav +  E(X, z), then they will
depend only on the deviation of E from its average value Eav. l-'here
\\'ill be further cancellations after these deviations are multiplied by
oscillating functions of x. l-'he same may be said about the impedances.
'rhus it is possible to obtain approximate solutions by the method
eXplained in Section 9.2. l'irst we neglect the coupling and solve the
equations for the individual modes. \Ve then use these solutions to
obtain their effects on other modes.
Equations (9.44) are called the generalized telegraPhist's equations.
For certain physical systems additional terms appear on the right

300
Electromagnetic fields
side. In the equation for a typical dV n/dz there appears a linear
function of the V m as well as of the 1m. Similarly, the equation for
dI n/dz \vould include a linear function of the 1m.
9.6 Waves between uniformly bent planes
In the preceding problem the modes were coupled by the variations
in the dielectric constant. In the case of uniformly bent planes,
Figure 9.4, they are coupled by the curvature. Let us use bent cylin-
x
FIGURE 9.4 Bent parallel planes.
drical coordinates as indicated. The y coordinate will be measured
along the axis of bending, the x coordinate from the cylindrical surface
half-way bet\\Teen the uniformly bent planes (cylinders, of course),
and the z coordinate in this surface. If R is the radius of the mean
cylindrical surface, then the differen tial elemen t of distance is
ds 2 = dx 2 + dy2 + (1 +  y dz 2
since the di5tances along the z lines for equal increments dz are pro-
portional to their radii of curvature R + x. Hence, l\laxwell's equa-

Generalized telegraphist's equations
301
tions in uniformly bent cylindrical coordinates may be obtained from
the general equations in curvilinear coordinates (Appendix II) by
letting
el = e2 = 1,
e3 =
x
1 +-.
R
Considering again fields independent of y and with magnetic
in tensity parallel to the axis of bending, \ve have
a: x = -jWJl (1 + : ) ll + :x [(1 + : ) E.]
(9.45)
ally (X )
a; - -jwE 1 + R Ex,
1 ally
E z = -.
jWE ax
Had \ve defined x as the distance from one of the bent planes, we
would have expressed the transverse field components by a cosine
series (9.32). As it is, one should replace in these series x by x +
(aj2). rrhus
Vo(z) 00 1n7r ( a)
Ex = + N L V m (z) cos - x + - ,
a m.....1 a 2
(9.46)
[o(z) rD 1nr ( a)
II II = + N L I m (z) cos - x + - ,
b m-I a 2
\\.herc we have taken into consideration that N m is independent of 1n.
1"he generalized telegraphist'8 equations are obtained exactly as
in the preceding section. 1"'hey will be of the form (9.44) except for
the first equation which will be more general, that is,
dl 1 0 (z)
dz
.
JWJJa ex)
- -- lo(z) - L ZOm l m(z),
b m-l
where
jWIJ.N J a /2 m7r ( a)
ZOrn = X cos - x + - dx.
R -a/2 a 2
(9.47)

302
Electroma gnetic fields
The remaining impedances and admittances are
jWEbN ja/2 m7r ( a)
YOm = xcos- x+- dx, m=0,1,2,...,
aR -0,/2 a 2
1 ( mn7r 2 ) ja/2 [ (X 1)].
Znm = - jw}J. + . x cos (m - n)7r - + - dx
aR JWEa 2 0,-/2 a 2
(9.48)
1 ( mn7r 2 ) fa/2 [ (X 1)]
+ - jwJJ. - . X cos (m + n)7r - + - dx,
aR }WEa 2 -0,/2 a 2
Y nm = 2jWE fa/2 X CDS m7r (x + ) CoS n7r (x + ) dx.
aR -0,/2 a 2 a 2
The last two equations are valid when neither n nor m is equal to
zero.
9.7 Waves between imperfectly conducting parallel planes
Let us return to parallel planes and assume that they are not perfect
conductors. The fields which between the planes are described by
equations (9.30) and (9.31) will be considered. At the boundaries
E z is proportional to H JI' Thus
EI:(O, z) = ZlH 71(0, z),
Ez(a, z) = -Z2HJI(a, z), (9.49)
where Zl and Z2 are the surface impedances whose real parts are
positive. The difference in algebraic signs is due to the fact that the
power flows downward into the lower plane, x = 0, Figure 6.1, and
upward into the upper plane x = a.
The transverse field components shall be expressed as cosine series,
equations (9.32). The longitudinal component E z between the planes
may be obtained from equation (9.31) by differentiation
N ex> m7r m7rX
E% = -:-  - Im(z) sin-,
JWf: m-l a a
o < z < a. (9.50)
At X = 0 and x = a it is given by equations (9.49) while the series
vanishes there. From the theory of Fourier series, and particularly
of sine series, it is known that a function which does not vanish at the
ends of a closed interval, such as (0, a), may be expressed by a sine
series in an open interval (0+, a - 0). The series is nonuniformly
convergent. That is, as x approaches zero or a, the series approaches

Generalized telegraphist's equations
303
the right values; but these values cannot be obtained simply by set-
ting x = 0 or x = a in the series itself. All terms of the series vanish
at these points. Furthermore, the derivative series does not converge.
Thus it is not permissible to substitute from equation (9.50) into
equations (9.30). However, in Section 9.5 we have explained a method
of obtaining the equations for the coefficients V n (z), In (z) which does
not involve such a substitution. For example, from equations (9.37)
and (9.30) we have
dVo(z) fa aE fa fa aE
=  dx = -jwJJ. H'/I dx + --.!. dx
dz 0 az 0 0 ax
.
JWJJ.a
- -- Io(z) + Ez(x, z)
b
.z-a
z-o
.
JWJJ.a . _
- -- Io(z) + Ez(a, z) - Ez(O, z).
b
Using the boundary conditions (9.49)
II 11( a, z) and H '/1(0, z), we obtain
dVo(z) (Zl + Z2 . wJJ.a)
= - + J - lo(z)
dz b b
and equations (9.32) for
00
L ZOmlm(z),
m-l
(9.51)
ZOm = lV[Zl + (-) mZ 2 ] ,
N = vTf(ib.
Similarly
dVn(z) fa 'n7rX fa aE '1t7rX
- -jwJJ.bN H'/I cos - dx + bN --.!. cos - dx.
 0 a 0 ax a
The first term on the right is calculated by using the series for H '/I'
The second integral may be integrated by parts
f a a E r, 1t7rX 1t7rX a
- cos- dx = E z cos-
O ax a a 0
n7r fa n7rX -Zl + ( - ) n Z2 00
+ E z sin - dx 10 (z) - L
a 0 a b m-l
(9.52)
1t7r fa 1t1fX
X[ZI + (_)n+mZ 2 ]Nl m (z) +- Ezsin-dx.
a 0 a
We could substitute from equation (9.50) for E" and integrate.
In this case we would obtain the correct result. However, the inte-

304
Electroma gnetic fields
gration of nonuniformly converging series should be performed with
care. This integration can be avoided by substituting from equation
(9.31) and integrating by parts
J a aH 11 n7rX n7rX CI n7r Ja n7rX
- sin - dx = H 11 sin - - - H 11 cos - dx
o ax a a 0 a 0 a
= !n7rNI,,(z).
Using this result and equation (9.52), we have
dV n(Z)
dz
- -Znn1n(Z) - L Znm1n(Z) ,
m
. n 2 7r 2 2 (Zl + Z2)
Znn = JWJJ + -:-- + ,
JWEa 2 a
(9.53)
2[Zl + (-) ,,+mZ 2 ]
Znm = ,
a
n, m = 1, 2, 3, ....
When n or m vanishes (but not both), Znm is given in equations
(9.51) .
No special difficulties arise in obtaining
dI,,(z)
= -jwEV,,(Z).
az
(9.54)
9.8 Generalized coordinates
The coefficients in such series as in equations (9.32) may be properly
called the generalized coordinates of the electromagnetic field. They
are similar to the generalized coordinates used for describing the
state of a mechanical system or to mesh currents in the network
analysis. There is no unique set of such coordinates for any given
problem. It is clear from the preceding sections that it is desirable to
choose them in such a way that the coupling coefficients are small.
It would seem that the optimum choice would be a set for which
all mutual coefficients vanish. However, this is not necessarily the
case. In the problem of uniformly bent planes a choice of cylindrical
coordinates and separation of variables would yield generalized
coordinates (the coefficients of certain Bessel functions) for which
all coupling coefficients vanish. Nevertheless when R is large, the
analysis by the method given in Section 9.6 is much simpler.

Appendix I
A.I Coordinate systems and vector components
rrhe most frequently used coordinate systems are the rectangular
or Cartesian, cylindrical, and spherical; in these systems a typical
point P is denoted by (x, y, z), (p, <P, z), (r, 0, 'P), respectively. The
meaning of these coordinates is eXplained in Figure A.1. Thus x, y, z
z
y
FIGURE A.l Coordinate systenls.
are the distances from three mutually perpendicular reference planes;
p is the distance from the z axis; r is the distance from the origin 0;
the "polar angle" () is the angle from the z axis to the radius 0 P;
the" longi tude" 'P is the angle from the half plane determined by the
z axis and the positive x axis to the half plane through P. Any vector
305

306
Electronzagnel'ic fields
z
y
x
FIGURE A.2 Vector components.
F can be resolved along three mutually perpendicular directions.
Figure A.2 illustrates such a resolution in a spherical frame of refer-
ence. The components Fr, Fs, and F f' are positive in the directions of
increasing r, 0, and <p, respectively.
A .2 Transformation of coordinates
From Figure A.1 we deduce the following relations between Cartesian,
cylindrical, and spherical coordinates:
x = p cos lp = r sin 0 cos <p; p = V x 2 + y2 = r sin ()
y = p sin <p = r sin (J sin <p; lp = tan- 1 (y/x) = <p
z = z = r cos o. z = z = r cos ()
,
r = V x 2 + y2 + Z2 = V p2 + Z2;
o == tan- 1 [(x 2 + y2)i/z] = tan- 1 (p/z);
<p = tan- 1 (y/x) = <p.

Appendix I
307
A.3 Elements of length, area, and volume
The square of the hypotenuse of a right triangle equals the sum of
the squares of its sides. From this theorem it follows that
ds 2 = dx 2 + dy2 + dz 2 ,
where ds is the distance between two neighboring points. Although
the theorem of Pythagoras does not hold for large curvilinear tri-
angles on the surface of a sphere and most other curved surfaces, it
does hold for infinitesimal right triangles and we have the following
.
expreSSIon:
ds 2 = ds + ds; + ds,
where ds u , ds l1 , and ds w are the infinitesimal sides of a right-angled
w line
e 2 dv
v line
FIGURE A.3 A n elementary coordinate cell in orthogonal curvilinear coordinates.
curvilinear parallelpiped (see Figure A.3) formed by mutually
perpendicular surfaces, such as spheres, cones, and half planes of
spherical coordinate system. Curvilinear coordinates may be defined
in various ways but invariably the differential lengths along co-
ordinate lines are proportional to the differentials of the coordinates,
ds u = el du,
dS l1 = e2 dv,
ds w = e3 dw.
Hence
ds 2 = ei du 2 + ei dv 2 + ei dw 2 .

308
llectronlagnetic fields
The clements of area and volume are
dS utJ = ds u dS I1 = ele2 du dv,
dS vw = dS I1 ds w = e2e3 dv dw,
d Swu ds w ds u = e3el dw du,
dV = ds u dS I1 ds w = ele2e3 du dv dw.
In spherical coordinates (r, 0, rp) ,ve have
el = 1,
e2 = r,
e3 = r sin O.
In cylindrical coordinates (p, rp, z)
el = 1,
e2 = p,
e3 = 1.
A.4 Gradient
Consider a function V(u, v, w) which depends only on the coordinates
of a poin t. Loci of equal values of this function
V(u, v, w) = constant
are called level surfaces or contour surfaces; in the two dimensional
cases we have level lines or contour li,z,es. Electric potential is an
example of such a function.
1"'he maximum rate of change of V is along the normals to contour
surfaces. This rate of change is called the gradient of V and is a vector
whose magnitude is
I grad V I
dV
ds'
where ds is taken along the normal to the contour surface. l"'hc com-
ponen t of this vector along the 'it line is obtained if we multiply this
magnitude by the cosine of the angle between the normal and the 11-
line, dsjds u . Hence
dV
grad u V = -,
ds u
dV
g rad V =
11 d '
SI1
dV
g rad V = -
w d.
Sw

Appendix I
309
In Cartesian, cylindrical, and spherical coordinates we have
( av aV av)
grad V = , , - ,
ax oy az
grad V = ( aV ,  av , av ),
op p acp OZ
( av 1 aV 1 av)
grad V = -, - -, . -.
ar r ao r SIn 0 ocp
A.5 Circulation of a vector and curl of a vector
Circulation of a vector ft round a closed curve is the line in tegral
of its tangential cOlnponen t round the curve. Thus the electro-
motive force round a closed curve is the circulation of E, and the
magnetomotive force the circulation of /1.
Consider an infinitesiJnal element of area centered at!L point P.
For a certain orientation of the area the circulation of F round its
edge per unit area is maximum. 'Ihis vector quantity is called the
curl of ft and is denoted by curl fl. I{efcrring to l\faX\vell's la,vs, \ve
observe that electric current density is the curl of II. and lnagnetic
curreJ: t density is the negative of the curl of E. 'rhus for time-har-
monic fields
curl E = -jwp.II,
curl II = (0" + jWE) E.
l'\he cOlnponents of curl F in orthogonal coordinates may be
calculated from its detinition. Consider };'igure A.4 ,vhich shows an
inflnitesiJnal curvilinear rectangle l1BeD in a surface formed by v
and w lines when It is constan t. The con tribution to the circulation
from the sides BC and DA is
a a
(Fw ds u ') dv = - (e3Fw) dw dv.
ov av
'rhe contribution from AB and CD is
a 0
- (F" ds,,) dw = -- (C2F,,) dv dw.
ow ow
rThe area of the rectangle .is C2e3 dv dw. Hence
- 1 [0 0]
curl u F = - - (C3Fw) - (e2F,,).
e2e3 av ow

310
Electronzagnelic fields
A
ds v
FIGURE A.4 A ss-isting in the calculation of the 1t cotnpOllellt of the curl of a
t'ec/or.
The remaining components may be obtained by cyclic permutation
of u, v, wand 1,2,3.
A.6 Flux of a vector and divergence
The flux of a vector F through a closed surface is the surface integral
of its normal component. If \ve choose such a surface around some
point P and let it shrink to zero, the limit of the flux per unit volume
is called the dilergcnce of vector ft and denoted by div P.
Consider an elementary coordinate cell about P (see Figure A.3).
1"he area of a It surface through P intercepted by the cell. is e2e3 dv d'll)
and the flux across it is e2e3F u dv dw. The rate of change of this flux in
the 1l direction is
a
(C2e3F u) d' dw
aU.
and the residual flux through the 1(, faces of the cell is
a
- (C2e3Fu) dl£dvdw.
aU,

Appendix I
311
By a cyclic permutation ofzt, 'lJ, wand 1, 2, 3 we obtain the residual
fluxes through the remaining pairs of faces and
div F =  [ (e2e 3 F..) + a (e3 e 1 F.) + a (e 1 e 2 Fw)].
elC2e3 au av aw
A.7 Laplacian
The Laplacian is defined as the divergence of a gradient
v = div grad V.
1'hus
1 [a (e 2 e 3 a V) a (e 3 e l a V) a (e 1 e 2 a V)]
v-- - -- +- -- +- -
- ele2e3 au Cl au av e2 av dW e3 aw ·

Appendix II
Maxwell's differential equations
In Cartesian coordinates
aE z aE y
- - - - -jwp,Il z ,
ay az
aE z aE z
- - - - -jwp,Hu,
az ax
aE u aE z
- - - - -jwp,H z ,
ax ay
In cylindrical coordinates
aE z aEf{)
- - p - = -jwp,pIl p ,
acp az
aE p aE z
- - - = -jwp,H",
az ap
a aE p
ap (pE.,) - alp = - jWlJ.pH "
In spherical coordinates
a
- (sin OEf{)
ao
aE 8
---
acp
a
- (sin OIl f{)
ao
aII 8
---
acp
all z all 11
- - - - (0" + jwe) Ex
ay az
aIl z aH z (0" + jwe) E y
---=
az ax
ally aIlz; (0" + jwe) Ez.
-
ax ay
aH z a/If{)
- - p - = (0" + jwe)pE p ,
dcp az
aH p aIl z
- - - = (0" + jWE)Ef{)'
az ap
a
- (pil rp)
ap
all p
--=
acp
(0" + jwe) pE z .
- jwp,r sin OH"
(0- + jwe) r sin OE r ,
aE a
 - sin 0 - (rEf{)) - -jw/Jr sin 011 8 ,
acp ar
aII r a
- - sin 0 - (rIIf{)) - (0- + jwe)r sin OE 8 ,
acp ar
a
- (rE 8 )
ar
aE r
--=
ao
a
all r
-=
ao
rH 8 )
dr
l"
312
-jwJ.LrII f{),
(u + jwe) rEf{).

Appendix II
313
}, curvilinear coordinates
In genera
a (e3Ew) a (e 2 E.) _ _ jwp. e 2 e JI u,
av aw
a (c:JI w) a (c211 fJ) (0" + jWE) C2C3Eu,
-
av aw
a (cIEu) a (e3Ew) - jWjJ. c 3 c l II fJ,
-
aw au
a (elll u) a (eJlw) (0" + jWE)e3eIEfJ,
-
aw aU,
a(c 2 E v ) a cIEu) - jWjJ. e le 2 H w,
-
au av
a (e2IIfJ) a (clll u) (0" + jWE) cle 2 E w .
-
au av
1
.
.
..

Appendix III
Laplace's equation
In Cartesian coordinates
iJ2V a 2 V a 2 V
- + - +- = o.
ax 2 a y 2 az 2
In cylindrical coordinates
a (av) a 2 v a 2 V
p - p + + p2 - = o.
iJp ap acp2 az 2
In spherical coordinates
sin2 0 !... (r2 av ) + sin e a (Sin e aV) + a 2 V = o.
ar aT iJO ao acp2
In general curvilinear coordinates
iJ (e 2 e 3 iJ V) a (e:el a V) iJ (e 1 e 2 a V)
- - + - - + -- = o.
aU, el au iJv e2 dV dW e3 aw
314

Problems
1.2-1 Two homogeneous spheres of mass 1n are equally and uni-
formly charged. Express the charge q on each sphere in terms of m
when the force of repulsion just annihilates the force of gravitational
a t traction.
Answer:
q = 86 n1, micro-microcoulombs.
1.4-1 Show that the following equations are the differential
equations for the electric lines of force:
dx dy dz
-- -- in Cartesian coordinates
Ez, EJI E,'
dp p dcp dz
--,-.-.-- in cylindrical coordinates
- - E,,'
E p Erp
dr r dO r sin 0 dcp
= - = in spherical coordinates.
Er E B Erp
See Appendix A.I for the definitions of " p, 0, and cpo
1.10-1 A copper sphere of radius a is submerged in sea water
far below its surface. An insulated wire conveys to it current [.
Calculate the following:
(a) the power P dissipated in sea water;
(b) the voltage V from the sphere to infinity along a radius or a
line made up of segments of radii and circular arcs concentric
with the sphere.
Answer:
p = /2/47rua,
'= /2/207ra,
v = I/47rua
= / /207ra.
1.11-1 Consider a current element of moment It in a conducting
medium. Calculate the voltage from point (ro, (0) to infinity along
two paths:
(a) along the radius I) = 1)0,
315

316 1leclronlag1lelic fields
11'1.s-,oer:
(b) along the meridian from (ro, 8 0 ) to (ro, 8 1 ) and then along the
radius 8 = 0 1 .
II cos 0 0
V= .
47r<1r5
1.15-1 A charged ring (thin circular wire) of mean radius a is
in the xy plane and coaxial ,vi th the z axis. The total charge is q.
Find the electric intensity on the axis of the ring.
Ansu'cr:
Er, = (QZ/47rE) (a 2 + Z2)-3/2.
1.15-2 A sphere of radius a is imbedded in free space. Let q be
the charge on the sphere. Find the voltage along a radius from the
sphere to infinity.
A'1lsu'cr:
v = q/47rEoa.
In circuit theory the quantity C - 47rEoa is called the capacitance
of the sphere.
1.15-3 Assume that in llroblem 1.15-2 we bring in inflnitesimal
quantities of charge, dq. Work is done against the field of the charged
sphere. Calculate the \vork done in raising the charge on the sphere
fronl q ql to q = q2 > ql.
.il nS'i.l'cr:
v = (q - qi) /87rEoa.
1.15--4 Consider a. metal sphere of radius a with a negative charge
-q. Suppose that an electron, originally at rest, starts moving. Find
its speed at inflnity (assuming that this speed turns out to be a
small fraction of the speed of light) .
Ansu'cr: v = vq(e/nt)/27rEor
r» a.
1.16-1 Consider an electric dipole along the z axis at the origin
of the coordinate system. Obtain the equation for an electric line
of force passing through the point r = "0,8 = 7r, cp = 'Po.
,llus(1}cr: r = ro sin 2 0, cp = 'Po.
1.17-1 i\ssume that a uniform magnetic field of flux density
Bo is parallel to the z axis. Suppose that at time t = 0, an electron

ProblclllS
317
(charge - e, mass l1l) is at the origin and is moving in the direction of
the y axis with a speed vo.
(a) Show that the equations of motion are
d . y
- = - (e/1n)Bov y , = (e/ln) Bov z .
dt dt
(b) !J rove that thereafter the speed of the electron remains con-
stant.
( c) Show that the electron is moving in a circle \\1 hose cen ter is
on the negative x axis and \vhose radius equals vaBa l (e/nt) -1,
and that the angular frequency is w = (e/1n)Bo.
1.21-1 Find the magnetic intensity in the interior of a cylinder of
radius a if the current in the cylinder is J and is distributed uniformly.
Answer: II JP = 1 p/27ra 2 .
1.22-1 Calculate the magnetic intensity on the axis of a thin
circular ring of mean radius a. Let I be the current and z the distance
from the plane of the ring.
Ansu'cr:
II z =a2/ (a 2 + Z2) -3/2.
1.22-2 Using the result of J)rohlcm 1.22-1. calculate the magnetic
-field on the axis of a solenoid of radius cl, extending froln z = 0 to
z = t. Assume that the current in the \vinding is I and the nUl11her of
turns per unit length is n. Assume that the \vinding is so thin that the
current can be smoothed over the cvlindrical surface of the solenoid.
oJ
In particular  conlparc the magnetic intensities at the center of the
solenoid and at its ends when 1 » 2a.
Ansu'cr:
II z = llI[ (l - z) rjl + zr-IJ, \vhcre
r = (£1,2 + Z2)!, rl = [a 2 + (1 - Z)2J1I2
and
II z(I/2)  n/  2/1 z(O).
1.22-3 For the solenoid in Problem 1.22-2 discuss the" end
effects" by computing II z at z = 0, a. 2a, .... Assume that l is large
compared \vith a. Show that if z =nza and 111- > 1, then
II z = Itl (1 -  +. -  + · ..).
41n 2 16nz 4 96m 6

318
Elec/ro11lagnetic fields
1.22-4 A thin \vire of radius b is bent in to a circular ring of mean
radius a. Assume that a. uniform magnetic field is created at a uni-
form rate in r seconds from t = 0 to t = 'T. Assume that magnetic
flux is perpendicular to the plane of the ring and its final density
is Bo. Calculate the current in the ring and the energy dissipated in
the ,,ire.
A nS7.ocr:
I = 7ruab 2 Bo/2r,
S = 7r 2 ua 3 b 2 B5/2 r.
O < t < T
1.22-.5 Consider a long solenoid of length l and cross sectional
area S. Assume that the current is I and the number of turns per
unit length is '11. I"ind the voltage between the ends of the winding.
.
.l4nswcr: V = Jl.of£2SII. In circuit theory the coefficient of
proportionality, L = JJ.on 2 St, is called the inductance of the solenoid.
1.22-6 Consider a curren t filamen t I, extending along the z axis
from z - 0 to z = 00. Superimpose on it another infinite filament
-I from z I to z = 00. j\ current clement of moment It is thus
obtained. Derive equation (1.57) from equation (1.56).
1.227 I{cfcrring to Section 1.22 and the equation preceding
equation (1.56), sho\v that \ve can also write
hr sin OIl <II = - f" j2" J,r 2 sin 0 dO dC{).
B 0
On the right the integration is performed over the spherical cap
"con1plementary" to the one used in the text. 1'\his is a variant
method of calculating the magnetic intensity of the. field generated
by a semi-infinite current filament. (See Figure 1.8).
1.26-1 A conducting cylinder of radius (], is coaxial \vith the z
axis. It is uniforn1ly charged. 'fhc charge per unit length is q. From
symmetry considerations and equation (1.79) flnd the field
.11 llsu'cr:
E p = q/27rf.p,
p > a
= 0, p < a
E(/J = E z = 0
1.26-2 i\ssumc that in Problem 1.26-1 there is another conducting
cylinder, also coaxial ".ith the z axis. Let b > a be its inner radius

ProblenlS
319
and -q the charge per unit length. Find the field and the voltage V
between the cylinders.
Answer:
E p = 0, p > b
= Q/27rEP, a < p < b
= 0, p < a,
Etp= E z == 0,
v = ( q /2 'IrE) In (b / a) .
1.263 1 4 rom equations (1.31) and (1.47) which define, respec-
tively, the dielectric cons tan t E and the pcrmcabili ty p. show that the
quantity c == (Jl.E) -1/2 has the physical dimensions of velocity and
'Y/ = (Il/ E) 1/2 has the dimensions of resistance [see equation (1.17) J.
rrhis means that 11 and 1111 have the same physical dimensions and
their magnitudes can thus be compared.
Show that in terms of these" secondary parameters" of a medium
1\;Iaxwell's equations for nondissipative isotropic media can be
written as
1 ELan ds = -  f (17/ 1 nor) dS,
r c at
1 a
,117l 1 tnn ds = - f Enor dS.
r c at
Note that for free space c equals approximately 3 X lOS In/sec.
1.26-4 In order to obtain an idea of the order of magnitude of
the interaction bet\veen electric and magnetic fields assume that
E(x, y, z; t) = E(x, y, z)e pt ,
II (x, y, z; t) = II (x, y, z)e Pt .
Note that the ratio of the tiInc derivative of it to E itself equals p
so that p is the relative rate of change or fractional rate of change of
11 (and /1). Show that for this time dependence lVlax\vcll'::; equations
become
f ELan ds = - (pic) f 7J/lnor) dS,
f 7J/ltan ds = (p/c) f Enor dS.

320
Elec/rOl1Ul£llclic ficlds
Suppose that p = 1 T so that in T seconds the lield increases in
the ra tio of e to 1 (that is, 2.718 ... to 1). Sho\v that the length c T
is indicative of the linear dimensions of the fIeld for which the electro-
magnetic interaction becomes significant. The same is true when
p = -1 T and the field is decreasing.
1.26-5 Assume that E and jj are varying \vith the angular
frequency w
E = Ec cas wt + E' sin wt,
- - -
II = lIe cas wt + II' sin wt.
Show that for nondissipative media ,Iax\vell's equations become
f Ean ds = - (w/c) f TJIlor dS,
f Ean ds = (wlc) f TJIlor dS,
f Illan ds = (wlc) f Eor dS,
f lILn ds = - (wlc) f Eor dS.
Note that w plays the same role as p in the preceding problem and
that T = 1/w.
1.27-1 Consider an infinitesimal rectangle in free space \vhose
'vertices are: A (x, y, z), B(x + x, y, z), C(x + X,)' + ill', z), and
D(x, :v + y, z). Let Ex, Ey, Ez' II z, II y' and II z be the Cartesian
components of 11 and II at point A. Apply l\Jax\vcll's equations
(1.76) and (1.77) to this rectangle and sho\v that
iJE y aE x aIl z
---- - J.L()
ax ay at '
all a/Ix aE z
y --- EO
ax av at
,
"fry some variations in the method of calculating electromotive
and magnetomotivc forces round the closed circuit ABC'DA in the

Problems
321
follo\ving \vays:
(a) Since l\x is infinitesimal, V AU = Ez; x and
a
V DC = Ez; x + (Ex x) fly
dY
except for inflnitcsin1als of higher order. Since V CD = - V DC,
these t\VO equations yield V All + V CD.
(b) !\Ioredirectly. VAll + VCl) = -(V nc - l'AB). The quantity
in paren theses is an incremen t in V A /J when y is increased by
1' and therefore is equal to the rate of change of V AB in the
y direction multiplied by y,
a
(E z Llx) y.
ay
Four other partial differential equations are obtained by applying
l\lax\vcIl's integral equations to infinitesimal rectangles in planes
parallel to the remaining coordinate planes. On account of symmetry,
ho\vever. these equations are readily obtained by cyclic permutation
of x, y. and z.
Similar equations may be obtained in cylindrical coordinates.
Choose a curvilinear rectangle A (p, cpo z), B (p + p, cp, z),C (p + p,
cp + t1cp. z) fJ p, cp + 6.cp, z) and show that
a aE p aH z
(pEl()) - - = - }J.oP .
ap acp at
1.27-2 Apply equation (1.79) to an intinitesimal parallelopided
formed by three coordinate planes passing through point (x, 1', z)
and three planes passing through (x + x, y + 6.1', z + z) and
sho\v that
aDz aD JI aDz
-+-+- = qv,
ax ay az
\vherc qv is the volume density of charge.
1.27-3 Consider a perfectly conducting sphere of radius a im-
bcddccl in an imperfect dielectric lnedium of conductivity u and
dielectric constant E. i\t time t = 0 an electric charge qo is placed on
the sphere. Sho\v that (see equation 1.78) at subsequent time the

322
]:lcctrn1Ha1!.lletic fields
radial electric in tcnsity satisfies the folIo\\Ting equa tion
aE r
(fEr + f- = 0
at
and that
qoe-ut/ f
E ' -
r - .
47rEr 2
Note that the first equation implies that the displacement current is
equal and opposite to the conduction current density. 'rhis equality is
possible because the clcctric fleld decays extremely fast (substitute
the constants of sea water or soil).
1.28-1 Consider two circular metal plates of radius a and coaxial
\vith the z axis, one in the plane z = 0 and the other in the plane
z Iz « a. LctJl 0 be the voltage from the lower plate to the upper.
Show that approximatcly Er, = Vo/h, Dr, = fVo/lz, and that the
charge on the lo\ver plate is
q = (-Jrfa 2 /h) V o .
Note that in circuit theory the coefficient C = 7rfa 2 /1z is called the
capacitance of the pair of plates.
Sho\v that if l"J is varying slo\vly \vith time, the magnetic intensity
due to the vertical displacement current is IIf{) = E p f ' o /21z = Iop/211"l2.
Compare this intensity with that in a circular wire (see I>roblem
1.21-1) .
1.28-2 Consider two uniformly and oppositely charged planes,
z = 0 and z = h. Let qs be the surface density of charge on z = 0
and -qs on z = h. Explain \vhy the field exists only bct\veen the
planes and show that Er, = qS/E.
2.1-} Show that the field of an infinitely long uniformly charged
linear fllamcnt or cylinder (sce Problem 1.26-1) can be expresscd
as the gradient of the following logaritll1nic potential
v = - (q/21rf) In p + A,
(1)
where A is an arbitrary constant. 'fhis constant may be chosen to
make V equal to zero at some reference distance p = Po. 'l'hcn
v = - (q/21rf) In (p/ po).
(2)
The form of equation (2) is quite proper since p/ Po is dimcn5ionlcss.

Problems
323
The form of equation (1), on the other hand, is ambiguous since In p
depends on the unit of length. The ambiguity, however, is absorbed
in the arbitrary constant A. Neither the field nor potential differences
depend on this constant. For this reason A is often dropped from
equation (1).
Show also that a uniform electric field of in tensity Eo parallel to
the z axis may be obtained from the potential function
V = -EoZ + A.
If the field is parallel to the x axis, then V = - EoX + A.
2.1-2 Consider two infinite uniformly and oppositely charged
filaments parallel to the z axis in the xz plane. Let the positively
charged filament pass through point x = s and the negatively
charged through x -s. Assume that s is infinitesimal. Express
the potential and the field in terms of the positive charge q per unit
length.
A llswer: V = (qs cas foP) /27rEp.
E p = (qs cos 'P) /2 7rEp 2.
Eip (qs sin 'P) /2 7rEp 2.
2.1-3 Obtain the potential difference between t\VO spheres of
radius a ,vhen the charge on one is q and on the other -g. Assume
that the distance s between their centers is rather large in com-
parison with 2a.
A JlSU)Cr:
q
VI - V 2 = - (1 - a/s).
27rEa
2.1-4 Obtain the potential difference bet\vcen t\VO infinitely
long parallel \vires of radius a ,vhen the distance s betwccn their
axes is rather large in comparison with 2a.Let q be the charge per
unit length on one wire and -q on the other.
./1 J1Swcr:
q
VI - V 2 = - In (s / a) .
7rE
2.1-5 Obtain the potential difference V bct\vecn concentric
spheres of radii a and b > a ,\then thcy arc equally and oppositely
charged.
Answer:
q(b - a)
v= .
41f'Eab

324
l/ec/r0l111lJ!.lle/ic fields
2.1-6 Sho,v that the potential difference in Problem 2.1-5 may
be exprcssc(l as
q Iz ( Ii'!.. 11 4 11 6 )
V= 1+ +-+-+...
47rEC 2 4c 2 16c 4 64c 6 '
,vhcrc Iz = b - a and c = (a + b)/2. ote that S = 47rc 2 is the
area of the sphere of nlean radius c so that \\rhen (11/2C)2 « 1 \ve have
a sinlple approximate fOrlTIula
v = qhjES.
2.1-7 ()btain an expression for the potential difference in ]>rohlem
1.26-2 similar to the one in })roblcm 2.1-6.
Ansu'cr:
qlt 00 1 ( Iz )2n
V=-L -
27rEC n=-O 2n + 1 2c .
2.1-.8 Consider a thin conducting circular disk of thickness h.
Let its inner radius be a anci the outer b. Let the potential difference
bct\vccn the edges be 1f. Find the curren t I.
A JZS7.ocr:
I = ev  G = 27rulzjln (bj a,).
2.1-9 Consider a thin conducting spherical shell concentric with
the origin. Let there be t\VO holes in this shell. 'rhe edge of one is the
in tersection of a cone () = ()l vvith the sphere. 'fhe edge of the other is
the intersection of a coaxial cone () = ()2 ,vith the sphere. Find the
relation betwecn the currcnt I and the voltage V betwccn the
edges. Let Iz be the thickness of the shell.
27rult
Ansu'cr: I = GV, G = .
In tan (()2) - In tan (()l)
2.1-10 Consider t\VO uniformly charged spheres of radius a.
Let the distance s bct\vccn their cen ters be ra ther large in comparison
,vith 2a. Show that the force between thcnl (in vacuum) equals
approxima tely
F = 47rEo(a/r)2V2[1 + (ajs) J-2,
where l' is the poten tial of each sphere.
Calculate the potential \vhich is just sufficient to neutralize the
gravitational force bct\veen the spheres ,\.hcn als is negligible in
comparison \vith unity.
Allsu'cr:
v = O.77(1nja.) volts.

Problems
325
2.1-11 Consider direct curren t flow in a square mesh plane
screen of conducting \vires. Choose some junction of two wires as the
origin of a Cartcsian system of coordinates. and the wires passing
through it as the axeg. l"'hc coordinates of a typical junction will
then he (11t, n), \vhere nt and n arc integers. Since the voltage between
any t\VO junctions is independent of the path connccting them, ,ve
can ascribe a potcntiall'(nt. n) to each junction (111, n). l"he voltage
from junction (11t.n) tnjunction (P,q) will then be V{nt,n) - l'(P,q).
Assume that the conductance betwecn any two adjacent junctions
is G 1 . Suppose that the current from junction (1n,n) to (1n + 1, n)
is 1(1n,n; 1n + 1, n). 'fhen
f(m, 1; In + 1, u) = G1[V(nt, 11,) - V(l1t + 1, n) J.
Sho\v that
V (1nn)
= [V(nt-l.n) + V(nz+ 1,n) + V(nz.,1t-l) + V(nt,n+ 1)J
so that the potential of each junction is the average of the potentials
of the four adjaccn t junctions.
Discuss the analogy bet\veen the condition imposed on the currents
leaving a junction and equation (1.76) for the case of time-invariable
fIelds. Discuss the analogy bctween
l ' ( l1t. n; 1n + 1, n,) = V ( 11l, n) - 11 (1n + 1, n),
where the left side is the voltage from junction (m, }t) to junction
(1n + 1, n), and equations (2.6), (2.7), and (2.8).
Sho\v that for a three dimensional lattice of wires the potential of
any junction is the cJ/i)crage of the potentials of the six adjacent junctions.
2.1-12 Using the result of Problem 1.27-2, sho,v that the po-
tential satisfies Poisson's equation
a 2 1 1 a 2 V a 2 V
- + - + - = -qv/ e
ax 2 a y 2 az 2
,,,,here q" is the volume densi ty of charge.
In the special case of charge free regions this equation is called
Laplace's equation.
2.1-13 l"hc potential function of the dipole type of field, equation
(2.10) varies as cos fJ \vhcrc () i the angle bct\vccn the axis of the dipole
and a line to a typical poin t in spa.ce. Arc there other poten tial func-
tions ,vhich vary as cos 8 but have a different dependence on r,
.

326
Electronzaglletic fields
R(r) let us say? Using Laplace's equation in spherical coordinates
(Appendix III) sho\v that \Vc n1ust have
d ( dR)
,2 = 2R
dr dr
and that the general solution is
R = Ar- 2 + Br.
1"he first term gives the dipole type I)otential. The second term gives
11 = Br cos 0 = Bz.
This is the potential in a uniform field.
2.1-14 Show that there is no potential function which varies as
cos 28 for a.ll values of r.
2.1-15 Sho\v that if Ee = 0 for a. < r < b, then the field in this
region is either radial or zero.
2.2-1 Three equidistant and equally charged metal spheres of
radius a are cen tcred on the z axis at poin ts z = 0 and z = :f::s. Let q
be the charge on each sphere and assume that s » a.. Obtain the
approximate potentials of the spheres.
q ( 2a)
V o = - 1 + - , for the middle sphere
41rEa s
Answer:
q ( 3a)
l'l = - 1 + - , for an outside sphere.
47rEa 2s
2.2-2 An electric tripole is on the z axis: charge - 2q at z = 0
and q at z = :i=s. Assume that s is infInitesimal and obtain the po-
tential and the electric intensity of the field.
qs2
V = - (1 + 3 cos 28),
81rEr 3
Answer:
3 qs 2 . 3 qs 2 sin 20
Er = - (1 + 3 cos 28), Ee = .
81rEr" 41rErt
2.2-3 Assume that a uniformly charged filament extends on the
z axis from z = -!l to z = l. Show that as 1 increases the

ProblenlS
327
potential approaches
1 1 = (qI27rE) In (lip) = - (qI21rE) In p + (qI21rE) In l
in an increasingly large volume surrounding the origin. This yields
the logarithmic potential introduced in Problem 2.1-1.
2.2-4 Find the poten tial of a uniformly charged thin circular
ring of radius a on its axis. IJet q be the total charge.
A 11su'er:
v = q/41r E'v a2 + Z2.
2.2-5 Find the potential at the center of a thin, uniformly charged
square loop. Let q be the total charge and a the side of the square.
Compare this potential with the corresponding potential of a circular
ring of diameter 2a.
Answer:
l ' = q/21rEa In (1 + V1).
V (square) = 0.88 X V (ring).
2.2-6 Assume that a thin \vire of radius a and length l » a is
uniformly charged and that the charge per unit length is q. Find the
average potential.
Allsu'er:
q (2l ) q (l )
V = - In - - 1 = - In - - 0.31 .
21rE a 27rE a
2.2-7 A square loop, length l on each side, is made of a thin
\vire of radius a (a « l). Assume that it is uniformly charged and
that q is the charge per unit length. Find the potentials at the corners
and in the middle of a side.
A llswer:
11 !L. (In £ + In (2 + 2V1)) = -L (In £ + 1.57),
21rE a 27rE a
l' !L. (In £ + In 7 + 3y1S ) = !L. (In £ + 1.93).
21rE a 2 21rE a
lVote: Since the average potential of a thin wire whose length equals
the perinzeter of the square is
-L (In £ + 1.08),
21rE a

328
Elec/ronra1!.llc/ic fields
it appears that the average potentials of a uniformly charged wire,
a square loop, a regular polygon, and a circle are approximately equal
if their perimeters are equal; in terms of the perimeter p the po-
ten tialis approximately
g (p )
V = - In - - 0.3 .
21rE a
For polygons and circles a better value would be obtained by taking
the average potential round the square.
2.2-8 Consider two parallel thin wires of radius a and length l,
disposed as in Figure 2.33 (Section 2.18). Let the interaxial distance
s be fairly large in comparison with 2a (at least twice as large) and l
be considerably larger than s. Assume that the wires are uniformly
and oppositely charged. Express the transverse potential difference
in terms of the distance 11-S from one end of the pair (ns < l/2) and
find how quickly the" end effect" disappears.
q s q ns + V (ns ) 2 + a 2
V = - In - + - In .
7rE a 27r€ ns + s y n 2 + 1
Answer:
Calculate the values of the second term for n = 1, 2, 3.
2.2-9 Suppose that in Problem 2.2-8 the wires are similarly
charged. Obtain a simplifIed expression for the potential, not too
near the ends. Assume that the charge per unit length on each wire
is q. Express the result in a form which can be compared with
equation (2.12).
Answer:
q 4z(l - z)
V = -In .
47rE as
2.2-10 Consider two infinitely long thin parallel wires, arranged
symmetrically about the axis of a cylinder of large radius b in one
of its radial planes. Let a be the radius of each wire and s « b the
distance between their axes. IJet the charge per unit length of each
,vire be g and that on the cylinder -g. Obtain the potential differ-
ence V between the wires and the cylinder and compare it with that
of a single wire coaxial with the cylinder.
g b
Answer: V = -In .
27rE vas
A single wire of radius vas would be at the same potential.

Problems
329
2.2-11 Consider two uniformly and oppositely charged thin
wires of radius a and length l, forming a V, as shown in the Figure
A
+
+ i'J+
+
s
FIGURE 2.2-11
2.2-11. Let the charge per unit length on OA be q. Show that the
potential at any point P is
q piTl (1 + cos 0) (1 + cos ( 1 )
V=-ln
47rE p1T2 (1 + cos 0') (1 + cos ( 2 )
and the potential difference betwecn the clemen ts of the wires,
equidistant from the apex and not too near thc ends, is
q 4p(1 - s)
V. UN = -In
27rf a 2 r2 (1 + cos 1/1) (1 + cas ( 2 )
q d q 2 (l - s)
= - In - + In ,
7rf a 27rf r2 + I - s cas 1/1
,vherc 1/1 is the angle bct\veen the \vires and d is thc distance bet\veen
the clements.
2.2-12 Consider 2n + 1 mctal spheres of radius a, centercd on the
z axis at z = 0, :i:c, ::I=2c, ... ::I=nc. Let ql be the charge on each
sphere. Sho\v that thc potential of the sphere centered at the origin
is approximately
ql ql m-n 1
Va = - +- -.
41rfa 27rfC m-l 1ft

330
Electromagnetic fields
Let l be the distance bet\veen the centers of the extreme spheres and
q the charge per unit length of the array. Noting that for large n
1
1 + . + ! + ... + -  In It + C,
'n
where C = 0.577 ... is Euler's constant. show that
V o = £ + ..i.. (In..!.. + C).
47rfa 27r€ 2c
Note that if this formula is used for spheres almost touching each
other, then the potential
V o = ..i.. (In: + 0.2)
27r'f a.
differs only a little from the potential at points half way bet\veen
the ends of a thin cylinder coaxial \vith a uniformly charged fIlament
[see equation (2.12) J.
2.2-13 In Problem 2.2-2 \ve obtained the potential function of a
tripolc and its field. Sho\v that there is another potential function
which has the same dependence on 8, namely,
1 1 = Ar 2 (1 + 3 cos 20).
2.2-14 Using the results of Problems 2.2-2 and 2.2-13 sho\v that
if we have a distribution of charge of density
qs = P(l + 3 cos 2(J)
on a spherical surface r = a, then the potential function of the
field is
v = P(aj5f) (a/r)3(1 + 3 cos 20),
= P(a/Sf) (r/o.)2(1 + 3 cos 28),
r > a,
r < a.
2.3-1 Consider a thin \vire of radius a on the z axis from z = -l
to z = I. Assume a point charge q on the x axis at x = s. Calculate
approximately the charge distribution on the wire on the assun1ption
tha t the net charge is zero.
In line with one method in Section 2.3 the following procedure is

Problems
331
suggested. Write the condition
Vi + Vr = V o = constant,
\vherc Vi is the potential of the poin t charge and Vr is the potential
of the charge displaced on the \vire; both are calculated on the axis
of the \vire. Take advantage of the thinness of the wire and express
the preceding equation in an approximate form. Show that to this
order of approximation the potential Va equals the average impressed
potential
1 I l dz 1 l + v s 2 + l2
V o = - = -In .
47rEl 0 vi S2 + Z2 47rEI s
Finally, obtain the approximate displaced charge per unit length
on the \vire
Va
q(z) - A
q
47rEA vi S2 + Z2
where
1 (4l )
A = - In - - 1
27l'"E a
is the average value of A1(z) in equation (2.18).
2.3-2 Consider a uniform flow of current in a dissipative medium
such as soil or sea \vater. "Then a metal wire is inserted in the medium
so tha t the \vire i paralIcl to the lines of flow, the current takes
advantage of the relatively lo\v resistance of the wire and enters the
wire along one half and leaves along the other half. rrhe disparity
between the conductivities of the \vire and the surrounding medium
is such that the \vire may be regarded as a perfect conductor. Sho\v
that the situation is analoguous to an electrostatic problem considered
in Section 2.3 [see :Figure 2.8(b) ] and that the current per unit
length leaving the \vire is given by equation (2.18) provided the
dielectric cons tan t E in Al (z) is replaced by the cond uctivit y q of the
medium.
Aumc no\v that the \vire is insulated from the medium except
at the ends and that the ends arc connected to small metal spheres
of radius b. Calculate the foIIo\ving:
(a) the curren t in the \vire \vhcn the spheres and the wire are
perfectly conducting and E is the electric intensity impressed
on the \virc;

332
lle(tronUlllet;( jirlds
(h) the current in the \vire \\'hen the spheres are perfectly con-
ducting but the resistance of the \\Tire is Rw;
(c) Rw for \vhich the po\vcr dissipated in the wire is maximum.
.,1 } lS7.Dcr :
(a) I = Ell R, \vhere R = 1/27l'"ub and l is the length
of the 'irc; or more accurately,
l = (1/27l'"u) (ljb - lll).
(b) I = Elj(R + Rw);
(c) Rw = R.
2.3-3 1"'wo thin wires of length l are connected to a direct current
generator as shown in Figure 2.8(a). Assume that the surrounding
medium is conducting and that the wires are insulated from it except
at the ends where they are connected to t\VO spheres of very high
conductivity and of radius b small compared with t. Let the resistance
of the wires be negligible and the generator voltage be V. Show that
the power delivered to the medium (and dissipated in it) is
P = V2/ R,
where
R = 1/27l'"ub
or more accurately
1 (1 1 )
R = 211"0" b - 2l ·
Consider a similar pair of wires, parallel to the 1irst pair and per-
pendicular to the line joining the centers. Let this pair of wires
be connected to a "load resistance" R L rather than to a generator.
Assume that the distance d 'between the t\VO pairs of wires is large
in comparison with l. Show that the po\ver delivered to the load is
V 2 l4RL
p -
ree - 7r 2 q 2 d 6 R2(R + R L )2'
where R is given above.
Obtain the ratio of the received power to the power delivered to
the medium and show that the maximum power transfer ratio is
(P rccl P) max = l41 41r 2 U 2 d 6 R2.

Problems
333
2.3-4 Assume that the potentials of the mctal sphcres in Problcm
2.2-1 are equal. f""ind the charges.
Ansu'cr:
41rfaV[1 - (3a/2s) ]
qo = . , on the middle sphere
1 + (a/2s) - (2a 2 /s 2 )
1 - (a/s)
ql = qo , for an outside sphere.
1 - (3a/2s).
2.3-5 Show that except for small quantities of the order (a,js)2
the charges in Problem 2.3-1 are inversely proportional to the po-
tentials in Problem 2.2-1. That is,
ql/qO = Vo/l'l = 1 + (a/2s).
rrhis is not surprising since the poten tial of a sphere of radius a,
under the conditions assumed in these problems. is dctermined largely
by the charge on it. lIenec \ve can equalize approximately the po-
tentials in J)roblem 2.2-1 by assuming that the charge on the middle
sphere is
qo = q / (1 + 2;: )  q (1 _ 2 s a )
and on an ou tside sphere is
/( 3lL) ( 3a)
ql = q 1 + 205  q 1 - 2s ·
2.3-6 Considcr three infinitc, coplanar, parallel, thin wires of
radius a. Lct q be the charge per unit length on one outside wire and
-q on the remaining two wires which arc kept at the same potential.
Let s » a and 2s be the distances bct\vccn the axis of the first wire
and the axes of the other t\vo. Find the ratio of the charge densities
on the latter \vires.
.Answer:
In ( s / a) + In 2 .
, the larger charge IS on the central
In (s/a.) - In 2
'VIreo
2.3-7 Consider t\VO parallel thin \vires of radius a and length l
shorted at one end as shown in I"igurc 2.3-7 (a). Let s » a be the
distance between their axes. Find the transverse voltage V. UN due

334
Electroma gnetic fields
...
2a
z
fez)

I
(a)
l(z)

v
...
--...
z
l
(b)
FIGURE 2.3-7
to the displaced charge, the charge q(z) per unit length on the lower
wire, and the current 1 (z) in the following two cases:
(a) The open loop is stationary in a slowly varying uniform
magnetic field perpendicular to the plane of the loop. The
time rate of change of the magnetic flux density is B.
(b) The loop is moving parallel to the long wires in its plane with
a speed 'V in a uniform magnetic field perpendicular to the
plane of the loop. The flux density is B. See Figure 2.3-7 (b).
Answer: (a) V. MN = -Bsz, q(z) = CV,MN where
c=
In (s/a)
I(z) = CBS(Z2 - l2)
(b) V.\lN = BS1J, q(z) = CBsv, f(z) = CBsiJ(l - z).
2.3-8 Answer the question in the preceding problem when the
loop is shorted at the other end. Let R be the resistance per unit
length of each long wire. Neglect the resistances of the short wires.
7rE
Answer:
(a) fez) = 10 = -Bs/2R, V ftfN = 0, q(z) = o.
(b) V JlfN = Bsv, q(z) = CBsv,I(z) = CBsiJ(!l - z).

Problems
335
2.3-9 Consider a closed loop as in Problem 2.3-8 except that the
resistance per unit length is Rl for z < 1/2 and R 2 for z > l/2 as
R. R 2
A
I V(zJ fez)  ! I
1/2 1/2
-- - - ..
... , .. --
c
D
B
FIGURE 2.3-9
indicated in Figure 2.3-9. Ans\\rer the questions in Problem 2.3-7
and compare them with the case Rl = R 2 = R.
A,lswer:
(a) fez) = 10 = -Bs/(R 1 + R 2 ),
V (z) = I3sz(R l - 12) (R 1 +R 2 ),
q(z) = CV(z).
(b) Same answer as in Problem 2.3-8(b).
Note: 1"he ans\vcr in case (a) is exact when 13 = constant. Other\vise,
it is the tirst approximation. If f3 varies \vith time, V(z) and q(z)
also vary with time and charging or displacement currents should
be added to I (z). l'hese currents in turn will affect V (z) and q(z).
rrhe step-by-step calculations may be continued.
2.3-10 Suppose that the thin wire sho\vn in Figure 2.8(b) is
moving to the right, with a. speed v, in a uniform magnetic field
perpendicular to the paper. Let B be the magnetic flux density
pointing into the paper. Obtain the density of displaced charge.
Anszoer: 1'he answer is given by equation (2.18) if \ve let
Eo= Bv. 1"0 compare with the results in Problem 2.3-7 note that
A (z) corresponds to the reciprocal of C. '-The calculation of current
is more difficult unless A (z) is approximated by its average value.
2.3-11 Let the plane z = h be the surface of an ocean of uniform
depth II » It and conductivity u. Consider t\VO copper spheres of
radius a, connected with insulated copper wires to a de generator.
Let the centers of these spheres be at points (0, -l, 0) and (0, l, 0)
where l > 2a. Assume that two other copper spheres arc centered
at (0, d - l, 0) and (0, d + l, 0) and are connected with insulated

336
Electr01Jlagnetic fields
copper \vires to a resistance R L (" load"). Let h « d « II. Find the
foIIo,ving:
(a) thc current through the gcnerator when the voltage bet\vecn
its terminals is 1'0'
the po\ver dissipated in the ocean,
the electric intensity at the second pair of spheres,
the current through R L ,
the maximun1 po\vcr \vhich can be abstracted from the field
by a proper luad resistance R L .
27ru(], Va
(b)
(c)
(d)
(e)
Answer:
(a)
1 - (a/2l)'
(b)
21l"u(], V o
,
1 - (a/2l)
(c)
E =
y [1
4alV o
(aj2l) Jd 3 '
(d)
r 1 - (a/21) . ]-1
8al 2 d- 3 [1 - (CL/21) J-l . + R L V o ,
 27ru(/,
327rua 3 l 2 d- 6 [1 - (aj21) J:q ".
(e)
2.3-12 Consider fIve metal spheres of radius a, centered on the
z axis at z = 0, =f=c, =f=2c. Let the intensity of the incident field be
Ez = Eo. Suppose that the spheres have been 11101ncntarily con-
nected \vith conducting \vires. Sho\v that if (jl and q'!. are the charges
on the spheres centered at z c and::; = 2c. respectively, then
q2/ql = 2[1 - (5lL/6c) J/[1 - (19a/12c) J.
Note that as a approaches zero. the ratio approaches 2.
2.3-13 l"herc is a fourth method of obtaining an approximate
solution for the static configuration sht)\vn in F'igurc 2.8 (a). It is
based on an assumption that the electric lines of force run along the
meridians on sphercg concen tric \vith the midpoin t bet\veen the ,vires.
l"his assumption enables us to obtain 1)8 and the voltage bct\,,"cen the
wires along a typical n1eridian. ']"hus \ve fInd
q(z) = 1rEl'o/ln (2z/a.)
,vhich agrees ,vith equation (2.16) except for the last term in the
expression for A (z). l"his last term represents the" end effect" due to
departure of clectric lines of force from meridians as z approaches t.

ProbJcnls
337
2.3-14 Using the result of the preceding problem, sho\v' that an
approximate charge per unit length of a thin wire, Figure 2.8 (b),
in a uniform electric field of intensity Eo is
q(z) = 21rEEoZ/ln (2z/a),
 27rEEoZ/[ln (2l/ a)
z > 0,
1J,
where in the second approximation the slowly varying (except in the
vicinity of z = 0) denominator has been replaced by its average
value.
Since this result. as well as equation (2.18), sho\vs that q(z) is ap-
proximately proportional to z \ve can improve our approximation by
assuming that
q(z) = Eoz f(z)
\vheref(z) is a slowly varying function. The equation which precedes
equation (2.18) would then become
I l Eouf(u) du
EoZ = -I 47rE Va 2 + (u - Z)2 '
\Ve no\v take advantage of the facts that f(u) is a slowly varying
function and that the greatest contribution to the integral comes
from the vicinity oflt = z: we replace f(u) by f(z), integrate, and
obtain
j(z)  27rE / [In l - 1 +  In (1 - :) J.
If the denominator is replaced by its average value, then
f(z)  27rE/[ln (41/t],) - 2J.
Since In 2 = 0.693 and In 2 - 1 = - 0.307. the nc\v val ue for the
charge density is somc\vhat larger than that given by the preceding
approxima tion.
2.3-15 Consider t\\"o thin ,vires of length I on the z axis, one
going up\vard from a point ..4 slightly above the origin and the other
going down\vard from a point B slightly bclo\v the origin. Let a
uniform field of intensity E z = Eo be imprcssed on the \vires. Since
the ilnprcsscd potential is a linear function of z, the potcntial due to
the displaced charge, on the surface of each \vire, is also a linear
function of z. Hence the density of displaced charge is approximately

338
1I('clr011laJ?lleli( .fields
a linear function. Shc>\v that 1/ nA is then approximately equal to
Eol and that
q(z) ==7rEEo( -I + 2z) [In (2/ja) - IJ.
and that q( -z) = -q(z).
z > 0,
2.4-1 Consider an electric dipolc of nl0ment ql along the z axis
at z 0 and a concentric conducting sphere of radius b. Find the
field and the surface density of charge displaced on the sphere.
l1nsu'cr: In thc interior of the sphere the field is the sum of the
original l1cld of the dipole and a "reflected" field due to electric
charge displaced on the sphere. '1'he reflected field is unifornl and
paralIc I to the z axis.
E = ql .
47rEb 3
Outside the sphcre the lleld is zero. The surface density of charge on
the sphere is
qs =
3ql cos ()
47rb 3
2.4-2 Suppose that instead of the dipole in Problem 2.4-1 there
is a point charge q on the z axis at z = l « b. Obtain the field and
the charge distribution on the sphere.
./1 llswer: 1'he interior field is the sum of the original fIeld of the
poin t charge and the reflected field equal to that in the preceding
problcnl. 1'he exterior field is radial and Er = q / 47rEr 2 . l'hc den-
sity of charge displaced on the sphere equals that in the preced-
ing problem. f.lote:Use t\VO methods:
(1) Expand the potcn tial of the given poin t charge in a
sui ta ble power series.
(2) I<eplacc the given point charge by a suitable point
charge in the center of the sphere and a suit,ble
dipole.
2.4-3 Solve the preceding problem \vhen the sphere is grounded.
Answer: 'I'he interior field is the same as in the prcccd ing
problen1, the exterior field is zero, and the charge density is
q ( 3l )
qs = -- 1 + cos 0 .
47rb 2 b

Prohlcn15
339
2.4-4 In }>roblem 2.4-3 assume that the charge q is on a sphere
of radius a « b "rith the center at z = t. 11'ind the potential of the
small sphere.
II = -3- _ --L (1 + l2 ).
47rEa 47rEb b<:.
2.4-5 Suppose that the sphere in Section 2.4 is perfectly con-
ducting and the conductivity of the medium around it is u. In \vhat
way does the sphere affect the field?
Ansu'cr:
2.4-6 Calculate the effect of a spherical airholc of radius a in a
conducting medium on a uniforn1 impressed field of intensity Ez Eo,
Ex = E y = O. 1\&';Un1C that the center of the hole is at the origin.
A nS'ioer: 1'hc field reflected from the hole is
E = - (a/r) 3 Eo cos 0,
E8 = -(a/r)3Eo sin 0
2 .4-7 Using the ans\vcr of the preceding problem obtain the field
in the holc.
Allsu'er:
Ez; = 1.5E o .
E ....-E-O
z - 11 - .
2.48 Consider a metal cylinder of radius a, coaxial \vith the z
axis, and assume that a uniform field
E = Eo, E = E; = 0
is impressed on it. Calculate the reflected field. Hin t: Take advantage
of the results obtained in Problem 2.1-2.
.Answer:
E; = (a,/ p) 2 Eo cos ({), E; = (a./ p) 2 Eo sin cpo
2.4-9 Consider a thin cylindrical metal shell of radius a, coaxial
\vith the z axis. Assume that along the z axis there is a double line
source of moment qs per unit length (see Problem 2.1-2). Obtain the
ficld and the charge density on the cylinder. Sho\v that the reflected
field is uniform and normal to the z axis.
qs cos cp (1 1 )
E p = 2 --:; + --:; ,
7rE p- a"
A JlS71.Jcr:
p < a
E", = qs sin If' ( _ ),
27rE p2 a 2
E p = Ef(J = 0, p > a,
qs cos ({)
p < a
qs =
7rEa 2

340
Elec/ro1Jzagnelic fields
2.4-10 i\ssunlc that instcad of the source in IJroblem 2.4-9 \V'e
have a charged filament parallcl to the z axis \vhich passes through
point x = s, )' = z = O \vhere s « a. Find the field \vhcn thc charge
per unit length is q.
A llsu'cr: In the in tcrior of the cylinder the field is the sunl of
the ticld of thc line charge and the reflcctcd field E qs/27rEa 2 ,
IJ: = E O. Outside the cylinder the field is radial and E p =
Q/27rfp.
2.4-11 Suppose that in ]>roblem 2.4-10 the cylinder is grounded
(at zero potential). Obtain the potential in the interior.
Allsu.'cr:
v = - (q/27rE) [In (PI/a) + sx/a 2 J,
V\'herc
PI = (p2 - 2sp cos 'P + S2)!.
2.4-12 Suppose that in the preceding problem the charge is on a
thin \vire of radius b « a rather than on a Jine. I"'ind thc potential of
the \virc.
A llS7,oer:
(q/27rE)[ln (a/b) - (s/a)2J.
2.4-13 Assume that the mument of the electric dipole in Problcm
2.4-1 is varying slo\vly and 5ho\\' that
ql sin () qlr sin ()
II", = + .
47rr 2 87rb 3
Suppose that b is vcry large; let a mctal sphere of radius a be cen-
tered on the z axis far from the dipole so that the intensity Eo of the
incident field is sensibly uniform. Sho\v that
II", = (Eoa 3 /ri)Eo sin 8 1 + EOrlEo sin 8 1 ,
\vhcre (rl, 8 1 ) are the spherical coordinates with reference to the
center of the snlall sphere. provided the lattcr is not too close to the
surface of the large sphere. \'Then the small sphere is close to the
surface of the large sphcre the field of the charge displaced on the
small sphere \vill in its turn displace a charge on the large pherc and
thus \vill generate an additional field.
If the small sphere is in a scnsibly uniform time-varying field Eo
off the z axis, then the magnetic intcnsity due to the induced currents

ProhlcnlS
341
is still given by the first term in the above expression. 'fhe magnetic
intensity of the incident field, ho\vcver. is no longer given by the sec-
ond term.
2.4-14 Suppose that a solid metal sphere of radius a is nloving
\vith a constant speed Vo in the x direction in a uniform magnetic
fIeld parallel to the y direction, 13 11 = Bo. Calculate the intensity of
the electric field outside the sphere and the density of charge on the
sphere.
l1nswer: If the origin of the spherical coordinate system is at
the een ter of the sphere, then
Er = 2voBo{a/r)3 cos 0, Ee = voBo(a/r)3 sin 8,
qs = 2EovoBo cos 8.
2.4-15 Show that if the solid sphere in the preceding problem is
replaced by a thin spherical shelL the external electric field \vill be
unaltered while an internal field E z = -'oBn will appear. Show also
that the charge density will be qs = 3Eovo13o cos ().
2.5-1 Consider an infinite medium of conductivity 0'0, supporting
a uniform electric field of intensity Eo. parallel to the z axis, and
therefore an electric current of density J o = uoEo. Suppose that
within a sphere of radius a, concentric with the origin of the coordinate
system. the medium is replaced by another medium of conductivity
u. l"'hc problem is to obtain the modification in the originally uniform
field.
lxp]ain \vhy the solution is essentially the same as that for a
dielectric sphere in a dielectric medium see Section 2.5 and that we
need only substitute Uo and u for Eo and E, and J for ]J. In particular,
sho\v that the fleld inside the sphere is uniform, parallcl to the im-
pressed field, and equal to
3 3u
E z = Eo, ] z = J o .
2uo + U 20"0 + 0'
'rhus a;;; u/uo  O. \ve have E z  1.5E o , Jz  1.5(u/uo) J o . That is,
the electric in tensity is not very differen t from the original intensity
in the homogeneous medium. 1"'hc current density, on the other hand,
bcconlcs vastly different. l"'his is to be expected intuitively since
the curren t tcnds to avoid en tcring the highly resistive sphere and
tends to flo\v around it. Similarly, the electric flux tends to avoid a
sphere \vhose dielectric constant is small in comparison \vith the
dielectric cons tan t of the surrounding nledium.

342
Electronlag'lletic fields
On the other hand, if u I Uo  00, then
Ez  3 (uolu) Eo, J,  310.
The electric intensity approaches zero. The current tends to seek
the highly conducting sphere but only from a limited area. From
regions more distant from the sphere the current takes longer paths
in the original medium in order to reach the sphere and thus en-
coun tcrs too great a resistance to benefit from the lo\v resistance of the
sphere.
'[hese considerations have an important bearing on the classical
definitions of E and jj in solid dielectric media. In vacuum, E is
defined as the force acting on a unit stationary charge and it can be
measured. In vacuum, tJ can be defined simply as foE since EO is a
universal constant. Thus fJ is also related simply to the force on a
unit charge. In fact, in classical electrostatic units the two quantities
are equal. In a solid dielectric medium the charge cannot move and
the force on it cannot be measured directly. For this reason, the
classical physicist defined E and iJ with the aid of Kelvin cavities.
One of them was a long thin tunnel in the medium, [Figure 2.5-1 (a) ]
and the other a thin pancake-like cavity [Figure 2.5-1 (b) J. Noting
that the tangential component of E and the normal compo:1ent of D
are continuous. one can argue that E in the dielectric can be measured
by E in a thin tunnel (a) and iJ in the dielectric by foE in a pancake
type of cavity (b). As far as the thin tunnel is concerned we are on
a solid ground. Intuitivcly. \ve feel that the field is not distorted
much by the tunnel. In the case of the pancake cavity, however. it is
evident that the field \vill be distorted considerably and that the
electric flux \vill tend to avoid the ca vi ty to an increasing extent as E
increases. Of course, by making the cavity sufficiently thin we can
eventually make the distortion negligible; but the thickness will
defin i tel y depend on the ratio EI EO (where Eo is the dielectric constan t
of vacuum) and will have to 'be halved \vhen E is doubled. These
conclusions can be verified by obtaining exact solutions for thin
prolate and oblate spheroids. Thus if a and b are semi-major and
semi-minor axes of a thin oblate spheroid, the quan tity bEl afo must be
very small in conlparison \vith unity to make fJ in the cavity ap-
proximately equal to jj in the medium before the cavity has been
hollo\vcd out.
1he definition of jj in Section 1.14 of this hook uses a thin l1zetal
Plate [Figure 2.5-1 (c) J. It is evident intuitively that tJ is not
appreciably affectcd by the plate. This also can be verificd by solving
the problem for an oblate metal spheroid.

Problems
343
(a)
E
EI
(b)
(c)
E
E
  l  A l A       4
Dt
Dt
F)(URE 2.5-1
2.5-2 Consider a dielectric cylinder of radius a, coaxial with the z
axis, in a uniform field of intensity Ex = Eo, Ell = Ez = O. Let the
dielectric constant of the cylinder be E and that of the surrounding
mcdiunl EO. Obtain the field inside the cylinder and the field reflected
by the cylinder.
A nsu'er:
E -
x -
2EQ
Eo,
E + EO
Ell = E z = 0,
p < a,
E - EO (a)2
E: = Eo - COS cp,
E+Eo P
p > a.
r E - EO (G)2
EI(J = Eo - sin <p.
E+EO P

344
Electronla gnelt'c fields
2.5-3 Calculate the field of an electric dipole of moment ql at the
center of a dielectric sphere of radius a. Assume that e and EO are
dielectric constan ts of the sphere and the surrounding space and
choose the coordinate system \vi th the origin at the cen ter of the
sphere and the z axis along the dipole.
A 1lSU)er:
ql CDS 8 (eo - e)ql CDS ()
E,= +
211'"Er 3 211'"e( 2E o + e)a 3 '
r < a,
3ql cos ()
-
,
211'" (2eQ + e) r 3
r > a,
ql sin 8
Ee=
411'"Er 3
(EO - E) ql sin ()
,
211'"e( 2e o + e)a 3 c
r < a,
3ql sin 8
-
,
411'" (2Eo + e) r 3
r > a.
2.5-4 Suppose that in Problem 2.5-3 the dipole moment is varying
slowly with time. Obtain the magnetic intensity (assuming that the
electric field is already kno\vn) .
ql sin () (EO - e) qlr sin ()
Hrp = + ,
411'"r 2 47ra 3 (2eo + e)
Answer:
r < a,
3Eoql sin ()
-
,
411'" (2eo + e) r 2
r > a.
2.5-5 In Problem 2.5-3 replace the solid dielectric sphere by a
spherical shell of radii a and b > a. Calculate the field.
Answer:
(e - eo) (a. / b) 3
Let k =
2eo + E
3ql
i.}f = ,
2 (1 - k) + (1 + 2k) (EO/E) .
,,\1 [ ( 1 - k) - (1 + 2k) (eo/E) ]
./1 =
6 7rE oa 3
3e o Jf
p = .
2Eo + e

Problclns 345
Then
,
ql cas 8
Er = + A cas 8, r < a,
27rEor 3
}.;{ cos 8 kA,f cas 8
- , a < r < b,
27rEr 3 27rEa 3
P cas 8
- , b < r
27rEQT3
E 8 = ql sin 8 A sin 8, r < a,
47rEor 3
Jf sin () kAf sin ()
+ ,
47rEr 3 27rEa 3
a < r < b,
P sin (}
,
47r E or 3
b < r.
2.5-6 A dielectric sphere moving in a magnetic field is polarized.
The polarization P equals (E - EO) E, \vhere E is the total force per
unit charge. Calculate the polarization. and the electric fIeld outside
the sphere, under the conditions specified in Problem 2.4-14.
2 (E - EO) 'oBo
Pz= ,
(E/Eo) + 1
Ansu'cr:
2 (E - EO) 'voBo (a)3
Er =  cos 8,
E + EO r
(E - EO) ""oBo (a)3 .
E8 = - sin 8.
E + EO r
2.5-7 In Section 2.4 V'C obtained the density of charge displaced
on the surface of a conducting sphere of radius a by a uniform elec-
tric ficld. 'T'his charge produces a.n exterior flcld of the dipole type,
,vhich ,vas also calculated. Since the total intcrior field must equal
zero, the displaccd charge must prod uce an interior field, equal and
opposite to the impressed field. \'erify this conclusion by direct
calculation. That is, assume that on the surface of the sphere r = a

346
Electromagnetic fields
there is a thin layer of charge of density
qs = 3EOEo cos ()
and calculate the field produced by this charge.
2.5-8 Consider 1wo parallel conducting plates, uniformly and
oppositely charged. Explain why a neutral metal sphere, introduced
between the plates, will reduce the voltage between the plates, and
hence, increase the capacitance between the plates.
Suppose now that the plates are connected to a source of constant
voltage. EXlllain why a neutral metal sphere, introduced between
the plates, will cause an additional displacement of negative charge
from the positively charged plate to the negatively charged plate.
Thus the capacitance between the plates will be increased.
Suppose that small metal spheres are scattered sparsely and
uniformly between the plates. Let N be the number of spheres per
unit volume and a be the radius of each sphere. Show that on the
average the displacement density between the plates is
D = EoEo + 47rN a3EOEo,
where Eo is the electric intensity, that is, the impressed voltage per
unit length between the plates. Thus the effective dielectric constant
of the medium between the plates is
E = (1 + 47rNa 3 )EO.
2.6-1 Three conducting spheres of radius a are centered on the z
axis at z = 0, z = l, and z = 2l. The first sphere has a charge q and
potential V. The second and third spheres are at the same potential
- VI and the total charge on them is -q. Find the approximate
charges on each of these two spheres when a « l and note the" prox-
imity" effect.
Answer: The approximate charges on the spheres centered at
z = land z = 2l are, respectively,
1 - (a/2l)
- q .  - !q[l + (aj2l)],
1 - (all)
and
1 - (3aI2l)
-!q  - !q[l  (aI2l)].
1 - (all)

Problems
347
2.9- 2 Solve the preceding problem for three infinitely long
conducting wires which are parallel to the z axis and whose axes pass
througp points (O 0, 0), (l, 0, 0), and (2l, 0, 0). Assume that q is
the charge per unit length.
Answer:
The ratio of the charge densities on the last two wires is
In (l/a) + In 2
In (lja) - In 2
and the charge densities are, respectively,
In (l/a) + In 2
1
-2q In (lja) ,
and
In (lja) - In 2
1
- 2q In (lja) ·
2.7-1 Show that the magnetic intensity of an infinitely long
current filament I along the z axis can be obtained from a potential
function
U = -I'P/27r + constant..
2.8-1 Consider a straight electric current filament I of length l.
Calculate the vector potential. In what way does it differ from the
scalar poten tial of a similar filament of electric charge [see eq ua tion
(2.11)]?
2.8-2 Show that the magnetic flux density around an infinitely
long electric current filament I along the z axis may be obtained
from the following logarithmic vector potential
A p = At' = 0,
A. = - (p./ /27r) In p + constant.
Note the resemblance to the scalar potential of an infinitely long
filament of charge.
2.9-1 Consider a square current loop, length b on each side,
made with a very thin wire of radius a. Find the magnetic intensity
at the center of the loop.
Answer:
2Y2I/7rb.

348
Electromagnetic fields
2.9-2 Find the fields at the center of a current loop in the shape
of a regular polygon with n sides, inscribed in a circle of radii a or
circumscribed about it.
Answer:
1£1 tan (7r / n )
27ra
and
nI sin (7r/n)
27ra
2.9-3 Consider two thin parallel wires of radius a and length l,
shorted at both ends to form a loop. Let s « l be the distance between
the axes of the wires. Suppose that the loop is perpendicular to a time-
varying uniform magnetic field. In Problem 2.3-9 it was assumed
that the time rate of change of magnetic flux density, that is, the
magnetic current density 13, was constant. It was found that current
in the loop is
I = iJs/2R,
where R is the resistance of the wire per unit length. This current
generates a magnetic field which is superposed on the field given by B.
As long as 13, and hence I, are constant, the added field does not
affect I. Otherwise, it will affect it. For instance, if the loop is brought
into the field, the current I can not suddenly assume a value different
from zero since this would mean an instantaneous creation of magnetic
flux associated with the cJlrrent, and therefore, an infinite electric
field. Hence, the above equation can be true only in the steady state
even when 13 is constant. Obtain the more exact equation for the
current by taking into consideration its magnetic field. Assume that
if the loop is in the plane of the paper, B is coming out of it and
that consequently the positive direction of electric current in the
above equation is clockwise.
Answer:
d .
2RI =  (Bs - LI)
dt '
where
p.
L = - In (s / a) .
7r
2.9-4 Note that if the loop in the preceding problem is perfectly
conducting and is brought into the magnetic field from a field free
region, no net magnetic flux will be linked with the loop and the
current is determined by B and not by 13,
I = Bsj L.

Problems
349
Show that if the magnetic flux density is generated at a constant
rate i3 from t = 0 and on, then
Bs
I (t) = (1 - e- 2R t I L) .
2R
2.10-1 Consider a thin solenoid extending from z = 0 to z = t.
Let C be the circulating current per unit length of the solenoid and S
the cross-sectional area of the solenoid. Using equation (2.45) show
tha t the magnetic potcn tial outside the solen oid is
CS CS
u=---
41J'"rl 41J'"r
where rand rl are the distances from the origin and the point at
z = t. To simplify thc integration take advantage of the fact that the
potential as given by equation (2.45) is the derivative with respect to
z of the poten tial of a certain poin t source.
Show also that if <I> is the magnetic flux through the solenoid, not
too near its cnds, then the above equation becomes
u= --
4> <I>
41J'"rl 41J'"J,tr
as one would naturally expect on intuitive grounds.
2.10-2 Consider a thin spherical shell of mean radius a, thickness
Iz, and conductivity u. Let the center of the shell be at the origin.
Assume that at t equal to zero a uniform magnetic field, parallel to
the z axis is impressed on the shell at the rate 13 0 . This time-varying
magnetic field will generate an electric field of intensity Erp. The
latter ,viII start circulating curren ts of density J rp in the shell, which
in their turn will generate a magnetic field, both inside and outside
the shell. On account of the boundary conditions and symmetry, this
magnetic field must conform to the original field. Following this line
of thought derive the equations for the subsequent field, first for an
arbitrary slowly varying impressed field Bo and then for a field in-
creasing at a constant rate. Calculate the current density in the shell.
Obtain the field when the conductivity of the shell is infinite.
Answer: If Bz = A is the magnetic flux density in the interior
of the shell (r < a) due to the circulating currents in the shell, then
Ii + (3/JJouha)A = -13 0 . (1)
For r < a
B. = Bo + A,
EfP = -}(Bo + A)r sin 0,

350
Electronzagnetic fields
and for r > a the field is the sum of the impressed field B is = Bo,
Ef/J = - }Bor sin 0 and the induced field
Br = A (air) 3 cos 0, B, = !A (a/r) 3 sin 0,
Erp = - !Aa (air) 2 sin (J.
The current density in the shell is
Jrp = -ua(Bo + .4) sin (J.
If u = 00, one obtains directly from equation (1)
A + Bo = constant = 0, (2)
which means that in this case there is no field inside the shell.
The general solution of equation (1) is
A = -exp (-3tjJl(}f1ha) [ .8 0 exp (3tjJl.ouha) dt.
o
If 13 0 = constant, then
A = -lJlouhaB o [1 - exp (-3t/Jlouha) J, (3)
so that ultimately
A = - 'lJlouhaB o
unless u - 00. Derive equation (2) from equation (3).
2.11-1 Consider a sphere of radius a, centered at the origin. Let
Jl be the permeability of the sphere, and Jlo that of the surrounding
medium. A magnetic field is created by a closely wound solenoid,
coaxial with the z axis and almost touching the sphere. Let C be the
circulating current in the tp direction per unit length of the solenoid.
Show that the magnetic intensity in the interior of the sphere is
3p.oC
ll =
2JJ.o + Jl
and that the intensity just outside the solenoid in the xy plane is
(JlO - J.L) C
11" = .
2JJ.o + JJ.
Show that the sphere becomes a magnet of moment
41l"a 3 ,uo (JJ. - J.Lo) C
2J.Lo + J.L

ProbJcnls
351
The sphere acts as if there were a layer of positive magnetic charge
on its upper hemisphere and a layer of negative charge on the lower
hemisphere \vhich create the exterior field of the magnet and produce
a downward magnetic intensity
(J1. - J1.o) C
2J1.O + J1.
opposing the field of the solenoid inside the magnet (" demagnetizing"
intensity) .
l\Jagnetic polarization is the dipole moment per unit volume
3J1.o (J1. - J1.o) c
p= .
2J1.O + J.I.
Sho\v that inside the sphere
Bz = J.l.ollz + P.
\\T e can also write
Bz = J.l.o(Il z + M),
where M is called the magnetization of the sphere. Here M is the
area 1noment per unit volume of amperian currents induced in the
sphere by the field of the solenoid, or the magnetization of the sphere.
2.11-2 Consider a spherical shell a < r < b of permeability Jl,
imbedded in a medium of permeability J1.o. Assume a current loop of
area moment I S centered at the origin and in the xy plane. Explain
why the magnetic intensity may be obtained from the results of
Problem 2.5-5 if E and EO are replaced by J1. and J.l.o, and ql by JJ,oI S.
2.11-3 Suppose that a uniform magnetic field of intensity II z = Ho,
Hz = Hu = 0 is impressed on the shell in the preceding problem
(without the current loop). Calculate the field inside the shell (where
r < a).
Answer:
Hz = H1I = 0, II,., = kIlo,
\vhcre
k = 9 (Jlo/ Jl)
(2 + J.l.o/J.I.) (1 + 2J.1.o/J.I.) - 2(a/b)3(1 - JJ.O/J.I.) 2.
The factor k represen ts magnetic shielding. The same factor repre-
sents the ratio of the external field in the preceding problem to that
which would have existed there without the shield.

352
Elcclro1na?1zclic fields
2.12-1 Obtain the potential of a charge q at point (0, O 0)
between conducting planes z = - a and z = a.
A 1lS"ti..'er:
q q 00 (1 1 )
v= . +- (_)n -+-,
41rE V p2 + Z2 47rE n-l r n Rn
where
Tn = y p 2 + (z - na)2,
Rn = V p2 + (z + na)2.
2.12-2 Find the potential of a uniform line charge on the z
axis between conducting planes y = -a/2 and y = a/2. Let q be the
charge per unit length.
Answer:
q q co 1 1 2n R 2n
V = -- In V x 2 + y2 - - In I ,
27rE 27rf n-l r2n- 1 R 2n - 1
where
Tn = yx 2 + (y - na)2,
Rn = yx 2 + (y + na)2.
2.12-3 Consider a metal cylinder of radius a, coaxial with the z
axis, and a uniform line charge of density q, parallel to the z axis and
passing through point (l, 0, 0) \vhere l > a. The charge on the surface
of the cylinder will be displaced in order to make its surface an
equipotential surface. Show that a hypothetical image line charge of
density -q, also parallel to the z axis and passing through (a 2 /l, 0, 0)
actually accomplishes this effect and thus produces outside the
cylinder the same field as the charge displaced on the cylinder.
Hence, the density of the displaced charge equals Dp(a, cp, z) where
D p is due to both the given line charge and the image charge.
Note that the result depends on the symmetry in the expression
(a 2 - 2a.l cos cp + l2) 1/2
for the distance between the given line charge and a typical generator
of the cylinder.
Show that if the logarithmic potential of a line charge q is taken as
q
V = -- In Po
27rE "
where Po is the distance from the line charge, then the potential of the
metal cylinder is
q l
V 2 = -- In-.
27rE a

Problems
353
Show that the potential of a cylinder of radius b « 1 - a, coaxial
with the line charge, is approximately
q b
VI = --In
21rE l - (a 2 /l)
and that
q [l2 ( a 2 )]
VI - V 2 = - In - 1 - - .
27rE ab l2
Show that if the metal cylinder is electrically neutral, its potential
is - (q/2rrE) In l and that the potential of the cylinder of radius b is
q 1 ( a 2 )
2'11'E In b 1 - l2 ·
Show that if charge of density per unit length, q, is placed on the
metal cylinder, its potential will become - (q/27rE) In (al) and that
the potential of the cylinder of radius b will be
(q/27rE) In [l-lb-1(1 - a 2 /l 2 )]
and that the potential difference between the latter and the former is
..L In [ (1 _ a 2 )].
27rE b l2
2.12-4 Consider, as in the preceding problem, a line charge of
density q and a metal cylinder with an equal and opposite charge.
Show that the surface density of charge on the cylinder is
q[(l2/a) - a]
qs = -
21rpl
where PI is the distance from the line charge to a typical generating
line of the cylinder.
Hence if the cylinder is neutral, the density of charge is
qs = !L [ + a - (l2/a) ].
27r a PI
Show that the density of charge is zero w hen
PI = -y! l2 - a 2 ,
that is, where the planes tangent to the cylinder pass through the
line charge.

354
Electromagnetic fields
When the cylinder is charged, as in the preceding case, the charge
density at points of tangency is -q/27ra, the same as for uniform
distribution of charge. When l » a, PI does not vary much with rp
and the charge is almost uniformly distributed. But as l approaches a,
the charge tends to concentrate on the side of the cylinder facing the
line charge. Show that the densities at points nearest the line charge,
at points of tangency, and at points farthest from the line charge are
proportional to
l+a l-a
l - a ' 1, l + a '
Show that the charge per unit length of the cylinder between the
planes rp = - rpo and rp = ({)o is
q(cpo) = - 2q tan- 1 ( l + a tan cpo).
7r l - a
Thus the charge on that half of the cylinder which faces the line
charge is
q(!7r)
2q 1 + a
- -- tan- 1
7r 1 - a
2q[l-a (l-a)3 ]
- -q + - - + ... .
7r l + a 3(1 + a)3
2.12-5 Consider two parallel metal cylinders of radii a and b.
Let q be the charge per unit length on the first cylinder and -q
that on the second. Assume that the distance between the axes is d.
When d is very large, there is little interaction between the charges.
On each cylinder the charge distribution is essentially uniform. As d
diminishes, the interaction becomes more important. Thus if the
charge on cylinder B (the one of radius b) is uniformly distributed,
its exterior field equals that of a line charge on its axis. Cylinder A
will then be equipotential if its field equals .that of a line charge at
distance SI = a 2 /d from its axis. This field acts on cylinder B which
will become equipotential if its field is not that of initially assumed
axial line charge but that of a line charge at distance Cl = b 2 / (d - SI)
from its axis. To make cylinder A equipotential, the image charge
must be shifted to distance S2 = a 2 / (d - Cl). Similarly, the image
charge inside cylinder B must be shifted to distance C2 = b 2 / (d - S2).
This iterative process can be continued indefinitely and the limiting
positions of the image charges
s = lim Sn,
c = lim en

Problems
355
can be expressed as periodic continued fractions. Show that the
limi t of. the potential difference between the cylinders is
q (d - s)(d - c)
VI - V 2 = -In .
27rE ab
Show that sand c are roots of a certain quadratic equation. Thus
s2d + s(b 2 - a 2 - d 2 ) + a 2 d = o.
This equation has two roots, one of which must be rejected on physical
grounds. What is the condition which the correct root must satisfy?
Once proper values for sand c are obtained, VI - V 2 is determined.
A simpler expression for this potential difference can be obtained if
we establish
a 2
s =
,
d-c
b 2
c =
d-s
and note that
(d - s) (d - c) = d 2 - sd - b 2 .
In this manner show that
q d 2 - a 2 - b 2 + V (d 2 - a 2 - b 2 )2 - 4a 2 b 2
VI - V 2 = - In .
27rE 2ab
Show that approximately
q d q ( a 2 + b 2 )
VI - V 2 = -In - - In 1 - .
7rE Vab 27rE d 2
2.12-6 Show that a sphere of radius a, centered at the origin and
in the presence of a point charge q at point (0, 0, l) becomes an
equipotential surface if we introduce a charge -qall at point (0, 0,
a 2 /l). Hence, this sphere may be replaced by a thin metal sphere
without disturbing the field. Show that the surface density of charge
on the metal sphere which produces the same exterior field as the
image charge - gall is
qs =
q[(l2Ia) - a]
47rTi
where TI is the distance between a point on the sphere and the point
charge.

356
Electrol1zagnet-ic fields
If a unifor'ln distribution of charge qall is superposed on this charge
distribution, the sphere becomes electrically neutral. Show that the
poten tial of the sphere is
v = qj47rEl.
Show that the charge density vanishes \vhen
( a2 a 4
rl = 3 y l(l2 - a 2 ) = l 1 - - - - -
31 2 91 2
..-).
The locus of these points on the sphere divides the sphere into posi-
tively and negatively charged parts.
2.12-7 Show that if a charge -q is placed on the neutral sphere
in the preceding problem, then the potential of the sphere and the
charge density will be
q (1 1)
V = - 411"E  - I '
qs = _ q(l - a) ( I + a + ).
47ra r1 al
Show that the charge on the surface of the sphere intercepted by
the cone fJ = fJo is
q(fJ o ) = _q [1 _ I - a cas 8 0 _ 1 2 - a 2 ].
2l 2l V a 2 + l2 - 2al cos 8 0
The charge on the hemisphere facing the point charge is
q(lI") = -q [1 - 21; :2 a 2 }
This equation may be used for studying the proximity effect. Thus if l
equals two diameters, the charge on the hemisphere facing the point
charge is - (18/33) q, which is about 4.5 percent in excess of one-half of
the total charge. If l = 3a, the charge on the hemispheres is distributed
in the ratio of 11 to 8. When l - a« a, only -O.35q(l - a)ja is on
the hemisphere looking away from the point charge.
2.12-8 Show that if the charge on the metal sphere equals that
of the poin t charge, the poten tial of the sphere is
V = -.f. ( + ).
47rE a l

Problems
357
2.12-9 Consider two metal spheres of radius a, with a charge q
on one and -q on the other. Let I be the distance between their
centers. The potentials of these spheres may be obtained by suc-
cessive approximations, based on the results of Problems 2.12-6 and
2.12-7. Thus starting with the assumption that in the first approxi-
mation the charges are distributed uniforn11y so that their fields out-
side the sphercs arc the same as the fields of equal charges located
at the centers of the spheres, one obtains the first image charges and
the compensating uniformly distributed charges. 1"he potential of
the positive sphere \vill then be
v =  ( - ).
47rE a l
Note that this agrees with the equation follo\ving equation (2.31)
except for the term of ordcr (a,jl) 4.
Applying the method of images for the second time, one obtains
v = 4:E [ - : - T ( - D]'
where 1 1 = l - (a 2 jl). Hence
q [1 1 a 3 ]
V = 41rE  - I - [2(12 - a 2 ) ·
rrhis agrees with the result in Section 2.6 as far as the term of order
(ajl)4.
rhc method of successive images can be applied indefinitely.
After the first term, the terms in the expression for V will be of the
.
same sIgn.
2.12-10 Show that if the spheres in the preceding problem are
equally charged, then
q [1 1 a 3 ]
1 1 =--+-- +....
47rE a l l2 (l2 - a 2 )
2.12-11 Sho\v that the next application of the method of images
will in troduce inside the brackets in the expression for V for two
equally charged spheres (Problem 2.12-10) the following term:
2a 6
l3(l2 - a 2 ) (l2 - 2a 2 )

358
l/ec/ronla1!.Jletic fir/ds
l"hc same term ,vith the negative sign ,viII appear in the potential
of t\VO equally and oppositively charged spheres.
2.15-1 Show that the resistance bet,vcen t,vo perfectly conducting
spheres of radius a imbedded in a conducting nlcdium is approxi-
mately
1 [1 1 a 3 J
R = 211"0" _ - l - l2(l2 - a 2 ) ,
where l is the distance bet,vecn the cen ters.
2.15-2 Show that the pO\\Tcr contributed to the conducting
medium by the generators bet,vecn conducting bodies Kl and K 2
(Section 2.15) and a point at infinity or ground is
P = rllli + 2rI2/1/2 + r22/i.
2.15-3 Sho\v that the currents in two resistive rods, connected to
a generator in parallel, arc so distributed that the dissipated po,ver is
minimum, and that a greater current is in the rod \vith the snlallcr
resistance.
2.15-4 Consider two perfectly conducting spheres of radius a,
imbedded in a conducting medium. Let their centers be on the z
axis at z = a + hand z = -a - h where Iz « a. lrom the results
of Problem 2.12-7 (as well as the principle established in Problcm
2.15-3) we conclude that current ,viII flo,v largely between the
hemispheres facing each other. One should be ablc, therefore, to
obtain the approximate conductance bct\veen the spheres by as-
suming that the lines of flow bct\veen the hemispheres are straight
lines parallel to the z axis. rrhe greatest deviation from straightness
will occur in the region where the current is small so that the resulting
effect on the conductance is small. Jascd on this approxinlation
show that
G = 11"0" [(a + 2/z) In a  2/z - (a + /z) J.
An estimate of the conductance bct\veen the remaining hemispheres is
G 1 = 1.57rua. rThis should be added to the preceding expression for a
bet ter a pproxima tion .
2.15-5 ]xplain \vhy the mutual resistance coefficient '12 of two
conducting bodies is approximately independent of their size and

Problems
359
shape as long as the mean distance l between them is large in com-
parison with their linear dimensions.
2.16-1 Show that the work done by the generators when placing
charges ql and q2 on the conducting bodies Kl and K 2 (Section 2.16) is
W = Pllqi + P12qlq2 + P22qt
2.16-2 Show that the work done by the generators when raising
the potentials of the conducting bodies Kl and K 2 from zero to VI
and V 2, respectively, is
IV = CllV1 + CIZV 1 V 2 + C22Vt
2.16-3 Show from the results of Section 2.2 that the total ca-
pacitance of a prolate spheroid whose focal distance is l, and whose
major and minor axes are, respectively, 2a and 2b, is
27rEl
C, =
coth (2a/l)
47rEl
In (2a + l) - In (2a - l)
27rEl
In (2a + ll2b)
Thus, for thin spheroids
27rEl
C t  In (l/b) + (bll)2'
Show from equation (2.12) that the approximate capacitance of a
thin cylinder of radius b and length l is
27rEl
C,  . .
In (lib) - 1 + In 2
Show that as l approaches zero, prolate spheroids approach a
sphere and
C t  47rEa,
where a is the radius of the sphere.
1vlore advanced methods arc needed to obtain the capacitance of
an oblate spheroid
27rEl
C, = ,
cot- 1 (1/2b)

360
Electromagnetic fields
where l is the focal distance and 2b is the minor axis. The oblate
spheroid of zero minor axis is a disk and the above formula gives
C t = 8Ea,
where a is the radius of the disk.
2.16-4 Suppose that the conducting bodies Kl and K 2 have equal
and opposite charges so that all tubes of displacement originating on
Kl end on K 2 . Consider two tubes of displacement whose capacitances
are C 1 and C 2 so that the charges intercepted by the tubes on Kl are
ql = C1V, q2 = C 2 V,
where V is the voltage from Kl to K 2 0 Show that the charge ql + q2
intercepted by both tubes is so distributed that the work done in
placing these charges on the bodies is minimum.
2.17-1 Show that the work needed to establish currents II and
1 2 in two perfectly conducting loops, Figure 2.30. is
V = Ll]li + L l2 1 1 1 2 + !L22I.
2.17-2 Using the results of Section 2.8 sho\v that the inductance
of a square loop is
J.ljaja [ 1
L=-
7r 0 0 V c 2 + ( - z) 2
where c is the radius of the wire and a is the side of the square.
Integrate and show that
2p.a [ 1 + Vl + (c/a)2 
L = - In (a/c) + In - 1 + v2
7r 1+V1
1 ]
d dz
v a 2 + a - Z)2  l
- Vl + (cla)2 + (cia)]
or approximately
L = (2p.a/7r) [In (a/c) + In 2 ("1 - 1) - 2 + v2J.
Using the result of Section 2.9, sho\v that
2p. jaja Z
L = - V---'- " dp dz.
7r c c P p- + z-
If the terms of order cia are neglected, \ve shall obtain the same
result as above.

Problems
361.
2.17-3 Consider a tube of magnetic flux <I> between two equi-
potential end surfaces. If U is the mmf between these ends, the ratio
R = U 14> is called the reluctance of the tube. In the analysis of
magnetic circuits it plays the same role the resistance does in the
analysis of direct current electric circuits.
Sho\v that the reluctance of a toroidal solenoid of length land
radius a is R = II J.L7ra 2 .
Assume that the solenoid is divided into two sections, one of
length II and the other of length l2 = 1 - ll. Assume that the fIrst
section is flllcd with substance of permeability JJ.l and the remaining
section with substance of permeability JJ.2. Let I be the total circu-
lating current. Obtain the magnetic flux cI>, the magnetomotive force
U 1 across the flrst section, and the magnctomotive force U 2 across
the second.
A llswer:
<1>=
7ra 2 J.1.1JJ. 2 1
,
1J. 2 l1 + JJ. l l2
J.l. 2 l 1 1
U 1 = ,
JJ. 2 l1 + J.L 1 12
U 2 = I - U 1 .
2.17-4 Consider a solenoid of radius a, coaxial with the z axis,
extending from z = 0 to z = l » a. In its interior the magnetic flux is
substantially parallel to the z axis and distributed uniformly except
near one end where the flux leaks out into the outer space and near
the other where the flux re-enters the solenoid. The magnetic intensity
on the axis of the solenoid was calculated in Problem 1.22-2 and may
be taken as indicative of the flux passing through the corresponding
cross section of the solenoid. Thus only about one-half of the total
flux cI>o actually comes from one end turn and en ters through the
other. Show that the flux <t>(z) in the solenoid at distance z from the
lo\ver end varies approximately as follows:
cI> (0) = O.SCPo,
<I> ( 3 a ) = 0 .97 CPo,
1> (a) = O.85c1>o,
cI> (4a) = 0.98ScI>o.
cI>(2a) = O.95cI>o
Sho\v that thc magnetomotivc force betwccn z = N a and z = l - iVa
when lVa « l is
u  e(l - 2Na),
\vherc C is the circulating curren t per unit length of the solenoid.
In view of the foregoing results we may subdivide the solenoid into an
inner" air-corc," extending from z = 4a to z = l - 4a, and two
"poles" of the solenoid at the ends. Show that the reluctance of the

362
Electrornaglle/ic fields
core is (l - 8a) /7rJJ.oa 2 and that the cxternal reluctance between the
poles is 8/7rJJoa.
If the solenoid has a magnetizable core of pcrmeability Il » JJo,
extending the en tire length of the solenoid, the flux density will be
substantially uniform all the way to the ends z = 0 and z = t. In this
case the external reluctance between these ends (constituting two
faces of two circular disks) may be obtained by analogy with the
capacitance of a circular disk (Problem 2.16-3). Show that this
rcluctance equals approximately Ij2J.Loa, assuming that l » a. Show
tha t a more accura tc reluctance is (1 /2JJoa) - ( 1/ 7rJJol). Show that
the flux through the solenoid is
7rJJa 2 C
1 + (7rJJa/2p,ol)
This means that the magnetic intensity inside the core is
4>=
II, = C[l + (7rJJa/2/Jol) J-1
as contrasted with the value C in abscnce of the corc.
2.18-1 Consider a direct current generator imbedded in a dis-
sipativc medium of conductivity ao and insulated from the medium
except at its terminals. Suppose that these terminals are highly
conducting spheres of radius b. Assume that the distance s betwecn
the centers of the terminals is equal to or larger than two diameters.
Two infinitely long parallel rods of radius a < b and conductivity
u » ao arc connected to the terminals. "fhese rods arc not insulated
from the medium. The distance between their axes is also s. The
problem is to find the curren t in the rods.
Since a is much larger than 0"0 the current entering one rod slowly
leaks out essentially at right angles to the rod, enters the other rod,
and returns to the generator. Assuming that I (z) is the current in
onc rod at distance z from the terminal and that V (z) is the trans-
verse voltage from this rod to the other rod show that
 V = -2RaeI z,
I = -G bh V z,
whcre
Rae = 1/7raa 2 ,
G ah = 7ruo/In (s/a) ,
so that as z approaches zero
dV
- = -2R I
dz Be ,
dI
- -GahV.
dz

Problems
363
Examine the solution for infinitely long rods when a - 10 7 mho
per meter (iron) and Uo = 0.01 mho per meter (soil). Assume that
a = 1 em, b = 2 em, S = 8 cm.
Show that if V o is the voltage between the terminals, the current
entering one rod is (G sh /2Ree)IV o . Compare this current with direct
leakage curren t between the terminals.
2.18-2 In the preceding problem it was assumed that the shunt
current emerging from a rod (or converging to it) is essentially per-
pendicular to the rod. This is true to the extent that the ratio, Ez/ E p ,
of the longitudinal to the radial component of electric intensity at
the surface of the rod is small in comparison with unity. Using the
solution obtained on this assumption show that
E,/ E p = [2 (uo/ a) In (s/a) J1/ 2 .
2.18-3 Consider two solid conducting cones, whose boundaries
are () = iJ and () = 7r - , with their apexes at the origin. Let their
conductivity be a. Assume that the conductivity of the medium
between the cones is ao « u. Show that the approximate differential
equations for the current J (r) in the upper cone and the transverse
voltage V (r) along the meridians from the upper cone to the lower are
dV
- = -2R J
dr Be ,
dJ
dr
-GehV,
where
Rae = 1/ arlr 2 ,
G eh =
In cot (/2)
7r U o
and
o = 27r ( 1 - cos t1)
is the solid angle of the upper cone.
Show that
d 2 J
r 2 - = kJ
dr 2 '
k =
27r a o
un In cot (il/2)
Show that the solution is
J = Ar n1 + Brn2,
nl.2 = ! ::i: v i + k.
For infinitely long cones J must vanish at infinity and therefore A
must equal zero. If the medium beyond r = l is nonconducting, the
ratio AIB will be obtained from the condition I(l) = o.

364
Elec/rornaglle/ic fields
2.18-4 A rod of radius a and permeability J.I. is bent into a long
rectangular loop ABC!) (see };igurc 2.18-4) of mean dimensions
sand l » s. A solenoid is ,vound round one short side of the loop
D C
... .
...
I
I
 <1>
 .
A B
FIGURE 2.18-4
AD. Assuming the direction of current as shown, the magnetic flux in
the solenoid will be from D to A. If the permeability J.l.o of the sur-
rounding medium is much smaller than J.I., this flux will follow the
highly permeable loop but gradually will leak out from the lower
rod, substantially at right angles to it, and enter the upper rod to
return to the solenoid. Sho\\r that if cI> (z) is the magnetic flux in the
lower rod at distance z from the solenoid and U (z) is the magneto-
motive force from the lower rod to the upper, then
dU
- = -2R se el>,
dz
del> U
dz Rljh
where Rse = 1/ p:lra 2 is the reluctance per unit length of the rod and
In (s/a)
RSh = ,
7r Jl.o
is the reluctance per unit length bet\veen the lo\vcr and the upper
sections of the loop.
Show that in general
eI> (z) = Ae- r , + Be r " r = V 2Rtie/ Rah,
U(z) = KAe- rz - KBe rz , K = V 2RscRsh.
Thus
 2 1lo
fa = Illn (sja)'
K = -.!..  2 In (sja) .
7ra J.l.J.l.o

Problems
365
Show that
K - Rt
B = ,,4e- 2fl
K+ R/
where Rt = s/7rJJ.a 2 is the rclcutancc of BC.
Sho\v that if I is the current in the winding and N is the number
of turns, then
1(0) = lVI - R.eI>(O),
,,,here R. = s/7rjJ.a 2 is the reluctance of AD. Hence, show that
IVI K - R.
.11 = + B.
K + R IJ K + R.
2.18-5 If the winding is extended to AB and DC and if n is the
number of turns per unit length, show that the flux transmission
equations become
dU
del>
u
dz
-2Rso4? + 2nl,
dz
RSh
Solve these equations and sho\v that if the winding is over the
entire loop (n turns per unit length over AD and Be as well), then
4> (z) = nI/ Rse,
U(z) = O.
3.4-1 Consider a circularly symmetric perfectly conducting
cavity \vhose cross section and dimensions are sho\vn in Figure 3.4-1.
Obtain the 1irst approximation to the natural frequency of oscilla-
tions.
A nS7.ocr:
w = (L t C t )-1/2
\vhere
JJ. (It - s) b JJ.S b
Lt = .... In - + In -,
27r C 27r a
E7ra 2
C t -
S
3.8-1 Consider t\VO perfectly conducting spheres, of radii a and
b > a, concentric \vith the origin. Assunle that a/b is not too small.
Suppose that segments of these spheres inside the cone (J = {} are
removed and a voltage is impressed uniformly round the periphery

366
Electromagnetic fields
- ...
... 
A
.... -- 2c
- -
h
,  
s
 2a
 ....
- -
2b
FIGURE 3.4-1
between the edges of the spheres. Show that the low-frequency
equivalen t circuit for this structure is a capacitance
27rEab (1 + cos {})
C e =
b - a
in series with an inductance
2p.(b - a) (In csc t? - ! cos 2 {})
Lt = .
1r (1 + cos t?) 2
As {} approaches 0, Lt approaches
p.(b - a) (In 2 _ ).
27r t? 2
3.9-1 Consider a circular parallel plate capacitor as shown in
Figure 3.5 (Section 3.8) and assume that the plates are connected
with a coaxial conducting rod of radius c. Show that the equivalent
circuit for this structure is an inductance Lt = (p./21r) In (a/c)
shunted by a capacitance
21rEP
C t = ,
h[ln (a/c) J2
where
P = i (a 2 - c 2 ) - C2 In (a/c) - C2 [In (a/c) J2.

Problems
367
3.9-2 Consider the structure in Problem 3.8-1 and assume that
the spheres arc shorted \vith a section of a perfectly conducting cone
(J = 7r - 1/1. Sho\v that the low-frequency equivalent circuit is an
inductance
Jl. (b - a) co t (l/I /2 )
Lt = In
27r tan (/2)
is parallel \vith a capacitance
27rEabP
C t =
(b - a) [In cot (1/1/2) + In cot (/2) J2'
P = r'- [In cot (1/1/2) + In cot ((}/2) J2 sin 2 () d(}.
"
To evaluate the integral, substitute cot (0/2) =It. The integration is
laborious.
3.12-1 Consider a perfectly conducting cylindrical cavity of
radius a and height h. This is a special case of the cavity in }:tigure
3.2 (a) (Section 3.4) in which s = It and b = 2a. If we use equation
(3.26), the natural frequency is given by way'"J;; = 3.4; the exact
natural frequency is given by wa = 2.40 .... Equation (3.26)
becomes more accurate as s becomes smaller. When s is smaller one
region of the cavity is occupied primarily by an electric field and the
other primarily by a magnetic field; this was the assumption in the
derivation of equation (3.26). This equation is even more accurate
for the cavity described in Problem 3.4-1.
Obtain an improved approximation for the cylindrical cavity by
using the low-frequency cquivalen t circuits for the cen tral portion
of the cavity, given by a = O.Sb, and for the remaining" slotted
toroid."
Answer:
2.40 · · · ) .
wby'"J;; = 2.36 (as compared with the exact value
4.2-1 Consider two parallel perfectly conducting planes, z = 0
and z = h, with circular holes extending from p = 0 to p = po. Assume
that a voltage V (po) is applied uniformly bet\vccn the edges of the
planes and that the resulting radial current in the plane z = 0 is
I (po). Obtain the integral equations for the voltage and current in

368
llectronlaglletic fields
this disk transmission line at distances P > Po. Compare these equa-
tions with equations (4.11) and dra\v conclusions about Land C:.
4nsu'er:
v (p) = V (Po) - (jw/J.Jz/2-rr) fP p-II (p) dp.
Po
1 (p) = 1 (po) - (2?rjWE/Iz) fP p V (p) dp.
Po
4.2-2 Consider two perfectly conducting spheres, concentric \vith
the origin of a spherical coordinate system. Let a be the radius of the
inner sphere and b the radius of the outer sphere. Assume that
b - a « (b + a) /2. Suppose that there are circular holes in these
spheres, concentric with the z axis, and that the edges of the sphere
are given by 8 = 8 0 . Let a voltage V (80) be applied uniformly between
the edges and assume that the resulting current along the meridians
on the inner sphere is 1(0 0 ). Show that the integral equations for
V (8) and 1(8), \vhen 8 > 80, are approximately
8
V(O) = V(Oo) - jw f L 1 (0)/(0) dO,
8 0
8
1(0) = 1(0 0 ) - jw f C 1 (0)V(0) dO,
8 0
where the inductance LI (e) and capacitance C I (e) per radian are
LI(e) = p.(b - a)/211" sin 0,
C I (8) = (27rEab sin e) (b - a).
Show that if s is the distance from the z axis along a meridian of
radius c = y{ijj [which is approximately equal to (a + b)/2], then
the transmission equations become
V(s) = V(so) - jw t L(s)/(s) ds,
'0
I(s) = I(so) - jw t C(s) V(s) ds,
"0
where the inductance L(s) and capacitance C(s) per unit length
along the mean meridian are
L ( s ) = jJ. (b - a) /2 'lrC si n (s / c) ,
C(s) = [27rE sin (slc)]/ (b - a).
What happens \vhen sic « I?

Problems
369
4.23 Consider a perfectly conducting cylinder of radius b and a
closely \vound coaxial solenoid of mean radius a < b. Assume that
the \vire with which the solenoid is wound is very thin and perfectly
conducting. Let c be the radius of the wire and l the thickness of the
insulation. Find an approximate value of Ez(a, z) in equation (4.5).
What is its effect on the parameters in equation (4.12)?
Answer:
Ez(a, z) = jwLll (z), where Ll = J,L7ra 2 /4(c + t)2,
4.2-4 Consider two parallel solenoids whose dimensions are the
same as those in the preceding problem. Assume that the interaxial
distance s is larger than 4a. Show that approximate equations for the
transverse voltage between the solenoids and the current I (z) are
equations (4.11) in which
y = }W7rE
In (s/a)'
where Ll is given in the answer to the preceding problem and L -
(}.L/7r) In (s/a).
Z =jw(2Ll + L),
4.2-5 Suppose that the outer cylinder of a coaxial pair (Figure
4.1, Section 4.1) is perfectly conducting and that the inner cylinder
is made of thin wafers which are alternately perfectly conducting
and dielectric. Let h be the thickness of a conducting wafer and s the
thickness of a dielectric wafer. Assume that the dielectric constan t
of the wafer is El and that of the medium between the cylinders is E.
Show that approximate equations for the transverse voltage and
longitudinal curren t are still cqua tions (4.11) \vi th
1
Z = jwL + . ,
}wC sc
Y = jwCs h ,
\vhere
J.L b
L = -In-
,
27r a
27rE
G' -
sh - In (bla) ,
El7ra2(S + 11)
Cae = ;
s
more accurately
J.L b J.LS
L = -In - +
27r a 87r (s + Iz)
4.4-1 Consider two perfectly conducting circular disks, coaxial
with the z axis, in the planes z = 0 and z  h. Assume that a voltage

370
Elcctr0l11l1R11C/ic fields
is applied uniformly bet\vcen the edges of these disks. Show that in
this case the ans\vcr to Problem 4.2-1 becomes
V(p) = V(O) - (jwp.hj27r) jP p-I/(p) dp,
o
I(p) = - (27rjwEjh) jP pV(p) dp.
o
Apply the step-by-step method for calculating yep) and T(p)
and sho\v that
V( ) = V(O) [1 _ ({3 /2)2 + (/3pj2)4 _ ({3p/2)6 + ...J
p , P (1.2)2 (3!)2 '
I( ) = - V(O) ( 'WE7r 2/11) [1 _ ({3pj2)2 + ({3pj2) 4
p J p 2 (1.2) (2.3)
(p/2)6 J
- (1.2.3) (2.3.4) + ... l
where (3 = w.
4.4-2 Suppose that along the axis of the structure in the preceding
problem there is a perfectly conducting \vire of radius a, connecting
the plates. Express V (p) and I (p) in terms of the current 10 in the
¥nre.
Answer:
V(p) = - (jwp.hj27r) fP p-1J(p) dp,
a
I(p) = 10 - (27rjwE/h) fP pV(p) dp.
a
4.6-1 Sho\v that the differential equations for the disk trans-
mission line are
dV
--
dp
dI
--
dp
- (jwJJ./t/21rp) I
- (27rjWEpjlz) V.

Problems
371
Show that V and I satisfy the following second-order differential
equations
d 2 V dV
p - + + 2p V = 0,
dp2 dp
{32 = W 2 JJ.E,
d 2 / dl
p - - - + 2pI = o.
dp2 dp
The first equation is Bessel's equation of order zero; the second may
be reduced to Bessel's equation of order one if we choose a new
dependen t variable u = lip.
4.6-2 Show that propagation of voltage along the meridians in
Problem 4.2-2 is described by
d 2 V dV
- + c- 1 cot (sic) - + 2V = 0, {32 = W 2 IJ.E.
ds 2 ds
If this equation is expressed in terms of the polar angle 8, it becomes
Legendre's equation
d 2 V dV
sin (J - + cos (J - + {32c 2 sin oV = O.
d8 2 dO
Usually, the constant (32c 2 is written as n,(n + 1), where n is called
the order of Legendre's equation.
4.6-3 Suppose that a distributed voltage, Ei(Z) per unit length,
is impressed on the conductor carrying curren t / (z). Show that the
transmission line equations (4.22) become
dV M
- - -Z/ + Ei(Z),
dz dz
Similarly, if Ci(Z) is the current per unit length, forced against
the transverse voltage V (z) from one conductor to the other, then
dV dI
- = -ZI, = - YV + Ci(Z).
dz dz
- -YV.
4.6-4 Derive from equations (4.22) the following relation
V(O)I*(O) - V(l)I*(l) = { [Z/(z)/*(z) + Y*V(z) V*(z) ] dz,
o
where the asterisks denote conjugate complex numbers.

372
Elec/ronUl1!.1 1ct ;eft clds
Sho\v that if the line is either open or shorted at z = /, then
Zinlinnn = YtnV in vtn = { [ZI/* + y*VV*] dz,
o
\\here Zin, fin, and V in arc. respectively. the input impedance, the
input current and the input voltage.
4.7-1 Consider a transmission line extending from z = 0 to z l.
Let it be terminated into its characteristic impedance at both cnds.
Lct V o be the voltage impressed in series with the line at z =  so
that there \vill be a discontinuity in the transverse voltagc. V ( + 0) -
Tl( - 0) = 110. Sho,v that at othcr points
I (z) - ( V o /2K) e- i(j(z-) , Z > ,
- (V o /2K) e-jIlCE-z), z < ,
V(z) - 1 V e- itHz-) Z > ,
20,
- - 1 Voe- iPCE- z) Z < .
2 ,
4.7-2 Suppose that in the preceding problem 10 is the shunt
current forced from one conductor into the other. Show that
fez) - ] I e- ifJ( z-E) Z > ,
20,
- -lfoe-j{Jc-z) z < ,
2 ,
V(z) - lKloc-j{J(z-E> Z > t,
2 ,
- ,1 K f oe- it3(E-z) z < .
1 ,
4.9-1 Obtain the propagation constant and the characteristic
impedance of the coaxial structure described in Problcm 4.2-5.
Discuss their behavior as the frequency increases from zero to infinity.
Discuss the conditions for the existence of waves on the one hand,
and for the concentration of energy in the ncighborhood (small or
relatively large) of the source on the other hand.
4.12-1 Suppose that in Problem 4.7-1 the line is shorted at z = 0
and tcrminated into its characteristic impedance at z = t. Assume
proper forms for the current and voltage distributions on either side

ProblenlS
373
of the source, determine the arbitrary constants from the given
conditions, and sho,v that
1 (z) - (Vol K) e-jJ3 cos f3z, z < ,
- ( Vol K) cos f3 e- j{Jz, z > ,
V(z) - -jV o e-jfJ sin f3z, z < ,
- V o cos 13 e- iPz , z > .
Alternatively, derive these exprcssions from those in Problem
4.7-1 by noting that the wave traveling in the negative z direction
is totally reflected from the shorted end.
4.13-1 Show that the input impedance of a nondissipative line
may be expressed as
. Zt C05 {31 + jK sin (3l
Z = K. .
I K cos f31 + jZt sin (3l
\Vhat are the input impedances when l = X/4 and l = X/2? Show
that if Zt is a pure resistance and t = X/8, then the magnitude of the
i input impedance is K. Show also that as f3l varies, the locus of Zi
in the complex plane is a circlc of radius . I Rt - K2R"t 1 I centered
at  (lt + K2R"t 1 ) on the real axis. Even if Zt is complex, the
locus is still a circle; but the proof is less simple.
4.16-1 Consider two perfcctly cond ucting coaxial cylinders of
radii a and b > a. Let the medium between them be nondissipative
from the generator at z = 0 to z = I and highly dissipative from
z = I to z = l + h. Show that if It is infinite. the field in the dissipative
n1edium decays at the ra te a n epcrs per meter, \vherc a = V7rf/J.u ,
and that 1 1 (1)11(1) = Y jW/J./u(1/21f') In (bla).
Sho,v that for a finite It,
If(l)/I(l) = y jW/J./u(1/21f') In (b/a) coth Iz y jw/J.u.
4.16-2 Consider three thin parallel wires of radius u. l\vo of
these wires are shorted at frequent intervals and serve as a return
path for the current in the third wire. Let s be the interaxial distance
from the latter to the wire nearest to it and It the interaxial distance

374 Electroma gn etic fields
D C
: -41( :
h 12
A  B
11
S
..
..
I = II + 1 2
FIGURE 4.16-2
to the remaining \vire as shown in the figure. Apply the Faraday-
l\Jaxwell law to the rectangle ABeD and sho\v that
[ jwp. Iz (s + Iz)] [ jwlJ. IzS]
R + - In /2 = R + - in II
2 as 2 (s + h)a
where R = 1/O'a2. Thus at very low frequencies the resistances of
the wires control the distribution of current and /2 = 11. At high
frequencies the inductive effects con trol the distribution and
1 2 In (ilia) - In (s + his)
- -
II In (Izja) + In (s + his)
These equations express the proxi1nity effect. The skin effect is evi-
dently related to it.
4.17-1 The disk transmission line, considered in Problems
4.2-1 and 4.6-1 is nonuniform. Apply the transformation introduced
in Section 4.17 and sho\v that if
f} = (jp,
V()
I()
K() = TJ(3IzI,
- (TJ{3IlI)IV(iJ),
- ({)It/(3/z)tl(iJ) ,
then
dV
d{}
dl
-- -jV - (1/2iJ)1.
dtJ
- -jf + (1/2iJ) V,

Problems
375
Hence, show that when {3p » 0.5, that is, when piX » 1/41r,
approximate expressions for the voltage and current in the disk
transmission line are
I(p) - (pl71h)i[.l1e-ip + BeipJ,
V (p) - (71Iz/ p)[Ae-i,9p - BejpJ.
For {3p « 0.5, an approximate solution may best be obtained in
the form of power series in (3p by using the step-by-step method.
The intermediate section, on both sides of I1p = 0.5, can be similarly
treated.
Of course, in the present case, a complete solution may be expressed
in terms of Bessel functions whose properties are now well known
and tables are available. Here, this case has been used for two reasons:
(1) to illustrate the phase integral method of Section 4.17, and
(2) to show by a concrete case that when the parameters of a
transmission line vary slowly, the line acts as an impedance
transformer.
4.17-2 Show that when the inductance and capacitance per
unit length vary in the same manner, that is, when L = LoF(z)
and C = CoF(z), then equation (4.56) can be solved exactly and
that the solution is
I (z) == Ae- j " + Be i ",
v (z) = K (Ae- i " - Be i "),
\vhere
K = V Lo/C o and t'J = w V LoCo l' F(z) dz.
o
4.17-3 Find an approximate solution for wave propagation along
meridians (see Problem 4.2-2), \vhen 8 is neither small nor near 7r.
\Vhat is the criterion of "smallness" of 8 and of its" nearness" to 7r?
Answer:
1(0) = /27r(ab)t sin 0 (Ae-ik8 + Be ik8 ) ,
'i 1](b - a)
V(O) = / 7J(b - a) (Ae-ikB - Be ik8 ) I
'i 21r (ab)4 sin (J
where k = wv;;oJ;. For these approximations to hold we should
have kO» 0.5 and k(7r - 8) »0.5.

376
llectronlaglletic fields
4.17--4 Consider a section of length l of a nonuniform trans-
mission line and assume that it is shorted at both ends. Show that
if K' /2K « 1, then the natural frequencies are given by
W n { v'L(z) C(z) dz = n?r,
o
11, = 1, 2, ....
4.17-5 In Section 4.17 ,ve obtained an approximate solution of
nonuniform transmission line equations \vhen K' « 2K. 1\n improved
approximation may be obtained from the results of problems 4.7-1
and 4.7-2 as follo\vs: Start with the first approximation
1 = Ae- i "
,
v = Ae-j{J
for the progressive wave propagating in the direction of increas-
ing fl. 1'he last terms in equations (4.61) may be interpreted as
the impressed series voltage - (K' /2K) Ii per unit length and shunt
current (K'/2K)I, also per unit length. Thus at a typical point
fl = cp we may assume an impressed infinitesimal series voltage
- (K'/2K)Ae- ifIJ dcp and an impressed shunt current (K'/2K) Ae- ifIJ .
The response to these may be obtained from the results of Problems
4.7-1 and 4.7-2. Then we integrate over the given length of the line
from cp = 0 to cp = 0, for example. Following this procedure show
that the improved approximations are
e
let'}) = Ae- j " - Ae j " 1 (K'j2K)e- 2j v>dcp,
"
e
Pet'}) = Ae- j " + Aei" 1 (K'j2K)c 2j v> dcp.
"
The second terms represent reflections which are inevitable ,vhen K
is not a constan t.
4.17-6
Prove that
K ' (fJ) K' (z)
-
2K(tJ) 2wL(z)'
\vhcre K(z) is, of course, K[t1(z)J. "fhis shows that K'(t1)/2K(fl)
decreases \vith increasing frequency.
4.18-1 l\S w approaches zero, the ratio K'(fJ)/2K({}) increases
and the approximations involved in equations (4.62 become invalid.

Problems
377
Sho\v that in this case approximate values of the image parameters
are (for a nondissipative line)
Kl = K 2 = vi Lt/C t ,
tan e  8 = w v LtCt,
where
Lt = II L(z) dz,
o
C t = t C(z) dz.
o
4.18-2 Sho\v that improved approximations for the image
parameters (at" lo\v frequencies") may be obtained from
1 ( l [' ]2
Yl,sh = :-- + jwLt 2 in C(.z) f L(z) dz dz,
JwLt 0 z
1 I l [l ]2
ZI,oP = -;-- + jWC t 2 L(z) f C(Z) dz dz,
JwC t 0 z
Y2,8h = - + jwLt211 C(Z) [I' L(z) dZ]2dZ,
JwLt 0 0
Z2.oP =  + jWC t 2 II L(z) [I' C(Z) dZ]2 dz.
JwC t 0 0
5.1-1 In Section 2.10 expressions were derived for the magnctic
intensity of an infinitesimal circulating current. In Section 2.11 the
expression for the clectric in tcnsi ty \vas obtained on the assunlption
that the circulating current varies slo\vly. Show that if this current
varies rapidly, then the field componen ts must satisfy the following
differen tial equations:
a
(sin OEtp) = -jwjJ.r sin OII r
ao
a
- (r E ) = J.wjJ.rII 8 ,
ar f(J
a aR r
- (rIle) = J.wErE +-.
ar tp ao
5.1-2 In Section 1.16 it 'vas found that for a static intinitesimal
electric dipole, along the z axis, a t the origin, the field components
Er and Ee arc proportional to cos () and sin O. respectively. In Section
1.23 it \vas further found that if the charge is flowing slo\vly bet\veen
the ends of such a dipole, there is ainagnetic field of intensity If f(J
\vhich is proportional to sin 8. 'The results so obtained should be

378
Eleclronlagl1elic fields
approxin1ate solutions of equations (5.1), (5.2), and (5.3) when w
is "sufficicn tly small." \\7 e are now in a position to look for exact
field intensities at all frequencies. Show that if
Be (r, 0) = Ee ( r) si n 8,
II f{) (r, 8) = II tp (r) si nO,
Er(r, 0) = Er(r) cos 0,
then
.!!: (r E,) = - (jw/J. + ) (r11 tp) ,
dr }WEr-
(1)
d _ _
- (rIItp) - -jwE(rE e ).
dr
Comparing \vith the transmission line equations (4.22) \ve find that
wi th the dipole (curren t elemen t), at the origin, space acts as a
radial transmission line \\'ith distributed parameters for r Ee and
rII rp:
Lf!1e = p"
Cab = E,
C -] .cr 2
8C - 2 ·
The radial current in this line is entirely a displacement current.,
Show that if I (r) is the total radial displacement current in the
northern hemisphere, then
I (r) = 27rrll rp.
If V (r) is the voltage along the merirlians from the northern axis
to the sou thern, then
VCr) = 2r£8.
'fhe transmission equations \\'ill then become
dV
_ _( jw ll + . 2 . ) I,
dr 7r J WE1rr-
(2)
dI
dr
-jwerrV
so that in this case
Lse = Jl/'Tr,
C ah = 7rE,
Cee = 7rEr2.
Note that if the radial displacement currents were distributed
uniformly in each hemisphere, the series capacitance per unit length,

Problems
379
in each hemisphere, would be E27rr 2 [see equation (2.33) ] and the
capacitance of both hemispheres in series \\"ould be 'TrEr 2 . 'The factor!
is the result of the cosine distribution of the radial current density.
Also, for a uniform current distribution, the series inductance of
each hemisphere \voulcl be p./87r [see equation (3.36) ] and that of
both hemispheres in series JJ./ 411'". l'he factor 4 is also the effect of
nonuniform current distribution.
Of course, \ve would not have made such assumptions unless we
were completely ignorant of the fact that in the static case Dr varies
as cos fJ. Even if we were ignorant of this fact, we would have thought
of better approximations. But once we know that Dr is proportional
to cos 0, the exact values of transmission line parameters are obtained
immedia tely.
5.1-3 \Vhen (3r» 1.41 in equations (1) of the preceding problem,
we have differential equations with constant parameters and the
general solutions are
rif9 = Ae-i/Jr + Be iIJr , (3 = w,
rE8 = 17Ae-i/Jr - 1}Be j {3r, 17 = .
For outward bound waves, B is equal to zero. \Vc also know that as
(3r  0, rli tp approaches Il/411'"r. These facts should help in flnding the
exact solution of equations (1). 'fentatively assume that
_ II
r1I9 = - e- jfJr + Ae- jpr .
47rr
The reason for including the exponen tial factor in the first term is
this: If this term does not have the exponential factor \:vhilc the
second does, there can be no hope to satisfy equations (1). "Te need
only try the substitution. l"here is no risk in the above tentative
assumption since the next step is to substitute rI19 in
d 2 _ _ 2_
- (rII rp) = - (32(rII rp) + - (riI tp).
dr 2 r 2
This \vill enable us either to express A in terms of It or disprove the
conjecture. Find .11 and note the agrcen1cnt \vith equation (5.23).
5.3-1 j\ssume a perfectly conducting sphere of radius I, concentric
with the apexes of the cones in Figure 5.4 (Section 5.3). Show that

380
llec/ronlaf!.lIeti( fields
the current in the cones and the transverse voltage bet\veen them are
I (r ) = [ I (0) cos fJ (I - r) J/ cos ,8t,
1'(r) = [V(O) sin ,8(l - r) J/sin ,8t,
\vhere the ratio of the input voltage V (0) to the input current is
v (0) /1(0) = jK tan {3l,
\vhere K is given by equation (5.12).
\\!hat is the magnetic intensity between the cones?
5.3-2 Obtain the natural frequencies of the cones in the preceding
problem:
(a) \\then the apexes of the cones arc insulated and
(h) when they arc shorted.
A 11su'cr:
(a) W n = (2n + 1)7r/2l,
(b) W n n7r/I,
n = 0 1 2 ...
, " ,
11, = 0, 1, 2, ....
5.3-3 Consider t\VO perfectly conducting coaxial cones, 0 = 0 1
and (} - (J2 > o. l\ssume that a voltage is impressed between their
apexes and that consequently there is a current I (r) in the cone
8 = 0 1 at distance r from the source. Since there is no radial displace-
ment current, \VC have
27rr sin 81/ f() = I (r) .
Using this equation show that the inductance per unit length of the
cone IS
L = J.L In [tan (8 2 /2) cot (lh/2) J.
27r'
Similarly assuming a charge per unit length, q(r), on the cone 0 = 01
calculate !)(J. Ee. the voltage V (r) along the meridians and show that
the capacitance per unit length is
21rE
c-
- In [tan (0 2 /2) cot (8 1 /2) J'
5.8-1 Using the results in Section 2.10, the equations in Problem
5.1-1, and the analogy between the fields of a current clement and

ProblcnlS
381
an intinitesimal circulating current. show that the exact field of the
1a t ter. for all f rcqucn cirs is
13'2/ S ( 1 )
Ef(J = 7'J 1 +  e-)(jr sin 8,
47rr J{3r
{32[ S ( 1 1 )
118 = - 1 + - -_ c-j{Jr sin 8
4 . 'l ., ,
'Trr J fJr 1Jr-
j{3! S ( 1),
II, =  1 +:- e-){Jr cos 8,
27rr- J{3r
,rhere [ is the current and S is the area of the loop.
Show that the field of an infinitesimal solenoid of moment Vl,
that is. a 1nagnetic current elel11,ent, may be obtained from the above if
I S is replaced by Vl/jwJ..L. (Here V is the magnetic current in the
solenoid and can be measured by the voltage induced in a single
turn of the solenoid.) Show that the distant field in this case is
jweVI
lIe = e- iP , sin (),
47rr
Ef(J = -71 H e.
5.8-2 Show that equations (5.1). (5.2), and (5.3) are invariant
under the following transformation:
r-)o -r,
II f(J -)0 -II rp,
Er -)0 Er,
Ee  Ee.
U sc this fact to obtain the following expressions for the fIeld of a
converging spherical ,vave
A'71 ( 1 )
E = - 1 - - e i13r cos 0
r 2". ,
7rr- J{3r
jW}J.A ( 1 1),.
Ee = - 1 - - - e):8 r SIn 8
47rr j I3r {j2r 2 '
. 11 A 1
H rp =  (1 - :-) e iIJr sin 8.
47rr J{3r
5.8-3 Use the expressions for the diverging and converging
spherical waves to obtain the following expressions for standing

382
J/ec/rorl111i:lletic fields
wave:
(a)
7]./1 [COS t:3r ( 1 ) ]
JiB = - + 1 - - sin I3r sin 8,
"Ar i3r {j:!r:!
.11 (sin pr )
j . - cos pr sin 8,
'Ar pr
Iff/)
17A (sin pr )
lr = --:; . - cos pr eos 8,
7rr" 13'
and
(b)
TJB [( 1 ) sin /1']
E, = - 1 - - cos fjr - sin (),
A' {32r'! fjr
. B ( . cos 13r) .
II f/J = J - sIn {3r + sIn (),
"Ar {3r
1}B ( cos 13')
Er = f) sin {3r + . cos (J.
r" fjr
'fhc jicld given by (a) is finite at r = O. The field given by (b)
is singular at r = O. Show that the latter implies a current clement
of moment Il = 2jB at r = O.
5.8--4 Using an appropriate equation from the preceding problem,
obtain the equation for natural frequencies of oscillations of electric
charge in a cavity bounded by a perfectly conducting sphere of radius
a, centered at the origin, \vhcn the charge density on the sphere
varies as eos 8.
il11su'cr:
W n = knia,
where k n is the nth root of cot k = k - k- J .
5.8-5 Calculate the smallest root of the equation in the pre-
ceding problem.
./t lls'U'cr :
k 1 - 2.74371.
5.8-6 Obtain the natural frequency and the damping constant'
of external oscillations on a perfectly conducting sphere of radius a
when the charge distribution on it is proportional to CDS (J.
Answer:
w == v3/2ay'J;;,
 = 1/2a.

Problems
383
5.8-'7 Using the energy n1cthod and the low-frequency field in
the spherical cavity (that is. the static tield distribution for E and
the effect of its time derivative on 1/), sho\v that the lowest natural
frequency is given by
w = vf6/aVE  3.16/l£.
l"'his is 11.6 per cen t higher than the exact natural frequency obtained
in Problem S.R,,5. i\ better approximation \vould have been obtained
if the sphere \vcrc divided into two parts by a cylindrical surface
p a '2 and the lo\v-frequency equivalent circuits for each part
obtained hy the energymethocl.
5.8"8 Sho\\T that the n1ethod suggested in the preceding prohlem,
\vhen applied to external oscillations on a perfectly conducting
sphere yields w a = V2 = 1.414 ..., as compared \vith the exact
value wV Jl.E t1 0.866. · .. l'hc reason for a much poorer result in
this case is that a very substantial loss of po\vcr by radiation is
neglected \vhen \Vc apply equation (3.17) \vith lt = O.
5.8--9 Sho\v that the natural frequencies of gyrations of currents
circulating in a perfectly conducting sphere of radiu=; a arc given by
W1LV a = k u , tan k = k, if we assume that these gyrations have
heen excited from inside the sphere in such a \vay that the density of
circu la ting curren t is proportional to :;in (}.
Shc)\v that k 1 = 4.4934 · · · .
5.8'lO Circulating currents, proportional to sin (J. may be excited
OIl a ptrfectly conducting sphere by an appropriate outside source.
Show that they \vould be exponentially dan1pcd at the rate 1 aV E
nepcrs per second.
5.8-11 F'ill in the details in the solution of the following problem.
In Section 5.2 \Vc obtained the iicld of a scn1i-infinite progressive
curren t \va '.l' along the = axis and issuing f ron1 the origin. I"'rom
equations (5.9) \\"t' Jind that OIl the positive: axis, lie and flip arc
infinite and that lr is tinite (exccpt at the origin). 'rhis means that
as \'-'C approach the po:-;itivc :; axis. the ratio Er 18 approa.ches zero
and it \vould appear that \\'C have a limit tield of a progressive current
wave on a thin l)(,'rfectly conducting \virc as the radius of the \\Tirc
approaches zero.
Considcr no\\. a thin perfectly conducting cone (J = l/1 « 1 and a.
current wave on it. 110w can we obtain the field? On account of sym-
metry Er, E8, anu II.p must satisfy '1ax\vcll':; equations (5.1). (5.2),

384
Eleclron1aJ!.lletic fields
and (5.3). In addition, the radial electric intensity must vanish on
the surface of the cone,
Er(..p) = O.
'\Te would expect that if 8 is large in comparison \\'ith if; , Er \vill be
approximately the same as in the case if; = 0 [see equation (5.5) J.
This suggests that Er for the ne\v arrangement may be given by
, R(r)8(8)
Er=
. ,
)Wfr
\vhcre e (0) is almost unity for 0 » '" and reduces to zero for 8 = if;,
(1)
eel/;) = o.
(2)
Function R (r) \vould probably be almost sinusoidal [see equation
(5.8) J. Our problem now is to find out \vhcthcr the assumption (1)
is consistent \vith l\Jax\vell's equations and calculate R(r) and
e (0) if it is.
To obtain the answer to the question we substitute Er from equa-
tion (1) into equation (5.1), integrate from 8 = 7r to 0 = 0, note
that H cp (r, 7r) = 0, and find
R(r) j9
II cp = . sin oe (0) dO.
SIn 0 ...
We substitute this in equation (5.2) to obtain rI9. l"'hen we sub-
stitute E 9 , II f{)' and Er in equation (5.3), multiply the result by jWE sin 8,
differentiate \vith respect to 0, and rearrange the terms to fInd
[ d 2 (rR)] d ( de)
+ 132(rR) sin 0 e = -Rr 1 - sin 8 -- .
dr 2 dO dO
\V c observe that if \VC divide by Rr- 1 8 sin 0, the left side of the equa-
tion \vill be a function of r only and the right side a function of ()
only. l'he equality is possible only if each side reduces to a constant,
k let us say. 'rhus, we find
d ( de)
- sin f) - = - k sin (J e
dO dB '
(3)
d 2 (r I)
= - f32(r I) + (k/r2) (r R). (4)
dr 2
The consistency of assumption (1) \vith }\ilax\vcll's equations has
now been demonstrated and it remains to solve the preceding equa-
tions.

Problems
385
\Ve know that in the limiting case'" = 0, the constant k is zero.
The general solution of equation (3) is then
8(0) = M In tan (0/2) + N,
where At and N are constants of integration. For our physical ar-
rangement J.lf must equal zero since Er must be finite at 0 = ?T. Indeed
in equation (5.5), 8(0) = 1 [the constant 1.'7 may be absorbed in
R(r) J. When l/I docs not vanish but is small, k must be small. I-Ience
the right-hand side of equation (3) is small. If we replace e by unity,
on the right-hand side, the error will be the product of one small
quantity k and another small quantity, namely, the deviation of e
from unity. Thus we have an approximate equation
 (sin 8 de ) = -k sin O.
dB dB
Integrating from (} = 7r to (} = 8, etc., we find
e = 1 + 2k In sin ( (} 12) .
rrhe boundary condition (2) is satisfied if
k =  In csc (l/I /2) ,
tha t is, for 1/1 « 1
k =  In (211/1).
Function e is
In sin (0/2)
e=l- .
In sin (1/1/2)
\\' e observe that as ,1/1 approaches zero, 0 approaches unity.
\Vhen k = 0, function rR is given by equation (5.8). If k  0,
then equation (4) ShOVlS that the fractional deviation from equation
(5.8) depends on klfJ 2 r 2 . This quantity approaches zero as I3r in-
creases, and, of course, k itself is small. It is only \vhen (3r « k that
the last term in equation (4) becomes dominant and \ve have ap-
proximately
r 2 d 2 (rR)
= k (r R).
dr 2
The general solution is
r R = Pr n1 + Qrn2,
where
1Z1.2 =! ::l:: v 1 + k  1 + k, -k.

386
Elec/ronlal1e/ic fields
This solution may be compared \vith equation (5.8) if we expand
the exponen tial function in po\ver series in r.
5.10-1 Obtain the average po\ver contributed by the generator
to the field produced by a circular current loop of area nZOl1zent IS.
Answer:
p =  {34S2II*.
127r
5.10-2 Calculate the power crossing a sphere of large radius,
concentric with the current loop in the preceding problem. Compare
this power \\'ith the average power contributed to the field.
5.10-3 Sho\v by the method used in the preceding problem that
the po\ver radiated by a magnetic current clement of dipole 111011zcnt
Vi is
1
]:> = - ({3l)2VV*.
127r17
5.10-4 Consider two current elements of moment It, coaxial with
the z axis, one at the origin and the other separated by distance r.
Show that in addition to the self-radiated power, equal to 2P, \vhere
P is given by equation (5.27), there is a mutual radiated power
'1]l2 11* (Sin (3r )
P AI = - cos {3r .
27rr 2 {3r
rrhis formula can be derived most readily from equation (5.24).
It can also be obtained by integrating equation (5.29) over a sphere
of large radius (in fact, of any radius, except that the integration
will be more complicated). 1"'he mutual power may be either positive
or negative.
5.10-5 Show that if one current clement is at the origin and
the other at (r, 7r/2, 0), then the mutual power is
1]l2 I 1* ( . cos (3r sin f3 r )
Pm = SIn {3r + -
2Xr {3r {32r2
assuming that the elements are parallel.
5.10-6 Show that if the currents in the preceding problems are
in quadrature, then Pm = O.

Problems
387
5.10-7 A charged particle q is oscillating about the origin. along
the z axis, \vith an angular frequency w. Let a be the anlplitude of
oscillations. Show that the radiated po\ver equals 10w 4 a 2 q2/c2, \vhcre c
is the velocity of light.
5.12-1 Apply the approximate equation (5.32) to obtain the
inductance of a square loop. First, imagine that the generator is
inserted at some corner and obtain the total inductance
2p.a
Lt = - [In (a/c) - 1 + ,! In 2J
7r
2p.a
= - [In (a/c) - 0.65J,
7r
where c is the radius of the \vire and a is the side of the square.
Next imagine that the generator is inserted in the middle of a side
of the square and obtain
2p.a
Lt = - [In (a/c) - 0.5J.
7r
Of course, the difference between the constan ts in the brackets is due
to somewhat different methods of approximation. The exact value of
the constant is -0.78 ...."" (See l>roblem 2.17-2).
5.12-2 Obtain an approximate value of the average character-
istic impedance 9f two thin wires of radius a and length I. diverging
from a common apex and making an angle tJ.
Answer:
Kav = 1J [In (2l/a) - 1 + In sin (tl/2) J.
1('
5.12-3
5.3-3.
Derive equations 5.32 by the method suggested in Problem
5.14-1 Show that at large distances the magnetic intensity of
a full-wave antenna (l = A) is
j/c,c-i8 r [1 COS (7r cos 0) ]
II" =
27rr sin ()

388
Electronzagl1ctic fields
5.14-2 Show that the power radiated by a full-wave antenna in
free space is
1 211' 1 - cos t 1 4 11' 1 - cos t
P = 60ft dt - 1515 dt
o tot
= (60 Cin 21l" - 15 Cin 41l")/o/t.
6.1-1 Consider direct current in a homogeneous medium. Show
that the Cartesian components of electric intensity and current
density satisfy Laplace's equation. Show that
aE x aE"
ay = ax '
aE"
c1E z
,
ay
aE z aE x
c:: -
az
ax az
whether the medium is homogeneous or not. If the medium is homo-
geneous, similar cquations connect the components of current density.
6.1-2 Consider a thin conducting plate bounded by planes
z = 0, z = It, y = O. and y = b» h. Suppose that current I is injected
into the plate uniformly by an infinitely thin blade of vvidth s picrcing
the plate. Let the edges of the hlade pass through points (0, )'0 - !s)
and (0, Yo + s). Sho\v that if x > 0
I 00 2/ . 1Z,7r S n7r)'O 1'l7rY
lx = - + L - sIn - cos - cas - e-nrxlb,
2blz ,,-1 1l1l"lzs 2b b b
00 2f 1t1l" s 1t7r'VO 1t7rY
] II = L sin cos  sin e-nrzlb,
n-I u1l"hs 2b b b
and that
Jz( -x, y) - - J:&(x, y)
J II ( -x, y) - J II (x, y).
Obtain J z and ]11 when s = o.
6.1-3 Consider two parallel resistive \vires, shunted at cqual
distances by other resistive ,vires as sho\vn in 14igure 6.1-3. Let lse
be the resistance of each section of the long wires and Rsh the resist-
ance of each shunting \virc. rrherc are t\VO sin1plc types of current
distribution in such a \vire net\vork. If equal currents arc injected
into the terminals .11 and B. there \vill be no current in the shunting
wires and the curren t in all series sections \vill he the same. See part
(a) of the ligure. In this case there \vill be no voltage either between

Problems
389
B
 10
DDD [- =
! 10
2
A
(a)
B
I 0 -411
II[B@][-:=
10
A
(b)
FIGURE 6.1-3
A and B or across any shunting wire. Another possibility is shown in
part (b) of the figure \vhere curren t To en ters the terminal A of the
lower wire and leaves through the terminal B of the upper wire. Show
that in this case the following relation exists between the currents
in the successive meshes
In-l - 2pln + 1"+1 = 0,
p = 1 + (RfW/R sh ).
Show that it is possible for the ratio k = I n+l/ I" of the successive
mesh currents to be constant, and that
k L2 = P :i: V p2 - 1.
(1)
Show that the product of k 1 and k 2 is unity and that therefore one
value is smaller than unity and the other greater than unity. Sho\v
that the current in the nth mesh, given by
In = Ak + Bk = Ak + Bk 1 n ,
(2)
\vhere .11 and B are arbitrary constants, satisfies the foregoing relation
bet\,reen mesh currents. Explain \vhy A must equal zero, \vhcn the
net\vork is continued indefinitely. In this case show that if current
lin is injected only into the .r1-tcrminaL then the currents in the nth
sections of the upper and lower wires are, respectively,
!Iin(l + kIn) and !Iin(l - kIn).

390
Electronlaglletic fields
! 10 .
3 _ Flr=:9Fl ==
!/o UUU
;[0 UUU_=
....  .
(a)
a C  .c
10 · [B[B[B-
a ttB
  ---
.-  ....
(b)
/  [B -----
4/:0 5J :
(c)
DDD
lin DDD =: =
---
(d)
FIGURE 6.1-4

Problems
391
6.1-4 1"1hree long \vires shunted across at regular intervals are
sho\vn in J4'igure 6.1-4. If \ve assume that all series resistances are
equal and all shunt resistances are also equaL \ve anticipate from
symmetry considerations the follo\ving three regular types of current
distribution [see parts (a), (b), and (c) of the figure]: type (a) in
\vhich the input current 10 is equally distributed bct\vccn the \vires
and there are no shun t curren ts; an ti-symmetric type (b) in \vhich
the curren t J en ters the lo\ver \virc and leaves the upper; there is no
net current in the middle \vire; the symmetric type (c) in ,vhich the
current J enters the middle \vire and leaves the screen by \vay of the
other t\\'o. Any other current distribution is a linear con1bination of
these three types. Show, for example. that if the current [in enters
the Io\ver wire as sho\vn in part (d) of the figure, then the resulting
current distribution lnay be obtained by superposition of these three
types \vi th
10 = fin.
f a - II
o - '2 ip,
f 8 - II
o - - 3 in.
Sho\V that the current in the nth mesh for the anti-symmetric
distribution is given by equation (2 of the preceding problem in
\vhich k1. 2 are given by equation (1) with
P = 1 + (J8P/2R8h).
Sho\v that for the synlD1etric type
p = 1 + (3RS(./2Jsh).
In the case of four parallel \vires there are four types of current
distribution for each uf \vhich the curren t in the nth mesh in the
longitudinal direction is given by equation (2 of I>roblem 6.1-3.
Only tv,.o of these types are readily obtained from symmetry con-
siderations. 'rhe remaining t\VO can be determined by solving the
three mesh equations for the nth section of the screen.
6.1-5
Sho\v that the t\VO dimensional Laplace's equation
a 2 V a 2 1f
+ =0
ax? iJ y 2
may be transformed into
iJ2ll'
cJrar*
\,'here r is the complex variable
r = x + j)'
=0
,

392
Electro1 n agnetic fields
and r* is its conjugate
r* = x - jy.
Therefore the general solution of Laplace's equation is
1 1 = fer) + g(r*),
wheref and g are arbitrary functions. Of course, V can always be made
real by a proper choice of g in relation to f.
6.1-6 Show that the real and imaginary parts of an arbitrary
function of the complex variable, W(r) = Vex, y) + jU(x, y),
satisfy Laplace's equation. The function W is called the complex
potential.
6.1-7 Show that the real part of the complex logarithmic potential
W = - (qj27rE) In (r - ro)
is the true potential of a line charge of density q, parallel to the z axis
and passing through point (xo, yo, 0).
Since we are dealing here with essentially two-dimensional prob-
lems, we may concentrate our attention on the complex plane where
points are represented by a single complex number r.
6.1-8 Show that the complex potential of a line charge passing
through S = So in the presence of a conducting plane y = 0 is
W = - !L In r - r; .
27rE s - So
The point sri is the image of roo
Show that if the conducting plane is x = 0, then the image point
is - rri and
w = _ q In r - ro .
27rE r + rci
6.1-9 Suppose that there are two perfectly conducting planes,
x = 0 and y = 0, and that the line charge is in the first quadrant
(xo > 0, Yo > 0). Show that the complex potential is
W = _ q In (r - ro) (r + fo) _ _3- In r 2 - r: .
27rE (r - rci) (r + rri) 27rE!;2 - <r 0 ) 2

Problems
393
6.1-10 Suppose that the conducting half-planes are 'P - 0 and
f{J = 1r/3, so that they form a 60° wedge. Show that the complex
potential of a line charge, passing through point to in the interior of
the wedge (0 < cp < 7r /3), is
W = _ In sa - s .
21TE t 3 - (sri) 3
6.1-11 Show that if the \vedge is formed by half-planes cp = 0
and <p = 1r/n, where n is an integer, then the number of images is
2n - 1 and
q r n - r
W = --In .
211"f sn - (sri) ft
6.1-12 Show that the complex logarithmic potential in Problem
6.1-7 may be expressed as
OJ
W = (qI27rE) :E (11m) (s Iso) m - (qj27rE) In (- so),.
m-l
when I s I < I So Land
OJ
W = - (q/27rE) In r + (q/27rf) :E (11m) (ro/r)m,
m-l
when S I > so. Show that consequently the true potential may be
expressed as
CD
V = (qI27rE) :E (11m) (p/ Po) m cas m(cp - cpo)
(q/27rE) In Po,
m-l
p < Po,
<Xl
- - (qj27rE) In p + (q/21TE) :E (l/1n) (pol p) m cosm( cp - <po),
m-l
p > Po.
6.1-13 Show that the complex potential in I)roblem 6.1-11 may
be expanded in two power series, one converging \vhen p < Po and the
other when p > Po. From these series show that the true potential

394
Elec/rOl1ulgllelic fields
may be expressed as follows:
(X)
yep, qJ) = (ql1rE) L (ll11t) (p/po)mn sin (l1lllqJ) sin (l1znr.po)
m-I
(X)
- (q/1rE) L (1/111) (po/p)mn sin (nzllcp) sin (1nncpo)
m-I
depending on whether p < Po or P > Po. Since V (p, 0) = V (p, 1r /11,)
= 0, irrespective of the value of n, the complex potential in Problem
6.1-11 must hold for a \vedge of any angle. A half-plane is a
"\\l'edge" of angle 21r, for which n = 1/2.
6.1-14 Sho\v that in the vicinity of S = ro the complex potential
derived in Problem 6.1-11 is
q nr-lLlr
lV = --In
21rE r - (rri) n
and that consequently the capacitance per unit length of a thin wire
of radius a inside the wedge and parallel to its edge is
27rE
c=
In (2poa- 1 n- 1 sin 1tr.po)
6.1-15 Given a conducting cylinder, coaxial with the z axis, and a
uniform line charge parallel to it. Let a be the radius of the cylinder.
Assume that the line charge is passing through the point (l, 0, 0)
where 1 > a. Using the result of Problem 6.1-12, show that the
poten tial of the charge displaced on the cylinder is
(X)
Vr = - (q/27rE) L (1/11x) (a 2 /lp)m cos mcp,
m-I
p > a.
Show that the density of displaced is
ex>
q. = - (q/7rl) L (a/l)m-I cos "lr.p.
m-I
6.1-16 By referring to Problems 2.7-1 and 2.8-2 show that the
real part of
w = - (J.I./ /27r) In (f - ro)
is the vector potential of an infinitely long current filament passing
through point So and that the imaginary part is the scalar potential,

Problems
395
multiplied by Jl.. Note the resemblance of the former to the scalar
potential of a line charge and use this resemblance to show that the
real vector potential of a curren t filamen t in a wedge of angle 1/1
formed by perfectly conducting half-planes, is
Jl.I [ r" - r ]
A = -- re In
21r t" - (rri) n '
where 11, = 1r IV;.
Further, show that if the radius of a thin current filament is a,
then the inductance per unit length, L, is
L = (Jl.127r) In (2poa- 1 n- 1 sin nq;o).
Explain why in two dimensional situations the product of the
inductance and capacitance per unit length is independent of the
geometry and that
LC = Jl.E:
6.1-17 Consider two parallel conducting planes given byim(r) = 0
and im(r) = jlt. Show that the image of any point r in the plane
r = jll is r* + 2jh. Use this result to show that the complex
paten tial is
q Q) [(r - to) 2 + 4n 2 1z2J
W = --In :E
21rf n-o [<r - rri) 2 + 4n 2 1z2J
q sinh [7r(r - ro) f21t]
- --In .
21rE sinh [7r(r - rri)/2hJ
6.1-18 Let a third conducting plane, rc (r) = 0, be added to the
two planes in the Problem 6.1-17. Show that the complex pOLential
will be the sum of the complex paten tial given above and
q sinh [7r(r + rri)/2h]
-In .
27r£ sinh [11" (r + ro) 12h]
6.10-1
text.
Derive equations (6.102) and (6.104) as suggested in the
6.15-1 Consider a perfectly conducting rectangular waveguide as
described in Section 6.10. At point (xo, yo, 0) there is a current

396
Electro111U1:lletiC fields
clement of moment II. parallel to the z axis. Show that only T1\l
,"'aves are excited and that for the (tn,11,) mode
2Ilxn sin (m1rxo/a) sin (n7rYo/b)
E = e=frmnz
Z ,mn K.. 2 2 b '
mnW f a
where the upper sign corresponds to z > O. Show that the power
radiated into each mode is
I I*l2xn sin 2 (m1rxo/a) sin 2 (n7rYo/b)
Pm = .
KmnW2f2ab
7.1-1 Consider a transmiSSIon line whose characteristic im-
pedance is KI' Suppose that a section of it from z = 0 to z = l is
replaced by a line with the characteristic impedance K 2 . Assume that
.
a progressIve wave
Ii(z) = loe- ifJu
is coming from z = - 00. Find the reflected and transmitted waves.
Answer:
If u = (K 2 - K 1 )/(K 2 + K 1 ) and k = ue-2ifJ2l, then
(k - u) Ioe ifJu
1 - uk
z < 0,
]r(z) _
I(z) = Ae- ifJ2z + Be j {32 z ,
o < z < l,
_ Ce- ifJ1 (z-l),
z > l,
where
A = (1 - u)I o
1 - jJ.k '
(1 - u 2 )Io
1 - uk
B = kA
,
c-
7.1-2 Equation (7.3) shows that there is no reflection at the
junction of two transmission lines with equal characteristic impedances
even though their phase constants are different. Similarly, equation
(4.61) shows that there are no reflections in a nonuniform transmission
line if the characteristic impedance K is constant. Suppose now that
K is not constant and that a section of this line from z = 0 to z = l is
sandwiched in between two uniform lines whose characteristic im-

ProblenlS
397
pcdances arc K (0) and K (l). Assume that a progressive wave is
coming from z = - ex>. Find the voltage and current in the interval
(0, I) taking into account the first-order reflections. Note the results
of Problem 4.17-5.
7.2-1 Suppose that instead of a lumped series impedance at
z =  of the line in Figure 7.2 we have a lumped shunt admittance
Y , . Discuss the problem of reflection and find voltage and current
distributions.
Answer:
AKY z
I (z) = Ae- iPz + e-2il3ejfJz
2 + KY , '
z < ,
2A
e- i13z
2 + KY , '
z > ,
AK2Y ,
V (z) = KAe- j13z - e-2i(JeifJz
2 + KY , '
z < ,
2KA
e- if3z
2 + KY , '
z > E.
7.3-1 Suppose that a wave given by equations (7.16) is incident
on a thin resistive sheet in the plane z = O. Let R be the resistance
between the opposite sides of a unit square in the sheet. Show that
(1]/R)Ho
H = Hoe- if3z + e if3z
II 2 + (TJ/R) ,
z < 0,
2Ho
e- jfJz
2 + (1]/R) ,
(1]2/ R) I/o
E = T]H oe- it3z - e if3z
z; 2 + (1]/R) ,
z > 0;
z < 0,
21]11 0 .
e-:ifJ z
2 + (11/ R) ,
z > o.
Compare this problem with Problem 7.2-1.

398
Electronlaglletic fields
7.3-2 Suppose that in the preceding problem we add a perfectly
conducting sheet in the plane z = X/4. Assume that R = 1]. Show that
there is no reflected wave in the region z < 0 and that between the
resistive sheet and the perfectly cond ucting plane
Ex = r, 11 o cos {3z,
lI y = -jilo sin (3z.
7.7-1 Discuss the situation in Problem 7.2-1 from the point of
view of scattering and obtain the expressions for the scattered waves.
Answer: ['(z) AKY l
- e-2jPei{Jz z < t,
2 + KY l '
2A
- e- jBz z > t,
2 + KY , '
V'(z) AK2Yl
- e-2jPceiPz z < t,
2 + KY l '
AK2Yl
- e-j(jz z > .
2 + KY l '
8.4-1 Show that equation (8.29) for the input adn1ittance of the
line shown in Figure 8.4(a) may be expressed as
1 00 jw
Y in = -=--L + L L (2 ')
JW 0 n-l n Wn - W"
1 00 jw
= · 2 + L 2 (2 2)
JW eo n-l en Wn - w
where: Ln is the inductance of the nth branch of the equivalent net-
work and en is the energy stored in the line a.t nth resonance when the
a.1nplitude of the current through the input terl1zinals is unity. Here eo is
the energy stored in the loop Figure 8.4(b), \vhen a direct current
of unit Inagnitude is circulating in it.
8.4-2 Consider a transmission line of length l shorted at both
ends, z = 0 and z = l, Figure 8.4(b). Assume that a voltage Vi is

Problems
399
impressed in series with the line at point z - . This introduces a
discontinuity in the transverse voltage, V ( + 0) - V ( - 0) = Vi.
To satisfy the boundary conditions at z = 0, 1 we expand V(z) into a
sine series, equation (8.19) in the text. We express I(z) by a cosine
series, equation (8.17) in the text. Show that
Vi  jw Vi cos (n7r/l) cos (17rz/l)
fez) = :--L +  . L (2 2) ,
JW 0 n-l JW n W n - W
where Ln and Wn are &till given by equations (8.27) and (8.28).
Show that in this case the inductance and capacitance in the nth
branch of the equivalent network, l"'igure 8.5. are
2el cos 2 (n7r/l)
C ' -
n -
L = !LlsCG(1't7r/l),
n 2 7r 2
8.4-3 Assume that the line in Figure 8.4 is open at z = l and that
a voltage Vi is impressed in series at z = . Show that the current in
the line may be expressed as
_ f jwVi COS [(21£ + 1)'/I"/21] cos [(21£ + 1)'/I"z/21]
n-O L fl (w - w 2 ) ,
I(z)
where
Ln = ! Ll, en = 8el/ (2,11, + 1) 27r 2 ,
W n = (2n + 1 1r/2lVLE.
rrhe corresponding equivalent network is of the type in Figure 8.5
with
L = ! Ll sec 2 [(21£ + 1) 7r/2lJ,
C = 8Cl cos 2 [(2n + 1)'/I"V 21 ] .
(211, + 1) 271"2
Here again L equals t\vice the energy stored in the line at nth reso-
nance \vhen the current through the source is unity.
8.5-1 Show that equation (8.38) for the input impedance of the
line in Figure 8.6(a) is equivalent to
00 JW
Zin =  2 2
n-o 2e n (w n - W )
where en is the e'uergy stored at Itth antiresonance when the voltage
amplitude across the input terminals is unity.

400
Electronlaglletic fields
8.5-2 Show that if the shunt current Ii is injected at z = ,
Figure 8.6 (a), rather than at z = 0, then the transverse voltage is
given by
V(z)
_ f jwl> cos [(2n + l)7r U ;lJ cos [(2n + 1)7rz/2lJ .
n-o Cn(W n - w 2 )
8.5-3 Show that if the line is open at both ends and the shunt
current is injected at z = , then
V(z) _   jwli cas (n7r/l) cas (n7rz/l)
- · C +  C (2 2) ,
JW 0 n-I n Wn - W
where
Co = Cl,
C n = !Cl,
W n = n7r/lVLE.
The corresponding equivalent network is of the type shown in Figure
8.7.
8.5-4 A shunt current Ii is injected at z =  of a line shorted at
both ends, Figure 8.4(b). Show that
 jw/ i sin (n7r/l) sin (nrz/l)
V(z) = 
n-l C n (w - w 2 ) ,
where
C n = !Cl and Wn = nr/lVLE.
Show that the input impedance seen by the shunt generator may be
expressed as
. CD jw 3 Lit.n
Z. = JwLo + ""
In .L...J L (2 2)
n-I n Wn - W
where
Lo = L(l - )/l
is the dc inductance of the two loops of lengths  and l -  in parallel
while
Ll sin (n7r/l)
Ln = !Ll and Lltl.n = .
1tr
The equivalent network is of the type shown in Fig. 8.8.
8.5-5 Solve the preceding problem when the line is shorted at
z = 0 and open at z = l.

Problems
401
Obtain the solution when the line is open at both ends and a volt-
age Vi is impressed in series at z = .
8.5-6 In the preceding problems W n is a typical natural frequency
of the corresponding section of the line when the line is nondissipative.
This suggests that the expressions for the input in1pedance and ad-
mittance may be generalized by substituting - n + jW n for jWn,
where n is the damping constan t of corresponding free oscillations.
This substitution is equivalent to the substitution of W n + jtn for
Wn or approximately w + 2jnwn for w. In the latter approximation
it is assumed that  is negligible in comparison with w. Another
approximation may and usually is involved in the substitution: the
coupling between various modes of oscillation, usually produced
by the resistance and conductance, is neglected.
The dan1ping constants may be calculated from energy considera-
tions as explained in Sections 3.5 and 6.13. Show for instance that if
R is the resistance per unit length of the line shown in Figure 8.4(a),
then the expression for the input admittance given in Problem 8.4-1
becomes
1 00 jw
Y in = 28 o [jw + (R/ L) ] + E 28n[w - w 2 + (jwnR/ L) 1
8.5-7 Consider a cavity bounded by planes x = 0, a; y = 0, b;
and z = 0, h. Assume that Iz is small in comparison with either a or b.
Around a point (xo, Yo, 0) there is a small hole through \vhich the
inner conductor of a coaxial line enters the cavity and joins the upper
wall at (xo, Yo, lz). Thus the curren t in the protruding post excites the
cavity. At an antiresonant frequency the current in the post is zero
and the post has a negligible effect on the antircsonant frequency.
Therefore the frequencies can bc calcula tcd as in Section 6.11 by
ignoring the post. The equation for the input impedance in Problem
8.5-1 is thus suitable for the present case. Show that
Wn = [(m1r/a)2 + (n1r/b)2J//.LE,
2e mn = Eab/4h sin 2 (m7rxo/a) sin 2 (n1rYo/b).
As w approaches zero,
co 1
Zin jw L 2
m,n-1 2W mn8mn
and the sununation represents the direct current inductance La of the

402
llcclronlaJ!.neti, fields
post in the cavity. This inductance may be calculated from static
considcra tions. 1"hen
co J.w 3 L 2
z. - · L "" !tl.mn
In - JW 0 + L-i 2 2 '
m,n-l Lmn{W mn - W )
v.rhere
Lif.mn
Lmn
4h sin 2 (11Z7rao/ a) sin 2 (ntryo/ b)
EabWn
1"'he equivalen t network is of the type shown in Figure 8.8.
8.5-8 Consider the cavity in the preceding problem and assume
that a given shunt current of density J z(x, y) is forced to flow from
the lower face z = 0 to the upper face z = h. Sho\v that in this case
l\laxwell's differential equations are
aE z
- -jwp.1I:r"
ay
aE z
- =}.wp.H
ax II'
ally aHJ; .
- - - = }wEE z + ]z(x, y).
ax ay
The electric intensity in the cavity may be represented by a double
:Fourier series.
E z = L: Emn sin (mtrx/a) sin (ntry/b).
m,n
Show that
Emn = - 4jw . {fa J.(x,y) sin (m7rxja) sin (n7ryjb) dxdy
Eab(wn - w..) 0 0
where W mn is given in the preceding problem. Fron1 these equations one
obtains the input impedance seen by the coaxial line in the pre-
ceding problem.
8.6-1 :From a purely mathematical point of view the method
presen ted in Section 8.6 is a method for solving transmission line
equations (8.43) and (8.44). Another such method is based on calculus
of variations. Suppose that we \\Oish to determine natural frequencies
of a nonuniform transmission line \vhich is either open or shorted at

Problems
403
its ends z = 0 and z = t. From equations (8.43 and (8.44) we then
have
d [ 1 dI]
- - - = -w 2 L(z)I
dz C(z) dz '
(1)
 [ LZ) : ] = -w 2 C(z) V.
l\Jultiplying equation (1) by I(z), integrating from z = 0 to z = l,
and using the boundary conditions, we find that the following integral,
(2)
P = fl [ ( dI )2 _ w 2 L(Z)I2] dz,
o C(z) dz
(3)
vanishes. Furthermore, taking the variation of P, we obtain
f l [1 dI dI ]
p = - 2 - {) - - 2w 2 L(z)llJ! dz
o C(z) dz dz
j l [ 1 dI d ]
= 2 - - - (oJ) - 2w 2 L(z)I fJ! dz.
o C(z) dz dz
Integrating the first term by parts and taking into account the boun-
dary conditions, we have
fJP = -2 fi [ { dI } + w 2 L(Z)I] fJ! dz.
o dz C(z) dz
Thus if fez) satisfies equation (1), then oP = o. Conversely, if
P = 0 for an arbitrary variation in 1(z) but subject to the given
end conditions, then I (z) satisfies equation (1). Thus, we have
p=O
and
oP = O.
(4)
The first of these equations expresses the physical fact that if the
oscillations ate free, the maximum electric energy in the line must
equal the maximum magnetic energy. Similar equations are obtained
for equation (2).
Equations (4) are suitable for approximate calculation of natural
frequencies. Suppose, for example, that the line is open at z = 0
and z = t. If the parameters Land C are independent of z, then we
know that for the lowest mode I (z) is proportional to sin (7rz/l). If

404
Electromagnetic fields
we use this function to approximate I(z) in the general case, we find
p = jl [ 1r 2 C05 2 (1rz/l) _ w2L(z) 5in2 (1rZ/l)] dz.
o l2C (z)
The condition P = 0 then yields
{ 1r 2 1- 2 [C(z) J-l C05 2 (1rz/l) dz
o
w 2 =
jl L(z) sin2 (1rz/l) dz
o
We could have used a different approximation to I (z), I - (z/l) 2,
for instance. The choice of the first approximation depends on the
physical insight in any specific case. If Land C are nearly independent
of z, the sine approximation would be quite good since it would yield
the exact answer when Land C are constants while other functions
would yield only approximate answers.
To improve the approximation we should introduce more disposable
parameters. Thus we could assume
I (z) = al sin (7rz/l) + a2 sin (21fz/l).
If we calculate P, then oP, and equate to zero the coefficients of oal,
and oa2, we obtain two linear homogeneous equations for the two
unknowns al and a,2. From these equations we can determine wand the
ratio a2/al. We can keep adding more sine functions for better ap-
proximations. We also can select a different set of approximating
functions.
9.4-1 Obtain the solution of equation (9.27) for the principal
mode, m = 0, when
jJ. =  J.l.o
E(Z) = Eo[I + k(z/Xo)J,
z > O
(J)
where k is a constant and Xo is the free space wavelength. At fIrst
use the approximate method of Section 4.17. The follo\\Ting substitu-
tion of the new dependent variable lV defined by
dZ
- = -jwE(z)fV
dz
(2)

Problems
405
will transform equation (9.27) into a pair of the first-order equations
of the type (4.56).
Answer: Z (z) = [K (z) J-1/2(Ae- i r} + Be i "),
where
K(z) = 170[1 + k(Z/AO) ]-1/2,
(z) = (47r/3k) [1 + k(Z/AO) J3/2 - (47r/3k).
9.4-2 The solution of the preceding problem is good to the extent
to \vhich K'(fJ)/2K(tJ) is negligible in comparison with unity. Show
that for equations (4.56)
K' (tJ) /2K (iJ) = K' (z) /2wL(z),
and use this result to show that the solution is good when k « 1 or
when k » 1 and Z is relatively large in comparison with Ao.
Show that
K'(iJ)/2K(iJ) = - (k/47r) [1 + k(Z/AO) J-3/2,
that the magnitude of this ratio for z = Zo is maximum when
k = 2 AO/ZO,
and that
\ [K' (iJ) /2K (iJ) ]max I = (Ao/zo) j67rV3.
If the approximation is considered good enough in the vicinity
of Z = Zo, it will be just as good or better for a.ll k when Z > Zoo
Note that the maximum ratio is relatively small even \vhen Zo = Ao/4.
Since the section from Z = 0 to Z = Ao/4 is relatively short, the ap-
propriate solution may be found by successive integrations as in
Section 4.4.
9.4-3 Show that the exact solution of the Problem 9.4-1 is
W(z) = AUl/ 3 J l13 (U) + BUl/ 3 iV 1 / 3 (U) ,
where
'It = iJ + (47r /3k) ,
and
dlV
Z(z) = - (l/jwJlo) -.
dz
Progressive waves arc obtained when B = =FjA. If u » 1, then
the Bessel functions may be approximated by the first terms of their

406
lle(/ronl(l1!.lle/ ic fields
asymptotic expansions and the approximate solutions given in Prob-
lem 9.4 - 1 arc obtained. The quantity 1(, is large \vhen either k « 1
or tJ » 1.
9 .4-4 If the stra tifica tion is at rightangles to the direction of
propagation and we have to solve equations (9.28) subject to the
boundary conditions (9.29), we may use either the method of Section
4.17 or the variational method. In the latter case sho\v that if
P _ fa [ dX 2 _ W2}J.X2 _  X2] dx,
- 0 E(X) dx E(X)
then
p=o
and
oP = O.
Since X(x) is proportional to cos (m7rx/a) when E(X) = const,
\ve may use these functions for obtaining the approximate values of
the propagation constants. Show that for the principal mode we have
jw
ro =
 (l/a) fa [E(X) J-l dx
o

List of Symbols
B magnetic flux density
C capaci tance _ linear curren t density
c speed of light in vacuum
D electric disphtcCn1cnt (or flux) density
E electric intensity
t energy
F force
J £rcftuency
G conductance
H magnetic intensity
I electric current
J electric current density
K characteristic impedance
L inductance
M magnetization
111, mass
p power, polarization
p moment
q electric charge
R resistance
S area
s distance
t time
U magnetic potential, magnetomotive force
V electric potential, electromotive force (voltage)
407

408
Electronzagne/ie fields
v
lV
X
Y
Z
a

r,,¥
E
EO
Er = E/EO
TJ
X
p.
Ilo
J..L, = p.1 Ilo
p
(f
'P
ell
'It
n
w
(x, y, z)
(p, f/J, Z)
(1, (J, f/J)
(U, 'V, W)
speed, volun1c
work
reactance
admittance
impedance
attenuation constant
phasc constant
propagation constant
dielcctric constant
dielectric constant of vacuum
relativc dielectric constant
in trinsic impedance
wavelength
permeability
permeability of vacuum
relativc permeability
charge density, cylindrical coordinate
conductivity
phase, cylindrical coorclina te
magnetic flux
electric flux
solid angle
angular frequency
Cartesian coordinates
cylindrical cQordina tes
spherical coordinates
general coordinates

Ampere's la\v, 41
Ampere-lVlax\\.ell equation, 51, 52
Ampere.l\rl ax\\.el1la\v, 44, 54
Area moment, 91
Attenuation constant, 162
Beats, 269
Boundary conditions, 55
Bre\vster angle, 249
Capacitance, 117
Capacitance coefficients, 117
Characteristic equation, 215
Characteristic functions, 209
Characteristic impedance, 161
Characteristic valuQS, 209
Circulation, 309
Conductance, 15, 114
Conductance coefficients, 115
Conductivity, 14
Conduction curcnt, 48
Conductors. 13
Con tou r Ii nes, 308
Contour surfaces, 308
Con vection cu rren t, 48
Coordinates, 305
Coulomb's law, 5
Coupled circuits, 268
Coupled transmission lines, 285
Coupling coefficient, 269
Curl of a vector, 309
Current element, 18
Damping constant, 136
Dielectric constant, 28
Dipole moment, 91
Dipole source, 18
Direct capacitance, 117
Direct conductance, 116
Directional coupling, 292
Directive radiation, 195
Discontinuities, 58
Displacement current, 27, 43, 44
Index
Di vergence, 310
Dominant ,vave, 232
Eigen function s, 209
Eigen val ues, 209
Electric charge, 6
Electric current, 13
Electric displacement, 27
Electric dipole, 29
Electric field. 7
Electric intensity, 8
Electric lines, 10
Electric net\vorks, 148
Electromotive force, 11
Equipotential Jines, 63
Equipotential surfaces, 63
Equivalent circuits, 139, 140, 141,
142, 159
Equivalent networks, 272, 276, 280
Faraday's law, 23
Faraday-rVIaxwell equation, 51, 52
Faraday-1Vlax\velllaw, 37, 54
Flux of a vector, 310
Free oscillations, 132, 166
Gaussian field strength, 32
Generalized coordi nates, 283, 304
Gradien t, 66, 308
Half-\\'ave antenna, 201, 202
I mage impedance, 170
J mage parameters, 170 .
Image transfer constant, 170
Impedance discontinuity, 165
I m pressed field, 77
Inductance, 117, 118
Input impedance, 165
Interference of ¥laves, 195
Internal impedance, 154
Intrinsic impedance, 146
Intrinsic velocity, 146
Kirchhoff's equations, 149
Laplace's equation, 314
Laplacian, 311
409

410
Electromagnetic fields
Level lines, 308
Leve I su rf aces, 308
Local potential, 66
Logarithmic decrements, 136
IVlagnetic dipole, 86
IVlagnetic field, 8
1\Iagnetic flux, 34
1\!lagnetic flux density, 31, 37
l\Iagnetic intensity, 39,40
l\Iagnetic point source, 86
1\1 agneti c ( scalar) potential, 86
l\rlagnetic vector potential, 86
1\lax\vell's equations, 312, 313
1\lesh currents, 149
iodes of oscillation, 166
l\iodcs of propagation, 287
N' atural oscillations, 132
IVlagnetomotive force, 41
1\'lode of oscillation, 270
K atural frequency, 136
Normal modes, 227
Ohm's la\v, 14
Orthogonal modes, 227
Orthogonality, 216
Permeability, 37
Phase constant, 162
Phase integral, 169
Phase velocity, 163
Point source, 17
Polarizability, 77
Polarization, 82
Polarization current, 45
Poten tial, 62
Poten tial coefficients, 117
Po\ver fio\\", 130
Poynting vector, 130
Pri mary parameters, 239
Primary wave, 253
Principal \vaves, 196
Propagation constant, 162
Proper functions, 209
Proper values, 209
Proximi ty effect, 83
Radiant energy, 192
Radiation, 191
H.adiation resistance, 200
Reactance, 155
Reciprocity, 115, 116
Reflected field, 76
H.eflected \vave, 164
H.eflection, 241
H.efiection coefficient, 165
Refracted \vave, 247
]{efractive index, 248
l{elative dielectric constant, 28
]{esistance, 15, 114
Resistance coefficients, 114
]{esonant frequency, 275
Retarded poten tial s, 202
Scattered \\-'aves, 252
Secondary parameters, 239
Separation constant, 208
Separa tion of variables, 207
Series impedance, 155
Short antenna, 198
Shunt admittance, 155
Ski n effect, 168
510\vly varying field, 147
Solenoid, 39
Spherical \vaves, 177, 182
Standing \vave, 166
Step-by-step method, 60
Surface impedance, 154
Susceptance, 155
l'El\/l \\'aves, 228
TIvl \vaves, 228
T'ime-harmonic quantities, 150
T'otal cu rren t, 43
Total reflection, 249
Transmission coefficient, 241
Transmission line, 124
Transmi tted field, 81
Transverse electromagnetic waves,
228
Transverse magnetic \vaves, 228
Tubes of current, 16
Tubes of fio\v, 17
Units, 3
\'cctor components, 305
Virtual source, 103
Voltage, 11
\Vave antenna, 250
\Vaveguides, 230
\\"ave inlpedance, 229
\V'avelength, 163
\V'eak coupling, 287

A Note About the Author
Professor Sergei A. Schelkunoff, born in Russia on January
27, 1897, emigrated to the United States in 1921. Having
majored in mathematics, he received both his B.A. and M.A.
degrees from the State College of Washington in 1923 and his
Ph.D. degree from Columbia University in 1928. Until 1960,
he was. with Bell Telephone Laboratories, doing research in
electroinagneic field theory. In 195 he became the Assistant
Director of Mathematical Research and in 1958, Assistant
Vice President. At various times he lectured on mathematics
and field theory at the State College of Washington, Brown
University, University of California at Low Angeles, New
York University, and Columbia University. In 1960, he was
appointed Professor of Electrical Engineering at Columbia
University.
Professor Schelkunoff is author of Electromagnetic Waves
(1943), ApPlied J.f athematics for Scientists and Engineers
(1948), ntennas: Theory and Practice (\vith Harald T. Friis,
1952), and Advanced A ntenna Theory (1952).
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