PRODUCTION MANAGER: Herbert Nolan The text o f this book was composed in Times Roman by Syntax International. Library of...
DAVID K. CHENG SYRACUSE UNIVERSITY
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ADDISONWESLEY PUBLISHING COMPANY
Reading, Massachusetts Menlo Park, California London Amsterdam Don Mills, Ontario Sydney
DAVID K. CHENG SYRACUSE UNIVERSITY
A vv
,
ADDISONWESLEY PUBLISHING COMPANY I 1
Reading, Massachusetts Menlo Park, California m on don' Amsterdam Don Mills, Ontario' Sydney
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This book is in the API)ISON7WESLEY SERIES IN ELERRICAL ENGINEERING
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SPONSORING EDIT& T o m Robbins PRODUCTION EDITOR: Marilee Sorotskin TEXT DESIGNER: Mednda Grbsser ILLUSTRATOR: Dick Morton , COVER DESIGNER AND ILLUSTRATOR: Richard H a n y s ART COORDINATOR: Dick Morton i.: PRODUCTION MANAGER: Herbert Nolan The text o f this book was composed in Times Roman by Syntax International.
Library of Congress Cataloging in Publication Data Cheng, David K. avid~(euni dqteField and wave e ~ e c t r o ~ ~ e t i ~ s . Bibliography: p. I 1. Electromagnetism. 2. Fiqld th :ory (Physics) I I. Title. QC760. C48 < 530.1'41 8112749 ISBN 0201012391 AACR2 3 I
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COPY%^^ O 1983 by AddisonW+ey
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Publishing Company, Inc. rights reserved. NO part of @is publication may be reproduqd, btored in a retrieval system, or tnnsm,ltted. in any, form or by any means. electmnic, mechanica~.photocopying, recordmg, or othenu~se, without the prior written permission df the publisher. Prmted m the Uolted States of Amenca. Published simultaneously in Canada. , ISBN 0201012391 ABCDEFGHIJAL89876543 I
11 12 13
Introduction The electromagnetic model SI units and universal constants Review questions
Introduction Vector addition and subtraction Products of vectors 23.1 Scalar or dot product 23.2 Vector or cross product 23.3 Product of three vectors Orthogonal coordinate systems 24.1 Cartesian coordinates 24.2 Cylindrical coordinates 24.3 Spherical coordinates Gradient of a scalar field Divergence of a vector field Divergence theorein Curl of a vector field Stokes's theorem 210 Two null identities 210.1 Identity I 210.1 Identity I1 21 1 Helmholtz's theorem Review questions . Problems
31 32
Introduction Fundamental postulates of electrostatics in free space J2 Coulonb's law 3 ?.: Electric fitla ciue LO a system of discrete charges 33.2 Electric field due to a contipuous distribution of charge 34 Gauss's law and applications 35 Electric potential I 35.1 Electric potential due to a charge distribution 36 Condugtors in static electric field 37 Dielectrics in static electric field 37.1 Equivalent charge distributions of polarized dielectrics 38 Electric flux density and dielectric constant 38.1 ~ielectric'strength 39 Boundary conditions for electrostatic fields 310 Capacitance and ~apacitors 3 10.1 Series aqd parallel connections of capacitors 31 1 Electrostatic enerpy and forces . 3 11.1 ~lectrost'aticenergy in terms of field quantities 3 11.2 Electrostatic forces Review. questions , Problems 6 I

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Solution of Electrostati~ Problems
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41 42 43 44
2
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Introduction i Poisson's and La~lace'sequations Uniqueness of eleptrgstatic solutions i Method of imageg' : : I 44.1 Point charge and conducting planes 44.2 Line charge and parallel .I conducting cylindpr . 44.3 Point charge aqd conducting sbhera, 1
45
,
46 47
Boundaryvalue problems in Cartesian coordinates Boundaryvalue problems in cylindrical coordinates Boundaryvalue problems in spherical coordinates Review questions ~roblems
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Steady Electric Currents
51 52 53 54 55 56 57
Introduction Current density and Ohm's law Electromotive force and Kirchhoff's voltage law Equation of continuity and Kirchhoff's current law Power dissipation and Joulc's law Boundary conditions for current density, Resistance calculations Review questions Problems
Static Magnetic Fields
1
,
61 62
Introduction Fundamental postulates of magnetostatics in free space 63 Vector magnetic potential 64 BiotSavart's law and applications 65 The magnetic dipole 65.1 Scalar magnetic potential 66  Magnetization ;~ntlcquivalcnt currcnt dcnsiiics 67 Magnetic field intensity and relative permeability I 68 Magnetic circuits 69 Behavior of magnetic materials I 610 Boundary conditions for magnetostatic fields 1 61 1 Inductances and Inductors . 612 Magnetic energy . 612.1 Magnetic energy in terms of field quantities 1
/
I
CONTENTS
xiii
1
1
. I
82.1 Transverse electromagnetic waves 82.2 Polarization of plane waves Plane waves in conducting media 83.1 Lowloss dielectirc 83.2 Good conductor 83.3 Group velocity Flow of electromagnetic power and the Poynting vector 84.1 Instantaneous ana average power densities Normal incidence at a plane conducting boundary Oblique incidence at a plane conducting boundary 86.1 Perpendicular polarization 86.2 Parallel polarization Normal incidence at a plane dielectric boundary Normal incidence at multiple dielectric interfaces 88.1 Wave impedance of total field 88.2 Impedance transformation wishmultiple dielectrics Oblique incidence at a plane dielectric boundary 89.1 Total reflection 89.2 Perpendicular polarization 89.3 Parallel polarization Review questions Problems
9
Theay and Applications of Mnsmission Lines
Introduction Transverse electromagnetic wave along a parallelplate transmission line 92.1 Lossy parallelplate transmission lines General transmissionline equations 93.1 Wave characteristics on an infinite transmission line
93.2
Tr:lnsrnissionli11c uurumctcl.~
93.3 Attenuation constant from power relations Wave characteristics on finite transmission'lines 94.1 Transmission lines as circuit elements 94.2 Lines with resistive termination
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1 I
312 314 317 318
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xiv
CONTENTS
1
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95
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94.3 Lines with G b j t i a q terminafian . , 94.4 ~ransmissioalinkcircuits . : The Smith chart 95.1, Smithchart ca1culations for losiy lines Transmissionline im&d&ce matching 1  . 9 4 . 1 Impedance rnatc$ng by quarter: , wave transformer 96.2 Singlestub matching 96.3 Doublestub matching Review questions 1 . , Problems ,
404 407 41 1 I
F
10
.
I i
(
Waveguides and Cavity R8so"ators
,
Introduction General wave behaviors along uniform ' guiding structures ,102.1 Transverse electromagnetic waves 102.2 Transverse magnetic waves 102.3 Transverse e1ecp;ic waves , Parallelplate waveg$de* t 103.1 TM waves petween yarahlel plates 103.2 TE waves between parallel plates 103.3 Attenuatioq in.~~rallelplate' waveguides . t Rectangular waveguides 104.1 TM waves in rectangular wqvebides 104.2 TE waves in recrangular waveguides 104.3 ~ t t e n u i i t i $ n ) r k t a n ~ u l a rwaveguides Dielectric waveguide6 : ! 105.1 TM wayes a dielectric slab 105.2 TE waves ;)ong ii dielectric slab ' > , Cavity resonators \ 106. ! TM,,,, moqes I I ,I 106.2 TE,,, modts 106.3 Quality factor &(cavity resonatbr Review questions Problems
lo^&
,
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420 422 423 426 43 1 435 437
11
An 111 11
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1.. . . ..,,., . . . ..: ,,  ... ;... .?;,; ...i . ', . ., ... ,:v.,,,,, :;. . , . ". '.. , ..,, . i , ~ ' Antekms and Radiating Systems. " ,. , . Introduction ~ ' .~ 1
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CONTENTS
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; .,
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,
I , .
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Radiation fields of elemental dipoles 112.1 The elemental electric dipole 112.2 The elemental magnetic dipole Antenna patiefh5 and at. tenna parameters Thi;. linear antennas 114.1 The haltwave aipde Antenna arrays 115.1 Twoelement arrays 115.2 General uniform linear arrays Receiving antennas 116.1 Internal impedance and directional pattern 116.2 Effective area Some other antenna types 117.1 Travelingwave antenna 117.2 YagiUda antenna 117.3 Broadband antennas Apeiture Radiators References Review questions ProbIems
S
,
*
{
.
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Amendix A Symbols and Units A1 A2
A3
Fundamental SI (rationalized MKSA) units Derived quantities Multiples and s~bmultiplesof units
Appendix B Some Useful Material Constants
B1 B2
Constants of free space Physical constants.of electron and proton
.
,. .
500 502
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\
B3 Relative permittivities (dielectric constants) .I ,.! i BQ ~onductivities , .., . . B5 Relative permeabilities ,
..I
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Back Endpapers
:
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Left: ., Gradient, divergence. cprl, and Laplacian opetations
Right: Cylindrical coordinates Spherical coordinates
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I
The many books on introductory electromagnetics can be roughly divided into two main groups. The first group takes the traditional development: starting with the experimental laws, generalizing them in steps, and finally synthesizing them in the form of Maxwell's equations. This is an inductive approach. The second group takes the axiomatic development: starting with Maxwell's equations, identifying each with the appropriate experimental law, and specializing the general equations to static and timevarying situations for analysis. This is a deductive approach. A few books begin with a treatment of the special theory ofrelativity and develop all of electromagnetic theory from Coulomb's law of force; but this approach requires the discussion and understanding of the special theory of relativity first and is perhaps best suited for a course at an advanced level. Proponents of the traditional development argue that it is the way electromagnetic theory was unraveled historically (from special experimental laws to Maxwell's equations), and that it is easier for the students to follow than the other methods. I feel, however, that the way a body of knowledge was unraveled is not necessarily the best way to teach the spbject to students. The topics tend to be fragmented and cannot take full advantage of the conciseness of vector calculus. Students are puzzled at, and often form a mental block to, the subsequent introduction of gradient, divergence. and curl operations. As ;I proccss for formu1;lting :in clcctroni:~gncticmodel, this approach lacks co,hesivonessand elegance. The axiomatic development usually begins with the set of four Maxwell's equations, either in differential or in integral form, as fundamental postulates. These are equations of considerable complexity and are difficult to master. They are likely to cause consternation and resistance in students who are hit with all of them at the beginning of a book. Alert students will wonder about the meaning of the field vectors and about the necessity and sufficiency of these general equations. At the initial stage students tend to be confused about the concepts of the electromagnetic model, and they are not yet comfortable with the associated mathematical manipulations. In any case, the general Maxwell's equations are soon simplified to apply t o static fields, which allaw the consideration of electrostatic fields and magnetostatic fields separately. Why then should the entire set of four Maxwell's equations ' be introduced a t the outset?
vi
I
PREFAC'
It may be argued tfiat'Coulomb's law, i h o u ~ hbased on experimental evidence, is in fact also a~postulate.'Cbnsider the tw6 stipulations of Coulomb's law: that the charged'bodies are very spa11 comparec/ with thcir distance of separation, and that the force between the qhafged bodies is isv'crscly proportional to t11c sclu;~rcof their distance. he question a~isesregarding the first stipulation: How small must the charged bodies be in ardqr to be considered "very small" compared with their distance? I,, practice the,charged bodies cannot .be of vanrs' ag sizes (ideal p o h t charges), and there is dificplty in determinirig the?'true" distance between two bodies of finite dimensions. Fgr given body sizes the relative accuracy in distance measurements is better when {he separation is largdr. However, practical considerations, (weakness o f force, existence of extraneous charged bodies, etc.) restrict the usable distance of separation in the laboratory, and experimental inaccuracies cannot be entirely avoided. This {ends to a more importa~ifquestion concerning the invcrsesquare relation of thc second stipulnlion. Even if thc clwrgcd bodies wcrc of vanishing . sizes, experiment?! measurements could not be of an infinite accuracy no matter how skillful and careful an experimentor was. ko+ then was it possible for Coulomb to know that the force,wns exactly inversely pr'pportional to thc syuure (not the 2.000001th or the 1.995j999th power) of the distance of separation? This question cannot be answered from an experimental1viewQoint because it is not likely that during Coulomb'$ time experiments could fiave been accurate to the seventh place. We must therefore conclud'e.that Coulombis lawsis itself a postulate and that it is a law of nature discovered and assumed on the basis of his experiments of a limited accuracy (see Section 32). This book builds tpe dcctromagnetic inode] using ap axiomatic approach in steps: first for static elecrric fields (Chapter 3), then. for siatic magnetic fields (Chapter 6), and finally for timemrying .fields leading to ~ a x k e f l ' sequations (Chapter 7). The mathematical basis:for each step is Halmhohr's theorem, which states that a vector field is determined to within an additive cbnst$bt if both its divergence and its curl are speciff?;devfrywhere. Thus, for the::'development of the electrostatic model in free space.. it is 'pnly necessary to define n singb vector (namely. the electric liclcl iulcuaily E) by spccil'~iug its clivcrg~.rdeand its cur! as postulates. All other relations in electrostati~sfor free space, ihchding Coul~mb'slaw and Gauss's !hq two rather simple postulates. Relations in material law, can be derivbmedia can be devclopedihydugh the concept of eguival&ntcharge distributions of r , f polarized dielectrics. Similarly, for fhe magphstatic model .in .fr& space it is necessary to define only a single magnetic fluyidensity vector B by .specifying its divergence and its curl as postulates all other formulas can be derived frpm these two postulates. Relations in rnater!al medip ban be developed t$ough.fhe concept of equivalent current densities. Of coyrsp' the validity o f the postulates lies in their ability to yield results that dpnfoqb pith experimentil evidince. For timevarying fields, 'the electric andYmagliftic field intensities are coupled. The curl E postula& for)hy electrostatic Godel rmst be modified to conform with Faraday's law. In addition, the curl B postdate the magnetostatic model must also be modified in'prder: to be consistent with of continuity. We have, '
E,
=
*
.
_ >
L
2'
8
: as.
I
\
PREFACE
bnental evidence, bb's law: that the hration, and that je square of their v small must the :d with their disizes (ideal point !ween two bodies ;istancc measurc
considerations
11
:strict the usable racies cannot be ling the inversevere of vanishing m c y no matter ble for Coulomb syuure (not the '! T . n p e s t i o n ; not . ~ ~ e that ly 2 >
.*'\ placu.
and that it is :nts of a'lirnited ie
~ t approach c
in
fields (Chapter ns (Chapter 7). :h states that a divergence and le electrostatic rly, the electric ates. All other v and Gauss's )us in material lihbutions of '
isary to define
rgenwd
its wo t .tulates. oi dent hei?,dity to
s are coupled. conform with ;c model must lity. We have,
vii
then, the. four Maxwell's equations that constitute the electromagnetic model. I believe that this gradual development of the electromagnetic model based on Helmholtz's theorem is novel, systematic, and more easily accepted by students. In the presentation of the material, I strive for lucidity and unity, and for smooth and logical flow of ideas. Many workedout examples (a total of 135 in the book) are included to emphasize fundamental concepts and to illustra~emethods for solving typical problems. Review questions appear at the end of each chapter to test the students' retention 2nd ur.dex;anding of the esse~tialmaterial in the chapter. Thc problcms in c;~chchap~crarc tlcsigncd rcinforcc >,;,!cnts1 comprchcnsion of the interrelationships btween the difTerent quantities in the formulas, and to extend their abilitywf appljling the formulas to solve practical problems. I do not believe in simpleminded drilltype problems that accomplish little more than sn exercise on a calculator. The subjects covered, besides the fundamentals of electromagnetic fields, include theory and applications of transmission lines, waveguides and resonators, and antennas and radiating systems. The fundamental concepts and the governing theory of electromagnetism do not change with the introduction of new eiectromaznetic devices. Ample reasons and incentives for learning the fundamental principles of electromagnetics are given in Section 11.. I hope that the contents of this book strcngthened by the novel approach, will pr'oiide students with a secure and sufficient background for understanding and analyzing basic electromagneiic phenomena as well as prepare them for more advanced subjects in electromagnetic theory. There is enough material in this book for a twosemester sequence of courses. Chapters 1 through 7 contain the material on fields, and Chapters 8 through 11 on waves and applications. In schools where there is only a onesemester course on electromagnetics, Chapters 1 through 7, plus the first four sections of Chapter 8 would provide a good foundation on fields and an introduction to waves in unbounded media. The remaining material could serve as a useful reference book on applications or as a textbook for a followup elective course. If one is pressed for time, some material, such as Example 22 in Section 22, Subsection 311.2 on electrostatic forces, Subsection 65.1 on scalar magnetic potential, Section 65 on magnctic circuits, and Subscctions 613.1 and 613.2 on magnetic forces and torqucs, may be omittcd. Schools on a quarter system could adjust the material to be covered in accordance with the total number of hours assigned to the subject of electromagnetics. The book in its manuscript form was classtested several times in my classes on electromr\gnetics at Syracuse University. I would like to thank all of the students in those classes who gave me feedback on the covered material. I would also like to thank all the reviewers of the manuscript who offered encouragement and valuable suggestions. Special thanks are due Mr. Changhong Liang and Mr. Bailin Ma for their help in providing solutions to some of the problems. '
Syracuse, New York January 1983
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11
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INTRODUCTION
!
Stated in a simple fashion, electromugnetics is the study of the effects of electric charges at rest and iq motion. From elementary physics we know there are two kinds of' charges: posithe ahd negative. Both positive and negative charges are sources of an electric field M b ~ i n ~ c h a r g eproduce s a current, which gives rise to a magnetlc field. Here we tentatively speak of electric field and magnetic ficld in a general way; more definitive meanings+willbc attached to these terms later. A j e l d is a spatial distribution of a q u a d y , which may or may not b;e,s function, of time. A timevarying electric field is accompanied by a magnetic field, and vice versa. In other words, timevarying electric and magn4iic fields are coupled, r$suiting i i rin electromagnetlc field. Under certain conditibns, timedependent electromagnetic fields produce waves that radiate from the sotlfce. The concept of fields and waves is essential in the explanation of action at a distance. In this book, Field mii Wave Electromugnetics, we study the principles a d applications of the laws of electromagnetism that govern electromagnetic dhenomepa. Eleciromagnetks is of fundamental importance to physicists and electr~cal engineers. ElEctromagnetic theory is indispensable in the understanding of the principle of atdm smashers, cathoderay oscillosco~s,radar, satellite communication, television reception, remote sensing, radio astronomy, microwave devices. optical fiber communication, instrumentlanding systems, electromechanical energy conversion, and s? on. Circuit concepts represent a restricted version, a special case, of electromagni?tic caflcepts. As we shall see in chap& 7, when the source frequency IS very low so that th&dimensions of a conducting ntfwork are much smaller than the wavelenglh, 3 e have a quasistatic situation, which simplifies an electromagnetic problem *circuit problem. However, we hasten to add that circuit theory is itself a highly deveroped, sophisticated discipline. It appHls to a different class of electrical engineering p r ~ b l e b s rtnd , it is certainly important in its own right. Two sitbations:illustrate the inadequacy of circuittheory concepts and the need of electroma~neticfieldconcepts. Figure !I depicts a monopole antenna of the type we sce on a wdkie4alkic. OH tnuwmit, thc sotlrcc at thc basc Cccds the antelm1 wlth P mcssagccarrying currcot d an appropriate carrier frequency. From a circuittheory
*
2
THE ELECTROMAGNETIC MODEL / 1
A monopole antenna.
Fig. 12 An electromagnetic problem.
point ofview, the source feeds into an open circuit because b b e r tip of the antenna is not connected to anything physically; hence no current would flow and nothing would happen. This viewpoint, of course, cannot explain why communication can be established between walkietalkies at a distance. Electromagnetic concepts must be used. We shall see in Chapter I1 that when the length of the antenna is an appreciable part of the carrier wavelengtht. a nonmiform current will flow :dong thc ol7cnended alltellna. This current radiates a tin~cv;lryiogelectrornil~neticfield in space, which can induce current in another antenna at a distance. In Fig. 12 we show a situation where an electromagnetic wave is incident from the left on a large conducting wall containing a small hole (aperture). Electromagnetic fields will exist on the right side of the wall at points, such as P in the figure, that arc not necessarily directly behind the aperture. Circuit theory is obviously inadequate here for the determination (or even the explanation of the existence) of the field at P. The situation in Fig. 12, however, represents a problem of practical importance as its solution is relevant in evqluating the shielding effectiveness of the conducting wall. Generally speaking, circuit theory deals with lumpedparameter systemscircuits consisting hf components characterized by lumped parameters such ar rcsisl:~~~ccs, i~~ductancbs, :111d ~ ~ a c i l s ~ ~VcOeI IsP .~ C S and currents are the main system variables. For DC circuits, the system variables are constants and the governing equations are algebraic equations. The system variables in AC circuits are timedependent; they are scalar quantities and are independent of space coordinates. The governing equations are ordinary differential equations. On the other hand, most electromagnetic variables arefunctions of time as well as of space coordinates. Many are vectors with both a magnitude and a direction, and their representation and manipulation require a knowledge of venor algebra and vector calculus. Even in static cases, the governing equations are, in general, partial differential equations. It
' The product of the w&elength and the frequency of an AC source is the vdocity of wave propagation.
i
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. in 0 , ) and a
!I'crence of two
identity. (Note
Fig. 28 Illustrating the backcab rule of vector triple product.
The expression above does not alone guarantee that the quantity inside the brackets to be D , since the former may contain a vector that is normal to D (parallel to All); that is, D .a, = E . a, does not guarantee E = D . In general, we can write
B(A,~.cj  c(A~, B) = D + k
~
~
~
,
where k js a scalar quantity. To determine k, we scalarmultiply both sides of the above e q u h n by All and obtain
The backcab ruk can be verified in a straightforward manner by expanding the vectors in the Cartesian coordinate system (Problem P.28). Only those interested in a general proof need to study this example.
I
18
.*
VECTOR ANALYSIS / 2
'
r
I
.
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' 1
1
1
Since All.D = 0, so k = 0 and
,
D=B(A;IIC)C(AIIiB),
.
which proves the backcab rule jnapmuch as All C = A
?
k
.
I,
ORTHOGONAL COORDINATE SYSTEMS
24
2
,
f
<

.. T
wh
and All B = A B.
Division by a vector is not $ejnt!d, a r d expr?ssions sluch as k/A and B/A are meaningless.
..
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the the
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,
We have indicated before that although the laws of'electromagnetism are invariant with coordinate system, solution of practical problert~srequires that the relations syshm ;~ppropsi:~~c t o lllc derived from thcsc laws he cspscwil in ;I cocrrdi~~.~lc geomctry of thc given problems. I:or cxampl~.,if wc are to determine the electric field at a certain point in space, we gt least need to describe the posilion of the source and the locatlon of this point in p coordinate system. In a threedimTnsiona1 space a point can be located as the intersection of three surfaceq. Assume that the three families of surfaces are described by u, = constant, u, = coqstant, and u, = constant, where the u's need not all be lengths. (In the familiar Cartesian pr rectangular coordinate system, u,, u,, and u, correspond to x, y, and z respectively.) When these three surfaces are mutually perpendicular to one another, we have an orthogonal coorriinate system. Nonorthogonal coordinate systems are not used because they complicate problems. Some surfaces represented byau,= constant (i = 1,2, or 3) in a coordinate system may not be plancs; they may b~ curvkd surhccs. Let a,,, a,,, and a,,, bc the unit vectors in the three coordinate directions. They are callqI the buse vectors. In a general righthanded, orthogonal, curvilinear coordinate system, the base vectors are arranged in such a way that the following relations ace satisfied: a,, x a,, = a,, a,,,
>j
,
I I
a,, = a,,
, aK3:xa,, = a,*.
I
(221a) (221 b)
(221c)
'
These three equations are not alliqde&4dent, as the spbificaiion ofone automatically I implies the other two. We have, qt cqube, ' . 1
and
.
Any vector A can be written as the sum of it$ cornpanents in the three orthogonal directions, as follows: " \!I 1
'
I
11 perfo
19
24 / ORTHOGONAL COORDINATE SYSTEMS
where the magnitudes of the three components, A,,, A,,, and A,,, may change with the location of A; that is, they may be functions of u,, u,, and u3. From Eq. (224) the magnitude of A is
=A.B. .nd
A = /A/ = (A;$
:
S/A are I a
I ,re invariant , he relations ricite to the clcctric iicld f the source ma1 space a at the three
/ I.
1.1
Solution: Firs? we writeA, B and C in the orthogonal coordinates (a,, u,, u3):
+
1
A A + A + aU3A,,, A= =a a,,,Aul f s %,Au2 %,A,,,
\ I
+
= AuIBul + Au2Bu2 Au3Bu3,
b, A x B = (%,A,,,+ ~ , , , A , , 4, %1/1,,1)x (a,,,B,, + a,,,B,,, + a,,,B,,,) sx %I( A112u143  A l l , ~ u42 ~~ ~ l , L ( ~ u L ~ ~ , ,~, ~ ~ Il j ,(A,,,LY~~ , , ~  Au2Bul)
~wtr

co~npl~cllw
mate system be the unit ecrors. In a :vectors are

(221a) (221b) (221c)
(226)
in view of Eqs. (222) and (223).
the? three
I
(225)
Example 23 Given three vectors A, B, and C, obtain the expressions of (a) A . B, (b) A x B, and (c) C (A x B) in the orthogonal curvilinear coordinate system ( ~ 1 ,U2,"3).
L
1
+ A ; ~+ A;~)'".
1
Equations (226) and (227) express, respcctivcly, the dot m d cross products of two vectors in orthogonal curvilinear coordinates. They are important and should be remembered.
f i
k
c) The expression for C (A x B) can be written down immediately by combining the results in Eqs. (226) and (227).
!
1 *
,
Eq. (228) can be used to prove Eqs. (218) and (219) by observing that a permutation of the order of the vectors on the left side leads simply to a rearrangement of the rows in the determinant on the right side.
In vector calculus (and in electromagnetics work), we are often required to perform line, surface. and'volume integrals. In each case we need to express the
(2 24) i
20
VECTOR ANALYSIS 1 2
1 %
I I
I I
r
I
differential lengthchange coriesponding to .a differe'ntial ~ h a n g ein one of the coordinates. However, some of the coordinatesy say u,. (i = 1, 2, or 3), may not be a length: and a conversion factor is needed to convert a differential change du, into a change in length dt, : (I/',= It, (Ill,,
( 2 20) where hi is called a metric coefic(en( a i d may itrclf L a fynction of u,, u,, and u,. For example, in the twodimensional polar coordinat& (u,, h)= (r, 4),a difrcrential change d4 (=du2) in 4 (= 7 l Z ) cor&onds to a differential lengthct~anged l = r rl$ (h2 = r = u , ) in the a + ( =  : ~ ~ , )  ~ ! i r ~ A~ cdirc~tc,l I i o ~ l . dillhc~~ii:~i I C I I ~ I ~ ~I  I ~ : I I , ~ c i l l :III arbitrary direction can be wrilkn i\s tbevec~orsum uflllc c.oo;pollmt l c t ~ ~ cIlanses:t tll
S1
Ill1
dt = a,,, dCI + aU2dt2 + a,, d&
. .
'pthe 1. = [ ( h , du!)' + (h2 h,)' + h, J u , ) ' ] ' ~ ~ . (232) he differential volume dv formed by differential coordinate changes du,, du,, and in directions a.,, a,, and a,, respect~velyis (dll dt2d!,), qr
a
Thi
24.1
Later we will have occasion to express the current or flux flowing through a differential area. In such cases the crdsssectional area perwndicular to the current or flux flow must be used, and it is convenient to consider the Pifferential area a vector with a direction normal to the surface; ihat is.
. I 1 II ds.= a, ds.
For instance, if current de;sity J ia ndt pcrpcndiculnr to a diLrcntinl area of a ,nagnitude ds, the current, dl, flowing through ds must be the c k p o n e n t of J normal to the area multiplied by the area. Usingthe notation in Eq. (234), we can write simply
( ~ r ; =J
JS
I
:I
,
.$
J a,&. (235) In general orthogonal curvilinear coordinates, the 4iff(rentiil area ds, normal to the unit vector a,, is . :. ' ds1 = a,,(dd, dt,) ' I
,_I
,
' This e is the symbol of the vector d.
J
9 ,
.
$ 4
,
I
(
! I
,I
.
,I
.I
1
1 .
,
1
Car
A
r
spec syst
,
me of the comay not be a nge dui into a
t 4t
,
. .
24 1 ORTHOGONAL COORDINATE SYSTEMS
21
or
r ,Z
. :$
F
Lf
c
(236)

1 ,;
Similarly, the differential area normal to unit vectors a,, and a,, are, respectively, u,, and u,. . a differential ge d l 2 = rd$ change in an gth changes:'
ds, = a,,(h,h, du, du,)
(237)
and
I ds, = au,(hlh2du, du,). I
(2 30)
Many orthogonal coordinate systcrns exist; but we shall only be concerned with the three that are most common and most useful:
I. Cartesian (or rectangular) coordinate^.^ 2. Cylindrical coordinates. 3. Spherical coordinates. These will be discussed separately in the following subsections. Cartesian Coordinates
24.1
ig through a
A point P(xl, y,, z , ) in Cartesian coordinates is the intersection of three planes specified by x = x,, y = y,, and z = z , , as shown in Fig. 29. It is a righthanded system with base vectors a,, a,, and a, satisfying the following relations:
the current area a yector
3
a, x a, = a, a, x a= = a, a, x a, = a,.
\
.ea of a magJ normal to write simply
n35) ormal to the
'
The position vector to the point P ( x , , y,, 2,) is
A v e w in Cartesian coordinates can be written as

The term "Cartesian coordinates" is preferred because the term "rectangular coordinates" is custornarlly associated with twodirnensio'na~geomztry.
I
I
22
VECTOR ANALYSIS 1 2
.
6
.
I
Fig, 29
Cartesian coordinates.

.
The dot product of two vectors A and B is, from E& (226), A t B = AxBx + AyBy
+ A&,
(242)
and the cross product of A and B i$ from Eq. (2271,
f
Since x, y, and z are lengihs themselves, all three etric coefficients are unity; that is. = h 2 = h, = 1. The expr&ons for the differe tial length, differential area, and differential volume ark Iroy E q s . (231). (236), $37). (238), and (233) respcctivcly, . .
ill
(245a) (245b) (245c)
ds, = a, dx dz ds, = a, dx d y ; ,
(246) t
,
1.: I E, t*
I.
P
Example 24
A scalar line integral of a vector field of the type
jp: F .dt' is of considerable importance in both physics and electromagnetics. (If F is a force, the integral is the work done by the force in moving from P1 to P 2 along a specified path; if F is replaced by E, the electric field intensity, then the'integral represents an electromotive force.) Assume F = a2;y + aY(3x y2). Evaluate the scalar line integral from P,(5,6) to P2(3, 3) in Fig. 210 (a) along the direct path @, PIP2; then (b) along path @, P,AP,.
Fig. 210 ' Paths of integration (Example 24).
.
s are unity; rentid area, ld (233) 
Solution: First we must write the dot product F d t in Cartesian coordinates. Since this is a twodimensional problem, we have, from Eq. (244), .
It is important to remember that dt' in Cartesian coordinates is always given by Eq. (244) irrespective of the path or the direction of integration. The direction of integration is taken care of by using the proper limits on the integral. Along direct path
aThe equation of the path PIP2is
This is easily obtained by noting from Fig. 210 that the slope of the line P I P 2 is f. Hence y = ($)x is the equation of the dashed line passing through the origin and parallel to PIP,.Since line inte intersects the xaxis at x = I, its equztion is that of the dashed line shifted one unit in the positive xdirection; it can bs
+
obtained by replacing x yith (x  I). We have, from E ~ S(247) . and (248),
SpyF .dP = Spy [ x y dx + (3x  y2) d y ] Path @
Path
t
In the integration with respect to y, the relatioq 3x = 2 y E q . (248) was usede<
+3
derived from
b) Along path @  This path has two straightline segments:
From P , to A : x = 5 , dx = 0.
From A t o P 2 : y = 3 , d y = 0 . I;. d t = 3x d x ,
Hence,
Path
We see here that the value of the line integral hepen+ on the path of integration. In such a case, we say that the vector field F is not conservative.
24.2
Cylindrical Coordinates
', .
I
' In cylindrical coordinates a point ~ ( r ,4,. , i,) is the intersection of a circular cylin
drical surface r = r , , a halfplane cpntaining the zaxis hnd making an angle 4 = 4, with the xzplane, and a pl ne frallel to the xyplane at z = 2 , . As indicated in 't Fig. 211, angle 4 is measured from the x  i u k and the base vcctor a, is tangential to the cylindrical !vrf&e. The following righthand relations apply: ,
d'"
'^'
'
I
I.
x
9 = 91plane '4
Fig. 211
Cylindrical coordmates.
Cylindrical wordinales are important for problsms wilh long line charges or curmnt,, and in places where cylindrical or circular boundaries exist. The twodimensional polar coordinates arc a special cusc at z = 0. A vector in cylindrical coordinates is written as
ri
The expressions for the dot and cross products of two vectors in cylindrical coordinates follow from Eqs. (226) and (227) directly. Two of the three coordinates, r and r (u, and u,), are themselves lengths; hence h , = h3 = 1. However, q5 is an angle requiring a metric coefficient h, = r to convert d$ to d 4 . The general expression for a differential length in cylindrical coordinates is then, from Eq. (231):
integration.
(251)
The expressions for differential areas and differential volume are rcular cylinngle 4 = 4 , indicated in v e r p a , is 1Pk 9
.
ja) (249 b) (249~) (
'
and
(
26
1
'
VECTOR ANALYSIS / 2
A typical differential volume element at a point (r, 4, 3 resulting from differential changes dr, d 4 , and dq in t& three orthogonal coordinate dir2tions is shown in Fig. 212. A vector given in cylindrical Fodrdinates can be lranlformed into one in Cartesian coordinates, and vice versa. Spppose we want to expresi A = a J r a d r n+ a,A, in Cartesian coordinates; that iq, w~ want to write A as a,A, ia,A, + a,A, and determine A,, A,, and A,. First ofall, we note that A,, the Icomponent ofA, is not changed by the transformation from cylindrical to Cartesian coor inates. To find A, we equate the dot products of both expressions of A with a,, Thus,
+
4 
' I
,
i
Ax
=.A. a, 5 Ara, ax
+ Ada,
,
A,.
i
The term containing A, disappei)rs here because a, a, = 0. Referring to Fig. 213, which shows the relative position's of the base vectors q,, a,, ar, and a+, we see that
'
Hence,
a4.p,=L"s(:+4)=:sis*. 1
*
i
I.
41 = A. cos 4  Ad sin 4 . :.
II
0
.
i

.
!'
I
!!
Fig. 213
Relations between
%,,a,, a,. and a*. 1.
I
ji. . : I'
I
I
.24 1 ORTHOGONAL COORDINATE SYSTEMS
27
f
Similarly, to find A,, we take the dot products of both expressions of A with a,: A, = A a, = Arar a,
. + A,a, .a,.
From Fig. 213, we find
.
a, a, = cos
(: 
$)
f
4
and a , . a, = cos 4.
It follows that differential is shown in
1
It is convenient to write the relations between the components oia vector in Cartesian and cylindrical coordinates in a matrix form:
Fig. 2 13, we see that
Our problem is now solved except that the cos 4 and sin 4 in Eq. (260) should be converted into Cartesian coordinates. Moreover, A,, A,, and A, may themselw be functions of r, 0,and z. In that case, they too should be converted into functions of x, y, and z in the final answer. The following conversion formulas are obvious from Fig. 213. From cylindrical to Cartesian coordinates:
y = r sin $J z = 2.
The inverse rilations (from Cartesian to cylindrical coordinates) are
%
28
VECTOR ANALYSIS I 2
,
2
,
1
:
1
:t
Example 25
.
r
,
2Express tbe yect& r
,
'
4 = a,/3 cos 4)  ia$r + azs in Cartesian coordinates.
,..
Solution: Using
0
or A = a,(3 cos2 q5
+ Zr
[email protected]) + a,(3 sin d, cos Q  2r cos 4) + n,5: 1
But, from Eqs. (261) and (262),
and sin 4 = , , i
Therefore,
1
.I
3x2
Y
.
Jq' i
, ,
I
1 ;
which is the desired answer.
i:
i
along the quartercircle showp in Fig. 2 14. I
"
1 ;
,
1 : I
J
i .I
rig, f t14 Path for line integ~pl (Exgmple 26). t

24 / ORTHOGONAL COORDINATE SYSTEMS
29
a
Solution: We shall solve this problem in two ways: first in Cartesian coordinates, then in cylindrical coordinates.
k
f.: C
I
II'
a) In ~ a r t e & z ncoordinates. From the given F and the expression for dC in Eq.
(244), we have
F . dC= x y d x  2x dy.
L
ti
+ y2 = 9(0 Ix , y I 3). Therefore,
The equation or the quartercircle ie x2
k
~
~
I

d
t 1
= +g3
=
~
 x2)3/2
1
~
~
x
~
=
~
d
0
[y~w+gsin'q 3
x

2
1:
b) In cylindrical coorrlinates. Here we first transform F into cylindrical coordinates. Inverting Eq. ( 2 5), wc Il;lvc
n
sin 4 cos (b
I
(4 i
sin 4 cos 4 0
>
i I
,
With the given F, Eq. (263) gives
L
cos 4
i
0 I
which leads to
T t 1 ; 0
F = a,(xy cos 4  2x sin 4 )  a,(xy sin 4
+ 2x cos 4 ) .
For ikpresent problem the path of integration is along a quartercircle of a radius 3. There is no change in r or z along the path (dr = 0 and dz = 0 ) ; hence Eq. (251) sinlplifies to
i
dC = a93 d4
t
:
sin 4 cos 4 0
and
~
~
~
30
VECTOR ANALYSIS I 2
.
. I
OVl
1: t
the
In this particular example, F is given in Cartesian coordinates and the path is circular. There is no compelling reasoli to solve tho problen~in onc or the other ~ co~lversin~l or v c c h ~ si ~ n dthe r o c c d ~ ~o(sollll\on rc coordinates We havc s l i o w ~tlle i n both coordinates.
r
over the surface of a closed cylinder about the zaxis specified by z = & 3 and r = 2, as shown in Fig. 215. S o l i o ~ : In connection with Eq. (234) wc noted that the direction of L is normal to the surface. This statement is actuplly~imprecisebecause a normal to a surface can point in either of two directions. No ambiguity would arise in Eq. (235), since the choice of a,, simply determines the reference direction of currebt flow. In the present case, where F . ds is to be integrated over a closed surface (denoted by the circle on the integral sign), the direction of ds is always to be taken qs that of the oir~~vard normal. Our problem is to carry out the surface integral ,
Fig. 215 jA cylindrical Surface (Exapple 2 7 ) .
24 1 ORTHOGONAL COORDINATE SYSTEMS
:ration. Along
31
over the entire specified surface. This integral gives the net outwardpw of the vector F through the enclosed surface. The cylinder in Fig. 215 has three surfaces: the top face, the bottom face, and , the sidc wall. So, ....
We evaluate,the three integrals on the right side separately. a) Top fa e. z = 3, a, = a,
F a,
= k2z = 3k2 ds = r dr d+ (from Eq. 252);
ci the path is o r the othcr c 01 mlution
lop
F a, ds =
C"So2
3k2r dr dq5 = 12nk2.
face
b) Bottom fnce.
2
=  3, all =  a z
I;?:
17 11,~ =
= 3k2
ds = r dr d 4 ;
F.
5;~,~l,,l,l i t , , ~=
12n/
which is exactly thc samc as the integral over the top face.
11or1n;il
surface can 51. since the the present he circle on .he out\c.rrrd
c) Side wall. r = 2, a, = a,
ds = r d 4 dz = 2 d 4 dz (from Eq. 252a); side wall
F .a, ds = f:3
So2"k, d 4 dz
= 12nk,.
Therefore,
$F
24.3
ds = 12xk2
+ 12nk2 + 12nk,
Spherical Coordinates
A point P(R,, dl, 4,) in spherical coordinates is specified as the intersection of the following three surfaces: aspherical surface centered at the origin with a radius R = R,; a right circular cone with its apex at the origin, its axis coinciding with the zaxis
'
VECTOR ANALYSIS / 2
Fig. 216
Spherical coordinates. 1 
and having a halfangle d = d l ;and q halfplane containing the zaxis and making an angle 4 = 4, with the xzplane. The pose vector aR at P is radialfrom the origin and is qzrite different /ram a, in cylindrical coordinates, the fatter being perpendicular to the zaxis. The base vector a, lies in the q5 F 4, plane and is t4pge&l to the spherical surface, whereas the base vector a, is the same as that in cylindrical coordinates. These are illustrated in Fig. 216. Far a righthanded system we have '
Spherical coordinates are importanr 'for problems :involving point sources and regions with spherical boundaries. When an observer isvery far from the source region of a finite extent. the latter could be considered as the origin of a spherical coordinilte system; and, as a rcsult, suitiihlc simplifying ;~ppn)xim:ttionscoi~J(lhc m;~clo.'l.ili5 i* the rcason that spllcrical ioordinatesnrc i~rcdin solving a n t c n ~ problems ~a in thc far field. A vector in spherical coordinates iswritten as '.
= a,AR ta,A,
+a A
Thc ent
ant
, ~ F O
mi
(265) or
The expressions for the dot and crgss products of two vectors in spberi~alcoordinates can be obtained from Eqs. (226) ~nd,(227). In spherical coordinates, only R(u,) is a kngfh. ?he other two coordinates, B and 4 (u2 and u,), are angles. Referring to Fig. 217, w t q e a typical differential volume element is shown, we see that m e t r i ~coekcients h, = Rand h3 = R sin B are required
(
Fig. 217 A differential volume element in spherical coordinates.
k n g an 'I.
urr
tc, .e
>her; hate3. 264a) 2 64b) 2 64c)
'df = a, d R
+ a,R dO + a,R
sin 0
(266)
The expressions for differential areas and d i k e n t i a l volume resulting from diflerentlal changes @, do, and dm in the three coordinate directions are
:s and region dilute This is he fhr
and
265)
FO: convenience the base vectors, metric coefficient$ and expressions for the differentlal volume are tabulated in Table 21. A ,VecfhgiWn in spherical coordinates can be transformed into one in Cartesian Or c~llndrlcalcoordinates, and vice versa From Fig. 217, it is easily seen that
n
nates es, e ume lired
Y = R sin 0 sin 4
(269a) (269b) (269c)
.
I
t
1
34
VECTOR ANALYSISIF
.
.
!
; ,
.
 ,
1

.. 1
\
2
.
I I
I
/
Table 21 Three Basic Orthogonal Coordinate Systems
b)
<
Cartesian Coordinates Coordinatesystem Relations
a .
( x , Y,4
P
.
4
Pz
h1 h, 3
dc
I
Spherical Coordinates (R, 8 , d
1 1.
1 dx d y &
a:
a,
1 r
1 R
r. dr
'
.
R sin 0 .
1 i
'
c;i
a~
a,
*
dl

a,
a~

Differential Volume
(r347
ax
 Yctors.
Metric Coefficients
Cylindrical Coordinates
I1
b!
'
d4' d;
R 2 sin 0 dR dU dd
Conversely, measurements in Cartesian coordinates can be transformed into those in spherical coordinates:
M
It coonl PC. cn~rd
n; ,
j
Esam ordin: Solutl! This i a poir all po functic definir in gcn produ
Example 28 The position of a point P in spherical coordinates is (8, 120°, 330"). Specify its location (a) in Cartesiali coordinates, and (b) in. cylindrical coordinatcs. Solution: The spherical coordinates df the given poipt are R
.
4 = 330".
a) I n Carresian coordinates. We use Eqs. (269a, b, c): .
'
r i
,
'4
..
* 8; 0 = 120°, and
*
!' : !
.
x = 8 s i n 1 2 0 ° c p ~ 3 3 0 " = 6 ,. . y = 8 sin 12 " &'330°=  2 8 ' ' z = 8 cos 120' F .4.
P
1
1
.
,
Hence, the location of the point is t ( 6 , 72J?, 4), and the position vector (the vector going from the origin to the point) is ,
Recgl; vector 069,
i;!
*::. ..{;;Y,$'',.:,$,
; , .., ,.
where the approximate sign (  ) ovd; 'the equal sign 'has been left out for simpliaty. !I 8
I)
'
33 1 COULOMB'S LAW
75
'
If the dipole lies along the zaxis as in Fig. 34, then (see Eq. 277)

(319)
p = aZp= p(a, cos 8

 a, sin 8)
..
R . p = R p cos 0 ,
(325)
'
(376)
and Eq. (324) becomes 2
write
E=
4m0R
( a , 2 cos 8
+ a, sin 6)
(V/rn).
(327)
Equation (3127) gives the electric field intensity of a n electric dipole in sphcri~n!. coordinates. We see that E of a dipole is inversely propoirional to the cube of the distance R. This is reasonable because as R increases, the fields due to the closely spaced q and  q tend to cancel each other more completely, thus decreasing more rapidly than that of a single point charge.
+
7
33.2
Electric Field due to n Continuous Dlstrlbution of Charge
The electric field caused by a continuous distribution of charge can be obtained by integrating i5kpcrposing) thc contribution of a n element of c h a r p over the charge clislrihulio~l.licfcr to 1 . i ~ . 3 5. whurc ; I voluinc cL;irp: tlisfrih~itioi~ is s l ~ o w i T ~ .~ I C V O I U I I I ~ ~ * C , ~ I : I II ,~K~ ( :I I . I ,II I ~I (
/ I I I I ) 11
:I I I I I I L I
i011
0 1 ~ I I CC O O I ~ ~ I I IS: II I~~CL:I: C .iIilli~c1111:11 .
clement ul' charge behaves like a point charge, the contribution of the charge p 10' in a differential volume element du' to the electric field intensity at the field point P is p
dv'
dE = a, . 4n.5,R2 We have
. h.
Fig. 35 Electric field due to a continuous charge distribution.
.I >
76
STATIC ELECTRIC FIELDS I
7
I
I
! !
or, since a, = R/R,
.,
i
I
i.
.
'
I *
I
, 3
(330) ( I
Except for some especially sikp14 ases, the vectdr triple ibtegral in Eq. (329) or Eq. (330) is difficult to carry out because, in general, all'three quantities in the integrand (aR,p, and R) change with the location of the differential volume dv'. If the charge is distributed on a,surface with a ~ u r f a c echarge density p, (C/m2), then the integration is to be carried out over the surface (not necessarily flat). Thus,
For a line charge, we have
8
1 . .
where p, (C/m) is the line charge 'density. and L' the line (not necessarily straight) along which the charge is distributed. I
Example 33 Determine the electric field inten& i line charge of a uniform density p h air.
o f I a n infinitely long. straight.
4'
Let us assume that tlje.line charge li& alopg the zfaxis as shown in Fig. 36. (We are perfectly frce. to dd . this because t\le field obviously docs not depend on how wc clcsign;~tcthc linc; 11 ?,>'irrr u r w p / ( * d cvrricir~ri/i,;nf r r r i : i ~p r i t n ! d r~r~orrlitrrc~c~.v jbr sotircc / ~ r , i t ~ /wid s ~ ~ t ~ ~ ~ r i ~ ~ ~/ ; , I~* , /~I v / ~ / /~) ! , ~~I I / .cS ~r W ~ rI 111m~ I~ ~ ~is ~(1 ~/ r rl r .~$ . r$ ~~/ ~~ ~ l i / y/ of conjusion.) The problem agks us to find the e ~ & ~ igeld c intensity at a point l', which is at a distance r from the I'ine. Since the p+blern has a cylindrical symmetry (that is, the electric field is in!dep&dent of the azlmutli angle 4). it woiild bc most convenient to work with cylijdriyd .  coordinates. h e re writ;.^^. (332) as Solution:
. ..
For the problem at hand p, is co&&nt and 5 line element dLr = dz' is chosen to be at an arbitrary distance z' from the,Grigin. It is mqit important to remember that R is the d~stancevector directed fr;h the source td'thejeld point, not the other way : . .i ii
I*
!'
:* A
,.i i :1
il 1:
I
~ ~ , s
33 1 COULOMB'S LAW
i
77
29) or
s in the
iL;', , [Cim2), t). Thus,
Fig. 36 An infinitely long straightline charge.
around. We have
R = a,,  a,:'. The electric field, dE, due to the difirential line charge element p , d t ' = p , d l is
where
dEr =
p,r dz'
4m0(r2
+ 2'2)3/2
and
dE, =
p,zf dz' 4neO(r2+ z'2)3/2'
In Eq. (335) we have decomposed d E into its components in the a, and a; directicns. It is easy to see that for every p, dz' at + z' there is a charge element p, d:' at  :', which will produce a dE with components dEr and dE,. Hence the a= components will cancel in the integration process, and we only need to integrate the dBr in Eq. ( 335a):
78
STATIC ELECTRIC FIELDS 1 3 ,
Equation (336) is an i m p o r t a ~ ~i.csult l for an inhiite linc chargc. Of course, no physical line charge is infinitely long; ncvcrtl~cless,Eq. (336) gives the approximate E field of a long straight.line charge at a point close to the line charge. 34
GAUSS'S LAW AND APPLICATIONS
!
Gauss's law follows directly from the divergence p&tulate of elect~ostatics,Eq. (34), by the application of the divergence theorem, 1t;has been derived in Section 32 as Eq. (37) and is repeated here on account of its impoytance:'
...,
I
1
Gtruss's Irr\\. osscrrs rlrtrr rlrc rortrl ool\\~rr~tl,/lrr.\(!I' tlrc~IG/icItl o r w t r ~ ~ c~/o,scd js~rr:/ircv in p e e sptrce is o q 1 ~ 1to1 the ruttrl C . / I ~ I I Y I C ~C I I ~ I O S C Y I ill tire S L I I $ I C ~ ~lil:idedby E ~ We . note that the surface S can be any hj~pothetical(nzathrr~atical)close&rJace chosen for come?lierlce: it does not have to be. and usually is not, a physical surface. Gauss's law is particularly usefui in determining he Efield of charge distributions with some symmetry conditions, such that the noynal c&lporreitt *f tlze electric ,field .srr~;luc.c.I n such c x c s tllc surfr~ccintcgral on thc irllcrlsil!~is c w l s ~ r r oocr ~ i ~ (111 cv~c~lo,sc~l left side of Eq. (337) would bc very easy to cval.u;~te,&nd Gauss's law would be a much more efficient way for fin'ding the electrik field intensity than Eqs. (329) through (333).On the ather hapd. whcn synimetiy conditions do not exist. Gauss's law would not be of much help. TLessence of applying Gauss's law lies first in the recognition of symmetry conditions, and second in thqsdtable choice of a surface over which the normal component of E resulting fro? a given charge distribution is a constant. Such n surface is referred to as a Ga~~ssiar? s u r f k c . This hasic principle s h outside the electron clouil. We obtain i h e same expression lor jL0E ds as in case (a). The total chargc e~closedis , h \.I
'
.
which follows the inverse qquare law and could have been obtained directly from Eq. (312). We observe that aurside the charged claud the E field is exactly the same as though the total chafige is concentrated on a single point charge at the center. T h ~ sis true, in general, for a spherically symmetrical charged rcglon even though p is a function of R. The variation of ER versus R is plotted in Fig. 39. Note that the formal solution s is not used. it is necessary of this problem requires only a fe% lines. If ~ a u s s law (1) to choose a differential volume dement arbitr&ily located in the electron cloud, (2) to express its vector distance R:to a field poi4t in a chosen coordinate,syitem, and (3) t6 perform a triple integration as indicated'in Eq, (329). This ir a hopelevly involved process. The moral is: T!) to apply Gdhssls.law if symmetry conditions exist for the given charge distributicn. i: , , ELECTRIC POTENTIAL:
35
'
t
h
I
.<
.
a
In connection with the null identhy in Eq. (2134 we poted that a curlfree vector field could always be expressed ,zs the gradient of'a sc, jar field. This induccs us to define a scalar electric potentiql, such that , .
1
c,
!'
(3 38)
, because scalar quantities are aasidr ;to handle thad vect r uantities. If we can determine V more easily, then E cap b&fbund by'a gradient operation, which is a straightforward process in an orthogonal~coordinatesystem. The reason lor the inclusion of a negative sign in Eq. (338) $ill be explained pr&eerltly,
5
t
a
34 / GAUSS'S LAW AND APPLICATIONS
,
81
Determine the E field caused by a spherical cloud of electrons with a volume charge density p = p, for 0 5 R i b (both po and b are positive) and p=OforR>b.
Example 36
,
Solution: First we recognize that the given source condition has spherical symmetry. The proper Gaussian surfaces must therefore be concentric spherical surfaces. We must find the E field in two regions. Refer to Fig. 39. A hypothetical spherical Gaussian surface Siwith R < h is constructed within i h s electron cloud. On this surface. E is radial and has a constant magnitude.
)In faces 'ig. 38. :d sheet
E = a,$,,
.
rls = a, d s .
The total outward E flux is
6,E .
ds = E, Js ds = ER4nR2.
The total charge enclosed within the Gaussian surface is
e = j', p du =

I
=

4n. 3
l;i&, 30 lilcclric licld Inlcnslty of' a sphcrical electron cloud (Example 36).
'
35 1 ELECTRIC POTENTIAL
,
aard the
83
Electric potential does have physical significance, and it is related to the work done in carrying a charge from one point to another. In Section 32 we defined rhe electric field intensity as the force acting on a unit test charge. Therefore, in moving a unit charge from point P , to point P , in an electric field, work must be done against ' the,jield and is equal to
I.
1
outside case (a).
'Many paths may be followed in going from P , to P , . Two such paths are drawn in P , and P , is not specified in Eq. (339). the question ~ i g310. . Since the ~ a t between h naturally arises, d o ~ iilc i work dcpund on the p t h i:lkcot! h liitlc il~ou$it i i i ; i lead Gs'to conclude fhat CV/q in Eq. (339) should not depend on the path; for. l f it did, one would be able to go from P1 to P, along a path for which W is smaller and then to come back to PI along another path, achieving a net gain in work or energy. This would be contrary to the principle of conservation of energy. We have nlrcndy alluded to the pathindependence nature of the scalar line integral of the irrotationni (conservative) E field when we discussed E q (38). Analogous to the concept of potential energy in mechanics, E q (339) represents the difference in electric potential energy of a unit charge between point P, and point P I . Denoting the electric potential energy per unit charge by V, the electric potenrial, wc have
f
11) ,..?rn L!LYl, c at the ion cvcn
\olutloll eccssary In cloud, 's)stem, ipclessly )nditions
Mathematically, Eq. (340) can be obtained by substituting Eq. (338) in Eq. (339). Thus, in view of Eq. (281),
Spy E .dP SP'' ( V V ) . =
:e vector 2es us to
(al d l )
,
r (3!I\ t l c l w ~trn~ghtluslon of
Fig. 310 Two paths Icading from P , to P , in an electric. field.
84
!
STATIC ELECTRIC FIELpS /:J' 4
:
Direction of
iwrcming V
, .
1
'
Fig. 31 1
u
of
Rclutivc direction\ and increasing V.
. ;j ,. . What we have defined in Eq, (340) is a ptentid difference (elcrrrorroric i.oltoge) between points P , and P I . It makes no moresense l;o talk about the absolute potential of a point than about the absolpte phasc of a, phasor ar thc absolutc altitude of a geographical location: a refercpceqropotential point. a reference zero phase (usmlly :I( r ; : 0). or :I rclixc~~cc zc1.11;~lli!h!c ( I I S I I : I;II I( ~S Y IwuI) I U \ I S ~lirst hb sl)chyiliccl.111 most ( 1 ~ 1not t i l l ) G I S ~ S ,tlic ~ c ~ ~ o  p i l;o1111 ~ l ~ i ~s l4Lc11 ~ ~ i ;I! ; ~i11Ii11ity. l W I I ~ (lie II I V I ~ ~ L  I I C ~ zeropotential poinl is 1101 ;\I inlini!y, it shoulri bcspcci/icully statctl. We want to make two mqre about Eq;(338). First, the inclusion of the negative sign is necessary in ordoe to copform with tile conve&on that in going oyoiirst the E field the electric potenlial V ina.eo.ser. For instance. when a DC battery of a voltage Vo is connected between two parallel qondu~tingplates. as in Fig. 31 1, positive and negative charges;cu$ulate, respectively,: 011 the top and bottom plates. The E field is directed from pdsitive to negative chitrges. whiic thc potential incrcnscs in the opposilc direction. Sccqn?.'dx know from &&tiqp 2 5 when we dclincd the gradient of a scalar field that the dirkction of V V isiiormyl to the surfaces of constant V . Hence, if we use directed flili'(iljir~e.sor .strennili+s~to indicate the direction of the E field. they are everywhere perp&diculitr to
[email protected] lines and ~qsiporoliid .SUI~~JC~'S.
,
,
.
.' I !
35.1 Electric Potential due to a Charge Distribution
: !'
1
I.
!
I
I'
The electric potential of a at 2 distance R fr* a point charge q referred to that at infinity. can be obtained rGdily from Eq. (343: , I , . d
.
which gives
I (342) '
I
r
,
This is a scalar quantity and on, besid~sq,~only the distance R. The potential difference between gny two p o i q t s ' ~ , and PI at bistalfes R, and R , , respectively.
, I
?
I'
.
!:
I. i; 1 , 'a !
i.
,
.i t : I
.
!4 ;I
8
35 / ELECTRIC POTENTIAL f
\$'
85
,,\ /'
/ 1
/
I I
I \
,/
/' I I\
\
\
9
\
'\'\
'/
'%.
I
R~
'\
1
\
I I
I
/
/'
/
I
/ $
, '/
/
Fig. 312 Path of integration about a point charge.
$
from q is
:I little siirprisi~~g a1 first, si~lcclJ2; ~ n d 1 1 ~ no1 ' l l ~ i srcsult m ~ ;~ppc:~r y y lie on tile same radial line through q, as illustrated in Fig. 3 12. However, the concentric circles (spheres) passing through P, and P, are equipotential lines (surfaces)and Vp2 1.,is tllc S : I I :IS ~ I i
$4
.
'.J 5
i

f1
i
r
whe
38 ELEC* DIELECTRIC
Bea
the
1
diffe ' mils
38 / ELECTRIC FLUX DENSITY AND DIELECTRIC CONSTANT
tensity surface
99
For a surface S bounding a volume V, the net total charge flowing out of V as
a result of polarization is obtained by integrating Eq. (387). The net charge remaining ' within the volume V is the negative of this integral. L
(383)+
0
Q = $sP.
a,, ds
I  .
(384)' 11ca 111
surjace field
which leads to the expression for the volume charge density in Eq. (384). Hence, when the divergence of P does not vanish, the bulk of the polarized dielectric appears ;; !x chdrged. However, since we started with an electrically neutral dielectric body, thc total charge of the body after polarization must remain zero. This can be readily verified by noting that
or
Total charge =
(3rx
llilldl.ly
:\I)
ll.ll;~cb:
2gallve et total .vhere 11 :ma1 to ld
(389)
.8 ELECTRIC FLUX DENSITY AND DIELECTRIC CONSTANT
Ih.c:~\~sc :I ~xil;~ri/.ctl tliclcc~ric~ i v c srise t i ) :I vol~~nlc c h ; ~ r ~tlcnsily c I),,. wc espcct ihc electric lield inicrlsity duc t o a given sourcc distribution in a ciiclcctric to be different from that in free space. In particular, the divergence postulated in Eq. (34) must be modified to include the effect of p,; that is, 1 . 5' v.32 = d P p,). (390) €0
(386) I
=$s~.a,,dsJvv.~d~=~,
ton$.
on\~der II
$ p,,, (1s + Jv p, du
where the divergence theorem has again been applied.
. ;I1 1111
'
+
,
Using Eq. (3841, we have
vector (387)
Wc now define a new fundamental field quantity, the electricJlux density, or e1ectr.i~ displacement, D, such that
p,'
r~Ii\ti\
9 and a
olve only
The use of the vector D efiiiblcs us to write il divergence rc1:rtion hctwccn the electric field and the distribution of j i v e charges in any medium without the necessity of dealing explicitly with the polarization vector P or the polarization charge density p,. Combining Eqs. (391) and (392), we obtain the new equation
!:i
I
;
!'
/
i'.. {f'
,.
1
1
,
I r. .i where P is the voiume depsity C$ file; char& ~ ~ u a k i &13s;!I and (35) are the f in any medium. two fundamental governing diffqreqfial equation9 ' f ~ &ctrosta;ics Note that thel~ermittivityaf fred space, ro, does not! appaar e$licitly in these two equations. $ The corresponding integral fyrrq ok Eq. (393) is obtaiqed by taking the volume integral of both sides. We have : '
*
;
' i
So CO I
sol >fin, I
'
pk
Equation (3951, another f o p of G&SS'S law, states! that ;he total outward ql the e/ecfric d i s p l a c o n e ~(or, ~ simply, tlfe totol of~t~varif e 1 e e ; ~ i c ~ uouer s ) ally closed suguce is equul to the total pee charg4 +enclosedin the~surjaqe.Asb,s been indicated in Section 34, Gauss's law is rqos{ju eful in dete&jninajthe elect& field due to :' charge distributions under symmetry c nditions. When the dielectric p r o p e r f p the medium $re linear and isotropic, the polariztion is directly proporticjpa] to the electric geld iqensity, and the proportionality constant is independent of the direction of the field. We write
6 b(
:i
,
, # = EOXA
,
(396)
where x i is a dimensionless is linear if X. is
dielectric medium of space
is a dimensionless constant know as t& relotiye or the dielectric constunt of the medium. The coefiicient c the absoluIe
[email protected] jvity (often called simply permittiuit~)of the medium and 1s p,+sured in meter (F/m). Air has a dielectric constant of l.WQ59; habcei,tb ennittivity h us$lly taken as that of free space. The dielectric constants gf s f ! t t h e r ma.r&i a r included in a table in :' . Appendix B.
gsdil
I
..
!4
,
.
f L
: I
; ! .!, '. , ' A tensor would be required to represent 1the d&tric swceptibility~the 't 'f i , :, . : 111
I
i
>
is anisotropic.
v1
,the 5t
the
38 / ELECTRIC FLUX DENSITY AND DIkLECTRIC CONSTANT
..
:irc 111c ledium. cse two
ha
'
volume
( 394)
(395)
flu of close~f t11c;llod d u i p IIC.
t'
)rap
Nolo 1h;il r , can fur~clionofspnm coordinnlcs. Ilr, is indcpcndcnt ofposition, the medium is said to be homogeneous. A linear, homogeneous, and isotropic medium is called~asimple medium. The relative permittivity of a simple medium is a constant. I
Example 311 A positive point charge Q is at the center of a spherical dielectric shell of an inner radius Ri and an outer radius R.. The dielectric constant of the shell is c,. Determine E, V,D, and P as functions of the radial distance R.
Solution: The geometry of this problem is the same as that of Example 310. The conducting shell has now been replaced by a dielectric shell, but the procedure of solution is similar. Because of the spherical symmetry, we apply Gauss's law to . 5nJ E and D inthiee regions: (a) R > R.; (b) Ri I R 5 and (cJ X R,
The situation in this region is exactly the same as that in Example 310. We have, from Eqs. (368) and (369),
(3 96) I ~ ~ I U J I I
f spacc
1
v1 = . Q
4nc,R From Eqs. (397) and (399), we obtain
(397) and (398) )rlstallf
sin* ILL
of fsw ~ b l e1.
101
.
,
P,, = 0 .
The applicaiion of Gauss's law in this region gives us directly
Dielectric shell
(4
:
. .t .
:
, 7
Fig. 520 'Vicld yuri:itions of a point charge +& at the center of a dielectrjc shell (Example 31 1,) a
..
I
38 I ELECTRIC FLUX DENSITY AND DIELECTRIC CONSTANT
1 J3
c) R < Ri Since the medium in this region is the same as that in the region R > R., the application of Gauss's law yields ;he 'same expressions for E,, DR,and PR in both regions:
To hnd v,, we must add to V2 at R = Rithe negative line integral of E,,:
The variations of e,E, and DR versus R are plotted in Fig. 3?O(b). The difference ( D R  roE,) is PR and is shown in Fig. 3PO(c). The plot for V in Fig. 320(d) is a composite graph for V,, V,, and V, in the three regions. We note that DR is a continuous curve exhibiting no sudden changes in going from one medium to another and that P R exists only in the dielectric region. It is instructive to compare Figs. 320(b) and 320(d) with, respectively, Figs. 318(b) and 318(c) of Example 3! 1. From Eqs. (383) and (384) we find
on the inner shell surface;
have a
,m.
Q
.
on the outer shell surface; and
=
 i .I j
('
aK.(K'l'R2) = 0.
(3lW)
Equations (3107), (3108);and (3109) indicate that there is no net polarization volunle charge inside the dielectric shell. However, negative polarization surface charges exist on the inner surface; positive polarization surface charges, on the outer
....
!i 1
i
<
3
1
t
\
I
1t 1
104
STATIC ELECTRIC ~ 1 @ \ 0 ~ 1 ' 3
' :,
II
3' I1
.
,
Ir
qic14lwf!Strengths of ~ o ~ c ! ~ & nMaterials ~on
Table 31

+
t.
.
Material li )
Air (atma~pher/cjyessure) Mineral oil I Polystyrene ., Rubber
~ i c l k c t r i c~ t r c n ~(tVl /~I ~ ) .._I . 3 x 10"
2
i 1
i
L
Glass Mica
; l$ix lo6
,
20x10~ 25 )( lo6 30 x 106 200 x' 10"
7
I
,
surface. These surface charges produce an electric fisld intensity that is directed radially inward, thus reducing tbe.E field in region 2 due to the point charge + Q . at the center. I
_
>.
3
t
38.1
Dielectric Strength
We have explained that an g\ecf$c field causes small Pisplacements of the bound charges in a dielectric material, teiulting in pola&ation. If the electric field is very strong, it will pull electrons ~oppletelyout of the m lecules, causing permanent dislocations in the molecular strtidure. Free char'ges w 11 appear. The material will become conducting, and larle cbirents may res$lt. Tbis phenomenon is called a dielectric breakdown. Thi mqiqhA electric figld hten$Ity that a dielectric material can withstand without breakdqwn is the dielecttic st ength of the material. The approximate dielectric streng hs f some common8'$ubst~ce~are given in Table 31. f 9 The dielectric strength of a mqtendmust not be c&nfustid with its dielectric constant. A convcnicnt number t o ~ c ~ k i n h ci \ r that tlsc iliu cctric strcnpth of ; i ~ ra t thc atmospheric prcwirc 1s. 3 k Y / i ~ j ~ nWIJCII . L ~ I C S I C C ~ I ICI,IICIC& I J I L C I I ~ A L Y cx~ccclsLll~s value, air breaks down.;,+4ass'ivivg~isnization taked'place, and sparking (corona discharge) follows. Charbe tendglto concentrate at sharp ~ o i n t s In . view of Eq. (367). the electric field intensity in t l ~ cimmediate vicinitxd shgrp points is higher than that at points on a surface with i j small curvature. This is 'the principle upon which a lightning arrester works. Disgbay$e through the s h a ~ p oinL of a lightning arrester fact that the electric prevents damaging discharg~ethmugh nearby ~ b i e c t  The , field intensity tends to be of a charged conductor point near the ~ s u with a larger curvature is in the follo$ng
4 1
.
1
.
:
7
i*
I
I
Example 312 Consider twp sph:erical conductsjrs~wjfh rqdii b, and b2 (b, > b,), which are connected by a coducdng wire. The ~$stange of,separation between the conductors is assumed to be,yeqIarge cornpafed to ' sothat the charges on the spherical conductors may
[email protected] as u a i f ~ ~ ~ l y ~ i s t r i b uAt etotal d . charge Q
P I
J
II (
I
! !
*
ii ; \ .
39 6 ELECTF
39 1 BOUNDARY CONDITIONS FOR ELECTROSTATIC FIELDS
Fig. 321 Two connected conducting spheres (Example 312).
is deposited on the spheres. Find (a) the charges on the two spheres, and (b) the electric field intensities at the sphere surfaces.
directed rg,: + Q
Solution a) Refer to Fig. 321. Since the spherical conductors are at the same potential, we have
r
:b ~ , , , d ! is vfry i'nlancnt :rial will cailcd a
Inatcrial id. The :ble 31. :onstant. ir at the :cds this ona dis. (367), han that which a arrester : e l e p inductor
bz > bl), ween the :s on the :harge Q
Hence the charges on the spheres are directly proportional to their radii. But, since we find
b2 b1 Q and 2Q. bl b2 b l + b2 b) The electric field intensities at the surfaces of the two conducting spheres are QI=
+
The electric field intensities are therefore inversely proportional to the radii, being higher at the surface of the smaller sphere which has a larger curvature. 39 BOUNDARY CONDITIONS FOR ELECTROSTATIC FIELDS
Electromagnetic problems often involve .hedia with different physical properties and require the knowledge of the relations of the field quantities at an interface between two media. For instance, we may wish to determine how the E and D vectors
wh
;
ill!(
L ,, 0
/
t I
Fig. 322 'An inttrface between two m e d i ~ .
or
1
l
ady know the bouqdary conditions that must interface. These copditi~nshave been given ln Eqs. (366) and (367). w e now consider an interface bl)tween&general media shown in Fig. 322. Let us construct a smqll path abc% with sidesab and c m media 1 and 2 respectlvely, both being parallel to t& inte'face and equal tb AW; ~ ~ u i t i (38), o n whlch is assumed to be valid for regions containing discoqtihuouq~medja,is applied to this path.' If we let s~desbc =ilo Ah ,Tpproach zero, .their pontribufions to the line integral of E around the path can beiqglected. We hbve ',6" *f
 whe
I
i
4
$
ebcde
'
E.~~=~~.Aw+~~.(Aw)=E~,Aw~E~,I;w=O. i I,
Therefore i
(31 10) i
, .
1 ;=.
if1
4,
1,
,I ,.
I
)
(31 1 1 )
€2
In order to find a relation betdecn the norrnal.com~pnemsof the fields at a boundary, we construct a small illldx with its top ?ace ip medium 1 and bottom face in medium 2, as was ia Fig. 322. Thdfaces ave an area AS, and the height of the pillbox Ah is vanish nglj( small. ~ ~ ~ lGayfs's ~ i nlawk Eq. (395) to the
:i ,, I
See C. T Tal. On the presentation of A U ~ U S1972. ~
$1
., . I
r,
we 1 or
i
I( cc elec
ii
\
Ii
 ,xa
elec
fi L
I
axwe$ theory: ~ r o c e e & o~f (he IEEE, vol. 60. pp 936945.
"1
iI
wh~
I
which states that the tanyenrid cQppotaent of un E je4d is c~ntinuousacrosr an interface. Eq. (3110) simplifies to &. (366) if one of the media is a conductor When media 1 and 2 are dieleftr&'$ith'.permittivities r; and r, respectively, we have '
,
t!
t
t
lnte
the becc
'1
1
.
Soil1 on: intc necc
39 / BOUNDARY CONDITIONS FOR FLECTROSTATIC FIELDS
137
4 i
pillbox, we have
,
I
+
$Dado = (Dl .an2 D, .anl) AS = a n 2 . ( D l  D,) AS
i
'
= pS As,
(3112) where we have used the relation a,, = a,,. Unit vectors a,, and a,, are, respectively, outward unit normals to media 1 and 2. From Eq. (3112) we obtain
1or
(3113a) 1
. . .
t
(3113b) lat must .n given :I
[email protected]$ respech1c i to this the line \\
infurr. When we have
a
where the reference unit nornml is omvard from rnediurn 2. Eq. (31 13) states that the norntal component of D field is discontinuous across an interface where a surface charge existsthe amount of discontinuity being equal ro the surface charge density. If medium 2 is a conductor, D, = 0 and Eq. (3ll3b) becomes Dln = 6lEln = P s ' (31 14) which simplifies to Eq. (367) when medium i is free space. When two dielectrics are in contact with no free charges at the interface, p, = 0, we have
Recapitulating, we find the boundary conditions that must be satisfied for static electric fields are as follows:
tin
Tangential components, El, = E,,; Normal components,
(3111) /
:Ids a bottom .and (j)to the
~
I
I
936945,
a,,
(Dl  D,) = p s .
Example 373 A lucite sheet ( E , = 3.2) is introduced perpendicularly in a uniform electric field E, = a,E, in free space. .Determine Ei, Di, and Pi inside the lucite. Solution: We assume that the introduction of the lucite sheet does not disturb the original uniform electric field E,. The sitnation is depicted in Fig. 323. Since the interfaces are perpendicular to the electric field, only the normal field components need be considered. No free charges exist.
.
*
t
108
.
\.d\
lr
STATIC ELECTRIC F Lw;/ 3 r
i
I
(
' .
1,
. . i
:I,
it
1
Boundary condition E ~(31;14) :
at the left ikt&rfagcgives
Di = a,D, = a,D, or . 1
$
D, = axcoEo:
There is no change in electric )
[email protected] density rcrpss thp i n t & & The electric field intensity inside the ludte shbet iv
$3. 4
',I 1 , Fi=Di=Di=fl
,e
cOE,
.
+2
The polarization vector is zbro oqtside the lucitd sheet(Po = 0). Inside the sheet, 11 1 ~ ,  ~ , i ~ ~ ~ ~$j)LO~O ~ = a . ( l
=p a,b.k875r0~, I
(q/rn2)* 2
!
Clearly, a similar applicati~ndffhe boundqry @ondipon Eq, (3114) on the right interface will yield the prigifid 4 and D,, in ihdfree /paceon the right of the lucite sbcct. Dacs thc solution ofi(hi$ firohlcrn clli~n if tb d p i n a l clcctric field i v n o t uniform, that is, if E, = q,E y)') , :t ,
$
.
, * *
.
i
,
?' , ;:. f
1
i
r
<
.
: . ?
;. . ( (;
*
* .
1%
:
I
.
ii
: 1
,
"[; i ' ;. Fig. 324 ~dund+ conditions at the interfwe betw two dielectric media (E~a$~lc.?/!cl). , 1 4
t
,
:. II
,I
It
r
81
It1 t
1:
I
,
,i
.i t. 4
; .
!
r
310 /
I
C~PACITANCEAND CAPACITORS
109
I
,I,!,
Example 314. do dielectric media with Permittivities c, and c, are separated by a chargefred boundary as shown in Fig. 324. The electric field intensity in mediuni 1 at the point F, has n magnitude E l and makks an angle cr, with the normal. Determine the magnitude and direction of the electric field intensity at point P , in medium 2. 1
Solution: ~ w o , k ~ & t i o nare s needed to solve'for two unkpowns E2, and E,,,. After E2, and E,,, have been found, E, and will follow directly. Using Eqs. (31101 and (3 115), we have . . . $ ,. E , sin a, = El sin ai (3117) and I n (3118) c2E2COS at = €,Elcos b l . ," '
I,
h
i
Division of Eq. (31 17) by 5q. (31 18) gives 
tan a , The magnitude of E, is
hc ight J'. luclte d is not
By examinink Fig. 324, can you tell whether r , is larger or smaller than s,?
7~
From Section 36 we understand that a conductor in a static electric field 1s an equipotential body and that charges deposited on a conductor will distribute themselves on its surface in such a way that the ilectric field inside vanishes. Suppose the potential due to a charge Q is V. Obviously, increasing the total charge by some factor k would merely increase the surface charge density p, everywhere by the same factor, witlroutaffecting the charge distribution because the conductor remains an equipotential body in a static situation. We may conclude from Eq. (357) that the potential of an isolated conductor'is directly proportional to the total charge on it. This may also bd seen from the fact that increasing V by a factor of k increases E =  V Y by a factor o f t . But, from Eq. (367), E = anp,/c,; it follows that p, and consequently the total charge Q will also increase by a factor of k. The ratio Q/V therefore
110
STATIC 'ELECTRIC FIELDS 1 3
remains unchanged. We write
where the constant of proportionality C is called the capacitance of the isolated conducting body. The capacitance is the electric charge that must be added to the body per unit increase in its electric potential. Its SI unit is coulomb per volt. or farad (F). Of considerable importance in practice is the capacitor which consists of two conductors separated by free space or a dielectric medium. The conductors may be of arbitrary shapes as in Fig. 325. When a DC voltage source is connected'between the conductors, a charge transfer occuqs, resulting in a charge + Q on one conductor and  Q on the other. Several eiectric field lines originating from positive charges and terminating on negative charges are shown in Fig. 325. Notc tha! the field S. arc cq~iipotcntiillsti~f;lces. lines are perpendicular to the conductor S L I ~ ~ ~ I C Cwhich Equation (3121) applies here if V is taken to mean the potential ditrerence between the two conductors, V, ,. That is,
The capacitance of a capacitor is a physical property of the twoconductor system. It depends on the geometry of the conductor6 and on the permittivity of the medium between them; it does not depend on either the charge Q or the potential difference V , , . A capacitor has a capacitance even when no voltage is applied to it and no free charges exist on its conductors. Capacitance C can be determined from Eq. (3122) by either ( I ) assuming n V,, and dotermining Q in terms of V,,. or (2) assuming a Q and determining V , , in terms of Q. At this stage, since we have not yet
Fig. 3125 A twoconductor
. capacitor.
'
i
1
(3121) '
isolated d to the volt, or of two may be 3etween nductor ch wgcs he field urfaccs. )elween
.r
I,
I I
i
3
studied the methods lor solving boundaryvalue problems (which will be taken up in Chapter 4), we find C by the second method. he procedure is as follows: 1. Choose an apprdpriate coordinate system for the given geometry. 2. Assume chargks 4 Q and Q on the conductors. 5 3. Find E from Q by Eq. (3114)' Gauss's law, or other relations. 4. Find V I 2by evaluating
!
r
( 3  1 ~
lductor
from the condbctor carrying  Q to the other carrying 5. Find C by taking the ratio Q/V,,.
.
+ Q.
.
Example 315 A parallelplate capacitor consists of two parallel conducting plates of area S separated by a uniform distance d. The space between the plates is filled with a dielectrlc of a constant permittivity E . Determine the capacitance.
Solution: A cross section of the capacitor is shown in Fig. 326. It is obvlous that the appropriate coordinate system to use is the Cartesian coordinate system. Following the procedure outlined above, we put charges + Q and  Q on the upper and lower conducting plates respectively. The charges are assumed to be uniformly distributed over the conducting plates with surface densities + p s and  p , , where
y of the
~tcnrial :d to It d from . or (2) not yet
"
From Eq. (31 14), we have
which is constant within the dielectric if the fringing of the electric field at the edges of the plates is negiectcd. Now
P
Cross sectlon of a parallelplate capacitor (Example 315).
Fig. 326

Therefore, for a parallelplate capacitor,
.
I
I
which is independent of Q or V,,. For this problem wecould have started by assuming a potential difference V12 between the upper and lower plates. The electric field intensity between the plates is uniform and equals
The surface charge densities at the upper and lower conducting plates are + p , and  p , , respectively, where, in view of Eq. (367),
,
. 1'; 2 p., = e b , = c. d Therefore, Q = psS = (cS/(i)Vl2and C = Q/V12= sS/d. as before. .Example 316 A cylindrical capacitor consists of an inner conductor of radius a and an outer conductor whose inner radius is b. The space between the conductors is filled with a dielectric of permittivity c, and the length of the capacitor is L Determine the capacitance of this capacitor. Solutioti: W e use cylindrical coordinates for this problem. First we assume charges Q and  Q on the surface of the inner conductor and the inner surface of the outer conductor, respectively. The E field in the dielectric can be obtained by applying Gauss's law to a cylindrical Gaussian surface within the dielectric a c r < b. (Note that Eq. (31 14) gives only the normal comporlent of the E field at a conductor surface. Since the conductor surfaces are not planes here, the E field is not constant in the dielectric and Eq. (31 14) cannot be used to find E in the u < r < b region.) Referring to Fig. 327, we have
+
Fig. 327. A cylindrical capacitor (Example 3 16).
310 1 CA~ACITANCEAND CAPACITORS
113
i
Again we neglect the fringing effect of the field hear the edges of the conductors. The potential difference Between the inner and outer conductors is
Therefore, for a cylin+ical capacitor,
tp;
and
We could not soive this problem from an assumed Vuhbecause the electric field is not uniform between the inner and outer conductors. Thus we wouid not know how lo cxprcss li and Q in lcrms of Vuhuntil wc l c m c d how to solve such a boundaryvalue problem. Example 317 A sphcriwl capacitor consists of an inncr conducting spllcrc nl radius R iand an outer conductor with a spherical inner wall of radius R,. The space inbetween is filled with a dielectric of permittivity s. Determine the capacitance.
:hargcs c outer ~ p i j i n g' . (Note ,urfacc. in the ferring
Solution: Assume ciiarges + Q and  Q, respectively, on the inner and outer conductors of the spherical capacitor in Fig. 328. Applying Gauss's law to a spherical Gaussian surface with radius R ( R i < R < R,,), we have
'Fig. 328 A spherical capacitor (Example 3 17). '
>.
114
i STATlC ELECTRIC FIELDS 1 3
tiC?+2yfq =
Fig. 329
+

4

capacitors.
Series connection of
Therefore, for a spherical capacitor,
310.1 Series and Parallel Connections of Capacitors
Capacitors are often combined in various ways in electrkhcuits. The two basic ways are series and parallel connections. In the series, or headtotail. connection shown in Fig. 329.' the external terminals are from the first and last capacitors only. When a potential difTerence or electrostatic voltage V is applied, charge cumulauons on the conductors connected to the external terminals are + Q and  Q. Charges will be induced on the internally connected conductors such that +Q and Q will appear on each capacitor independently of its capacitance. The potential differences across Ihc individual capacitors arc Q / C , , Q/Cz,. . . , (LIC,,.anti
where C, is the equivalent capacitance of the seriesconnected capacitors. We have
In the parallel connection of capacitors, the external terminals are connected to the conductors of all the capacitors as in Fig. 330. When a potential difference V is applied to the terminals, the charge cumulated on a capacitor depends on its capacitance. The total charge is the sum of all the charges.

' Capacitors, whatever their actual shape, are conventionaily represented in circuits by pairs of parallel
bars.
,I 310 / CAPACITANCE AND CA'PACITORS I
F
on of
(3 127) 1Y
b
+
I
d
Fig. 330 Parallel of capacitors.
connection
Therefore the equivalent capacitance of the parallelconnected capiicitors is /
\LO
>IC
lnnef 'n mrs ",.,y. nuldt~ons rirgc\ w111 Q will ~ITtrences
\\it
Iiavc
C,,= C ,
+ C2 + . . . +C,,.
We note that the formula for the equivalent capacitance of seriesconnected capacitors is similar to that for the equivalent resistance of parallelconnected resistors and that the formula for the equivalent capacitance of parallelconnected capisitors is similar to that for the equivalent resistance of seriesconnected resistors. Can you explain this?
Example 318 Four capacitors C , = 1 pF, C2 = 2 pF, C, = 3 pF, and C, = 4 / I F are connected as in F i g 331. A DC voltage of 100 V is applied to the external terminals ab. Determine the following: (a) the total equivalent capacitance between terminals ub: (b) the charge on each capacitor; and (c) the potentla1 difirence across each capacitor.
( 3  128)
nected to krence V cis l r r s
of pdrallel 100 ( V )
Fig. 331 A cornblnatlon of capacitors (Exxmple 3 18).
\ '
116
STATIC ELECTRIC FIELDS I 3
Solution
31 1
AND F
a) The equivalent capacitance C,, of C , and C , in series is
The combination of C 1 2in parallel with C3 gives C l Z 3= C 1 2+ C 3 = 9 (PF). The total equivalent capacitance C, is then
b) Since the capacitances are given. the voltages can be found as soon as the charges have been determined. We have four unknowns: Q,, Q,, Q,, and Q,. Four equations are needed for their determination. Series connection of C , and C 2 : Q I = Qz. Kirchhoff's voltage law, Vl + V2 = V3: Kirchhoff's voltage law, V3
+ V4 = 100:
 +  .
Qi
I.
,
Qi'Q3 =
c, cz c3 Q3  +  = 100. c, c4

Q4
Series connection at d : Q2 + Q 3 = Q4. Using the given values of C,, C,. C,. and C4 and solving the equations. we obtain
c) Dividing the charges by' the capacitances, we find

\
d
Q4
V4 =  = 47.8 (V).
c4
These results can be checked by verifying That V, 100 (V).
+ V2 = V, and that
V3 + V4 =
1
I
4
In Section 35
I
11 L t
117
311 1 ELECTRdSTATlC ENEI'GY AND FORCES
i
+.ihdicated that electrlc potential at a polnt in an eiectrlc field 1s ,.
I
the work req$& rb bring a unit positive charge from mfinlty (at reference reropotential) to that bolnt. In order to bring a charge Q, (slowly, so that kinetlc energy and radiation effects may be neglected) from lnfinlty againrt the field of a charge Q, in free space to a distance R12, the amount of work required is
~ h charges c Four equa
r
This work is stored in the assembly of the two charges as potential energy. Combining Eqs. (3 130) and (3131), we can write
Now suppose adother charge Q, is brought from.infinity to a point that is R , , from Q1 and R,, from Q,; an additional work is required that equals ,. \\ r: obtain
hq.
(3133) and W, in Eq. (3130) is the potential energy. The sum of AW in stored in the asserhbiy of the three charges Q,, Q,, and Q,. That is.
W,,
We can rewrite W, ifi the following form:
.
("
+ Q 3 4 7 r ~ ~ R ~, ~ T ~ E ~ R ~ ~ = f(Qlv1
+ QZV2 + Q3V3).
(3 135)
III Eq. (3135), V,, tHe potential at the position of Q,, is caused by charges Qz and Q,; it is different frbm the V, in Eq. (313n in the twocharge case. Similarly. V2 and V, are the potentials, respectiyely. at Q2 and Q, in the threecharge assembly. Extending thii procedure of bringmg in ddditional charges, we arrive at the followins general expression for the potential energy of a group of N discrete point
,
I
charges at rest. (The purpose of the subscript a on of an electric nature.) We have
CC; is to denote that the energy is
1I I
where V,, the electric potential at Q,, is caused by all the other charges and has the following expression:
.
V,=
(3137)
(I*hl
Two remarks are in order here. First, We can be negative. For instance, W2 in Eq. 0130) will be negative if Q, and Q, are of oppos~tesigns. In that case, work is done by the field (not against the field) established by Q , in moving Q, from Infinity. Second, &, in Eq. (3 136) represents only the lnteractlon energy (mutual energy) and does not ~ncludethe work requlred to assemble the ~ndividuaipolnt charges themselves (selfenergy). Example 319 Find the energy required to assemble a uniform sphere of charge of radius b and volume charge density p. Because of symmetry. it is simplest to assume that the sphere of c h a k e is assembled by bringing i ~ pa si~cccssionof sphsric;lI i:~yersof thicl\i>ess, / R , ~ c t l!~ili)llnv o l ~ l l Cl ~I I ; I ~ ~ dcnrity C he p, I\[ :I r:ldila H s h o r n in Fig. 332, the potential is
Solution:
Q VR =.!L.
47i~~R where QR is the total charge contained in a sphere of radius R: Q , ~= pinlhc\for r > b'!
1
a) Determine V d;d E at a distant point P ( R , 0, (b). b) Find the equations for equiporential surfaces and streamlines C) Sketch a famil? of equipotential lines and streamlines. (Such an arrangement of three charges is called a linear electrostatic qimdrupolr.) I*
P.313 A finite line chsrge of length L carries a uniform line charge density p , a) Determine V iil thr plane bisecting the line charge. b) Determine E from p, directly by applying Coulomb's law. c) Check the answer in part (b) with VV.
measurable 1
atom, as
P.314 A charge Q is distributed uniformly over an L x L square plate. Determine I/ and E at a point on the axis perbendicular to the plate, and through its center.
). Find the
1, 1, 0) will
P.315 A charge Q is distributed uniformly over the wall of a circular tube of radius b and height h . Determine V and E on its axis 4)
)pended at pi.iced on !r redchcd 1
'
D r tn ~ le quliaL .I ~terof the
at a point outside the tubs, then
b) at a point inside the tube.
.
*ti
,
r
P.316 A simple clasdcsl model of an aton1 consists of a nucleus of a posit~vecharge k ~ surrounded by a spherical electron cloud of the same total negative charge. ( N is the atomic number and e is the electronic charge.) An external electric field E,, will cause the nucleus to be displaced a distance r,, from the center of the electron cloud, thus polarizing the atom. Assuming a uniform c h ~ ~ e  d i s t r i b u t i owithin n the electron cloud of radius b, find r,,. P.317 Determine the work done in carrying a Z(pC) charge from P,(2, 1.  1) to P,(8. 2, in the field E = a,y + a,,x a) along the parabola s = ~ J J ~ , b) along the straight line joining P, and P 2
 1)
r ~
130
STATIC ELECTRIC FIELDS I 3
P.318 The polarization in a dielectric cube of side L centered at the origin is given by P = P,(a,x + a,y + a,z). a) Determine the surface and volume boundcharge densities. b) Show that the total bound charge is zero.
P319 Determine the electric field intensity at the center of a small spherical cavity cut out of a large block of dielectric in which a polarization P exists. P.320 Solve the folloiving problems: a) Find the breakdown voltage of a parallelplate capacitor, assuming that conducting plates are 50 (mm) apart and the medium between them is air. b) Find the breakdown voltage if the entire space between the conducting plates is filled with plexiglass, which has a dielectric constant 3 and a dielectric strength 20 (kV,'mm). c) If a 10(mm) thick plexiglass is inserted between the plate.;, what is the maximum voltage that can be applied to the platCs without a breakdown?
P321 Assume that the z = 0 plane separates two lossless dielectric regions with E , , = 2 and 3. If we know that E, in region 1 is a,2y  a,.3.u a,(5 + z), what do we also know about E, and D, in region 2? Can we determine E, and D, at any point ip region 2? Explaln.
E , ~=
+
_
._ 
,
P.322 Dctcrtninc t i ~ cbound:~rycondirions for thc tangential and the normal components of P at an interface between two perfcct diclcctric tncdia with dielcctric constants E , , 2nd E , ? . P.323 What are the boundary conditions that must be satisfied by the electric potential at an interface between two perfect dielectrics with dielectric constants E,, and E,, ? P.324 Dielectric lenses can be used to collimate electromagnetic fields. In Fig. 334, the left surface of the lens is that of a circular cylinder, and the right surface is a plane. If E, at point P(r,, 45", 5 ) in region 1 is a,5  a,3, what must be the dielectric constant of the lens in order that E, in region 3 is parallel to the xaxis?
Fig. 334 Dielectric lens (Problem P.3 24). P325 The space between a parallelplate capacitor of area S is filled with a dielectric whose permittivity varies linearly frome, at one plate (y = 0) to E , at the other plate (y = d). Neglecting fringing effect, find the capacitance.
PROBLEMS
I
131
,by P =
P.326 Consider the earth as a conducting sphere of radius 6.37 (Mm). a) Determine its ~ a ~ a c i t a n c e . b) Detcrminc thimhximum charge that can exist on it without cawing a breakdown of the air surroundirlg it.
out of a
P.327 Determine the capacitance of m isolated cbnducting sphere of radius h that is coated . . with a dielectric layer ~f uniform thickness d The dielectric has an electric susceptibility %
nducting
P328 A capacitor cbnrists of two concentric spnerical shells of radii R, add Ro. The space between them is filled with a dielectric of relative pkrmittivity c from R, to b(R, < b < R.) a r d another dielectric of relative permittivity 26, from b to R,. a) Determine E and D everywhere in terms of An applied voltage I/.
,lit(\ with
h) Dctcrminc thc c:~pacitancc.
11:11j.
n vduge
=
2 and
.w about r
.>ncnt, f 1d E r * %ti31ar an 4. the left
, st point ; in
order
n xric whose Keglecting
P.329 Assume that the outer conductor of the cylindrical capacitor in Example 316 is grounded, and the inner conductor is maintained at a potential Vn. a) b) c) d)
Find the electiic field intensity, E(a), at thcsurface of the inner conducror. With the irrner radius, b, of the outer conductor fixed, find a so that E ( n ) is minimized. Find this minimum E(ul. Detcrrhine the capacitance under the conditions of part (b).
P.330 The radius of the core and the inner radius of the outer conductor of a very long coaxial transmission line are ri and r, respectively. The space between the conductors is filled with two coaxial layers of dielectrics. The dielectric constant4 of the dielectrics are E,, for ri < r < b and E , ~for b < r < ro. Determine its capacitance per uriit length. P.331 A cylindrical capacitor of length L consists of coaxial conducting surfaces of radii i i and r,,. Two diebctricmcdii~of dillcrcnl diclvctric copstants r,, and r,, fill the space between the conducting surfaccs as shown in Fig. 335. Detcrmihe its capacitance.
I


A
Fig. 33 5 A cylindrical capacitor with two dielectrlic media (Problem P.331).
P332 A capacitor consists of two coaxial
[email protected] cylindrical surfaces of a length 30 (mm) and radii 5 (mm) and 7 (mm). The dielectric material between the surfaces has a relative permittivity E, = 2 + (4/r), where r is measured in mm. Determine the capacitance of the capacitor.
,
132
STATIC'ELECTRIC FIELDS I 3
t
P.333 Calculate the amount of electrostatic energy of a uniform sphere of c h a m with radius b and volume charge density p stored in the following regions:
a) inside the sphere, b) outside thesphere. Check your rekits with those in Example 319. P.334 Find the electrostatic energy stored in the region of space R z 6 around an electric dipole oT moment p.
[
I
P.335 Prove that Eqs. (3149) for stored electrostatic energy hold true for any twoconductor capacitor.
,:
i
P.336 A parallelplate capacitor of width s,length L. and separation d is partially filled with a dielectric medium of dielectric constant t.?m shown in Fig: 336. A battery of I,, volts is connected between the piates. a) Find D. E, and p, in each region. b) Find distance r such that the electrostatic cnergy storcd in each region i r the same.
Fig. 336 A parallelplate capacitor (Problem P.336). P.337 Using the principle of virtual displacement, derivc m expression for the force between two point charges + Q and Q separated by a distance x in free space. 1
P.338 A parallelplate capacitor of width IV. length L. and separation d has a solid d~electric slab of permittivity r in the space between the plates. The capacitor is charged to a voltage Yo by a battery, as indicated in Fig. 337. Assuming that the dielectric slab is withdrawn to the position shown, determine the force acting on the slab a) with the switch closed,
b) after the switch is first opened.
\
.... . x .L
Switch
Fig. 337 A partially filled parallelplate capacitor (Problem P.338).
'
.
1
i "
with rad~us
<
I ,
I
an electric
)conductor filled w ~ t ha 011s IS con
INTRODUCTION
41

.ce between
d diclectric ~ItagcV, by
hi position
Electrostatic problems are those which deal with the effects of electric charges at rcst. These problem,d can present themselves in several different ways according to w h a t is initially kiwwn. T h c solutioli usually calls for ~ h determination c of electric potential, electric field intensity, and/or electric charge distribution. If the charge distribution is given, both the electric potentilll and the electric field inlensity can be found by the formulas developed in Cha ter 3. In many practical problems, however, the exact qharge distribution is not everywhere, and the formulas in Chapter 3 cannot be applied directly for finding.the potential and field intensity. For instance, if the charges at certaln discrete points in spaco and the potentials of some conducting boclics arc givcn, it is rathcr difficult LO find the distribution of surface charges on the conducting bodies and/or the electric field intensity in space. When the cobducting bodies have bohdaries of a simple geometry, the method of images may be used to great advantage. This method will be discussed in Section 44. In another type: of problem, the potentials of all conducting bodies may be known, and we wish to find the potential and field intensity in the surrounding space as well as the distribution of surface charges on the conducting boundaries. Differential equations must be solved subject to the appropriate boundary conditions. The techniques far solving partial aiiferctjai equations in the various coordinate systems.wil1 be discussed in Sect;& 45 through 47. *
.
.
,
42 POISSON'S AND LAPLACE'S E Q U A T I O ~ ~
In Section 38, we pointed out that Eqs. (393) and (35) are the two fundamental gpverning dificrcntirll cquations for clectrostatics in any medium. These equations , are repeated below for convenience. Eq. (393): Eq. (35):
134
SOLUTION OF ELECTROSTATIC PROBLEMS 1 4 "' ' '
'"?
The irrotational nature of E indicated by Eq. (42) enables us to define a scalar electric potential V, as in Eq. (338).
In a linear and isotropic medium, D = rE, and Eq. (41) becomes V.cE=p.
(44)
Substitution of Eq. (43) in Eq. (44) yields
where E can be a function of position. For a simple medium$ that is, for a medium that is also homogeneous, E is a constant and can then be taken out of the divergence , operation. We have

 .
.
In Eq. (4G), we have introduced a new operator, V2, the Luplnciun operutor, which stands for "the divergence of the gradient of," or V V. Equation (46) is known as Poisson's equution; it states that the Laplacian (the divergence of the gradient) of V equals  p/e ji,r u simple tnediutn, where e is the permittivity of the medium (which is a constant) and p is the volume charge density (which may be a function of space coordinates). . Since both divergence and gradient operations involve firstorder spatial derivatives, Poisson's equation is a secondorder partial difkrential equation that holds at every point in space where the secondorder derivatives exist. In Cartesian coordinstcs,
and Eq. (46) becomes
Similarly, by using Eqs. (286) and (2102). we can easily verify the following expressions for V2V in cylindrical and spherical coordinates. Cylindrical coordinates:
4
'
'r
% '
~ 4  2 '1
PO IS SON*^ AND LAPLACE'S EQUATIONS
135
I
(43)
. .
<
1
*
d'V
(49)
The solut~onof Poidon's equation in three dimensions subject to prescribed boundary conditions is,.ih beneral, not an easy task. At points ifi a kimple medium where there is no free charge, p = 0 and Eq. (46) reduces to
(43)
Y
(45) 1 mcdiLm livergence
which is known as Laplace's eqr~atiotr.Lapiace's equation occupies a very important position in electrom~netics.It is the governing equation for problems involving a set of conductors, sbch as capacitors, maintained at different potentials. Once V is found from Eq. (41 o), E can be determined from  VV, and the charge distribution on the conductor sdrfaces can be determined from p, = € E n (Eq. 367). .
(46) /
\, ..dl known as adient) of urn (which t l of space
tor.
:la1 Jerivaat holds at an coordi
!'I ; 
I Example 41 The .two plates of a parallelblate capacitor are separated by a distance d and mainiained at potentials 0 and V,, as shown in Fig. 41. Assuming negligible fZlnging effect at the edges, determine (a) the potential at any point between the plates, and (b) the surface charge densities at the plates.
a) Laplace's equation is the governing equatidn for the potential between the plates since p = 0 therq. Ignoring the fringing effect of the electric field is tantamount to assuming that tHi field distribution between the plates is the same as though the plates were infidtely large and that there is no variation of V in the x and z directions. Equation (47) then simplifies to
/
(47)
n Ilowh
where d2/dy' is used instead of d,;!'. 1 
k
(48)
since y is the only space variable.
't Fig. 41 A parallelplate capacitor (Example 41).
t
136
SOLUTlON OF ELECT .ubr''!C
i
PROBLEMS / 4
f
Integration of Eq. (411) with respect to y gives
;
dV
,
.
 = C1,
dy where the constant of integration C, is yet to be determined. Integrating again. we obtain v = Cly C,. (412)
+
I
I
1
Two boundary conditions are required for the determination of the two constants of integration:
I
Substitution of Eqs. (413a) and (413b) in Eq. (412) yields immediately C , = V,/d and C, = 0. Hcnce thc potmllid at m y point I. bctwern 111~. plates is. from Eq. (412),
The potential increases linearly from y = 0 to y = d.
b) In order to find the surface charge densities, we must first find E at the conducting
,
plates at y = 0 and y = d. From Eqs. (43) and (414), we have
\
which is a constant and is independent ofy. Note that the direction of E is opposite to the direction of increasing V. The surface charge densities a t i h e conducting plates are obtained by using Eq. (367)' I(,
>
. 11;  1'
:I,,
*
E
At the lower plate,
'7
At the upper plate,
Electric field lines in an electrostatic field begin from positive charges and end in negative charges.
1
t
Ekample 42 D e t e h i n e the E field both i n ~ i d ehnd outside a spherical cloud of R 5 b and p = 0 electrons with a unibrm volume charge density p =  p , for 0 i for R > b by solving~Poisson'sand Laplace's equations for V. I
g again,
(412) mstants
Solution: We re$ that this problem was dblveh in Chapter 3 (Example 36) by applying Gauss's aw. We now use the sam problem to illustrate the solut~onof onedimensional Poisson's and Laplace's equ tions. Since there are no variations in 0 and 4 directio?s,'tue are only dealing with functions of R in spherical coordinates.
3
a) Inside the cloud,; 1.
(4 13a)
(44 3b) .ly C, = IS, from
J
t O S R I b , p y PO.
In this region, Poisson's equation ( V 2 y terms from kq. (49), we have
a/@
 p h o ) holds. Dropping i l Z O and
which reduces to
(@PI)
Integration of E4, (416) gives ~duct~ng The electric field intensity inside the electron cloud is (415)
opposite nductmg
Since Ei cannot be infinite at R = 0, the integration constant C, in Eq. (417) must vanish. We obtain
b) Outside the cloud,
E i =  a ,   RP,0 360
01Rsb.
R2b,p0.
Laplace's equaiion holds in this region. We have V2 Y. = 0 or
n \
; and
Integrating Eq. (4191, we obtain end
(418)
138
i
SOLUTION OF ELECTROSTATIC PROBLEMS 1 4
The integration constant C , can be found by equating Eo and E, at R = b, where there is no discontinuity in medium characteristics. c2 Po b ,
b2
t
i i
kO
from which we find pOb3 C2 = 3e0
and
(422)
4 3 ELECT
P
Eo = a
R ' O b 3
3c,R2 '
R R ~ .
(423)
Since the total chargc coniaincd in the clcctron cloud is
Equation (423) can be written as
1
.
which is the familiar expression for the electric field intensity at a point R from a point charge Q. Further insight to this problem can be gained by examining the potential as a function of R. Integrating Eq. (417), remembering that C , = 0, we have
It is important to Aote that C; is a new integration constant and is not the same as C,. Substituting Eq. (422) in Eq. (420) and integrating, we obtain
However, C; in Eq. (426) must vanish since V,is zero at infinity (R + a). As electrostatic potential is continuous at a boundary, we determine C; by equating l/i and Voat R = b: p0b2 ; b'+C  pob2 1  3c0 660
1
43 1 UNIQUENESS Q+ ELECTROSTATIC
..or
. .*
'
5
.
'and, from Eq. (425),
( 4  23)
n
(424) t R from
itial as a
(425)
same as
(426)
r'
elec~ anu
:T/T
r
: podZ. = ,
,

(427)
2%
C: in &q. (428) is the same as V in Eq. (3i42), with p P
(422)
..
c;
C
We see that
139
i'
4
3
SOLUTIONS
= po.
I
11
UNIQUENESS OF irh ELECTROSTATIC SOLUTIONS
43
111 L I I C Lwu rcla~ivcl~. siniplc cxa~nplcs~n t l ~ eIds[ section, wc ob1.1' ' lned t i x solutions by direct integration. In more complicated situations other methods of solution must be used. Befoie these methods are discussed. it is important to know that n solutioti o j Poissoft's eq~ution(of which Laplace's equation is s special cxci r h r rurisjies rile giaen hoiiniiaq. c~oiiilirioi~s is ii wiipzrr .solsiioti. This statement is cnlied the u~liqtienesstheorem. The implication of the uniqueness theorem is that a solution of an electrostatic problem with its bou~ldsryconditions is tile o ~ l i !poisihle . s ~ / ~ i t i o t i irrespective of the method by which the solution is obtained. A solotion obt:lined even by intelligent guessing is the only correct sollrtion. The import:lnce of this theorem will be appreciated when we discussthe method of ima;~ e ins Section 41. To prove the uniqueness theorem. suppose a volume r i s bo~indedoutside by a surface So, which may be a surface at infinity. Inside the closed surhce So there are a number of charged conducting bodies with surfaces S,, S,, . . . , S n at specified potentials, as depicted in the twodimensional Fig. 42. Now :lssumr th:ll. contrnry to the uniqueness theorem, there are two solutions, V, and V,, to Poisson's equation in 5 :
.
140
SOLUTION OF ELECTGO7FTIC PROBLEMS I 4
..,
\
Also assume that both Vl and V, satisfy the same boundary conditions on S , , S,, . . . , S , and So. Let us try to define a new difference potential
From Eqs. (4L29a) and (429b), we see that V, satisfies Laplace's equation in
s
O n conducting boundaries the potentials are specified and I/, = 0. Recalling the vector identity (Problem 218),
V . (,/'A) = ,/'V . A and letting f = V, and A
=
+ . Vj'. I\
(432)
VV,, we have
where, because of Eq. (431). the first term on the right side vanishes. Integration of Eq. (4 33) ovcs n vnlumc r yiclds
.
where a, denotes the unit normal outward from r. Surface S consists of So as well as = 0. Over the large surface S,, S,, . . . , and S,. Over the conducting boundaries, So,which encloses the whole system, the surface integral on the left side of Eq. (434) can be evaluated by considering Soas the surface of a very large sphere with radius R. As R increases, both V , and V, (and therefore also I/,) fall off as 1/R; consequently. VV, falls off as 1/R2, making the integrand (5VV,) fall off as 1/R3. The surfacc area So, however, increases as R2. Hence the surface integral on the left side of Eq. (434) decreases as 1/R and approaches zero at infinity. So must also the volume integral on the right side. We have
v,
Jr
..
jVv,12 do = 0 .
(435)
Since the integrand IVV,I2 is nonnegative everywhere, Eq. (435)can be satisfied only if IVI/,I is identically zero. A vanishing gradient everywhere means that T/, has the same value at all points in z as it has on the bounding surfaces, S,, S,,.. . . , S,, where V, = 0. It follows that I/, = 0 throughout the volume z. Therefore V, = V2, and there is only one possible solution. It is easy to see that the uniqueness theorem holds if the surface charge distributions ( p , = E E , = E JV/Jn), rather than thc potentials, of the conducting bodics are specified. In such a case, VV, will be zero, which in turn, makes the left side of Eq. (4 34) vanish and leads to thc same conclusion. In fact, thc uniqucncss thcorcrn applies even if an inhomogeneous dielectric (one whose permittivity varies with position) is present. The proof, however, is more involved and will be omitted here.
L
'
F
i 44

$ ! I1
44
,
141
t
b METHOD OF I M A ~ E S. There is a cladi(o[%l~ctro~tatic problems with boundary conditions that appear to be difficult to satis? if the governing Laplace's quation is to be solved directly, but the conditions oh the bounding surfaces in these can be set up by appropriate image (equiva1ent)kharges and the potential distributions can then be determined in a straightforward danner. This method of replac{hg bounding surfaces by appropriate image charges in lieu of a formal solution of Laplace's equation is called the method of images. . Consider the case of a positive point char&, Q, located at a distance d above a . large grounded (zeropotential) conduc~ingplane, as shown in Fig. 43(a). The problem is to find thk. potential at every point above the conducting plane ( y > 0). 'IIlc formal proccdurc for doing 50 would b~ L O sdvc Ldplacc's c q ~ ~ u t l oinn C:rrtca~,m . coordmates .I\*
,
1 METHOD OF IMAGES
i
which must hold for i, > 0 except at the point charge. The solution V ( s ,J., z ) ahould satisfy the following conditions: 1. At all points on the grounded conducting plane, the potential is zero: that is, ,IS [cell as ge burface Eq. (434) radius R. sequently, rfa& area Eq. (434) le integral
2. At points very close to Q, the potentialapproaches that of the pomi charge alone; that is
v+~ ?0C E' ~ R
i
where R is the distance to Q.
3. At points very far from Q(x proaches zero.
 + cc, or
jCQ,y
z
& a),the potential ap
e satisfied hat V , has (2:.
. . ,s,,,
Grounded plane conductor
e V, = V,,


\
(a) Physical arrangement.
eft
SL..
0
of
.
;s theorem
x i e s with ~ittedhere.
Fig. 43
(Image charge) I
Point charge ;rnd grounded plilne condt~ctor.
(b) Image charge and field lines.
142
SOLUTION OF ELECTROSTATIC PROBLEMS / 4
c
.
I*.
i 4. The potential function is even with respect to the x and z coordinates; that is,
It does appear difficult to construct a solution for V that will satisfy all of these conditions. From another point of view, we' may reason that the presence of a positive charge Q at y = d would induce negative charges on the surface of the conducting plane, resulting in a surface charge density p,. Hence the potential at points above the conducting plane would be #
; ;
:
where R , is the distance from ds to the point under consideration and S is the surface of the entire conducting plane. The trouble here is that p , must first be determined fro111tlic houndary condilion I.(\.. 0. ): =: 0. Mmcovcr, the indic~tcdsurface intcp,il is dillicult to cvaluatc cvcn after p , ha> bccn tictcrmlncd at cvcry point on tlic conducting plane. In the following subsections, we demonstrate how the method of images greatly simplifies these problems.
44.1
Point Charge and Conducting Planes
The problem in Fig. 43(a) is that of a positive point charge, Q, located at a distance d above a large plane conductor that is at zero potential. If we remove the conductor and replace it by an image point charge  Q at y =  d, then the potential at a point P(x, y, z ) in the y > 0 region is
where R+ = [ x 2 + ( y  d ) 2 + z 2 ] l i Z , R = [x2 + ( ~ + d + ) z~2 ] l i 2 .
It is easy to prove by direct substitution (Problem P.45a) that V ( x , y, z ) in Eq. (437) satisfies the Laplace's equation in Eq. (436). and it is obvious that all four conditions listed after Eq. (436) arc satislicd. Thcrcforc Eq. (437) is a solution of this problem; and, in view of the uniqueness theorem, it is the only solution. Electric field intensity E in the y > 0 region can be found easily from  V V with Eq. (437). It is exactly the same as that between two point charges, iQ and  0,
I
.
$; 8
that is,':
.
/
i .  .
,'
!
'
2
1 of these
1
pos~tive mducting nts above, 1
+Q (a) Physical arrangemeht.

ie surface Termined e Integral tb3nlet,. of
Irkl~~ce
onductor .t a point
(437)
Fig. 44
vv with illd Q, I
1
4Q
(b) Equivalent imagecharge arrangement.
//
+t?
I
I Q
(c) Forces on charge Q.
Point chard and perpendicular conduct& planes.
spaced a distance ?d apart. A few of the field lines are shown in Fig. 43(b) The solution of this elec ostatic problem by the method of images is extremely simple; but it must be asized that the image charge is located ourside the region in which the field is to be determined. In this problem the point charges + Q and  Q cutmor bu used to ca cula~cthe V or E in the y < 0 region. As P matter of fact, both V and E are zero in he y < 0 region. Can you bxplain that? It is readily seed that the electric field of a line charge p, above nn infinite conducting plane can be found from p, ;ind its imiigc  p , (with tile condticting plane removed).
t
: I  3 A pusilric p o i ~ ~d1i11.g~ l Q is I O C : L L C ~ .LL d~hliillcc~ d l 1111d i f i , reipectivel~,from two grounded perpendicular conducting halfplanes, as shown in Fig. 44(a). ~ e t e r m i h ethe force on Q caused by the charges induced on the planes.
Solution: A formal solution of Poisson's equation, subject to the zeropotential boundary condition at the conducting halfplahes, would be quite difficult. Now an image charge  Q in the fourth quadrant would make the potential of the horizontal halfplane (but not that of the vertical halfplane) zero. Similarly, an image charge  Q in the second uadrant would make the potential of the vertical halfplane (but not that of the brimntal plane) zero. But if a third image charge + Q is added in the third quadrant, we see from symmetry that the imagecharge arrangement in Fig. 44(b) satisfies the zeropotential boundary condition on both halfplanes and is electricallyeqyivaknt to the physical arrangement in Fig. 44(a). Negative surface charges will be induced on the halfplanes, but their effect on Q can be determined from that of the three image charges. Referring to Fig. 44(c), we have, for the net force on Q,
2
q. (437) )n+n?s 3rotr&. <
I
I I p•
1144
SOLUTION OF El ';tTRCJ$TATIC PROBLEMS 1 4
where
I
F, =

F2 =
ax
Q 4ns,(~d,)~' Q2
4~6~(2d~)~'
F, = 4rrc0[(2d,)'Q2+ (2d2)2]31' (a"2d1
+ a,2d2).
Therefore,
F=
Q2
d2
'
The electric potential and electjic field intensity at points in the first yuodtat~tand the surface charge density induced on the two halfplanes can also be found from the system of four charges.
14.2 Line Charge and Parallel Conducting Cylinder
1 .
We now consider the problem of a line charge p, (Clm) located at a distance d from the axis of a parallel, conducting, circular cylinder of radius a. Both the line charge and the conducting cylinder are assumed to be infinitely long. Figure 45(a) shows a cross section of this arrangement. Preparatory to the solution of this problem by the method of images, we note the following: (1) The imagcmust be a parallel line charge inside the cylinder in order to make the cylindrical surface at r = u an equipotentiai surface. Let us call this image line charge pi. (2) Because of symmetry with respect to the line OP, the image line charge must lie somewhere along OP, say at point P,, which is at a distance di from the axis (Fig. 45b). We need to determine the two unknowns, pi and d,.
,I /
I
\
(a) Line charge and Parallel conducting cylinder.
\
(b) Line charge and its Image.
Cross section of line charge and its image in a parallel conducting circular cylinder.
Fig. 45
.
,:
.
.
i ,
!r
::
44 1 METHOD OF IMAGES
i
:, . * *
145
. .* 8
.i
let u$ assume that
i I
"i
(438)
At t h i ~stage, Fq. (4181 is just a trial solution (an intelligent guess), and we are not sure that it will hold tbue. We will, on the one hand, proceed with this trial solution until we find that it $ils to satisfy the boundary conditions. On the other hand, if Eq. (438) leads to a iolutian that does sitisfy dl1 boundary conditions, then by the uniqueness theored 14 is the only solurion. Our next job will be to see whether we can determine d;. . 3 The electric poiedtial at a distance r from a line charge of density p, can be obtained by integrating the electric field intensity E given in Eq. (336). c l r ~ f and nd from
.
:d from :charge
shows a n by thc c chargc mtcntial respect ?oint Pi,
Note that the refeihce point for zero potential, r,, cannot be at infinity becais: setting ro = in ~ q . ( 4  3 9 )would make V infinite everywhere else. Let ui leave r0 unspecified for the time being. The potential at a point on or outside the cylindricnl surface is obtained by addin: the contributions ofp, and pi.In particular, at a point .I on the cylindrical surface shown in Fig. 45(b),.we have
In Eq. (440) we have chosen, for simplicity, a point equidistant from p, and p , as the reference point for zero potential so that the In ro terms cancel. Otherwise. s constant term should be included in the right side of Eq. (440), but it would not affect what follows. Equipotential surfaces are specified by ri (441) r = , l ) , ihc If an equipotential surface is to coincide with the cylindrical surface (OX point P, must be louted in such a way as to mate triangles OMP, and OPM s~mllar. Note that ttwo triangles already have one common angle, L MOP,. Point P, should be chosen70 make ,L OMP, = L: OPM. We have
 = Constant.
rt =di  a . d  Constant. r
q
,
,, . *
) I r .
146
SOLUTION OF ELECTROSTATIC PROBLEMS I 4
,!
A%
From Eq. (442) we see that if
the image line charge p,, together with p,, will make the dashed cylindrical surface in Fig. 45(b) equipotential. As the point M changes its location on the dashed circle, both ri and r will change; but their ratio remains a constant that equals ald. Point Pi is called the inverse point of P with respect to a circle of radius a. The image line charge p, can thcn replace the cylindrical conducting surface. and V and E at any point outside the surface can be dcter~nincdfrom thc line charges p, and p,. By symmetry, we find that the parallel cylindrical surface surrounding the original line charge p, withadius a and its axis at a distance 0, to the right of P is also an equipotential surface. This observation enables us to calculate the capacitance per unit length of an openwire transmission line consisting of two parallel conductors of circular cross section. 1 .
Example 44 Determine the capacitance per unit length between two long, parallel, circular conducting wires of radius a. The axes of the wires are separated by a distance D. Solution: Refer to the cross section of the twowire transmission line shown in Fig. 4 6. Thc cquipotcnti:ll surfxcs of the two wircs can bc considcrcd to hnvc been generated by a pair of line charges p, and,p, separated by a distance (D  2 4 ) = d  di. The potential difference between the two wires is that between any two points on their respective wires. Let subscripts 1 and 2 denote the wires surrounding the equivalent line charges p, and p, respectively. We have, from Eqs. (440) and (4421,
Fig. 46 Cross section of twowire transmission line and equivalent line charges (Example 44).
i
t
1 t
I
r

'
and, similarly,

. 4
1
rc t
(443)
i'

.
V, =
t
 Lin

a
2nco ,
1'
I
,f..l t : . , ,.,; .
(
.
1
a
Fig. 415 Diagrams for Problem P.48.
(b) Line charge between grounded intersecting planes.,
P.49
Two infinitely long, parallel line charges with line densities pc. and  p , are located at . &  + b and z =   b f 2 2 respectively. Find the equations for the equipotential surfaces, and sketch a typical pair.

1i : i i
I
P.410 Determine the capacitance per unit length of a twowire transmission line with parallel conducting cylinders of dillkrent radii u , and o,, tllcir axcs being scp;wutcd by n distancc D (where D > a, + a,). 1 .
PAI 1 A straight conducting wire or radius tr is p;u;~llclto and nt height 11 from thc surbce of the earth. Assuming that the earth is perfectly conducting, determine the capacitance per unit length between the wire and the earth. P.412 A point charge Q is located inside and at distanced from the center of a grounded spherical conducting shell of radius b (where b > d). Use the method of images to determine a) the potential distribution inside the shell, b) the charge density p, induced on the inncr surface of the shell.
P.413 Two dielectric media with dielectric constants E , and e 2 are separated by a plane boundary at x = 0, as shown in Fig. 416. A point charge Q exists in medium 1 at distance d from the boundary. a) Verify that the field in medium 1 can be obtained from Q and an image charge Q,, both acting in medium 1.
(Image charge)
I
(Image charge) Medium 1 ( e l )
Medium 2 (62) x=O
Fig. 416 Image charges in dielectric media (Problem P.413).
i
I
i

Verify that the field in'medium 2 can be obtained $om Q and an image charge +Q,, both acting in mediud 2. c) Determine Q, and Q,. [ ~ i n t Consider : neighboring boints P, and P , In media 1 and 2 , respectively and requird the continuity of the tangential component of the Efield and of the normal coapon&t of the Dfield.) P.414 i n what way should we hodify the solution in E ~1491) . for ~ x a r n ~47 l e rfthe boundary conditions on the top, bottbm, and rlght planes in Fig. 410 are dV/an = O? PA15 In what way shollld i d m o d i f y the solution in Eq. (491) lor Example 47 if the top, bottom, and left planes in F& 4110 are grounded ( V = 0) and an end plate on the right is rnaintained at a constant potential to? P.416 Consider the rcctangulab region shown in Fig. 4. 10 as thc cross scction of an cnclosurc pli~t~?i. I'hc Icl't mid right plntcs arc yroundcd, and the top and bottom formcd I~y'rourcontlucli~~g pixto i ~ l rin;~imiiincd a1 constnht potentials v, and V, respstively. Determine the potential distribution insidc the enclosurq;
7f1ir.
P.417 Consider a metallic rektrihgular box with sidcs a and b and height c. The side walls and thc bottom surface are grounded, The top surface is isolated and kept at a constant potentla1 Vo. Determine the potential distribution inside the box.
r parall~l stance D
PA18 An infinitely long, thin, c nrluctinp circular cylinder of radius b is split in four quartcrcylinders, as shown in Fig. 417. he quartercylinderi in the second ; ~ n dfoilrtli L ~ Ii tI,
!
, .d 2 ' ,
. .
,
,.
.
.
.
,
,
.*,
.. '
. ,
.
."'
.. ., '. .L ,.;.. .,;.,.,~.!'l . , ' : , .. ' .
,;
1
?..,;.,*a:,>
'
.
; .
,
')
/.
.
, ' . . .
5 / Steady Electric Currents
51
F
.,A
,..
* . ,.>
1
,
',
.
I
'
,
,. .,! :, ,

v,, = E d
.,
.
,
..
or
E=.Vl 2
(510)
IP
The total current is .,
or
t I
,
I=JJ.~~=Js
.
I
1
,. ,
,*
_a
 , .k+rs 7
.,
,:
+,
< ,
J = ~. .
(51 1)
Using Eqs. (510) and (511) in Eq. (58), we obtain
which is the same as Eq. (59). From Eq. (512) we have the formula for the resistance of a straight piece of homogeneous material of a uniform cross section for steady current (DC).
We could have started with Eq. (59) as the experimental o h m s law and applied it to a homogeneous conductor of length C and uniform crosssection S. Using the formula in Eq. (513), we could derive the point relationship in Eq. (58). Example 51 Determine the DC resistance of 1 (km) of wire having a I(mm) radius (a) if the wire is made of.copper, and (b) if the wire is made of aluminum.
Solution: Since we are dealing with conductors of a uniform cross section, Eq. (513) applies.
a) For copper wire, a,, = 5.80 x lo7 (S/m):
G = lo3 (m),
S=7~(lO~ =) lo% ~ (m2).
We have
., .,
R,==G a,S
,~";;;
 . 
lo3 = 5.49 (n). 5.80 x lo7 x lo%
' We will discuss the significance of V,, and E more in dctail in Section 53.
I : b) For a l u k n u h i l r e , ual 1
II
= 3.54 x
*
.lo7
mi:.
.
1
I
The conductahcej{G, or the reciprocal of ri%istance, is useful in combining resistances in parallel : 1 b. S + : , G== (514) * R e ' (9. % I
(5li)
ft
d
From circuit theory %e know the following:
',
a) When resistakeb R , and R , are connectod in series (same current), the total resistance R is (5 15)
I
/
2
12)
b) When resistances R , and R , arc connected in parallel (same voltage), we have
rem~unce for steady J
(513) d applied it Using the mm) radius
53 ELECTROMOTIVE FORCE AND KIRCHHOFF'S VOLTAGE LAW .
In Section 32 we &inted out that static electric field is conservative and that the scalar line integral a! static electric i?+oGty around any closed path 1s zero; that IS, $~dt'=0.
(517)
For an ohmic material J = oE, Eq. (517) becomes
Equation (518) t1119hs that a steady current cannot be maintained in the same direction in a closed circuit .by an electrostaticjield. A steady current in a circuit is the result of the motion of c ~ a r g carriers, e which, in their paths, collide with atoms and dissipate energy in the circuit4 This energy must cdme from a nonconservative field, since a charge carrier completing a closed circuit in conservative field neither gains nor
178
STEADY ELECTRIC CURRENTS I 5
+1+
++02
+
+ &
.
.
,
2  

+ +E'Electric battery
Fig. 53 Electric fields inside an electric battery.
loses energy. The source of the nonconservative field may be electric batteries (conversion of chemical energy to ,electric energy), electric generators (conversion of mechanical energy to electric energy), thermocouples (conversion of thermal energy to electric energy), photovoltaic cells (conversion of light energy to elect'ric energy), or other devices. These electrical energy sources, when connected in an electric circuit, provide a driving force for the charge c~rriers.TllisJorcc mmifcsts itself as an equivalent irnpressed electric field irtterisity E;. Consider an electric battery with electrodes 1 and 2, shown schematically in Fig. 53. Chemical action creates a cumulation of positive and negative charges at electrodes 1 and 2 respectively. These charges give rise to an electrostatic field intensity E both outside and inside the battery. Inside the battery, E must be equal in magnitude and opposite in direction to the nonconservative E, produced by chemical action. since no current flows in the opencircuited battery and the net force acting on the cli;~rgccarriers must vanish. Thc line inlcgrnl of thc impressed ficld intensity E, from the negative to the positive electrode (from electrode 2 to electrodc 1 in Fig. 53) inside the battery is customarily called the electromotive forcet (emf) of the battery. The SI unit for emf is volt, and an emf is not a force in newtons. Denoted by V , the electromotive force is a measure of the strength of the nonconservative source. We have
.
Inside the source
The conservative electrostatic field intensity E satisfies Eq. (517).
Outside the source
' Also called electromotance.
Inside the source
combining Eqs. (5119) and (520), we have
.,
"
.
'
, = J : E ~ ir
)I
:rics (conw u o n of la1 energy ic energy), in electric ts itself as n
(519)
"=
v,,= v, v,.
(522)
In Eqs. (521) and @22) we have expressed the emf of the source as a line integral of the conservative & and interpreted it as a uqltage rise. In spite of the nonconservative nature of E,, the emf can be expressed as a potential difference between the positive and negative terml?als. This was what we did in arriving at Eq. (510). Whcn a resisto? in tile form of Fig. 52 i.p conncctcd bctwecn tcrm~nals1 '~nd2 . o l l l x battery, com~lclingthe circu~l. lie lotul c l c a r ~ cfield intcnbity (c1ectroat:rtic E caused by charge cumulation, as well as impressed Ei caused by chemical sctlon) must be used in the point form of Ohm's law, We have, instead of Eq. (58),

'
(521)
O u t s ~ d e' the source
or
a t d in char at IC field in:e equal in ;chemical me actlng 1 ldtensity [rode 1 in (emf) of ,. Denoted riservative
.
"
.
where E, exists inside the battery only, while has a nonzero value both inside and outside the source. From Eq. (523), we obtain
The scalar line integkal of Eq. (5 24) around the closed circuit yields, in view of Eqs. (517) and (519),
r = $(E
+ E,) .dP =
II . J
LIP.
Equation (525) should be compared to E q (518), which holds when there IS no source of nonconse~vativefield. If the resistor has a conductivity o, length /, and uniform crosssection S, J = I / S and the right side of Eq. (525) becomes RI. We havef =XI. (526) If there are more than x c soarc; illc!c~.;;,,.~otiveforce and more than one resistor (including the interdal resistances 01 the sources) in the closed path, we generalize Eq. (526) to
0)

' We assume the battery to have a degligible 1ntern:l
resistance; otherwlu is elfcvt must be lncluded in Eq. (526). An idcul witaye solrm in one whosc t c h i n a l Voltage is equal to ltr c n ~and l is , n d c p c n d c ~ of the current flowing through it. This Implies that an ideal voltage source has a zero internal resistance.
L
'
L
,
Equation (527) is an expression of Kirchhoff s voltuge l& It states that around a closed path in an electric circuit the algebraic sum of the emf's (volhige rises) is equal to the algebraic sum of the voltage drops across the resistances. It applies to any closed path in a network. The direction of tracing the path can be arbitrarily assigned. and the currents in the different resistances need not be the same. Kirchhoff's voltage law is the basis for loop analysis in circuit theory.
I
4
.
I
:
I,
:'
I
.
,
I i i
.
I
I
I
I
< .
1 \
L
The priwiplc ofmnrrruotion of cbergr is one of the fundamental postulates of physics. Electric charges may not be created or destroyed; all charges either at rest or in motion must be accounted for at dl times. Consider an arbitrary volume V bounded by surface S. A net charge Q exists within this region. If a net current I flows across Illc surfhce o i r l oS Illis region, the cll;~rgein 1I1e volume inus( ( / ~ ~ c ~;I( . ;I~ r;~le ~ r atI1;11 equals the current. Conversely, iSa net currunt llows across the surhce illto the region, the charge in the volume must i~~creusr at a rate equal to thtcurrent. The current leaving the region is the total outward flux of the current density vector through the surface S. We have
: i t

I
Divergence theorem, Eq. (2107), may be invoked to convert the surface integral of J to the volume integral of V .J. We obtain, for a stationary volume, S V v . ~ d v = S v z ap du.
t
.
54 EQUATION OF CONTINUITY AND KIRCHHOFF'S CURRENT LAW
,
(529)
7
:
In moving the time derivative of p inside the volume integral, it is necessary to use partial dinerentiation because p may be a function of time as well as of space coordinates. Since Eq. (529) must hold regardless of the choice of V, the integrands must be equal. Thus, we have.
I
; I i
h
S! t
(530) This point relationship derived from the principle of mnxrvation of charge is called the equation of continuity. For steady currents, charge density does not vary with time, ap/& = 0. Equation (530) becomes V.J=O. (53 1) Thus, steady electric currents are divergenceless or solenoidal. Equation (531) is a point relationship and holds also at points where p = 0 (no flow source). It means
u 1

t
i
7

#
:
I
.. ,
I .
"
.
T
C(
1
1
:
U
in
t
; 4
I
A 1
54 I EBUA~ON:?F CONTINUITY ,
,
urourid u
J~seyttul rtly closed jned, and s voltage
.
,
,
,
,
' I
,
r I
KIRCHHOFF'S CURRENT LAW
181
!
< . <
d
A/I
i
.4 that the field lines df stryimlines of steady cdrrents close upon themselves, unlike tho& of electrosiatit field intensity that orig!hate and end on charges. Over any enclosed surface, kq. (531) leads to the f o l l o ~ i n gintegral form: I $ J  d s  0 ,I (532)
which can be writted as (533) I
fphysics. est or in bounded h s across rate that ic rcgion, :current 0 ug'P,2 i
.
1
.
,
Equation (533) is a9exb;kssion of ~ i r c / & f l ' scurrrnr iuw. It states that rile o/~ehr?iic sum o/ull the out ($a junction ill on electric circuit is zero.' KirchholT's for node analysis in circuit theory. In Section 36 ~e stilted that charges introduced in the interior of a coiiductor will move to the c o d uctor surface and redistribute themselves in such a way as to make p = 0 and E ='! inside h d e r equilibrium conditions. We arc now in a position to prove this statem&nt and to calculate the time it takes to reach a n equilibrium. Combining Ohm's I&, Eq. (58), with the equation of continuity and assuming a constant n, we have
t 5 28)
ilegral of
(529)
In a simple medium, V E'= p i t and Eq. (5343 becomes
The solution of Eq. (334) is
ry to use
pace cotegrands
(530) is c 6 d
:qui
where po is the initial charge density at r = 0. Both p and po can be functions of the space coordinates, add Eq. (536) says that the charge density at a given location will decrease with time cxponentially. An initial charge density po will decay to l/e or 36.8% of its vdue in a time equal to
1
The time constant T is cdled the re/avo: 3,. : i r ~..or a good conductor such as coppera = 5.80 x 10' (S/m), r 2 so = 8.85 x lo'' (F/m) r equals 1.52 x 1019(s), a very short time indeed. The transient time is so brief that for all practical
(531) n (531) It means
'
This includes the currcnta of current generators at the junction. if any. A n ideal a n e n t genemior s one whose current is independent of its terminal voltage. This implies that an ideal current source h n an Infinite internal resistance.
'
I
purposes p can be considered zero in the interior of a conductorsee Eq. (364) in Section 36. The relaxation time for a good insulator is not infinite, but can be hours or days.
, 55 POWER DISSIPATION AND JOULE'S LAW
In section 51 we indicated that under the influence of an electric field, conduction electrons in a conductor undergo a drift motion macroscopically. Microscopically these electrons collide with atoms on lattice sites. Energy is thus transmitted from the electric field to the atoms in thermal vibration. The work An*done by it11 clectric field E in moving a charge q a distance A( is qE . (At). which co~.respondsL O ;Lpower
where u is the drift velocity. The tot;~lpower delivered to all the charge carriers in a volume dv is
which, by virtue of Eq. (57), is
Thus the point function E ..I is a power dm,siry rtndcr rtcztdycurrcnt conditions. For a given volume V, the total electric power converted into heat is
This is known as Joule's law. (Note that the SI unit for P is watt, not joule, which is the unit for energy or work.) Equation (539) is the corresponding point relationship. In a conductqr of a constant cross section, dv = ds d t , with d t measured in the direction J. Equation (540) can be written as P = S ~ E ~ / ~ J ~ ~ = V I , where I is the current in the conductor. Since V =,RI, we have (541) Equation (541) is, of course, the familiar expression for ohmic power representing the heat dissipated in resistance R per unit time.
.  .
:I>
I
. 2.
a
(6
1
j364) in
.
topically
.
'.
.
'
a
.
;ted from' n electric ) 3 power
I
)
8
t can be
nduction
~
I
.
i 1
I
f
I
/BOUNDARY
I
CONDI~~ONS F
183
I
.
I A
;
'
~ CURRENT R DENSITY
& 
56 BOUNDARY C O N D ~ T I ~ NFOR S * r: CURRENT DENSITY L
P
.A
I
k'
'
,
When current .obliqiaely posses an interface4;between two media with difierent conductivities, the cuhent density vector changis both in direction and in magnitude. A set of boundary cohditions can be der~vedfur J in a way similar to that used in Section 39 for obt~iningthe boundary conditions for D and E. The governing equations for steady currcpt density J in the absence of nonconservative energy I, sources are

dovernjng Equations for $tea& Current Density
Differential Form
lhtegral Form
(538) ricrs in a
'P
(539) ions.
or'
The divergence equation a the same as Eq. (531), and the curl equation is obtalned by combining Ohm'sd lau ( J = oE) with V x E = 0. By applying Eqs. (542) a d (543) at the interface between two ohmic media with bonductivities o, and oi, we obtain the boundary conditions for the normal Bnd tangential components of J. Without actually bonstructing a pillbox at the interface as was done in Fig. 322. we know from Section 39 that the normal rowq>onontof 'l.u diuergence1er.s oectarJe1~1 is continuous. Hence, ham V * J = 0, we have

. (540)
\~hichis tionship. ed in the
Equation (
block and defined in spherical coordinates by
R,IRSR, and 010~8,. Determine the resistance between the R = R , and R = R , surfaces.
+
P.513 Redo problem P.512, asguming that the truncated conical block is composed of an inhomogeneous material with a nonuniform conductivity a(R)= aoRl/R, where R , 5 R IR,. P514 Two conducting spheres of radii b , and b, that have a very high conductivity are immersed in a poorly conducting medium (for example, they are buried very deep in the ground) of conductivity a and permittivity e. The distance, d, between the spheres is very large compared with the radii. Determine the resistance between the conducting spheres. Hint:Find the capacitance between the spheres by following the procedure in Section 310 and using Eq. (567). P.515 Justlfy the statement that the steadycurrent problem associated with a conductor buried in a poorly conducting medium near a plane boundary with air, as shown in Fig. 59(a), can be replaced by that of the conductor and ~ t image, s both immersed in the poorli conducting medium as shown in Fig. 59(b).
Q o=o

u
*A.e
...
4.a.
d
0
.
I
A
%
i /I ' t
!
Boundary removed
.. . . , . . . . . . (a) Conductor in a poorly (b) Image conductor in conducting. conducting medium near medium replacing the a plane boundary. plane boundary. *a
I
I.U
I
Fig. 59 Steady current problem with a plane boundary (Problem P.515).
i
I !
P516 A ground connection is made by burying a hemispherical conductor of radius 25 (mm) in the earth with its base up, as shown in Fig. 510. Assuming the earth conductivity to be S/m, find the resistance of the conductor to faraway points in the ground.
Fig. 510 Hemispherical conductor in ground (Problem P.516).

P.517 Assume a rectangular conducting sheet of conductivity u, width a, and height b. A , ...
potential difference V, is applied to the side edges, as shown in Fig. 511. Find a) the potential distribution b) the current density everywhere within. the sheet. Hint: Solve Laplace's equation in Cartesian coordinates subject to appropriate boundary conditions.
1
i
j
r'
v=0
an =
sed of an iR5R1.
I
1
~mmerscd d) of conlared with spacitance 1.
I
I
a 7
Fig. 511 A conducting sheet (Problem P.517).
P.518 A uniform curreht density J = a,Jo flows in a k r y large block of homogeneous material ' .
conductor ig. 59(4, zonducting
/'
:ad y em with a lry
15).
%us 25 (mm) ~ctivityto be
fd heirL+ b. A $*.
's equation in
of conductivity a. A hole of radius b is drilled in the material. Assuming no var~ationin the : Laplace's =direction: find the nevJ current density J' in the cdllducting material. H ~ n tSolve equation in cylindrical &ordinates and note that V approaches (Jor/a)cos$ as r + a.
,
6 / Static M ngnetic Fields
61
INTRODUCTION f
In Chapter 3 we dealt with static electric fields caused by electric charges at rest. We saw that electric field intensity E is the only fundamental vector fikld quantity required for the study of electrostatics in free space. In a material medium, it is convenient to define a second vector field quantity, the electricJlx density D, to account for the effect of polarization. The following two equations'form the basis of the electrostntic model : V .U=,, (6 1) The electrical property of the medium determines the relation between D and E. If the medium is linear and isotropic, we have the simple cortstit~itiuerclurion D = EE. ' When a small test charge q is placed in an electric field E, it experiences an electric force F,, which is a function of the position of q. We have 6 2 MAGE
When the test charge is in motion in a magnetic field (to be defined presently), experiments show that it experiences another force, F,,,,which has the following characteristics: (1) The magnitude 'of I?, is proportional to q ; (2) the dircction of I;, at any point is at right angles to the velocity vector of the test charge as well as to a fixed direction at that point; and (3) the magnitude of F, is also proportional to the component of the velocity at right angles to this fixed direction. The force F, is a magnetic force; it cannot be expressed in terms of E or D. The characteristics of F, can be described by defining a new vector field quantity, the magnetic Jux density B, that specifies both the fixed direction and the constant of proportionality. In SI units, the magnetic force can be expressed as ?:J$.
.
,
(64)
.
1
I:
I
where u (m/sJ is thk vdocity vector, and B is mkasured in webers per square meter (Wb/m2) or t e d d 0,' The total eleetrohag$etie force on a charge q is, then, F = F e + F m ; t h a t i s ,; (65)
which is called Lotdm's jhrce equation. Its kilidity has been unquestionably cstabiishcd by experinlcnts. We may tonbider Fe/q for a small q as the definition for electric field intensity (as we did in Eq. 32) and Fm/q = u x B as the defining relation for magneiic t u x density B. Alternative y, we may consider Lorentz's force equation as a [undadental postulate of our elktromagnetic model; it cannot be derived from ofher postulates, We begin the stlldy of static magnetlc fields in free space by two postulates specifying the divergedce and the curl of B. From the solenoidal character of B. a vector magnetic potential is defined, which is shown to obey a vector Poisson's equation. Neit we deiive the BiotSavart law, *hich can be used to determine the magnetic field of a curf'enicarrying circuit. The postulated curl relat~onleads d~rectly to Ampire's circuital law which is particularly useful when symmetry exists. The macroscopic bffet of magnetic materials in a magnetic field can be studled by defining a magnetihation vector. Here we introduce a fourth vector field quantity, the magnetic field intensity H. From the relation between B and H, we define the permeability of the nl$terial, following which .we discuss magnetic clrcults and the microscopic behaviorlof magnetic materials. We then examine the boundary conditions of B and H at the ~pterfaceof two different magnetic media; self and mutual inductances; and magfieti; energy, forces, and torques. ,
,k
r s at rest. i quantity . it is con
o account .sis @he
(61) \
.2)
and E. If )n D = EE. :riences an
)
I
62 FUNDAMENTAL POST~JLCTESOF MAGNETOSTATICS IN FRE$ SPACE
tly), experi:characterI' F, at any s to a fixed to the com4 a magnetic :F, can be 3sityFthat ;I ur. , the
,
To study magnetostatjcs (;;teady magnetic fields) in free space, we need only consider the magnetic flux dendity iector, B. The two fundamental postulates that specify the divergence and the curl of B in free space are

' One weber per square meter or one I&
y u a l s la4 y u s s in CCS units. The earth rnagnetlc field about 4 gauss or 0.5 x lo' T. ( t ,weber is the same as a voltsecond.)
IS
198
STATIC MAGNETIC FIELDS 1 6
In Eq. (67), po is the permeability of free space
/
(see Eq. 19),'and J is the current density. Since the divergence of the curl of any vector field is zero (see Eq. 2 l37), we obtain from Eq. (67) which is consistent with Eq. (531) for steady currents. Comparison of Eq. (66) with the analogous equation for electrostatics in free space, V E = p/eO(Eq. 34), leads us to conclude that there is no magnetic analogue for electric charge density p. Taking the volume integral of Eq. (66) and applying divergence theorem, we have .
.
where the surface integral is carried out over the bounding surface of an arbitrary volume. Comparing Eq. (68) with Eq. (37). we again de$the existence of isolated magnetic charges. There are no magnetic Jow sources, and the magnetic J u x lines always close upon themeloes. Equation (65) is also referred to as an expression for the law of conservation of rnagnetic Jux, because it states that the total outward magnetic flux through any closed surface is zero. The traditional designation of north and south poles in a permanent bar magnet does not imply that an isolated positive magnetic charge exists at the north pole and a corresponding amount of isolated negative magnetic charge exists at the south pole. Consider the bar magnet with north and south poles in Fig. 6 l(a). If this magnet is cut into two segments, new south and north poles appear and we have two shorter magnets as in Fig. 6l(b). If each of the two shorter magnets is cut again into two segments, we have four magnets, each with a north pole and a south pole as in Fig. 61(E). This process could be continued until the magnets are of atomic dimensions; but each infinitesimally small magnet would still have a north pole and a south pole.
Successive division
flux lines follow er end outside the magnet, and then The designation of north and south ends of a bar magnet freely and south directions. in Eq. (67) can be obtained by integrating Stokes's theorem. We have
ics in free1 analogue ~PP~Y~Q
"
or
,
.
.
.
i ( V x B )  d ~ = ~ , , *Jf . d s !s
B  dP= poi,
(6 8) arb~trary >f ~srficd flux ..,ies
ession for I or .d ar magnet 1 polc and outh pole. magnct is vo shorter I into two as in Fig. msnsions; outh pole.
n
(69)
I
whcrc thc path C for !he line ihtcgral ia thc contour boundlng the iurlicc S, m d I is the total current t S The sense of tracing C and the dl~ectionof current flow follow the Equation (69) is a form of Ampere', c ~ r c s ~ t nlniv, l which states that the ~irculationof the lnayrietic flex doisltj! in jrrc space nroiinii oiijl closed path is equal toI po tunes the rotol current jawing through die u t j a c e bounded by the path. 4mpkre9Jcircuital law is very usefbl in determining the maqnrtlc flux density B caused by LI cuirent I when there is a closed path C around the current such that the magnitdtle cf R is constant over the path. Thc followihy is a'sunrmary of the two fundamental postulates of magnetostat~cs in free space: Free Space
Example 6 4 An infinitely long, straight conductor with a circular cross section ..of radius b carrles a 3teady current I. Determine the magnetic flux density both inside and outside the contiuctor. Solution: First we note tliat this is a problem with cylindrical symmetry and that Ampkre's circuital law can beused to advantage. If we align the conductor along the "axis, the magnetic fldx density B will be &directed and will be constant along any

t
!
circular path around the zaxis. Figure 62 shows a cross section of the conductor and the two circular paths of integration. C , and C,, inside and outside, respectively, the currentcarrying conductor. Note again that the directions of C , and C , and the direction of I follow the righthand rule. (When the fingers of the right hand follow the directions of C , and C,, the thumb of the right hand points to the direction of I.)
:
a) Inside the cortductor: 1.
Bl=agBgl,
de=a,,,r,drl,
$c, B ,  d f = So2" B,,r1
dg = 2 n r l B g i .
The current through the area enclosed by C , is
Therefore, from Ampkre's circuital law, f
Bl = .,B,l
liorlI = a, j &.
rl 5 b.
1t
(610)
C
,
b) Outside the conductor:
c
!
B2 = a,Bg2,
d t = a,r2 d 4
$B2dP=Znr,B,,. Path C 2 outside the conductor encloses the total current I. Henc?
Examination of Eqs. (610) and (61 1) reveals that the magnitude of B increases linearly with r , from 0 until r , = h, after which it decreases inversely with r,. Example 62 Determine the magnetic flux density inside a closely wound toroidal coil wilh an air core having N turt~sa n d cilrryillg ;I c~trrcI11 I. Tllc 101.0itl I I ~ I S a lncilrl radius b and the radius of each turn is u.
i 1
i I

62 1 FUI:
AMENTAL
POSTULATES
OF'MAG~~ETOSTATICS IN FREE SPACE :<
1
4 . conductor :spectively, C 2 dnd the dnd follow a i o n of I.)
n
1
.
I
.
41
1
1
(6 10) I
i1 i
Fig. 63
Currentcarrying toroidal coil (Example 62). I
Solution: Figure 63 drpicts the geometry of this problem. Cylindrical symmetry ensures that B has only a +component and is cbnstant along any circular path about the axis of the toroid; We constructa circular contour C with radius r as shown. For (b  a) < r < b a, kq. (69) leads diredtly to
+
$ B  d~ = 2nr~,= ~ , N I , where we have assumed that the toroid has an air core with permeability p,. Therefore, oNI, B = asB4 = a4 'I

2zr
(b
 o) < r < (b + a).
(612)
It is apparent that = 0 for r < (b  a) and r > (b + a), since the net total current enclosed by a contour constructed in these two regions is zero. Example 63 Determine the magnetic flux density inside an infinitely long solenoid with an air core having n closely wound turns per unit length and carrying a current I.
r'.
(61 1)
3 Increases Ir 2 . d toroidal as a mean
I
i I
1i 1
1
Solution: This problem can be s'olved in two ways.
a) As n dire&application ofAmpere'scircuita1 law. It is clear that there is no magnetic field outside of the holenoid. To determine the Bfield inside we construct a rectangular contour C of length L that is pdttially inside and partially outside , the solenoid. By reason of symmetry, the Bfield inside must be parallel to the axis. Applying Ampkre's circuital law,,we have
202
STATIC MAGNETIC FIELDS I 6
la Fig. 64
Currentcarrying long solenoid (Example 63).
The direction of B goes from right to left, conforming to the righthand rule with respect to the direction of tjle current I in the solenoid, as indicated in Fig. 64. b) A s a speciul cux of toroid. Thc straight solenoid may be regarded .as a special case of the toroidal coil in Example 62 with an infinite radius (b + cc). In such a case, the dimensions of the cross section of the core are very small compared with b, and the magnetic flux density inside the core isa~proximatelyconstant. We have, from Eq. (612).
which is;the same as Eq. (613). The &directed B in Fig. 62 now goes from right to left, as was shown in Fig. 63. 63
VECTOR MAGNETIC POTENTIAL

The divergencefree postulate of B in Eq. (66), V B = 0, assures that B is solenoidal. As a consequence, B can be expressed as the curl of another vector field, say A, su'ch that (see Identity 11, Eq. (2137), in Section 210)
The vector field A so defined is called the vector magnetic potential. Its SI unit is weber per meter (Wb/m). Thus, if we can find A of a current distribution, B can be obtained from A by a differential (or curl) operation. This is quite similar to the introduction of the scalar electric potential V for the curlfree E in electrostatics (Section 3S), and the obtaining of E from the rclation 15 =  V V . Ilowcvcr, the definition of a vector requires the s p e c i b t i o n of both its curl and its divergence. Hence Eq. (614) alone is not sufficient to define A; we must still specify its divergence. How do we choose V A? Before we answer this question, let us take the curl of B in Eq. (614) and substitute it in Eq. (67). We have
I
,
63. I V&TOR
MAGNETIC POTENTIAL
203
I
Here we digress t6 inhoduce a formula for the curl curl of a vector:
f
v2~=v(v.li)VXVXA.
(616b)
Equation (616a)' or (616b) can be regarded as the definition of V'A, the Laplacian of A. For Cartesian &ordinates, it can be rekdily verified by direct substitution (Problem P.610) that , 2 V24 = ax VIA, a , , ' v 2 ~ , az VIA=. (617) ;.! Thus for Cartesian coordinates, the ~ a ~ l a c i aofna vector field A 1s another vector field whose componehts are the Laplaclan (the divergence of the gmdlent) of the corresponding components of A. This, howevbr, is not true for other coordinate systems. We now expand V x V x A in Eq. (615) according to Eq. (616a) iind obtun
+
rule with Fig. 64. special 111such :mpared :onspt.
.
+
V(V.A)  V% = p o J .
(613) Wlth the purpose oidmpl~fyingEq. (618) ta the greatest extent possible, we choose:
/ V.~=O,/ les from
and Eq. (6 18) becomes
c~loidal.
This is a vector IJoissn's quutiotz. In Cartesian Coordinates, Eq. (620) is equivalent to three scalar Poisson's zquations:
A. such
unit is can be to the
V'A ,. =  11, J,, , (621b) V2A, =  p o J , . (621~) Each of these three equations is mathematically the same as the Poisson's equation, Eq. (46), in electrostatics. In free space, the equation
Jsq
.cr, ~..e rgence. rgel,. curl of
'
Equation (6Iba) can also be obtained heuristl~lllyfrom the vector triple product formula in E q (220) by $onsidering the del operator, V, a vector:
.
V X (V x ~j = V(V A)  (V.. V)A = V(V . A)  V2A. : Equation (619) holds 10; static magnetic fields. Modification is necessary for timevarylnp electm
magnetic fields (see Eq. 746),
,"
204
STATIC MAGNETIC FIELDS 1 6
.,
has a particular solution (see Eq. 356),
s
,
Hence the solution for Eq. (621a) is
We can write similar solutions for A, and A:. Combining the three components, we have the solution for Eq. (620):
Equation (622) enables us to find the vector magnetic potential A from the volume current density J. The magnetic flux density B can then beobtained from V x A by differentiation, in a way similar to that of obtaining the static electric field E from
 vv.
Vector potential A relates to the magnetic flux @ through a given area S that is bounded by contour C in a simple way: @=
bBds.
(623) The SI unit for magnetic flux is weber (Wb), which is equivalent to teslasquare Using Eq. (614) and Stokes's theorem, we have meter (T.m2).
(D = 1 7 ( V x A ) . d s = $ A .
dP
(Wb).
(624)
64 BIOTSAVART'S LAW AND APPLICATIONS
In many applications we are interested in determining the magnetic field due to a currentcarrying circuit. For a thin wire with crosssectional area S, dv' equals S dd', and the current flow is entirely along the wire. We have
J dv' = J S dP' = I d t ' ,
(625)
and Eq. (622) becomes
where a circle has been put on the integral sign because the current I must flow in
',
4
' L' 1 6101Si VART'S LAW AND APPLICATIONS
64
.'
v
.',
205
i:
'
N).is known as BiotSouart's law. It is a formula for determining B caused by a current I in a closed path C', and is obtained by taking the curl of A in Eq. (626). Sometimes it is convenient to write Eq. (631) in two steps.
(620) : flow in
' w e are now dealing with direct (nontimevarying).currehts that give rise to steady magnetlc fields. Circuits containing timevdrying sources may send timevarying currents along an open wire and deposit charges at its ends. Antennas are examples.
206
S T A ~ CMAGNETIC FIELDS 1 6
t, (6 32) with
which is the magnetic flux density due to a current element I dt". An alternative and sometimes more convenient form for Eq. (633a) is
=
/(,)I
(
tic'
3
x
N
)
mi
Comparison of Eq. (631) with Eq. (69) Iwill reveal that BiotSavart law is, in general. more difficult to apply than Ampere's circuital law. However. AmpCre's circuital law is not useful for determining from I in a circuit if a closed path cannot . be found over which B has a constant magnitude. Example 64 A direct current I flows in a straight wire of length 2L. Find the magnetic Hux density B at a point located at a distance r from the wire in the bisecting plane: (a) by determining the vector magnetic potential A first, and (b) by applying BiotSavart's law. Solution: Currents exist only in closed circuits. Hence the wire in the present problem must be a part of a currentcarrying loop with several straight sides. Since we do not know the rest of the circuit, Ampere's circuital law cannot be used to advantage. Refer to Fig. 65. The currentcarrying line segment is aligned with the ;axis. A typical element on the wire is dP' = a, dz'. The cylindrical coordinates of the field point P are (r, 0, 0).
Fig. 65 Currentcarrying straight wire (Example 64).
!
r (632)
64 I B I O T  ~ V A ~ T 'LAW S AND APPLICATIONS

a) By jinding B frod V x A. Substituting R
i
207
/,
into Eq. (6261, we have
I L dr' A = ~ Z K J ,/~
(633a) itive and
(03ib)
Cylindrical symmetry around the wire assures that BA,/d$ = 0. Thus, tiv
i\, in
'FS CJ. c .;
IhL
:lbeCilllg ipplying ~ o b l e m, :do not
vantage. axis. A
When r > b2, b 2 / ~ can ' be neglected in comparison with 1.
+
1 R
 (1
g
b +sin 0 sin 4' R
Substitution of Ey. (G42) in 13q. ( 6 . 41) yiclcls A = a,
*
Jni2
2nR
(1
b +sin 6 sin 4'
R
,
,
211
65 / THE MAGNETIC DIPOLE
I .C, ,+ I
oint: R in d the field 1 Fig. 68. angle 4 of
The magnetic flux density is B = V x A. Equation (2127) can be used to find
, which is our answer.
it point P.
lb2 O B/f = (a, 2 cos 0 + a, sin O), 4R3
(643)
At this point we recognize the sin~ilaritybetween Eq. (644) and the expression for the electric field intensity in the far field of an electrostatic dipole as given in Eq. (349). T o exanline the similarity further, we rearrange Eq. (643) as
(640) lement on !
the a, ( 0 .39)
1011
r
TL' ,is
1)
is defined as the rniy(t1eti~dipole inotncric, whlch is a vector whose ma,onltude 1s the product of the current in m d the area of the loop and whose direction is the direction of the thumb as the fingars of the right hand follow the direction of the current. Comparison of E q (645) with the expression for the scalar electric potential of an electric dipole in Eq. (348),
the
,
reveals that, for the fwo cases, A is analogous to V. We call a small currentcarrying loop a magnetic dipole. The analogous quantities are as follows: . 
Electric Dipole
Mngnet~cDipole
(642)
0 ,
(643)
In a similar manner we can also rewrite Eq. (644) as
212
STATIC MAGNETIC FIELDS 1 6
Except for the change of p to m, Eq. (648) has the same form as Eq. (349) does for the expression for E at a distant point of an electric dipole. Hence, the magnetic flux lines of a magnetic dipole lying in the xyplane will have the same form as that of the electric field lines of an electric dipole positioned along the zaxis. These lines have been sketched as dashed lines in Fig. 3 14. One essential difference is that the electric field lines of an electric dipole start from the positive charge + q and terminate on the negative charge q, whereas the magnetic flux lines close upon themselves.+ 65.1
Scalar Magnetic Potential In a currentfree region J = 0, Eq. (67) becomes The magnetic flux density B is tfxm curlfree and can bc exprcssed as thc gradient of a scalar field. Let B = 1'0 VT/;,,, (650) where I/;, is called the scular. r~ltrgrreticpolc~r~ti~il (esprcssed il~iqmperes).The negative sign in Eq. (650) is conventional (see the definition of the scalar electric potential V in Eq. 338), and the permeability of free space p, is simply a proportionality constant. Analogous to Eq. (340), we can write the scalar magnetic potential difference between two points. P 2 and P I , in free space as
If there were magnetic charges with a volume density p, (A/m2) in a volume V', we would be able to find V, from
The magnetic flux density B could then be determined from Eq. (650). However, isolatcd magnetic charges have never been observed experimentally; they must be considered fictitious. Nevertheless, the consideration of fictitious magnetic charges in a mathematical (not physical) model is expedient both to the discussion of some magnetostatic relations in terms of our knowledge of electrostatics and to the establishment of a bridge between the traditional magneticpole viewpoint of magnetism and the concept of microscopic circulating currents as sources of magnetism. The magnetic field of a small bar magnet is the same as that of a magnetic dipole. This can bc vcrilied expcrimcntally by obscrving'thc contours of iron Jilings around a magnet. The traditional understanding is that the ends (the north and south poles) QT
' Although the magnetic dipole in Example 67
was taken to be a circular loop, it can be shown (Problem P.613) that the same express~onsEqs. (645) and (648)are obtalned when the loop has a rectangular shape, w ~ t hm = IS. as given in Eq. (646).
66 1 MAGNET' ITION AND EQUIVALENT CURRENT DENSITIES
does for ~ a l fcux .; that of :se lines that the :rminate ~selves.'
.
.
$
213
a
of a magnet are the location of, respectively, positive and negative magnetic charges. For a bar mhgnet', the fictitious magnetic charges + q,, and  q,, are assumed to be separated by a pistance d and to form an equivalent magnetic dipole of moment
,
!
m = q;d = a,lS.
(653)
v,,
The scalar magnetic potential caused by this magnetic dipole can then be found by following the procedure used in subsection 35.1 for findin,0 the scalar electric potential that is caufed by an electric dipole. We obtain, as in Eq. (348), II
i o do) dicnt of
(6 5 0 ) ;eg;P:
;tend :on I ; (i~llcr
(65 1) ume V',
(652) owever, xust be xges in ,f some e estabgetism
.d i ~n , 4.
are\; '
h POL, Problem &kngular
Substitution of Eq. (6 54) i n Eq. ( 6 5 0 ) yiclds thc same R as given in Eq. (648). We note that th,c ex~rcssionsof the scalar magnetic potential V,, in Eq. ( 6 ~ 5 4 ) for a magnetic dipole are cxactly analogous to those for the scalar electric potential V in Eq. (647) for an electric dipole: the likeness between the vector magnetic potential A (in Eq. 635) and b' in Zq. (617) is nor as exact. Holvever, since magnetic ch:~rgcs do not exist in practical problems, I/;, rilust be determined from a givzn current distribution. This determination is usually not a simple process. Moreover. the curlfree nature of IZ indicatccl in Ey.(649), from which the scalar magnutic potential V,, is defined, holds only ~t points with 110 currcnts. In a region whcrc currents csis:. ce, scalar magnetic potential is not a singlethe magnetic field is rlot ~ c i l s ~ u ~ i i iand,the valued function; hence tile magnetic potential difference evaluated by Eq. (651) depends on the path of il~tegration.For these reasons, we will use the circularingcurrentandvectorpbten;ial approach, instead of the fictitious magneticcharge andscalarpotential approach. for the study of magnetic fields in magnetic materials. We ascribe the macroscopic properties of a bar magnet to circulating atomic currents (Ampkrian currents) caused by orbiting and spinning electrons. 66 MAGNETIZATION CURRENT DENSITIES
AN^
EQOIVALENT
According to the elerhentary atomic mcdel of matter, all materials are composed of atoms, cach with a positively chargcd nuclcus and a numbcr of orbiting negatively charged electrons. The orbiting electrons cause circulating currents and form microscopic magnetic dipoles. In addition, both the electrons and the nucleus of an atom rotate (spin) on their own axes with certain magnetic dipole moments. The magnetic dipole moment. of a spinriing I I L I C ~ C ~ IisS u~u;tIly negligible compared to that of an orbiting or spinqing electron hecause of the much largcr nx~ssand lower anp1:ir velocity of the nucleds. A camplete understanditig of the magnetic effects of materials requires a knowledge of quantum mechanics. (We give a qualitative description of the behavior of different kinds of magnetic materials later in Section 69.) In the absence of an external magnetic field, the magnetic dipoles of the atoms of most materials (except permanept magnets) have random orientations, resulting
214
STATIC MAGNETIC FIELDS I 6
in no net magnetic moment. The application of an external magnetic field causes both an alignment of the magnetic moments of the spinning electrons and an induced magnetic mom'ent due to a change in thc orbital motion of clcctrons. 111 ortlcr 10 obtain a formula for determining the quantitativc change in the magnetic llux dcnsity caused by the presence of a magnctic matcrisl. we kt in, hc thc ~nagaeticd ~ p u l s moment of an atom. If there are H atoms per unit volun~e,we define a inugi~etizution vector, M, as
M
= lim AWO
k= l 
AV
(A/m)
9
which is the volume density of magn'etic dipole moment. The magnetic dipole moment dm of an elemental volume dti' is'dm = XI da' that. according to Eq. (6451, will produce a vector magnetic potential
Using Eq. (375). we can write Eq. ( 6  5 6 ) a s
Thus,
where V' is the volume of the magnetized material. We now use the vector identity in Eq. (628) to write
.
and expand the right side of Eg. (657)hto two terms:
The following vector identity (see Problem P. 6 14) enables us to change ;he volume integral of the curl of a vector into a surfacc integral.
jv,P x F d d = $s,
F x ds',
(6 60)
where F is any vector with continuous first derivatives. We have, from Eq. (659)
66 1 MAGNETIZATION AND E ~ I V A L ~ NCURRENT T QENSITIES
215
I
\es both induced srdcr to . density ;. dipole 'ti:ation
.
where a: is the unit outward normal vector from ds' and St is thc surface bounding I,,. the volume V'. ,' I A comparison ofthe expressions on the right side of Eq. (661) with the form of A in Eq. (622), expdssed in terms of volume current density J suggests that the effect of the magnetizatio$vectqr is equivalent to b&th a volume current density
d
I
and a surface current density
(655) J,, = M x a,, noment Is), will
16 56) f.
(Ajm).
(663)
In Eqs. (662) and (66.i) we hare omimd the primes on V and a,, for simplicity, since it is clear that both refer to the coordinates of the source point where the mapnctization vector M exisis However. the primes should be retained when there is n possibility of confusing tile coordinates cf the source and field points. The problem of finding the magnetic flux density E caused by a given iaiume density of magnetic dipvls moment i\.i is then reduced to finding the equiiaicnt tnaqnerizoiion current iie.&rirs J, and J,,,, by using E q s (661) m d (6631, determining A from Eq. (6611, and then obtaining B from the curl of A. The externally applied magnetic field. if i t also exists, must be accounted for separately. The mathematical derivation of E q s (662) and (663) is straightforivord. Tbs equivalence of a volume density of magnetic dipole moment to a volume current density and a surface cuxent density can be appreciated qualitatively by rekrring to Fig: 69 where a cross scction of a magnetized material is shown. It is assumed that an externally applied magnetic field has caused the atomic circulating currents to align with it, thereby magnetizing the material. The strength of this magnetizing
volume
P (660)
Fig. 69 A cross section of a magnetized material.
,
216
STATIC MAGNETIC FIELDS 1 6
effect is measured by the magnetization vector M. O n the surface of the material, there will be a surface current density J,,, whose direction is correctly given by that of the cross product M x a,. If M is uniform inside the material, the currents of the neighboring atomic dipoles that flow in opposite directions will cancel everywhere, leaving no net currents in the interior. This is predicted by Eq. (662), since the space derivatives (and therefore the curl) of a constant M vanish. However, if M has space variations, the internal atomic currents do not completely cancel, .esulting in a net volume current density J,. It is possible to justify the quantitative relationships between M and the current densities by deriving the atomic currents on the surface a i d in the interior. But as this additional derivation is really not necessary and tends to be tedious, we will not attempt it here. Example 68 Determine the magnetic flux density on the axis of a uniformly magnetized circular cylinder of a magnetic material. The cylinder h i s a radius h, length L, and axial magnetization M. Solution: In this problem concerning a cylindrical bar rnagqt. let the axis of the magnetized cylinder coincide with the zaxis of a cylindrical~oordinatcsystem, as shown in Fig. 610. Since the magnetization 1CI is a constant within the magnet, J, = V' x M = 0, and there is no equivalcl;t volume current density. The cquivulent magnetization surface current density on the side wall is
Jms= M x a; = (aJ4) x a, = a,M.
(664)
T h e lnagrlet is then like a cylindrical sheet with a lined clrrrerlt density of M (A/m). There is no surface current on the top and bottom faces. In order to find B at P(0, 0, s), we consider a differential length dz' with a current a,M dz' and use Eq. (638) to
Fig. 610 . A uniformly magnetized circular cylinder (Example 68).
atcrial,
obtain
,
j
by that
of the where, e space s space ~g In a mhips surface 3 tends formly dius b, , o i the 3.~5tir
.?asnet, ! I \ d'
, i l o ~ bdzl 2
and
, '
,
dB = a, 2 [ ( z  z'JZ + b 2 F
+,
B =  J ~ B ' =a,
J 0L
p o ~ b dz' 2 2 [ ( z  z')' b2]3 / 2
+
67 MAGNETIC FIELD INTENSITY AND RELATIVE PERMEABILITY
Because the application af an txternal magnetic fieid ciloses both :In :~ligilmeorai the internal dipole moments and an induced magnetic moment in 3 magnetic material. we expcct that the resultant magnetic flux density in the presence of Y magnetic l different from its value in free space. The n~acroscopicetTect a i magmaterial ~ i i be netization can be studied by incorporating the equivalent volume current densit!. J, in Eq. (662). into the basic curl equation, Eq. (67). We have
i h 64)
( A m). (0,0, I), 38) to
n,
We now define a new f~mdamentalfield quantity, the inagnetic field intensity, H , such that
The use of the vector H enables us to write a curl equation relatlng the magnetic field and the distribthon of free'currents in any medium. There is no need to deal explicitly with the ninpnetization vector M or the equivalent volume current density .I,,,. ~ o m h i i i TEqs. i ~ ~ ((660) and 3667). we ohlain the ncw cqu:ition (668) where J (A/m2) is the volume density olfiee ci~rlrnt.Equations (66) and 1668) are the two fundamental governing differential equations for magnetostatics in any
218
STATIC MAGNETIC FIELDS 1 6
medium. The permeability of free space, pO,does not appear explicitly in these two equations. The corresponding integral form of Eq. (668) is obtained by taking the scalar surface integral of both sides.
or, according to Stokes's theorem,
where C is the contour (closed path) bounding the surface S, and 1is the total current passing through S. The relative diiections of C and current flow I follow the righthand rule. Equation (670) is another form of Ampc'.re's circuital law: It states that the circulation of the ntugneticjeld intensity urounil any closed path is equal to the J r c c c l i l  r e n r jlor\.inc/ T / I I . O L I ~~/ ~/ I CS I I I . / ~ K C 'h l l l 7 d d I.!. p ~ l t . / ! ~ ~  \ s \ve iildic,ltcd in Section 63. Amptre's circuital law is most useful in determi'ning the magnetlc field caused by a current when cylindrical symmetry existsthat is, when there is a closed path around the current over whichthe magnetic field is constant. When the magnetic properties of the medium are linear and isotropic, the magnetization is directly proportional to the magnetic field intensity: where zmis a dimensionless quantity called magnetic susceptibility. Substitution of Eq. (671) in Eq. (667) yields
(672b) where
is another dimensionless quantity known as the relative permeability of the medium.
b
<
I .) L
67 / M A ~ N E T ~FIELD C INTENSITY AND RELATIVE PERMEABILITY
219
Fig. 611 Coil on ferromagnetic toroid w ~ t h air gap (Example 69).
The parameter p = u,p. is the absoliite perazeu8ilit)~(or, sometimes. just pemiubilitj 1 of the medium and is measured in H/m; z,,,, and therefore p,, can be a function of space coordinarcs. For a simple mcdium  linear, isotropic, anct homogeneous z,,, m d p, are consttlnts. The permeability of nost materials is very close to that of free space ( p , , ) . Fdr ferromagnetic materials such as iron. nickel, and cobalt. p, could be very large (505000, and up to 106 or more for special alloys): the permeability depends no[ only on lhc magnitude ol 11 but also on liic prcvious hibtory of the material. Section 68 contains some quali~ativediscussions of the macroscopic behavior of magnetic materials. Examplc 69 Assunw illat iZl turns ol'wire arc wo~inciaround a roroid~llcare ol'.i ferromagnetic material with permeabiilty p. The core has a mean radius r,],3 circular cross section of radiUs u iu I< r,,), and a narrow air gap of length I,, as shown in Fig. 611. A steady currdnt I,,Rows in the wire. Determine (a) the magnetic flux denhity. B f , in the ferromagnetic core; (b) the magnetic field intensity, H,., in the core; and (c) the magnetic field intensity, H,, in the air gap.
Solution
a) Applying Ampkke's circuital law, Eq. (670), around the circular contour C, which has a mean radius r,, we have
will lloa iii 170111 ilia Scrron~ayctic lilliir lr:~k:~gais n c g l c ~ k ~i lil.r S : I ~wCh l 11~1s
.
corf ;md in ihc air g:lp: I f lhc fringing clkul ol. lllc llur i n the air gap is also ncglected, the magnetic Hux density B in both the core and the air gdp will also be the same. However, Zccause of the different permeabilities, the magnetic field intensities in both parts will be different. We have
* STATIC MAGNETIC FIELDS / 6
220
In the ferromagnetic core,
and, in the,air gap, Po Substituting Eqs. (675), (676), and (677) in Eq. (674), we obtain
and
B
IUOPNI~ + pi,
;' po(2ni,, 
'
b) From Eqs. (676) and (678) we get
c) Similarly, from Eqs. (677) and (678). wc havc
Since H,/H, = p/p0, the magnetic field intensity in the air gap is much stronger than that in the ferromagnetic core. Why do you think the condition a cc r, is stipulated in this problem? 68
MAGNETIC CIRCUITS
The problem in Example 69 is, essentially, one of a magnetic circuit in which the current applied to the winding causes a magnetic flux to flow.in the ferromagnetic core and thc air gap in series. We define the line integral of magnetic field intensity around a closed path,
as magnetomotiw force,* mmf. Its SI unit is ampere (A): but, because of Eq. (674), mmf is frequently measured in ampereturns (A t). An mmf is not a force measured in newtons. Assume 1'; = N1,denotes a magnetomotive force that causes a magnetic flux, 0,to flow in a magnetic circuit. If the radius of the cross section of the core is much
' Also called magnetomotance.
I
:
L
68 1 MAGNETIC CIRCUITS
221
.:
r 6 76) (677)
smaller than the mean radius of the toroid, the magnetic flux density B in the core is approximately co&ant,'and cf, E B S , (681) wlicrc S is thc crossscclionnl area of the cute. Combination of Eqs. (681) and (678) yields 4
Equation (682) carfbe rewritten
(678)
i0,Pl
( 6 SO)
.rongcr
with
where tl = 2nr,
 f gis the length of the ferrorhagnetic core, and
Both 3,and 2, have the rime form :is the formula. Lq. (51 3). for ihc DC res~st:incc o 1 s : p i c f 1 o 1 1 c 1 o si i I a i cross s i S. U l l h arc called rcliiaunr~:2 , of the fcrromiignctic corc; and r?,,,of the ;iir g i p . Tile SI unit for reluctance is reciprocal henry ( H  I ) . The fact that Eqs (684) and 1685] are as they are, even though the core is not straight, is a consequence of assuming that B is approximately constant over the core cross section. Equation (683) is analogous to the expression for the current I in an electric circuit, in which an idealvoltage source of emf Y is connected in series with two resistances Rf and R,:
ich the lgnetic .tensity
(6
p,
' J X I I , ,J
x Hu. s much (a) Magnctic circuit,
(b) Elcctric circuit.
Fig. 612 Equivalent magnetic circuit and analogous electric circuit for toroidal coil with air gap in Fig. 6 1 1.
.
222 ,
STATIC MAGNETIC FIELDS / 6
,
respectively. Magnetic circuits can, by analogy, be analyzed by the same techniques we have used in analyzing electric circuits. The analogous quantities are Magnetic Circuits
Electric Circuits
mmf, YC,, ( = N I ) magnetic flux, 0 reluctance, 94 permeability, p
emf, /'
electric current, I resistance, R conductivity, a
In spite of this convenient likeness, an exact analysis of magnetic circuits is inherently very difficult to achieve. First, it is very difficult to account for leakage fluxes, fluxes that stray or leak from the main flux paths of a magnetic circuit. For the toroidal coil in Fig. 611, leakage flux paths encircle every turn of the winding; they .partially transverse the space around the core, as illustrated, because the perrneabilit> means that for any arbitrary angle a, that is not close to zero. the magnetic field ~ almost parallel to thc intc;fa~c:Third, if medium I in a ferromagnetic n ~ e d i u nruns is ferromagnetic and medium 2 is air (p, >> p2), then a, will be nearly zero; that is. if a magnetic field originates in a ferromaghetic medium, the flux lines will emerge into air in a direction almost normal to the interface.
In currentfree regions the magnetic flux density B is irrotational and can be expressed as the gradient of a scalar magnetic potential Vm, as indicated in Section 65.1. B = $i'V,. (6105) Assuming a constant p, substitution of Eq. (6105) in V B = 0 (Eq. 66) yields a Laplace's equation in Vm: VZVm= 0. (6106) Equation (6106) is entirely similar to the Laplace's equation, Eq. (4lo), for the scalar electric potential V in a chargefree region. That the solution for Eq. (6106) satisfying given boundary conditions is unique can be proved in the same way as for Eq. (410)see Section 43. Thus the tcchniqucs (method of images and method of separation of variables) discussed in Chapter 4 for solving electrostatic boundaryvalue problems can be adapted to solving analogous magnetostatic boundaryvalue problems. However, although electrostatic problems with conducting boundaries maintained at fixed potentials occur quite often in practice, analogous magnetostatic problems with constant magneticpotential boundaries are of little practical importance. (We recall that isolated magnetic charges do not exist and that magnetic flux lines always form closed paths.) The nonlinearity in the relationship between B and H in ferromagnetic materials also complicates the analytical solution of boundaryvalue problems in magnetostatics.
. I
.
, i
?
i
611
(6103) :ofH, is
? can be i in Sec
I~DUCT~RS
I
.
:
233
t
$\ f
!
Consider two neighboring closed loops, C, add C, bounding surfaces S l and S2 respectively, as shown in Fig. 619. If a current I, flows in C,, a magnetic field B, will be created. Some of the magnetic flux due to B, will link with C , that is, wlll pass through the surf$ce S2 bounded by C,. Let us designate this mutual flux a,,. We have
i
ecs. This
INDUCTANCES AND
"
611 1 INO~CTANCESAND INDUCTORS
I
i
From physics we knok thiit a timevarying I , (and therefore a timevarying 81,) wll produce an induced e~ectromotiveforce or voltage in C 2 as a result of Foraday's low of electromagnetic i$uction. (We defer the discussion of Faraday's law untll the next chapter.) I Iowcdor, ( I , , , cxists cvcn ~f I , is a steady DC current. From ~ i o t  ~ a v a law, r ! Eq. (631), we see that B , is directly proportional to I, ; hence ( D l , is also proportional to I,. We write
where the proportionslity constant L,, is called the mutual inductance between loops C , and C,, with SI unit henry (H).In case C 2 has N , turns, thejux linkage A,, due to Q 1 2is
[email protected] (Wb), and Eq. (6108) generalizes to
1, for the . (6106) :ty as for
elhod of ~undary
:ry'PY ,s unci
.etostafic imf xtic flux :n B and mndary
11
Fig. 619
Two magnetically coupled loops.
,
234 '
STATIC MAGNETIC FIELDS / 6
The mutual inductance between two circuits is then the magneticjux linkage with one circuit per unit current in the other. In Eq. (6108), it is implied that the permeability of the medium does not change with I , . In other words, Eq. (6108) and hence Eq. (611 1) apply only to linear media. A more general definition for L , , is
Some of the magnetic flux produced by I , links only with C, itself, and not with C,. The total flux linkage with C, caused by I , is
j
!
f
t
v
!
1
The selfinductance of loop C, is defined as the magneticflux linkage per: unit current in the loop itself; that is,
! L
for a linear medium. In general,
,
The selfinductance of a loop or circuit depends on the geometrical shape and the physical arrangement of the conductor constituting the loop or circuit, as well as on the permeability of the medium. With a linear medium, selfinductancc does not depend on the current in the loop or circuit. As a matter of fact, it exists regardless of whether the loop or circuit is open or closed, or whether it is near another loop or circuit. A conductor arranged in an appropriate shape (such as a conducting wire wound as a coil) to supply a certain amount of selfinductance is called an inductor. Just as a capacitor can store electric energy, an inductor can storage magnetic energy, as we shall see in Section 612. When we deal with only one loop or coil, therc is no need to carry the subscripts in Eq. (61 14) or Eq. (61 15), and inductunce without an adjective will bc taken to mean selfinductance. The proccdurc for determining the selfinductance of an inductor is as follows:
1f I
I
' t
1 1 f 5'
'
1. Choose an appropriate coordinate system for the given geometry. 2. Assume a current I in the conducting wire.
3. Find B from I by Ampire's circuital law, Eq. (69), if symmetry exists; if nbt, BiotSavart law, Eq. (631), must be used. ;
1!
'
611 / INDUCTANCES AN3 INDUCTORS t
4. Find the flux linking with each turn, Q, from B by integration,
\vith orle
meability nd hence
B= I
S I
I not with
Ss
Bads,
where S is the area over which B exists and links with the assumed currenr. 5. Find the flux linkkge A by multiplying B by the number of turns. 6. Find L by taking !he ratio L = All. '
(61 12)
235
s
Only a slight modification of this proceduk is needed to determine the mutual inductance L I Z b e t d e n two circuits. After choosing an appropriate coordinate system, proceed as fd~lows:.AssumeI , + find B, find B,, by integrating B, over surface S2 .find fluk linkage A , , = N,(P,, ifind L , = A, ,/I,

,
(6113) 1t curreut
( ( 4) 0
Example 612 Assdme I? turns of wire are tightly wound on n toroidal frame of a rectangular cross section with dimensions as shown in Fig. 620. Then assuming the permeability of the medium to be ,uo. find the selfinductance of the toroidal coil.
Sulution: It is clear jhat the cylindrical coordinate system is approprlate for t h ~ s problem because the toroili is symmetrical about its axis. A s s u m q a currenr I in the conducting wire, we find. by applying Eq. (69) to a circular path wlth radius r ( u < r < h):
(6115)
e and the file11as on does not ardless of :r loop or
This result is obtained because bath B, and r a r e constant around the circular path C, Since the path encircles a total current NI, we have
re wound Just as rgy, as we !o need to . ad/\.e le sellinIr.
ts;
if not,
Fig. 620 A closely wound toroidal ,coil (Example 612).
236
and
STATIC MAGNETIC FIELDS / 6
,
.,,,
,.
.. .
,.
.
i
 B, = PoNI .2nr Next we find
The flux linkage A is
[email protected] or
Finally, we obtain
.
A=
poN21h b In . 2n u
A poN2h b L==InI
2
a
(H).
(6116)
We note that the selfinductance is not a function d?ffor a constant nmiium permeability). The qualification that the coil be closely wound on the toroid is to minimize the linkage flux around the iliriividual turns of the wire. Example 613 Find the inductance per unit length of a very long solenoid with air core having n turns per unit length. Solution: The magnetic flux density inside an infinitely long solenoid has been found in Example 63. For current I we have, from Eq. (613),
B = ,uonI, which is constant inside the solenoid. Hence, where S is the crosssectional area of the solenoid. The flux linkage per unit length is Therefore the inductance per unit length is Equation (61 19)is an approximate formula, based on the assumption that the length ol' Ihc solcnoid is vcry n~uchyraltcr th;m thc 1inc;lr dimcnsions of its crow scction.
A more accurate derivation for the magnetic flux density and flux linkage per unit length near the ends of a finite solenoid will show that they are less than the values given, respectively, by Eqs. (613) and (6118). Hence, the total inductance of a finite solenoid is somewhat less than the values of L',as given in Eq. (61 19), muitiplicd by the length.
.c\;
,,.,
.(:
$:,.;
I..'.
. . .
,, ;  3 :. '. ..,;";,,: ,, ,!:, , . , . . . I,..,
.
":
i
.
.
,
,
.;.:
1
q,;
.;
..+.:.
,
.
,.,.~ , ,
* :,
,:;
>
,
,,:
.
.
'
:
,
r!
, ,
,,;,
':r
, . f
.
J
'4
,

;;..:,
.
::.., ., . . .i.i
[,,
; .. .:
I
,
.
611 ,'I I ~ D U C T A ~ L CAND S INDUCTORS
.:
C
.,.
,!
.!a; ,
~
.
i:
,
237
,
,
4,
The following isia significant obser&ion hbout the results of the previous two examples: The self4 ductance of wirewound ihductors is proportional to the square ~f the number of tu ps. t
1
Example 614 ~ n ' k icdaxial r transmission lide has a solid inner conductor of radius a and a very thin outer conductor of inner rahius b. Determine the inductance per unit length of the lide.
!'
'
.
Solution: Refer to i g . 6721. Assume that a &rent I flows in the inner conductor and returns via the h e r donductor in the othet direction. Because of the cylindrical symmetry, B has onl) a $component with diherent expressions 111 ihc two resions: (a) ins~dethe inner cbndu;tor, and (b) between the inner and outer conductors. Also assume that the currlk!nt I is liniformly distributed over the cross section of the Inner conductor. (6116)
a) Inside the inner cbndu :tor,
n
: meuium
:oic'
O l r l a .
From Eq. (6lo],
to
d with air
b) Between the inner and outer conductors, u
~
r
~
h
.
From Eq. (61 11, een found
(61 17)
Now consider an:annclar ring in the inner conductor between radii r and r i cir. The current in a unit length of this annular ring is linked by the flux that can be obtained by integrating Eqs. (61 20) and (6121). We have
.t length is
(61'1 8)
the 1, lth m se~.~n. ;e per unit the values tance of a 19), multi
Fig. 621 Two views of a coaxial transmission line (Examplz 614).
'
238
STATIC MAGNETIC FIELDS / 8
But the current in the annular ring is only a fr&tion (2nr dr/na2 = 2r dr/a2)of the total current I.' Hence the flux linkage for this annular ring is 2r dr dA' = dW. a2
The total flux linkage per unit length is
(6123)
,
.
.
The inductance of a unit length of the coaxial transmission line is therefore
The first term po/8n arises from the flux linkage internal to the solid inner conductor; it is known as the internal inductance per unit length of the inner conductor. The second term comes from the linkage of the flux that exists between the inner and the outer conductors; this term is known as the external inductance per unit length of the coaxial line. .
Before we present some examples showing how to detkrmine the mutual indue
. tance between two circuits, we pose the following question about Fig. 619 and Eq. (6111): Is the flux linkage with loop C 2 caused by a unit current in loop C, equal to the flux linkage with C , caused by a unit current in C,? That is, is it true that L I Z= L 2 , ? (6125)
We may vaguely and intuitively expect that the answer is in the affirmative .because of reciprocity." But how do we prove it? We may proceed as follows. Combining Eqs. (6107), (6109) and (6Ill), we obtain
' .I
N2 L I Z= I, J
L2
.
B, ds,.
(6126)
.
It is assumed that the current is distributed unifohnly in the inner conductor. This assumption does not hold for highfrequency AC currents.
.
,.
4 '
n2) of the
. ij;'
>I
.I.,!: I
1' . I
(6123)
r
r ' i
a
But, in $ew of Eq.(6,14), B1 can be written as the curl of a vector magnetic potential A,, B1 = V x A,. Welhave <
:
'
:
f
'
.
!
N2 L12= 
(V x A l ) . ds2
11
I
In Eqs. (6127) and (6128), the contour integrals are evaluated only once over the periphery of the loogs C, and C , respectivelythe effects of multiple turns habinp been taken care of sebarately by the factors N , and N , . Substitution of Eq. (6128) in Eq. (6127) yields
\nductor; .?;'',!. 8 ,
,
.,
,
'\
4
259
1,
!
R636 What boundaryi condi!ion must the tangenVal components of magnetization satisfy at an intcrfacc? If regiorl2 i6 n:onmaynctic, what is the relation betwcen thc surface current and thc tanycntial c o m p b a c ~ tof M I ? . R.637 Define (a)'the mhtual j?ductance betweka two circuits, and (b) the selfinductance or a single coil. , .
,
. $
% :
R.638 Explain how the,.selfhd~lctanceof a wirewd~ndinductor depends on ~ t number s of turns.
b,
i
, would the answer be ;he dame if the outer conductor is not "very I
R.639 In Example 6f$, thin"? Explain. 1
i
4
R.640 Give an expression o i magnetic energy In terms of B dndlor H. I
R.641 Clve thc integral exprebslon for the force on a closed clrcult t h d c m x s In a magnetic field B. :S1 units
d
current 1
R.642 Discuss first the net force and then the net torque acting on a currentcarrying cllcult situated in a uniform mlgnetq field. R643 What IS the reladon tztween the force and the stored magnetic energy In system of currentcarrymg c~rcuitsdnder the condition of constant flux linkages? Under the condmon of constant currents? 1
t h P.
and
n mmf of nat in the
PROBLEMS P.61 A positive point charge q of mass m 1s Injected with a velocity u, = a,u, into the J > 0 region where a uniform magnetic fleld B = a,B, exists. Obtarn the equat~onof motion of the charge, and describe the pdth that the chargc follows. I
P.62 An electron is injected with a velocity u, = a,ilt, into a region where both an electrlc field E and a magnetic field B exist. Describe the motion of the electron if a) E = a,E, and B = a$,, b) E = a,E, and B = a,B,. Discuss the effect of the relative n~agnitudesof E, and B, on the electron paths of (a) and (b). I
A current I Ilows it) (lie inncr I A ~ I W I O I  ( , ~ : ! I I idillrlcly long co:i\i.~I Iinc atld roturnb wa ~ . , I & ~ I I I SoI' LIIC IIIIICT C L ) ~ ~ L I C 'is I O1T1, and the Illlicr and outer radil of the lhc 0 \ 1 1 ~L X~ ~ I I ~ U ~ I'I'hc O U ~ C I 'L ' O I ~ ~ I CL&~ W 1) I I I \ ~e ~ ~ c h ~ ~ c lI ;~~ ~ I I c111e ~ l yt,~ u ~ g l i cIhu k clcudy 11 Vor ,dl ucgionh ;mu plot IBI versus r. \
P.63
magnetic
P.64 Determine the magnetic flux denslty at a point on the axis of a solenold with radius h and length L,and with a current 1 in its N turns of closely Wound coil. Show that the result reduces to that given in Eq. (613) when t approaches infinity.
260
STATIC MAGNETIC FIELDS 1 6
.. ,
,
,
*
P65 Starting from the e ~ p r e s ~ i ofor n vector magnetic potential A in Eq. (622), prove that
.
Furthermore, prove that V B = 0. J
P.66 Two identical coaxial coils, each of N turns and radius b, are separated by a distance d, as depicted in Fig. 629. A current I flows in each coil in the same direction. a) Find the magnetic flux density B = axBx at a point midway between the coils. b) Show that dB,.dx vanishes at the midpoint. c) Find thc relation between h and (1 such that r12B,/rl.~2 also vanishes at the rnidnoint r Such a pair ufcuils arc used to obtain a n appruxi~l~ately uniform magnetic field in the midpoint coils. region. They are known as Heli~rholt~
d
(Problems P.66).
dP.67 A thin conducting wire is bent into the shape of a regular polygon of N sides. A currcnt I flows in the wire. Show that the magnetic flux density at the center is
.
B=a,
!JON[ n tan , 2nb N
where b is the radius of the circle circumscribing the polygon and a, is a unit vector normal to the plane of the polygon. Show also that as N becomes very large this result reduces to that given in Eq. (638) with z = 0.
P.68 Find the total magnetic flux through a circular toroid with a rectangular cross section of height h. The inner and outer radii of the toroid are u and b respectively. A current I flows in N turns of closely wound wire around thc toroid. Determine the percentage of error if the flux is found by multiplying the crosssectional area by the flux density at the mean radius. P.69 in certain cxpcriments it is dcsirshle to havc a rcgion of constant magnetic [lux density. This can be created in an offcenter cylindrical cavity that is cut in a very long cylindrical conductor
.
I
..
we that (6197)
,
, !,
' ' /
, 1
: '
tmce (i,
!,,
carrying a uniform current density. Refer to the cross section in Fig 630. The uniform i~rial currcnt density is .I :;I,J: Find [hc in;~yilil~~dc xnd dirCcii011 of U ill ihc vyiindiicll cavity who* axis is displaced from that of the conducting part by a distance d. (Hint: Use principle of superposition and consider B in thecavity as that due to t d o long cylindrical conducton with radii b and a and current densities J and  J respectively.)  ,
P.610 Prove the following:
p.612 Startrng from the t expressron of A In Eq, (634) for the vector magnet~cpotent~alat a point in the brsecting ~ l a f l eof 8 straight wlre of length 2L that carrles a current 1 a) Find A at pomt k x , y.0) in the blsectrng plane of two parallel r ~ r e seach of length 21, located at y = &!/2 and carrylng equal and obpos~tecurrents, as s h o r n in Fig. 631. b) Flnd A due to ecjbal and opposite currents m n very long twow~retransm~ss~on line. c) Find B from A in part (b), and check your answer against the result obtarned by ~pplying Ampkre's circuital law.
1 current
s
to that
il
density. .onductor A
I
Fig. 631 parallel wires carrying equal and apposite currents (Problem P.612).
262
STATIC MAGNETIC FIELDS I 6
P.613 For the small rectangular loop with sides a and b that carries a current I, shown in Fig. 632: a) Find the vector magnetic potential A at a distance point, P(.u,y,d. Show that it can be put in the form of Eq. (645). b) Determine the magnetic flux density B from A, and show that it is the same as that given in Eq. (648).


P.614 For a vector field F with continuous.first derivatives, prove that & ( v x ~ ) d u = $ , F ~ L , where S is the surface enclosing the volume V. (Hint:Apply the divergence theorem to (A x C ) , where C is a constant vector.)
P.615 A circular rod of magnetic material with permeability p is inserted coaxially in the long solenoid of Fig. 64. The radius of the rod, a, is less than the inner radius, b, of the solenoid. The solenoid's winding has n turns per unit length and carries a current I. a) Find the values of B, H, and M inside the solenoid for r < a and for a < r < b. b) What are the equivalent magnetization current densities J, and J,, for the magnetized rod'? P.616 The scalar magnetic potential, Vm,due to a current loop can be obtained by first dividing thc loop area into many small'loops and then summing up thc contribution of thcse small loops (magnetic dipoles); that is,
where
dm = a,I ds.
(6198b)
Prove, by substituting Eq. (6198b) in Eq. (6198a), that
where R is the solid angle subtended by the lbop surface at the field point P (see Fig. 633).
PROBLEMS
i i
I\
Fig. 633 Subdivided current loop for determination of scalaf magnetic potential (Problcm P.616).
P.617 DO the following by ilsing Eq. (6199): a) Determine the scalar magnetic potential at a point on the axis of a circular loop having radius h and carryinb'a current I. b) Obtain the maghetic flux density B from Po OK,, and compare the rcsult with Eg. (638). (A x C),
)
the long no~cl.The
i
P.618 A ferromagnetikpphere of radius b is magnetized uniformly wlth a magnetlation M = azM0. a) Determrne the equivalent magnetlzatlon current densltles J, m d ,J b) Dctcrmlnc the nlagnclic llux densrty at the center of the qhere.
P.619 A toroidal lron core of relative permeab~l~ty 3000 has a mean radlus R = 80 (mm) m d .i c~rcularcross section wlth rrd~ilsb = 25 (mm). An .ur gJp fU= 3 (I,,) exlatr, J I I ~.I cui rc11, i ilou \ in a 5Wturn wlndlng to p o d u c e a lnagnetlc flux of
(Wb). (See Rg. 634.) Neglecting leakage
dividing loops
:;lit
Pig. 634 A toroidal iron core with air gap (Problem P.619).
264
STATIC MAGNETIC FIELDS 1 6
and using mean path length, find a) the reluctances of the air gap and of the iron core.
b) B, and H,in the air gap, and B, and H, in the iron core. C) the required current I.
P.620 Consider the magnetic circuit in Fig. 635. A current of 3 (A) flows through 200 turns of wire on the center leg. Assuming the core to have a constant crosssectional area of (m2) and a relative permeability of 5000: a) Determ~nethc magnctlc llux in cach leg. b) Dctcrrninc thc magnetic liclcl intensity ill c;~chIcg of l l i c core i111tl ill the itlr gi~p.
Fig, 635 A magnetic circuit with air gap (P~oblemP.6XI), P.621 Consider an infinitely long solenoid with n turns per unit length around a ferromagnetic core of crosssectional area S. When a current is sent through the coil to create a magnetic field, a voltage v , =  1 1 dO/dt is inducod pcr unit length, which opposes the current change. Power P, =  v , I per unit length must be supplied to overcome this induced voltage in order to Increase the current to I. a) Prove that the work per unit volume required to produce a final magnetic flux density B j is W, = JOE' H dB. (6200) b) Assuming the current is changed in a periodic manner such that B is reduced from BJ to B, and then is increased again to BJ, prove that the work done per unit volume for such a cycle of chayge in the ferromagnetic core is represented by the area of the hysteresis loop of the core material. P.622 Prove that the relation V x H = J leads to Eq. (699) at an interface between two media. P.623 What boundary conditions must the scalar magnetic potential Vm satisfy at an interface between two different magnetic media? P.624 Consider a plane boundary ( y = 0) between air (region 1, p,, = 1) and iron (region 2, P a = 5000).
a) Assuming B, = a,0.5 b) Assuming B, = a,10 to the interface.
 aJ0
+ a,OS
(mT), find B, and the angle that B, makes with the interface. (mT), find B, and the angle that B, makes with the normal
P.625 The method of images can also be applied to certain magnetostatic problems. Consider a straight thin conductor in air parallel to and at a distance d above the plane interface of a magnetic material of relative permeability p,. A current I flows in the conductor.
I.,
1 turns
6
~
,
and these cuhents ~ i r cequidistant from the interface and situated in alr, ii) the magnet~dfieldbelow the boundary plane is calculated from I and  I , , both at the same lodtion! These currents a r i sitbated in an infinite magneuc material of relative perx&bility p,. < b) F o r a long m d b c t o r carrying a current I ahd for p? >> 1, determine the magnetic flux density B at thejoint P in Fig. 636.

' P k Y)
[email protected]
t
I
d C
0
Ferromagnetic rncd~pm / '
(h21)
,6200) rom L?, wlume I of the I
P.627 Refer to ~ x a r n & 613. Dctcrminc lhc inductmcc pcr unit icngtii of tile sir coaxial transmission line anumibg that its outer conqu&r is dot very thin but is of a thickness d.
111
density
Fig. 636 A currentcarrying conductor near a ferromagnetlc medium (Problem P 625).
I'.626 Dctcrmme the 811inductance of a toroidrl coil of N turns of wlrc wound on an ilr frame with mcan radius 'roand a circular cross section of radius b Obtain an rpproxlmate expression assuming b rill~~l is rcachcd. A t ccluilib~.iu~n. w l ~ ~ cisl itcnclicd vcry ropiclly, the ncl rorcc on tllc kcc cllargcs in the moving conductor is zero. To an observer moving with the conductor, there is no apparent motion and the magnetic force per unit charge F,/q = u x B can be interpreted as an induced electric field acting along the conductor and producing a voltage V2,
=J:
(U x
B ) . dt'.
(77)
If the moving conductor is a part of a closed circuit C, then the emf generated around the circuit is
This is referred to as a fluxcutting emf, or a motional emf. Obviously only the part of the circuit that moves in a direction not parallel to (and hence, figuratively, "cutting") the magnetic flux will contribute to V' in Eq. (78). Example 72 A metal bar slides over a pair of conducting rails in a uniform magnetic field B = a,Bo with a constant velocity u, as shown in Fig. 72. (a) Determine the opencircuit voltage Vo that appears across terminals 1 and 2. (b) Assuming that a resistance R is connected between the terminals, find the electric power dissipated in R. (c) Show that this electric power is equal to the mechanical power required to move the sliding bar with a velocity u. Neglect the electric resistance of the metal bar and of the conducting rails. Neglect also the mqchanical friction at the contact points.
F,ig.72 A metal bar sliding over conducting rails (Example 72).
a) The moving bar generates a fluxcutting emf. We use Eq. (78) to find the opencircuit voltage Vo: , V, = V,
on and the .n induced
 v2
;
gC
(U
x Bj
're
=
J2!(a,u x a , ~ , ).(a, 'it)
=
 ~rB,h
(V).
(79) b) Wlicn a rcsis~anccR ~i conneclcd between terminals 1 and Z,3 current I = icB,li R will flow from terminal 2 to terminal I , so that the electric power. P,, dlsslpiited in R is
c) The mechanical power, P,,,, required Lo move the sliding bar is ted around
P,=F.u
(w),
(71 1 )
where F is the mechanical force required to counteract the magnetic force, F,, which the magnetic field exerts on the currentcarrying metal bar. From Eq. (6159) we have ily. the part iguratively,
m mn t i c ermine,the I 'a . dissipated required to c mctol bar tact points.
F,.= 1
S2t'
dP x B = a,IBoh
(N).
(712)
The negative sign in Eq. (712) arises because current I flows in a direction opposite to that of dP. Hence,. F =  F , = axIBoh= a , u ~ ; h ' / ~
(713) (N). Substitut~onofEq. (7'13) in E q . (71 1) proves P,,,= P,., which i~pholdstlic principle of conservation of cncrgy. 1
E h m p l e 73 The Faroilny disk gerwrntor consists of n circular nietnl disk rot;ltmp with :l constant ongulilr vclocily (u in a ull~l'urmand constant magnetic lield of flux density B = a,B, that is parallel to the axis of rotation. Brush contacts are provided
274
1
TIMEVARYING FIELDS AND MAXWELL'S EQUATIONS 1 7
I'
a
2
Ld
Fig. 73
Faraday disk generator (Example 73).
at the axis and on the rim of thepdisk, as depicted in Fig. 73. Determine the opencircuit voltage of the generator if the radius of the disk is b. Solutiorz: Let us consider the circuit 122'341'1. Of the part 2'34 that moves with the Eq. (78). disk, only the straight portion 34 "cuts" the magnetic flux. ~ & a v e from ,
which is the emf of the Faraday disk generator. To measure V, we must ,lac n voltmeter or :I vcry high rcsisl;~nmso tI1i11 n o ;~pprcci:ihlecurrent Ilowr i l l llle circuit lo modify thc cxlcrn;~llyt~pplicdn1:tgnclic licltl. 72.3 A Moving Circuit in a TimeVarying Magnetic Field
When a charge q mdves with a velocity u in a region where both an electric field E and a magnetic field L1 cxist, the clcctrornugnetic brcc 17 on y, us mcasured by a laboratory observer, is given by Lorentz's force equation, Eq. (65), which is repeated below: F = q ( E + u x B). (715) To an observer moving with q, there is no apparent motion, and the force on q can be interpreted as caused by an electric field E', where
t
'
4il4
Hence, when a conducting circuit with contbur C and surface S moves with a velocity u in a field (E, B), w e m e Eq. (717) in Eq. [7$) to obtain
dB $~'?dt'* Ldt.ds+
sv:
\
(u x B).dt'
(V).
(718)
t
:e open
Equation (718) is tHe general form of Faraday's law for a moving circuit in a timevarying magnetic fidd. The line integral on tge left side is the emf induced in the moving frame of t'&fbence.Thc first term o n thd right sidc represents the transformer emf due to the time inriation of B; and the second term represents the motional emf due to the motioli of the circuit in B. The division of the induced emf between the transformer and the motimal parts depends on the chosen frame of reference. Let us consider P circuit with contour C that moves from C, at time t to C, at limo r + AI in a chilngi~igm;ignolic lield U. Thc motion may include trunslaticn, rotation, and distortibn in an arbitrary manner. Figure 74 illustrates the situaticn. The timerate of chatlge of magnetic flux through the contour is
B(r
+ At) .ds,  J. ~ ( t )d s , S1
(719)
B(r + At) in Eq. (719) can be expanded as a Taylor's series: e a volt2 circuit
B(i + At) = B(t)
aqt) +A t + H.O.T., at
where the highorder tkrms (H.O.T.) coniain the second and higher powers of (Ai). Substitution of Eq. (720) in Eq. (719) yields
!c'iicld E :ed by a repeated
on q can
Fig. 73 A moving circuit timevarying magnetic field.
.
276
TIMEVARYING FIELDS AND MAXWELL'S EQUATIONS I 7
where B has been written for B(t) for simplicity. In going from C, to C,, the circuit coven a region that is bounded by S,,S,, and S,. Side surface S3 is the area swept out by the contour in time At. An element of the side surface is
d~= , dP x u At. (722) We now apply the divergence theorem for B at time r to the region sketched in Fig. 74: where a negative sign is included in the term involving ds, because outward normals must be used in the divergence theorem. Using Eq. (722)in Eq. (723)and noting that V B = 0, we have
L,
B . h, 
k, B . ch,
= Ar$Ju
x B)dP.
Combining Eqs. (721)and (724), we obtain ti

tlr
C
0 * 'k =
('.s c'R .(1s  ., /; 
(f, (.
(a x R)
.
lit.

which can be identified as the negative of the right side of Eq. (718). If we designate 1' = $c El dP = emf induced in circuit C measured in the moving frame
Eq. (718)can be written simply as
which is of the same form as Eq. (76).Of course, if a circuit is not in motion. f ' reduces to Y 7and Eqs. (72.7)and (76)are exactly the same. Hence. Faraday's law that the emf induce? in a closed circuit equals the negative timerate of increase of the magnetic flux linking a circuit applies to a stationary circuit as well as a moving one. Either Eq. (718)or Eq. (727)can he used to evaluate the induced emf in the general case. IIil highimpedance .~oltmcteris insenal in a conducting circuit, it will read the opencircuit voltage due to electromagnetic induction whether the circuit is stationary or moving. We have mentioned that the division of the induced emf in Eq. (718)into trimsformcr and motioo;ll emf's is not stiiqac, but their sum is ;~lw;lys eqi~alto that computed by using Eq. (727). In Example 72 (Fig. 72). we determined tho opencircuit voltage V, by using Eq. (78).If we use Eq. (727),we have
72 / F: WDAY~S LAW
1:
6~ ELECTROMAGNETIC INDUCTION ..
277
I
6 1
and
in Fig.
,
'
4
'
which is the s a k e & h q . (79). Similarly, for thd ~ a r d d adisk ~ generator in Example 73, the magnetic Aux linking the circuit. 1?2'341'1 is that whish passes through the wedgeshaped area 2'342'.
. I
orn~nls . ' ;i. 'noting . ; . and
Example 74 An h by i v rectangular conducting loop is sltuatcd iil ;I changing magnetic field B = a,B, sin o)r. The normal of the loop initially n~;ikesan angle 1 w ~ l ha,, as shown in f)g. 75. Iiind thc induced emf in thc loop: (a) when the loop is at rest, and (b) when the loop rotates with an angular velocity w about the xaxis.
(727)
ion. Y
"
~ y ' slaw
rease of movmg fin the t, it f l :c i r ~ ! c1nf in
I
1
.&
alwi
i
6
I

! )y using
II '
I
.
.
(a) Pcrspectivc vicw.
I:ig. 75
A
rcvlllll~ullrc o ~ l d a c h yloop rolnli~lgill
b (b) Vicw from +x direction.
n cliu~lpiag~ ~ l i i ~ ~licld l c ~(Example ic 74).
,
278
TIMEVARYING FIELDS AND MAXWELL'S EQUATIONS / 7
Solution a) When the loop is at rest, we use Eq. (76). a , = J ~ . d ~
.
= (a,,Bosin wt) (anhw) = Bohw sin wt cos a. 
Therefore
where S = hw is the area of the loop. The relative polarities of the terminals are as indicated. If the circuit is qompleted through an external load, *Kuwill produce a current that will opposc thc changc in (I). b) When the loop rotates about the xaxis, both terms in Eq. (718) contribute: the first term contributes the transformer emf Y; in Eq. (728). and the second term contributes a motional emf r,': where
'
\ I 
=l[( ;)
.
w x (a,Bo sin 031 (a, dx)
an
s)
+s:L(
 a.  w
x (a$, sin cot)] . (a, dx)
Note that the sides 23 and 41 do not contribute to Y , and that the contributions of sides 12 and 34 are of equal magnitude and in the same direction. If a = 0 at t = 0, then cr = wt, and we can write
.
', = B,Sw sin wt sin w t .
f/"
(729)
The total emf induced or generated in the rotating loop is the sum of f'. in Eq. (728) and V:,in Eq. (729): \
,
which has an ;tngulnr frcqucncy 20). We can determine the total induced emf Y : by applying Eq. (727) directly. At any time t, the magnetic Run linking the loop is
.
@(t)= B(t) [a,(t)S] = B,S sin o t cds cr = BoS sin wt cos o t = i B o S sin 2 o t .
i
1
'
 ,
1
r,
33
I MAXWELL'S EQUhtONS
279
t
.
Hence
(' i
L
8
e
r:
<
:
,
*
[email protected] y:F = dl
!
.
ri(; BOs

sin 2wt
% I
. =  B,So
, I
cos 2 0 r
,
as before.
;ids are xoduce ~trlbute: . second
The fundamental dbstulate for electromagn~ticinduction assures us that a timevarying magnetic figld gives rise to an electric field. This assurance has been amply verified b y numerous npcrimcnts. The V x E = 0 equation in Table 71 must ' therefore be replacekl by Eq. (71) in the timevarying case. Following are the revised set of two curl and two divergence equations from Table 71.
7
In addition, we know that the principle of conservation of charge must be satisded at all times. The mathematical csprcssion of charge conservation is the equation of c o ~ ~ l i r l u i t ISq. y , ( 5 XI), which IS ~'cpc;~tctl Iwlow.
butio ions x=Oat (729)
of %< in (7
C ) 1
I Jir&
,.
1
The crucial question here is whether the set of four equations in (731a, b, c, and d) are now consistefit +,withthe requirement specified by Eq. (732) in a timevarying situation. That the Answer is in the negative is immediately obvious by simply taking the divergence of Ed. (731 b),
which follows from the null identity, Eq. (2137). We are reminded that the divergence of lhc cuslof;in$ MIhc11;wd vector field 13 zero. Sincc Eq. (732) :lsserts V . J ~ O C S11otv:111i~hill L I ' I ~ I I I C  V ; I ~S~ ~~ ~ I \L ~ I : I 1[i ~ 1 , O( 7I I: ,~ ) i,, s ~n ' gcllcr;~I,not true. How should Eqfi. (731a, b, c, and d) be modified so that they are consistent with Eq. (732)? First of all, a term ap/St must be added to the right side of Eq. (733):
\
Equation (7,361 indicates that a timevarying electric field will give rise to a magnetic field. even in the absence of a current flow. The additional term aD/& is necessary in order to make Eq. (736) consistent with the principle of conservation of charge. It is easy to verify that 2 D B t ha's the dimension ofa current density (SI unit: Aim2). The term SD/dt is called displacement e~rrrentdensity, and its introductibn in the V x H equation was one of the major contributions of James Clerk M;i\\vcll (18311579). In order to be cotisistent with tlie cqu;ltion of continuity in a timevarying situation. both of the curl equations in Table 71 muh'be generalized. The set of four consistent equations to replace the inconsistent equations, Eqs. (731a. b. c, and d), are
They are known as Maxwell's equations. These four equations, together with the equation of continuity in Eq, (731) and Lorentr's force equation in Eq. (65), form the foundation .of electromagnetic theory. These equations can be used to explain and predict all macroscopic electromagnetic phenomena. Although the four Maxwell's equations in Eqs. (737a. b, c, and d) are consistent, they are not all independent. As a matter of fact, the two divergence equations, Eqs. (737c and d). can be derived from the two curl equations, Eqs. (737a and b), by making use of the equation of continuity. Eq. (732) (see Problem P.77). The four fundamental field vectors E, D, B,H (cach having three components) represent twelve unknowns. Twelve scalar equations are required for the determination of these twelve unknowns. The required equations are supplied by the two vector curl equations and the two vector constitutive relations D = aE and H = B/p, each vector equation being equivalent to three scalar equations.
. 8

73.1 '
.
(736)
.
.
.
. *.,  . .. ,+. .
11
8
.
,
, "
.
i) I
I
. ,
and
.
:
i'
'i
7
.i
I:
lgnetic xssary large. 1/m2). in the dxwell L
'
iI 1 1
The four Maxwell's bq~a'lionsin (737a b, c,.and d) are differential equations that ,are valid at ever$ $oint;h space. In explain/ng electromagnetic phenomena in a physical environm t, we must deal with dhite objects of specified shapes and boundaries. It is bo ,venient to convert the differential forms into their integralform equivalents. We ta e the! sdrface integral of both sides of the curl equations in Eqs. (737a) and ( 7 ~ 3 7 bover ) an open surface'! with a contour C and apply Stokes's theorem to obtaiti I
" "I..
'.
,
f
?
j
3.
,
.Integral Form o ~ , ~ ~ x w e Equations; ll's ,
I
E.dP*
.•
s
at
ds
(738a)
,
d. The 73'

I
Taking the volume ihtegdl of both sides'of the divergence equations in Eqs. (737c) and (737d) over a ~ o l u m kv with a closed surface S and using divergence theorem, we have
!
7374 737b)
..
.
' 3
and
737c) 7 37d) ~ t hthe (65), !sed to
.
2
~,
sistent,
i .
.;iti~P
and u), 7). 7;' ?rese. ~f these or curl u, each
..
f , I .
.
. ,. *.
.
,
,
i
i,
;
. ,'
, '
I
.
i ,
. I
, /
The set of !OL& equat~onsin (738a, b. c and d) are the integral form of Ma~well's t liq. (738a) is the same as Eq. (72), which is an expression ctrarnrignetic induction. Equation (738b) is a generalization law given in Eq. (6701, the latter applying only to static magnetic urrent density J may consist of a convection current density of ;I ireccharge distribution, as well ;a ;;canduction current t h presence of,an electric field in a conducting medium. The s the:c~rrentI flowing through the open surface S. a n b e recognized as Gauss's law, which we used extensively icli remains the same in the timevarying case. The volume ihtegral ofApeqvals the total charge Q that is enclosed in surface S. No particular h w is associated with EQ (738d); but, incdmparlng it with Eq. (738c), we conclude that there are no isdlatrd magnetic charges and that the total outward magnetic
"


. .. _. ;
i
?
282
> .,',""'" . >
2 :
:
'
TIMEVARYING FIELDS AND MAXWELCS EQUATIONS / 7
rh
Table 72 Maxwell's Equations Differential Fonn
VXE=
:.+
Integral Form
Significance
dB at
Faraday's law.
dt
dD fC
.
H de = I +
6
dD
 ds
Amfirs's circuital law Gauss's law.
VD=p
V.B=o
r..fl
'$B.L=o
No isolated magnetic charge.
6
flux through any closed surface is zero. Both the differential and the integral forms of Maxwell's equations are collected in Table 72 for easy reference. Example 75 An AC voltage source of amplitude V, andangular frequency w, U. = V, sin wt, is connected across a parallelplate capacitor C,. as shown in Fig. 76. (a) Verify that the displaccmcnt current in t l ~ ccapacitor is the same us thc conduction current in thc wires. (b) Detcrminc the magnetic licld intensity at il distancc r from the wire. Solution
a ) The conduction current in the connecting wire is
i c = c l 3 = 1cv0 a cos at
(A).
dt
For a parallelplate capacitor with an area A, plate sepmtion d, and a dielectric medium of permittivity e, the capacitance is 0

4(
Fig. 76 A parallelplate capacitor connected to an AC voltage source (Example 75).
,
I
.I(c
ii .I
1
7
E2 0
,$26)
:ction, a,,
(829)
3
!or of the the plane e posit~ve wparate :ec:~on of ,P L
c;
which leads to the foilowing equation for an ellipse:
0 region. The incident wave is reflected, giving rise to a reflected wave (E,, H,). The reflected electric field intensity can be written as
E,(I) = a,Eroe+jl'l=, (S71) where the positive sign in the exponent signifies that the reflected wave travels in the  z direction, as tliscusscd in Section 82. The total electric field intensity in medium 1 is the sum of E and E,. E,(z) = E,(z)
+ E,(z) = a,(EloeJP1' + EroefJP1z).
ith a perfect ncidcnce and nirorm plane
(872) Continuity of the tangenti 11 component of the Efield at the boundary r = 0 demands that I;,(O) = a,(Elo E,,) = E,(O) = 0,
:Is in the + z t electric and
which yields Ero =
+
Aio. Thus, Eq. (872) becomes El(z) a , ~ , ~ ( ~  j Pl 'e+jP1z) f
j2Ei0 sin /I,:. (873,) The msirctic field il.tensity H, of the reflected wave is related to Er by Eq. = a,
(8 24).
ely, the phase the Poynting , which is the 1.
.. * .>
t
.
I.._.,.
k

"
,
I
Combining H,(z) with Hi(z) in Eq. (870b), we obtain the total magnetic field inten. '' sity in medium 1 : Hl(z) = Hi(z)
+ H,(z) = a,2 Eio cos P1z. 'I
(873 b)
1
It is clear from Eqs. (873a), (873b). and (869) that no average power is associated with the total electromagnetic wave in medium I, since El(z) and Hl(z) are in phase quadrature. In order to examine the spacetime behavior of the total field in medium 1. we first write the instantaneous expressions corresponding to the electric and magnetic field intensity phasors obtained in Eqs. (873a) and (873b): E,(z, t)
= .%'o[~,(z)ej"']
= a,2Eio sin
PI? sin r
~ .
(874a)
Eio H ~ ( zt), = .%e[H,(:)eJm'] = a,2 cos Plz cos wt.
(8 74b)
'11
Both El(i, t ) and Hl(z, t) possess zeros and maxima at fixed distances from the conducting boundary for all t, as foilows: Zeros of El(,, t)
1
i
Maxima of H ,(z, t) Maxima of El(;,
'.
i)
occur at /Il:
A
i.
nx, Or==
"z'
n = 0, 1.2,.
..
)
7i
occuratB,z= (2n+ I), 2 Zeros of Hl(z, t )
or:=
.;
(2n+ I), 4 n=0,1,2....
The total wave in medium 1 is not a traveling wave. It is a standing wave, resulting from the superposition of two waves traveling in opposite directions. For a given t. both El and H1 vary sinusoidally with the distance measured from the boundar) plane. The standing waves of E l = a,E, and H , = a,H, are shown in Fig. 88 for several values of wt. Note the following three points: (1) E l vanishes on the conducting boundary (E,., =  Eio); (2) H , is a maximum on the conducting boundary (H., = Hio = Eio/vl); (3) the standing waves of E l and H , are in time quadrature (90 phase dillercncc) and are shifted in spiicc b y a qu;iricr wavelength.
. .
Example 87 A ):polarized uniform plane wave.(Et, Hi) with a frequency 100 (MHz) propagates in air and impinges normally on a perfectly conducting plane at x = 0. Assuming the amplitude of Ei to be 6 (mV/m), write the phasor and instantaneous expressions for: (a) E, and Hi of the incident wave; (b) Er and H, of the reflected wave; and (c) E l and H, of the total wave in air. (d) Dcierminc tile location nearest to the conducting plane where E l is'zero.
icld intenI
, , .
;
(873h)
, 1
1
1+ {,I
 .It
'
.
1
issociated e in phase ium 1, we magnetic
i
I
I
(8744 i Y 74b)
Irom the ,Fig. 88
Standing waves of ;i, = r,E, md H I = r,H, for several values o f o ~ .
Solution: At the giveh frequency 100 (MHz),
'
I
I ) :. 4 7 ... ,
. resulting a given t , boundary 2. 58 for ! the conboundary uadrature
1encn0
:I .
.I
~d InstanI f , of the :I~~c'l~lclll
a) For the incident w i v e
((1
travel~ngwave):
i) Phasor expressions E,(x) = ily6 x 103eJ2nxi3 (v/m), 1 H,(x) = ax X El(.;) = a, eJ2"x'3 Y1 2n

(A/m).
336
PLANE ELECTROMAGN"ETIC
WAVES 1 8
b) For the rejected wave (a traveling wave): i) Phasor expressions Er(x) = ,a,6 x
[email protected]"/3 (v/m),
. ,. ..
ii) Instantaneous expressions
Er(x,r) = Wc[E,(x)eja'] =  5 6 x
cos 2n x 108t +  s 3
(
Hr(x, t ) = a=cos 271 , x 108t lo' . + 2n c) For the total wave (a standing wave): i) Phasor expressions
ii) Instantaneous expressions E,(x, t ) = ~%e[E,(x)e~"']= a 9 2 x H ( x , t) = a, 
(?)
sin x
cos (277 x 108t)
sin(2n x 10")
(A/m).
d) The electric field vanishes at the surface of thc conducting plane at x = 0. In medium 'I, the first null occurs at
86 OBLIQUE INCIDENCE AT A PLANE CONDUCTING BOUNDARY
Whcn a uniform plane wave is incidcnt on a plane conducting surfacc obliquely, the behavior of the reflected wave depends on the polarization of the incident wave. In order to be specific about the direction of Ei, we define a plane of incidence as the plane containing the vector indicating the direction of propagation of the incident wave and the normal to the boundary surface. Since an Ei polarized in an arbitrary direction can always be decomposed into two componentsone perpendicular and
Medium I (01 = 0)
I ,
'
*

2 4 0
Fig. 89 Planc wave inc~dcnt obliqacly 011 a plane conducting boundary (perpendicular polarization)
the other parallel to the pime of incidencewe consider these two cases sepnmtelj. The general case is ostilined by superposing the results of the two component cas~s. 86.1
Perpendicular ~ o l a r i r a t i o n ~
In the wsc of perpendicuiilr p d u r i ~ u t i o n Ei , IS p&+endicular to the plane of incidence. as illustrated in Fig. 89. Noting that
+
ani = a, sin Oi a, cos Oi, (8753 where Oi is the ungle of inci~lencemeasured from the normal to the boundary surface, we obtain, using Eqs. (817) and (823),

' Also referred to as horizontal poiari:ation
or Epolcrrlzatioh.
.*
338
PLANE ELECTROMAGNETIC WAVES / 8
,I
i
.
At the boundary surface, z = 0, the total electric field intensity must vanish. Thus,
In order for this relation to hold for all values of x, we must have Ero =  Eio and Or = Oi.The latter relation, asserting that the angle of reflection eqtlals the artgle of irtcideilce, is referred to as Snell's law oj' reJectiort. Thus, Eq. (878) becomes
I 3
The corresponding Hr(x, z ) is
The total field is obtained by adding the incident and reflected fields. From Eqs. (876a) and (879a) we have El(x, z) = E,(x, z) + E,(.Y,Z ) = a E (e~8~ZC~d e &PI: , cos8,
Jb~xsrnO,
)e =  a,j2Eto sin (P,z cos Ol)eJ"lX Y
10
(880a)
e l .
Adding the results in Eqs. (876b) and (879b), we get I
E;0 [a, cos Ui cos (/jl; cos Ui)eJI1~X"nol M,(.K,z ) =  2 '11
+ a,j sin 0, sin ( p l z cos Oi)ej51xsin1. o1
1880b)
Equations (88Oa) and (8;80b) are rather complicated expressions, but we can make the following observations about the oblique incidence of a uniform plane wave with polarization on a plane conducting boundary: 1. In the direction ( z direction) normal to the boundary; El, and H I , maintain standingwave patterns according to sin Plzz and cos filzz, respectively, where fil= = P I cos Oi.No average power is propagated in this direction since El, and HI, are 90" out of time phasc. 2. In the dircction (.K direction) parnllcl to thc boundary, E , , and If,, arc in both time and spacc phasc and propagate with a phasc vclocity Ulx
(I)
V)
u1
/3, sin Oi
sin 8,
===.
PIX
... .'
1
st vanish. ..: .
.
.
/
:i'
1
, ,I:.
., ... . *,, r,
b
:
I
,..':;:,
'
':
,,
,
,
.
,.' ,
!1
:, 1.. '
I
:.
'
 E,,
and unyle of
:s
3. The propagating wave in the r direction is a nonunijom plane wave because its amplitude varies witIi'z. I. Since E, = 0 for all x when sin (P,z cos OiJ = 0 or when
2n fllz cas oi=   cos 0, = ^
A
(879a)
)'1
 1117t,
nz=1,2,3 ,...,
a conducting plilte can be insertdd at
(879b)
:on1 f".
(
x  aoa)
issob) t
without changinh tht: field pattern that exists between the conducting plate and the conducting ~~~~~~~~~y at I = 0. A tiitn.queiac elearic (TE) IIYIC~ ( E l , = 0)will bounce back and forth between the conducting pinnes ilnd propagate in tne r direction. We hitVe, i I effect, a parallelplate waveguide. Example 88 A uniforrrl plilhe wavc ( E l , H,) of an angular Srequcncy w ir lncldent from air on a very ILrge. perfectly conducting wail a r a n angle of incidence 0, with perpendicular polarihllun. Find (a) thc current induced on the wall surface, and (b) the timeaverage Poynting vector in medium 1.
a) The conditions df this problem are exactly those we have just discussed: hence we could use the kormulas directly. Let r = 0be the plane representing the rurfxe of the perfectly cbnducting wall. and let Ei be polarized in the y direction. as was shown in Fig. 89. At z = 0, El(x, 0) = 0, and H,(r. 0) can be obtained from Lq. (880b):
call plane
\VC
t11
7 x 1infain
>.\\~llcl.o L,. apq in
.
Inside the perfectly conducting wall, both E2 and H, must vanish. There is then a discontinuity ih iho milgnktic field. Thc amount of discontinuity is equal to ilic surfacu currchl. Iqroin Lcl. (752b), wc have
.
The instantaneous expression for the surface current is Ei0 J,(x, t ) = a, cos 4 cos o 6 0 ~
(882)
It is this induced current on the wall surface that gives rise to the reflected wave in medium 1 and cancels the incident wave in the conducting wall. b) The timeaverage Poynting vector in medium 1 is found by using Eqs. (880a) and (880b) in Eq. (869). Since El, and HI, are in time quadrature. Pa,will have a nonvanishing x component. 
E2 a,2C sin Oi sin' /l,,r, '/I
where /IlZ = /3, cos 0,. The timeaverage Poynting vector in medium 2 ( a p r f c c t conductor) is, of course, zero. 86.2
1
Parallel polarizationt
We now consider the case of E, lying in the plane of incidence while a uniform plane wave impinges obliquely on a perfectly conducting plane boundary, as depicted in Fig. 810. The unit vectors a,, and a,,, representing, respectively, the directions of propagation of the incident and reflected waves. remain the same as those given in
Reflected
wave
Incident
wave
Fig. 810 Plane wave incident obliquely on a plane conducting boundary (parallel polarization).
' Also referred to as tiertical polarizatiotl or Hpolarizatidt~.
86 1 OBLlOUE !NCIDENCE AT A'PLANE CONDUCTING BOUNDARY , . f"', ,I
341
1
i? Eqs. (875) and ($I!). Boih Ei and Er now have components in x and z directions, whereas Hi and H, hdve only a y component. w e have, for the incident wave, (882)
t
i
1,
1
Er(x, 2) = EIO(axcos 0, ' a: sin Oi)ejbl(x'ln ' 1 :tcd wave
;. (880a)
*. .Pav will
H,(x, = ay
+ z cosUf),
5
e  ~ \ l ~ (sxt n o i + z tb\ 0 1 ) .
(883b)
"t 1
The reflected wave (E,, H:) have the following phasor expressions:
E,(x, k) = Ero(a, cos 0, +'a, sin Or)ejP1(xsln '
cos Or),
:
H , ( ~ i), =
(885a) (88jb)
BIEroe~~~(xsln8rzcos~r~.
YI 1
(8 83) (a perfect
(884a)
A1 thc surklcc ol'~licpclfcct conductur, := O, ~ h tanyentiti1 c compolicn~( t l x component) of the total electric field intcnsitymust vanish for a11 s, or E,,(,Y.0) 6,,(.u, 0) .= 0. From Eqs. (8.841) ant1 (8.X~:I), wc 1i;ivc

.Y
/?
n
11e
Cplc'l,.,
117
\)I
whlch requires Ero =  f;,, and 0, = 0,. The total electric field intensity in medlum 1 is the sum of Eqs. (8+84id and (885a):
cctlolls of L' given in
Adding Eqs. (884b) and (885b). we obtain the total magnetic field intensity in medium 1. Hl(s. z ) = H,(s, z) + Hr(x, z) El 0
= ay2cos (/jlz cos O , ) C  J D ~ ~ ' ~ ~ "4,
YI 1
(886b)
I
m
The interpretation oi' Eqs. (886n) and (886b) is similar to that of Eqs. (880a) and (8SOW b r the erpendicularpol:~rizi~tio~~ cclsc, cxcept that E,(\.z), ~nbtc;idof fl,(s.:), now 11;)s1x1 11 ; I I I Y :rnti :I :componc~it.Wc C O I I C ~ L I ~ I~ICI.L'~OI.C: C.
P
,1. In the direction
( 2 direction) normal to the boundary, E l , and H , , maintain standingwave pattens according to,sin P1,z and cos Dl,z, respectively, where j,, = Dl cos Oi. No average power ispropagated in this direction, since El, and H,, are 90" out of t i p e phase.
. .
342
*
,
..r,
4.
.
,
!
'
!$ . i'.
. \ . . . .. ; ..
,, ' . , ..
.'
,
*
..
: I '
, ,,<
?.
:,,. \
:!
373
92 1 TEM WAVE ALONG PARALLELPLATE LINE
At y = 0 (lower plate), a,, = a,: a, )n along a nal dimen:tidds the the homo, t w phasor
D = psc
aj, x H = J,,
At y
=d

or
psd = EE, = E
or
J,,
E
~
~

Eo eJ". a,H, = a, 
~
~(963) ~ (97a)
'7
(uppet plate), a, = a,:
hy
x H = J,,
or
E J,,,=azHx= azorjpi.
(97b)
r?
Equations (96) add (9 7) indicate thar surface charges and surface currents on rhc conducting planes var) sinusoidally with 2, as d o E, and H,. This is illustrated schematically in Fib. 93. Field phasors E and H in Eqs. (912)and (91 b) satisfy the two Maxwell's curl equations:
Since E = a,E, and H

a,EI,, Eqs. (98) and (99) become
and
.
Ordinary derivatives a p p ~ i labove r because phasors E, and H , are functions of only.
f
TEMmode fields, surface charges, and surface currents in parallelplate transmission line. '
Fig. 93
n line,
374
THEORY AND APPLICATIONS OF TRANSMISSION LINES 1 9
Integrating Eq. (910) over y from 0 to d, we have
'=
~Ll(:),
where
v(r)= 
:J
E, d y =
(9 12)
E , f z ) d
is the potential difference or voltage between the upper and lower plates; '
is the total current flowing in the
r
+:
direction in the upper plate: and
I is the inductance per unit length of the parallelplate transmission line. The dependence of phasors V ( z ) and I ( z ) on z is noted explicitly in Eq. (912) for emphasis. Siniilarly, wc intcgwtc Eq. (91 I ) nvcr .s from 0 to I\ to obtain
=j o C V ( z ) ,
where
. is the capacitance per unit length of the parallelplate transmission line. Equations (912) and (914) constitute a pair of timeharmonic trai~smissiorzline ec1ucrtiot1.s for phnsors V ( z ) and I(:). Thcy may bc coinhincd to yield secondordcr dilli.rcntial ccluutions ror V ( z )and for I ( = ) :
The solutions of Eqb. (916a) and (916b) ate, for waves propagating in the +: direction, V(Z)= Voejsz (917a) and I(z) = ~ , e  ~ ~ ' , (917b) where the phase conlltant
is the same as that given in Eq. (92). The relation between V, and I , can be found by using either Eq. (9 1.1) or Eq. (9 14) :
which becomes, in view of the results of Eqs. (913) and (915).
I
I
2,
which, again, is the same as that of a TEM plahe wave in the dielectric medium 92.1 Lossy ParallelPlate Transmission Lines
.
Wc l w c so far :i~sulhcdllic parallelpl~~lc tramnission linc to be lossless. In actual situations loss may arise from two causes. First, the dielectric medium may have a nonvanishing loss tahger.1; and, second, the plates may not be perfectly conducting. To characterize these two effects we define two new parameters: G, the conductance
' This statement will be proved in Section 93 (see Eq. 987).
376
Ti4EORY AND APPLICATIONS OF TRANSMlSSlON LINES 1 9 ,
.
per unit length across the two plates; and R, the resistance per unit length of the two plate conductors. The conductance between two conductors septlratcd by a dielectric medium having a permittivity E and a conductivity 0 can be determined readily by using Eq. (567) when the capacitance between the two conductors is known. We have
Use of Eq. (915) directly yields
If the parallelplate conductors have a very large but finite conductivity a, (which must not be confused with the conductivity a of the dielectric medium), ohmic power will be dissipated in the plates. This necessitates the presence of a nonvanishing axial clcctric field a,E, at the plate sarbces. such that the avera$hynting vector (924) = aspn =::.Xa(a,E, x a,Hz) has a y component and equals the average power per unit area dissipated in each of the conducting plates. (Obviously the cross product of a y E , and a,Hx does not result in a !. component.) Consider the upper plate where the surface current density is JSu= H,. It is convenient to define a sur/ace impedance of an imperfect conductor. Z,, as the ratio of the tangential component of the electr'c field to the surface current density at the conductor surface.
For the upper plate, we hive
E
=f = "SU
EA , Hx
= ~ C9
(926a)
where qc is the intrinsic impedance of the plate conductor. Here we assume that both the conductivity 0, of the plate conductor and the operating frequency are sufficiently high that the current flows in a very thin surface layer and can be reprcxnted by the surlacc current J,,,,, Thc intrinsic impcdvncc of ;l good conductor has been given in Eq. (846). We have

where the subscript c is used to indicate the properties of the conductor.
I
92 1 TEM ~ A V E ALONG PARALLELPLATE LINE
377
I
length oi'the two
.
Substihi$ of;Eq. (926a) in Eq. (624) gives
electric~medium lily by using Eq. . We have :'
(927)
I
,
(9 23)
The ohmic poker dissipated in a unit length of :he plate havlng a wldth which can be expressed in terms of the total surface current. I = WJ,., as
vj
IS
spc,
.
Equation (b28) is the power dissipated when a sinusoidal current ofdmplitude I flows through a resistance RJw. Thus, the effective serles resistance per unit length for both plates of a parallelplate transmission line of wldth iv is
~chvityo, (which m), ohmic powcr ):.ivanishingaxial ng vector
r'
(923)
ipat  each of ', doe3 not result s J,,
= H,.It
is
'. Z,, as the ratio nt density at the
.
Table 91 lists the expressions for the four distributed purut~~eiers (R,L, G. 2nd C per unit length) of s parallelplate transmission line of width L I and separation (1. We note from Eq. (926b) !hat surface impedance Z, has a positive renctancterm X, that is nun3erically equal to R,. If the total complex power (insrend of its real Part, the ohmic Power P.. only) associated with a unit length of the plate is considered. X ,will lend to an ititert?aiiniiesinductance pcr wit lengtll L, = x,,~,, = A,, ,.,. At hi& frquedcics; Li is negligible in comparison with the exterllal inductlncu ,r.
Table 91 Distributed Parameters of ParallelPlate Transmission Line (Width = W. Stparation = d )
\ve assume that 1g frequency are nd can be repred c o w t o r has
clor.
We note in the calculation of the power loss in the plate conductors of a finite conductivity o, that a nonvanishing electric field a,E, must exist. The very existence of this axial electric field makes the wave along a lossy transmission line strictly not TEM. However, this axial component is ordinarily very small compared to the transverse component E,. An estimate of their relative magnitudes can be made as follows:
I
(930) For copper plates [oc = 5.80 x lo7 (S/m)] in air [s = r0 = 10'/36n (F/m)] at a frequency of 3 (GHz), IEJ ~ ~ 5x. los/Eyl 3 ;.
&
., :.
*.I
>
,,'.
, .,I
: . ,: . . , R,. Figure 99 illustrates some typical standing waves for a lossless line with resistive terminaticn.
. The standing waves on an opencircuited line are similar to those on a resistanceterminated line with RL > R,, except that the IV(zl)/and Il(zl)l curves are now magnitudes of sinusoidal functions of the distance zr from the load. This is seen from Eqs. (9104a) and (9104b), by letting RL + co. Of course, I, = 0, but VL is finite. Wc have (9 1 lua) v(zt)(= VL lcos Vz'(
I
All the minima go t o zero. For an opencircuited line,
r = 1 and S
Voltage and current standing waves on resistanceterminated lossless lines.
Fig. 99
+
co.
L
'1
I V(z')I for openc~rcu~ted line. IMz1)l for shortcircuited h e , ll(z')l for openc~rcuitcdline. V(zl)/for shortcircuiml h e .
ossible. ion, z' = 0, :maximum. xima of the :located at rr.At voltage ,lace. Other wave) will ioltaqnd
Fig. 910 Voltage h d current standing waver on open and shortsircu~ted ,i . lossless lines.
i;
O n the other and, the standing waves on a shortcircuited line are simdar to those on a resistanceterminated line with RL < R,. Here R L = 0. VL = 0, but 1 , is finite. Equations (9104a) and (9104b) reddce to
r =
i. X
i It,.
d rcsistancere now magis seen from t V, is finite.
I V(zf)l= ILRo/sinPz'l
(91 1 la)
Il(z')/ = I , /cm bz'l.
(9lllb)
Typical standing Wzves for open and shortcircuited lossless lines are shown in Fig. 9 10. Example 96 The staxlingwave ratio S on a transmiss~online is an easily measurable quantity. (a) Show how the value of a terminating resistance on a lossless line of known charactetistic impedance Ro can b determined by measuring S. (b) What is the impedance of the line looking toward the load at a distance equal to one quarter of the operating wavelength? Solutio)~ :
a) Since the terminating impedance is purely resistive, Z L = R,, we can determine etc.) whether RL is greater than R, (if there are voltage maxima at z' = 0, i/2, i, or whether RL is less than R, (if there are voltage minima at '; = 0, i.12, 1,etc.). This can be easily ascertained by measurements. and (1.~ 1 occur at pz' = 0 ; and 1~ ,,,I First, if RL > R,, 0, = 0. Both /VmaXI and (1,.1 occuk at /Irt = n/2. We have, from Eqs. (9102a) and (91OZb).
404
THEORY AND APPLICATIONS OF TRANSMISSION
Second, if RL < R,, 8, =
/vmaxl and IImi,Ioccur at pz'
A.
Both (Vmi,I and /1,,;I occur at
pz'
= 0; a n d
= n/2. We have
b) The operating wavelength, A, can be determined from twice the distance between two neighboring voltage (or current) maxima or minima. At 2' = 214, 11': = nI2. cos /kt= 0, and sin /I:' = 1. Equations (9103a) and (91UbJ become
V(jL/4)= jILRo
(Question: What is the significance of thc j in thcsc cquations?) The ratio of V(1.14) to 1(1/4) is the input impedance of a quarterwavelength, resistively terminated, lossless line.
This result is anticipated because of the impedancetransformation property of a quarterwave line given in Eq. (994). 94.3
.
Lines with Arbitrary Termination
In the preceding subsection we not:d4hat the standing wave on a resistively terminated lossless transmission line is such that a voltage maximum (a current minimum) occurs at the terminatiorwhere z' = 0 if RL > R,, and a voltage minimum (a current maximum) occurs there if RL < R,. What will happen if the terminating impedance is not a pure resistance? It is intuitively correct to expect that a voltage maximum or minimum will not occur at the termination, and that both will be shifted away from the termination. In this subsection we will.show that information on the direction and amount of this shift can be used to determine the terminating impedance.
Let the termindting (or load) impedand be 2, = RL + jXL, and assume the voltage standing wdve on the line to look like that depicted in Fig. 911. We note that neither a voltage maximum noca voltage minimum appears at the load at z' = 0. If we let the standing wave continue, say, by an extra distance f,,,,it will reach a minimum. Thc volthge minimum is where it should he if the original termmating terminated by a pure resistance impedance ZLis replaced by a line section of lehgth l,,, R, < R,, as shown in the figure. The voltage distribution on the line to the left of the actual termination (where z' > 0) is not changed by this replacement. The fact that any complex impedance can be obtained as the input impedance of a section of loul&s line terminated in a resfstive load can be seen from Eq. (989). Using R, for 2, add dm for d, we have ' I
The rcll u d im:&ar) pails of Eq (9114) form two equations, Sroin w h ~ ht l x two unknowns, R,, hnd I,,,can bc solved (sce Problem P.934). Thc load impcd:ln~c % , can bc determined expcrimcntally by musunng the stand~ngwaveratio S and the distance 2: in Fig. 911. (Remember that 1:. + i, = ;,Q.) The procedure is as follows: 1. Find le ratio of brivel~fler
Sl S+l Use 0 , = 2P::,
lrl from S. Usu Il1
2. Find 0, from
6
= from Eq. (9106).
 n forjl
= 0 from Eq. (910s).
operty of a
rzmtirrared ti[,)
OF
'rent
1
is 1 m t x i or away from :direction mce.
.iIlCC
t
I t;n
'
Fig. 911 Voltage standing wave on line terminated by arbitrary im~edance and . equivalent line section with pure resistive 1. 7 load.
I
z'Z o
Ia
.
A
Find ZL, which is the ratio of Eqs. (9102a) and (9 102b) at I' = 0: ZL = RL +FL = Ro
+ ITlejer I  Irl ,or 1
(9115)
The value ;f R m that, if terminated on a line of length em,will yield an input impedance ZLcan be found easily from Eq. (91 14). Since R, < Ro. R,,! = R&. The procedure leading to Eq. (91 15)is used to determine Z, from a measurement of S and of 2;. the distance from the termination to the first voltage minimum. Of course, the distance from the termination to a voltage maximum could be used instead of 2:. However, the voltage minima of a standing wave are sharper than the voltage maxima. The former, therefore, can be located more accurately than the latter, and it is preferable to find unknown quantities in terms of S and &. Example 97 The standingwale ratio on a lossless 50(Q) trvnsmission line terminated in an unknown load impedance is found to be 3.0. Thc dintan& betwccn successive voltage minima is 20 (cm), and the first minimum is loc;~tcdi l l 5 ( ~ 1 1 1 ) fro111the load. D ~ ~ ~ I (:\I I I~ II I CI ~cllcctio~\ I ~ cocIlicic111l',: I I I ~(b) [IIC l ~ UII~CLIJIICC d Z,.In addition, find (c) the equivalent.length and
[email protected] of a line such that the input impedance is equal to 2,.
Solution a) The distance between successive voltage minima is half a wavelength. 
2x
2x
/?= =  = 5x (rad/m) . i, 0.4
A = 2 x 0.2 = 0.4 (m),
Step 1 : We find the magnitude of the reflection coefficient, lrl, from the standingwave ratio S = 3.
Step 2: Find the angle of the reflection coefficient, B,, from
0, = 2&
 n =2 x
5n x 0.05  n =  0 . 5 ~(rad)
r = (rider = 0.5ej0.5n = j0.5 b) The load impedance ZLis determined from Eq. (9115):
/
c) Now we find R, and C,, in Fig. 91 1. We may use Eq. (91 14) 30  j4O = 50
R m + j50 tan pt, 50 + jR, tan pt,
94 1
WAVE CHARACTERISIICS
Ohl FINITE TRANSMISSION LINES
40i
I
and solve the airndtaneous equations obtained from the real and imaginary = 112 and Rm = R,/S. Hence,' pans for R; and Pt,,,, Actually, we know i, t,,,
+
(9 IS] 5 ) an input RoIS. lsurement irnum. of 1 be used r than the the lattei,
n line terc bctwccn < i t 5 (crnr mpedancc :of 4 4 e
: standing
e m =1   zk 2
= 0.2
 0.05 = 0.15 (m)
and 50
Rm=
3
94.4
= 16.7 (R).
TransmissionLine Circuits
Our discussions on the properties of trmsrnission lines so far have been restricted primarily to the effects of the load on the input impedance and on the characteristics of voltage and current waves. No attention has been paid to the generator at the "other cnd," which is thc sourcc of Lhc wavcs. Just as the constraint (the boundary condition), V, = I&,, which the voltage V, and the current I , must satisfy at the 0), a constraint exists at the generator end where 2 = 0 and load end (z = l,z' i 2' = i.Let a voltage generator V, with an internal impedance 2,represent the source connected to a finite transmission line of length L that is terminated in a load impcdance Z,, as shown i l l Fig. 95. Thc additional constraint at := 0 will cnable the voltage and current anywhere on the line to be expressed in terms of the source characteristics (V,, Z,), the line characteristics (1,Z,, 4, and the load impedance iZ,). The constraint Bt z = 0 is = V,  I i Z , . But, from Eqs. (9i00aj and (9IOOb),
IL
= y (ZL
+ Z,)eiLII + re'?' 1
(9117a)
and
Substitution of Eqg. (9 117a) and (91 17b) in Eq. (9116) enables us to find
/7
1
'
Another set of solutiofls to pirt (c) is fm= t',,, why?

114 = 0.05 (m)and Rm = S R , = 150 (a).Do you see
408
THEORY AND APPLICATIONS OF TRANSMISSION
is the voltage reflection coefficient of the generator end. Using Eq. (9118) in Eqs. (9100a) and (9100b), we obtain
+'
Similarly,
f
Equations (9120a) and (9120b) are analytical phasor expressions for the voltage and current at any point on a finite line fed by a sinusoidal voltage s'ource V,. These are rather complicated expressions, but their significance can be interpreted in the following way. Let us concentrate our attention on the voltage equation (9120a); obviously the interpretation of the current equa&&(9120b) is quite similar. We expand Eq. (9120a) as follows:
where
V; = r(VMeyd)eyz' V z = rg(TVMe2Yd)eYz.

(9121b) (9121c)
The quantity
v
 v,zo + Zg z0
,
(9122)
is the complex amplitude of the voltage wave initially sent down the transmission line from the generator. It is obtained directly from the simple circuit shown in Fig. 912(a). The phasor V: in Eq. (9121a) represents the initial wave traveling in the
Fig. 912
Lrr thc voltsource Vg. interpreted &e equation 3b) is quite 2

A transffiission4necircuit and travelitlg waves.
+ z direction. Before this wave reaches the load impedance Z,. it sees Z, of the line as if the line were h n i t e l y long. When the first wave VT = Vb,e'' reaches 2, at 2 = d', it is reflected because of mismatch, resulting in a wave V ; with a complex amplitude T(V,,,e:IC)travelifig in the  2 direction. As the wave V ; returns to the generator at z = 0,it is again reflected for 2,# Z, giving rise to a second wave V: with a complex amplitude T,,ITV,,:,,C~"') traVclin~i n 4: direction. This proccss continues indefinitely with refections at both ends, and the resulting standing wave I/(:') is the sum of all the waves traveling in both dircctions. This is illustrated schematically in Fig. 9lZ(b). In practice, 7 = ct + j p has a real part, and the attenuation effect of r  " l diminishes the amplitude of rl reflected wave each time the wave transverses the length af the line. When the lirle is terminated with a nlatched load, 2, = Z,, r = 0, only V ; exists, and it stops at the matched load with no reflections. If 2, # 2, but Z, = 2, (if the internal idpedance of the generator is matched to the line), then r f 0 and r, = 0. As a consequence, both V : and V ; exist, and V ; , V ; and all higherorder reflections vanish,
Esnmple 98 A 100(Mllr,) gcnerutor with V, = 100 (V) and ~nternalresistance 50 (R) is connected to J lassless 50 (R) air line that is 3.6 (rn)long and term~narsdin a 25 + j25 ( R )load. Find (a) V(z) at a locatiod z Irom the generator, (b) I.; at the input terminals and V, idt tht load, tc) the vo!tage standingwave ratio on the line, and (d) the average power delivered to the load. 1 
sansr~~~ssion iown in Fig.
Ro = 50 (Q), ZL= 25 + j25
= 35.36/450 (R). t' = 3.6 (m).
. '
410
THEORY AND APPLICATIONS OF TRANSMISSION LINES 1 9
Thus, w
2n1O8
2n
p== (rad/m), p/ c 3 x lo8 3  
= 2.4n (rad)
(25 + j25)  50  25 + j25  35.36/135" r = ZL  Z o  (25 +'j25) + 50  100 + j25 Z,
+ Z,
103.1/14"
= 0.343/121" = 0.34310.672~
.
1, = 0.
a) From Eq. (9120a), we have V(z) = 
z, + z,;

e j2n:/3[l
100
+
0.343~j(0.672&8)nj 4 n z I 3 e I
We see that, because r, = 0, V(z) is the superposition of only two traveling waves, V : and V ; , as defined in Eq. (9121).: b) At the input terminals,
Vi = V(0) = 5(1 + 0.343eJ0.'28n) = 5(1.316  jO.134) = 6.611
At the load,
5.82" (V).
VL = V(3.6) = 5[ej0.4n + 0.343ej0.272*] = 5(0.534  j0.692) = 3.461 52.3" (V).
c) The voltage standingwave ratio is
d) The average power delivered to the load is
It is interesting to compare this result with the case of a matchcd load when Z L = Z O= 50 j O (a).In that case, l = 0,
+
IV,l =
V Iq ='I = 5 (V), 2
95 / THE SMITH CHART
41 1
/
and a maximum avirage:bower is delivered td the load:
I t
'I
Mlximum P,, =
v: 5 == 0.25 (W), 2R, 2 x 50
which is consideradly larger than the Pa. calculated for the unmatched load in part (d). ThE SMITH CHART
,
b
Transmissionline ~alculationssuch as the determination of input impedance by Eq. (989), reflection coefficient by Eq. (9101), and load impedance by Eq. (91 15)often invdhe tedious miinipulations of complex numbcrr This lcdl~rmc.in be alleviated by using a grilphicd mec!i~dof solut~on.The beit known m d most widely used graphicel chart is the Siriirh cbnrt devlsed by P. H. Smlrh.' Stated succinctly, a Sn~itllchilrt I: a graphical plot of norn~alizedresistance and reactance functions in the reflectioncoefficient plane. cs~ h e 15 In order to understlnd how the Smith chart for a l o ~ ~ l transmlision constructed, let us examine the voltage reflection coefficient of the load impedance defined in Eq. (9101):
Let the load impedance ZL be normalized with respect to the characteristic impedance R, = of the line.
where r and x are the normalized resistance and normalized reactance respectively. Equation (9101) cdn be rewritten as
where r, and r, are the real and imaginary parts of the voltage reflection coefficient r respectively. The inverse relation of Eq. (9124) is

P. H.Smith. "Transmissionlinc calculator," Electronlo. vol. 12, p 29. January 1939: m d "An improved transmissionline calculator," Electronics, vol. 17, p. 130, January 1944.
412
THEORY
.AND
APPLICATIONS OF TRANSMISSION LINES I 9
Multiplying both the numerator and the denominator of Eq. (9126) by the complex conjugate of the denominator, and separating the real and imaginary parts, we obtain
If Eq. (9127a) is plotted in the?,  Ti plane for a given value of r, the resulting graph is the locus for this r. The locus can be recognized when the equation is rearranged as
.
Fig. 913
Smith chart with rectangular coordinates.
.. I
,
.
; ,
,. ,
8 . :
i ..,
..
,
95 1 THE SMITH CHART
fi
413
1
;
!.
:complex
.
weobtain
..
/
'
1, ;
!
(YI&)
. ,
/
.
,
, I
. :
I
(9l27b) .
tion is re
,
,
.
:
:
.
:redulting
Several sa~ikntproperties of the rdircleslare noted as follows:
.
:
, , . ; I
.
'5
;, 1 ., :
1. The centers oh11 rhrcles lie on the'ir$xis. 2. Thc r = ~,.circ~c, hwiog ;I unity radi'us and ccntcrcd at thc origin, is thc largcst. 4' 3. The rcmles becoAe progressively 'smiller as r increases from 0 toward m,
(r,
ending at the = 1, Ti = 0) point. 4. All rcircles pdss through the (I, = 1, Ti = 0) point. Similarly, Eq. (9 l2lb) may be rearranged as
.
1
This is the equation for a circle having radius l/[x(and centered at T,= 1 and T,= l/x. Different values o f x yield circles of different radii with centers at different positions on thc r, = 1 line. A family of the portions of xcircles lying inside thc /I/= 1 boundary are shown in dashcd lines in Fig. 913. The following is a list of several salient properties of the ucircles. 1. The centers of ali scircles lie on the Tr = 1 line; those for x > 0 (inductive reactance) lie above the r,axis, and those for x < 0 (capacitive reactance) lie below the r,hxis. 2. The x = 0 circle bzcomes the rraxis. 3. The xcircles become progressively smaller as 1x1 increases from 0 toward a, ending at the [ r , = 1, Ti = 0) point. 4. All xcircles pass through the (T, = 1, = 0) point.
rl
A Smith chart is a chart of r and xcircles in the T,  Ii plane for Irl 5 1. It can be proved thi; the r and scircles are everywhere orthogonal to one another. The intersection of an rcircle and an xcircle defines a point that represents a normalized load impedance z , = 1. + ,js. Thc actual load impedance is Z , = R,(r + js).Since a Smith chart plots the normalized impedance, it can be used for cn1cul;~tionsconcerning :I Inssicss i~x~lklllissitl~l linc wit11 ; I I ~;~~.l>iilx~'y ~I;II.:IC~C ~ !S~~~X~~C~LLI ': I I ~ C C . As an illustration, point P in Fig. 9 13 is the intersection of the r = 1.7 circle and the x = 0.6 circle. Herice it represents 2,. = 1.7 + j0.6. The point P,, at (T,=  1, T i = 0) corresponds toi r = 0 and x = 0 nhd, therefore, represents a shortcircuit. The point Po, at (T, = 1, Ti A 0 ) corresponds to an infinite impedance and represents an opencircuit. The'Smith chart ic Fig. 913 is marked with T, and Ti rectangular coordinates. The same chart can be marked withpolar coordinates, such that every point in the rplane is specified by a magnitude I f 1 and a phase angle 8,. This is illustrated in Fig. 914, where several ]TIcircles are shown in dotted lines and some Orangles = 1 circler The Irlcircles are normally not shown on are marked aroufid the commercially available Smith charts; but once the point representing a certain
Irl
'
I
.
,
,. I :
414
THEORY AND APPLICATIONS OF TRANSMISSION
9n0
270'
Fig. 914
z, = r
Smith chart with polar coordinates.
+
jx is located, it is a simple matter to draw a circle centered at the origin through the point. The fractional distance from the center to the point (compared with the unity radius to the edge of the chart) is equal to the magnitude Irl of the load reflection coefficient; and the angle that the line to the point makes with the real axis is 0,. This graphical determination circumvents the need for computing by Eq. (9124). Each Irlcircle intersects thc real axis at two points. In Fig. 914 we designate the point on the positivercal axis (OfJ,,,)as I', and thc point on thc: ncgativcrcal axis (OP,,) as P,. Since x = 0 along the real axis, l', and P , both represent situations with a purely resistive load, Z , = R,.. Obviously RL > R, at P,,, where r > 1 : and R L < I=
0.60/21".
(rl= 0.60 circle intersects wit6 the positivereal axis OP.. at r = S = 4. Thus the voltage stwndi~~~wavc ratio is 4.
h) The
,
*.
,418
THEORY AND APPLICATIONS OF TRANSMISSION LINES / 9
Fig. 916
I
.
Smithchart calculations for Examples 99 and 910.
c ) To find the input impedance, we proceed as follows: 1. Move P; at 0.220 by a total of 0.434 "wavelengths toward generator," first to 0.500 (same as 0.000) and then further to 0.1%[(OSOO  0.220) + 0.154 = 0.4341 to P;. 2. Join 0 and P; by a straight line which intcrsects the Irl = 0.60 circle at P,.
;r
f
.,
1
L,
95 / THE SMITH CHART
'1 i
1
3. Read r = 0.64 and x = 1.2 at P,. Hencd,
..
; z;=
ROz, = lOO(q69
419
,
.
+j1.2) = 69 + j120 (R).
d) In going from Pi to P,, the lrl = 0.60 circle intersects the positivereal axis OPoc at P, where thd voltage is a maximum. Thus, a voltage maximum appears at (0.250  0.220)A or'0.030A from the load. Example 911
~ o h Ee ~ a m p l e97 by using the Smith chart. Given 4 w
1., ,, I
Ro .? 50 (Q) S=3.0
2 = 2 x 0.2 = 0.4 (m) First voltage minimum at rk = 0.05 (m),
find (a) a), (b) Z,, (c)'&, 2nd R,,, (Fig. 91 1). Solution
a) On the positivercill dris OP,. locale ~ h po~nt c I>,,, .lr wlilcb = s = 3.0 (see F , ~ . 917). Then DM = /r/ = 0.5 (mot = 1.0). We cannot find 0, until we have located the point that represents the normalized load imped;ince. b) We use the following procedure to find the load ihpedance on the Smith &art: 1. Draw a circle centered at the origin with radius which intersects with the negativer a1 anis OP,. at P,,, where there will be a voltage minimum. 2 Since &/i = J05/0.4 = 0.125, move from P,. 0.125 '6wav&ngths toward load7' in the counterclosewise direction to P;. 3. Join 0 and P; by a straight line, intersecting the (TI = O.j circle at P,.This 1s the point rebrescnling the normalized load impedance. 4. Read the angle iPOCOP;= 90' = n/2 (rad). There is no need to use a protractor, because L POCOP;= 4n(0.250  0.125) = n/2. Hence 6 , = @ ( n d ) , or r = 0.5140" = jO.j. 5 Read at PL,2 , = 0.60  j0.80, which gives
m,\,,
.
rcle at P,.

All the above results are the same as those obtained in Example 97, but no calculations with Complex numben are needed in using the Smith chart.

420
THEORY AND APPLICATIONS OF TRANSMISSION LINES 1 9
Fig. 917
Smithchart calculations for Example 911.
95.1 SmithChart Calculations for Lossy Lines
In discussing the use of the Smith chart for transmissionline calculations, we hav.e assumed the line to be lossless. This is normally a satisfactory approximation for we generally deal with relatively short sectioxis of lowloss lines. The lossless assumption enablcs us to say, following Eq. (9130), that the magnitude of the Te12"' term
?
S'
I
1
does not change with line length z' and that we can find z, from z,, and vice versa, by moving along the lrikircle by an angle eqhal to 2p.2'.
1
1I 8 I
i
For a lossy line of a sufficient length 8, sdch that 2 d is not negligible compared to unity, Eq. (9136) must be amended to read .r 1 + re 2 a ~ j2pz' ' ~ 2,
'
=
1  re2az' e 12pr'
l+(r(e2a'e~9
1
6 = or  2 w . (9132) ,r Hence, to find I , @om ,z, we cannot aimply move along the (TIcircle; auxiliary calciilations are necessary in order to account for the e'"' factor. The following example illustrates What has ta be done.

Example 912 Tlle input impedance of n shortcircuited loshy transmission line of length 2 (m) and characteristic impedance 75 $2) (approximately real) is 45 + j225 (0). (a) Find a and /3 of the line. (b) Determine the input impedance if the shortcircuit is replaced by a lo:ld impc~lilnccZ , = 67.5  j45 (Q). Solution a) The shortcircuit load is represented by the point P, on the extreme left of the Smith impedance chart. 1. Enter zil= (45 j225)/75 = 0.60 j3.O in the'chart as P , (Fig. 918), 2. Draw a straight line from the origin OthrouSh P , to P;. 3. Measure OP,/OP', = 0.89 = c". It follows that
+ 
+
4. Record that the arc P,P; is 0.20 "wavelengths toward generator." We have //A = 0.20 and 286 = 4n//A = 0 . 8 ~Thus, .
b) To find the input impedance for Z , = 67.5  j45 (R): I . h n k r ZL = %,/2, = (67.5  /45)/75 = 0.9  jO.6 on thc Smith chart as P,. 2. Draw a straight line from 0 through P, to P2where the "wavelengths toward generator" reading is 0.364. 3. Drawa/r(circlecentered at 0 with radius 4 Move P i alohg the perimeter by 0.20 "wavelengths toward generator.' to Pi at 0.364 0.20 = 0.564 or 0.064. , 5. Join P3 and d by a straight line, intersecting the /r/circle at P,. 6. Mark on line OP, a point Pi such that 0P1/CF3= e2a' = 0.89. 7. At Pi, read zi * 0.64 j0.27. Hence,
m2.
.
we nave on for we sumption 12i7:' 4 term
+
+
2, = 75(0.64
+ j0.271 = 48.0 + j20.3 (R).
\
422
THEORY AND APPLlcATloNs OF TRANsMlssloN LINES 1 9
e i
I
96.1 Quarter
/.
Fig. 918
Smithchart calculations for lossy transmission line
(Example 912). 96 TRANSMISSIONLINE IMPEDANCE MATCHING
Transmission lines are used for the transmission of power and information. For radiofrequency power transmission it is highly desirable that as much power as possible is transmitted from the generator to the load and as little power as possible is lost on the line itself. This will require that the loztd be matched to the characteristic
>
I:
96.1 Impedance ~ a k h l by h ~ QuarterWave Transformer '
Kb = ,/=.
(9133)
Since the length of the quarterwave line depends on wavelengrh, this matching is frequency~eniitive.as are a]! the other methods to be discussed. Example 913 A &pal generator is to feid equal power through a lossless air transmission line with a characteristic impedance 50 (Q) to two separate resistive loads, 64 and 23 (n).Quarterwave transformers are used to match the ]oa& to the 50 cn) line, as shouo in Fig. 9 1% (a) Determine the required charactenstic impedances of the quarlerwave lines. (b) Find the standingwave ratios on the matching line scctiohs. , .
fig. 919
impedance matching by quarterwave lines
(Ex:irnplr 9 13).
i
424
THEORY AND APPLICATIONS OF TRANSMISSION LINES 1 9
Solution
a) To feed equal power to the two loads, the input rehstance at the junction with the main line' looking toward each load must be equal to 2Ro. Ril = Ri2 = 2Ro = 100 (Q).
~b~ = J G 2=
~ = J i W Z i = 8 o ( n )
Jm 4TGEZ =
= 50 (Q).
b) Under matched conditions, there are no standing waves on the main transmission line (S = 1). The standingwave ratios on the two matching line sections are Matching section No. 1
.
Matching section No. 2
.
Ordinarily the main transmission line and the matching line sections are essentially lossless. In that case both Ro and Roare purely real and Eq. (9133) will have no solution if R L is replaced by a complex ZL.Hence quarterwave transformers are not useful for matching a complex load impedance to a lowloss line. In the following subsection we will discuss a method for matching an arbitrary load impedance to a line by using a single open or shortcircuited line section (a single stub) in parallel with the main line and at an appropriate distance from the load. Since it is more convenient to use admittances instead of impedances for parallel connections, we first examine how the Smith chart can be used to make admittance calculations. Let YL = l/ZL denote the load admittance. The normalized load impedance is
where
YL = YLP0 = YLIGo . = ROYL= g
+j b
(Dimensionless),
(9135)
ion with = Rtz =
smission .Sare
.
is the normalized load admittance having normalized conductance g and normalized susceptance h as its teal and imagin;~rypdrts respectively Equ;ltion (9134) suggcsts 111:11 ; I qtmrlcrw;tvt li:w will, ;I tt~l~ly ilornlilli~dchar:~~tert~tic impedance will l j ' m h  r n 2 , lo y,, i l t i U vice versa. On the Smith chart we need only to move the point representing T L along the IT(circle by a quarterwavelength in order to locate thc point r c p r e s ~ n t i h ~,. Since a 214change in line length (AzlN = i)corresponds to a change of n radians (2PAzf = x) on the Smith chart. the poir~tsrrpreset~titlgZ, and y, are then diametrically opposite to eachuther on the Jr/circle.This observation enables US to find yi. from r,, and z, from y,, on the Smith chart in a very simple manner.
re essenwill have mers are
, ubitrary sction (a from the r'parallel mittance
Solution: This problem has nothing to do with any transmission line. In order to use the Smith chart, we can choose an arbitrary normalizing constant; for instance, RO = 50 (R).Thus,
Enter z, as point P , on the smith chart in Fig. 920. The point P, on the other side of the line joining PI and 0 represents y,: 07, =
m1.
1 
Example 915 ~ i n the d input admittance of an opencircuited line of characteristic impedance 300 (a)and leapth 0.042.
426
THEORY AND APPLICATIONS OF TRANSMISSION LINES 1 9
a
Fig. 921 Finding input admittance of opencircuited line (Example 915).
Solution 1. For an opencircuited line, we start from the point Po, 6nYthcextreme right of the impedance Smith chart, at 0.25 in Fig. 921. 2. Move along the perimeter of the chart by 0.04 "wavelengths toward generator" to P 3 (at 0.29). 3. Draw a straight line from P 3 through 0, intersecting at P; on the opposite side. 4. Read at P; y, = 0 jO.26. Thus, 1 (0 j0.26) = j0.87 (mS). 300
+
x=
+
In the preceding two examples we have made admittance calculations by using the Smith chart as an impedance chart. The Smith chart can also be used as an admittance chart, in which case the r and x circles would be g and h circles. The points representing an open rind shortcircuit termination would be the points on the extreme left and t%e extreme right, respectively, on an admittance chart. For Example 915, we could then start from extreme left point on the chart, at 0.00 in Fig. 921, and move 0.04 "wavelengths toward generator" to P; directly. 96.2
SingleStub Matching
We now tackle the problem of matching a load impedance 2, to a lossless line that has a characteristic impedance R, by placing a single shortcircuited stub in parallel with the line, as shown in Fig. 922. This is the singlestub method for impedance matching. We need to determine the length of the stub, L', and the distance from thc load, z', such that the impedance of the pariillel combination to the right of points BB' equals R,. Shortcircuited stubs are usually used in preference to opencircuited
I
$ 2
stubs ~ C C ; I U S Ci l n inlinitc tormin:~lingi m a d i ~ n u IS : more difficult to re;~lim (hiin il zero t o d i n a t h i irdpcdancc for reasons of radiation from an open end and coupling effects with neighbbring,objects. Moreover, a shortcircuited stub of an adjustable length and a constant qharacteristic resistdhce is much easier to construct than an opencircuited one. Of course, the difference in the required length for an opencircuited stub and,that for a shortcircuited stub is an odd multiple of a quarterwavclcncth. The parallel cdhbination of a line terminated in Z, and a stub at points BB in Fig. 922 sugsddts that it is advantageous to analyze the matching requirements in terms of admitt~hces.The basic requirement is

>

ttance 915).
In terms of normalized Lidmittances.Eq. (9 136) becomes I
rght of
' r
eratc
tte stae.
1= + J*,, (9137) where y , = ROY, is for tne load sectloll and y , = ROY,is for the shortcircurted stub. However, since thdi input admtttance of a shortcircuited stub is purely susceptive. y, is purely imaginary. As a consequencg.Eq. (9137) can be satisfied only if
and
YII
=1
+ jb,
.
(9138a)
Y B=  J ~ B , (9138b) where b, can be either positive or negative. Our objectives, then, are to find the length d such that the admittance, y,, of the load section looking to the rlght of terminals BE' has a hnity r e d purt and to find the length fB of the stub required to cancel the imuginury purr.
y using i as an es. The 1111h fill l
l 1'111
0.W i n
n
ne th, 3ara" ' ledan, om the points rcuited
Y
Fig: 922 Impedance matching by ringicrtvb method.
428
THEORY AND APPLICATIONS OF'TRANSMISSION~LINES/ 9
Using the Smith chart as an admittance chart, we proceed as follows for single, stub matching: t
1. Enter the point representing the normalized load admittance y,. 2. Draw the Irlcircle for y,, which will intersect the g = 1 circle at two points. At these points y,, = 1 jbBl and y,, = 1 jb,,. Both are possible solutions. 3. Determine loadsection lengths dl and d2 from the angles between the point representing y, and the points representing y,, andly,,. 4. Determine stub lengths t,, and tB2 from the angles between the shortcircuit point on the extreme right of the chart to the points representing jb,, and jh,], respcctively.
+
+
The following example will illustrate the necessary steps. Example 916 A 50(R) transmission line is connected to a load impedaice Z, = 35  j47.5 (Q). Find the position and length of a shortcircuitcd stub rcquircd to march rhc linc. \
Solution: Given
R o = 50(Q)
2, z,
=
.,
35  j47.5 (R) = 0.70  j0.95.
= Z,/R,
1. Enter z, on the Smith chart as P , (Fig. 923). 2.. Draw a Irlcircle centered at 0 with radius 3. Draw a straight line from P, through 0 to point P i on the perimeter, intersecting the Il1circle at P,, which represents y,. Note 0.109 at P; on the "wavelengths toward generator" scale. 4. Note the two points of intersection of the Irlcircle with the g = 1 circle.
m,.
+
+
At P,: y,, = 1 j1.2 = 1 jb,,; AtP,: y,,= 1 j1.2= 1 +jbBz. 5. Solutions for the position 'of the stub: For P, (from P; to P;): d l = (0.168  0.109)A = 0.0591.; For P 4 (from P; to f i ) : d, = (0.332  0.109)A = 0.2231. ,
6. Solutions for the length of shortcircuited stub to provide y, = jb,: For P , (from P,, on the extreme right of chart to P;',which represents jh,, j1.2): For P 4 (from P,, to Pi,which represents jb,,
= j1.2):
=
r single
.. points. Jutions. e point
t .
t
i
<
:

sccting lengths
Fig. 923
Construction for singlestub matching.
In general,. the solution with the shorter lengths is preferred unless there are other practical constraints. The exact length, t'B,of the shortcircuited stub mey require fine adjustments in the actual matching procedure; hence the shorted matching sections are sometimes called stub tuners. The use of Smith chart in solving impedancematching problems avoids the manipulation of complex numbers and the computation of tangent and arctangent
_.
.?.
I
430
THEORY AND APPLICATIONS OF TRANSMISSIQN LINES / 9 1
functions; but graphical c~nstructionsare needed, and graphical methods have limited accuracy. Actually the analytical solutions of jmpedancematching problems are relatively simple, and ebsy access to a computer'may diminish the reliance on the Smith chart and, at the same time, yield more accurate results. For the singlestub matching problem illustrated in Fig. 922, we have, from Eq. (989),
where (9 140)
t = tanpd. The normalized input admittance to the right of points BB' is
where
and
A perfect match requires the simultaneous satisfarion of Eqs. (9138a) and (9l3Sb). Equating g, in Eq. (9142a) to unity, we lii~vc . (r:  I)t2  2sLt
+
(I.,
 1.;  .yl) = 0.
(9133)
Solving Eq. (9143), we obtqin
The required length d can be found from Eqs. (?140), (9144a), and (91 44b):
 tan't, (n
t20
+ tanIt),
,
t ?:rt~ti~ti/~t~r(~tio! l~,,l/l~,l) in t c m s of 2, and Z L ,

2' s the for a >JO(R), ie inner radius tor is 0.6 (mm).
+
P.919 A 75(R)lossless lj& is terminated in a load impedance Z , = RL jXL. a) What must be the relation between RL and X, in order that the standingwave ratio on !$e line be 31 b) Find X,, if dt = 150 (Q). c) Where does the voltage minimum &arest to the load occur on the line for part (b)?
. 1
440
THEORY AND APPLICATIONS OF TRANSMISSION LINES I 9
s. .. P.920 Consider a lossless transmission line. a) Determine the line's characteristic resistance so that it will have a minimum possible standingwave ratio for a load impedance 40 + j30 (Q). b) Find this minimum standingwave ratio and the corresponding voltage reflection coefficient. c) Find the location of the voltage minimum nearest to the load.
P921 A lossy transmission line with characteristic impedance Zois terminated in an arbitrary load impedance ZL. a) Express the standingwave ratio S on the line in terms of Z0and ZL. b) Find in terms of S and Zothe impedance looking toward the load at the location of a voltage maximum. C) Find the impedance looking toward the load at a location of a voltage minimum.
P.923 The standingwave ratio on a lossless 300(R)transmission linet'bnhated in an unknown load impedance is 2.0,and the nearest voltage minimum is at a distance 0.31. from the load. Determine (a) the reflection coefficient l of the load, (b) the unknown load impedance Z L , and (c) the equivalent length and terminating resistance of a line, such that the input impedance is equal to 2 , . P.924 Obtain from Eq. (9114) the formulas for finding the length t,,,and the terminating resistance R, of a lossless line having a characteristic impedance Rosuch that the input impedance equals Z i = R, j X i .
+
P.925 Obtain an analytical expression for the load impedance ZL connected to a line of characteristic impedance Zoin terms of standingwave ratio $ and the distance, zui., of the voltage minimum closest to the load. P.926 A sinusoidal voltage generator with V, = O . l p (V) and internal impedance Z , = Ro is connected to a lossless transmission line having a characteristic impedance Ro = 50 (R). The line is f meters long and is terminated in a load resistance R, = 25 (R).Find (a) I i , VL,and 1,; (b) the standingwave ratio on the line: and (c) the average power delivered to the load. Compare the result in part (c) with the case where R, = 50 (R).
v,
.
P.927 A sinusoidal voltage generator u, =, ll0sinwt (V) and internal impedance 2, = 50 (R) is connected to a quarterwave lossless line having a characteristic impedance Ro = 50 (R) that is terminated in a purely reactive load ZL= j50 (R). a) Obtain voltage and current phasor expressions V(zl)and I(zl). b) Write the instantaneous voltage and current expressions u(zi, t) and i(z1,t). c) Obtain the instantaneous power and the average power delivered to the load. P.928 The characteristic impedance of a given lossless transmission line is 75 (R). Use a Smith chart to find the input impedance a n 0 0 (MHz) of such a line that is (a) 1 (m) long and opencircuited, and (b) 0.8 (m) long and shortcircuited. Then (c) determine the corresponding input admittances for the lines in parts (a) and (b).
?
+
P.929 A load impedadce 30 110 (Q) is connected d(ba l~sslesstransmission line of length 0.1011. and characteristic im&danke 50 (a). Use a Smith. chart to find (a) the standingwave ratio, (b) the voltage re~ectlbncoefficient, (c) the jnput impedance, (d) the input admittance, and (e) the location of the $b~tagdm~nimum on thiline.$
ssible :don
~ 9  3 0Repeat probl+ P.9%29for a load i m p d a n k 30  jlO (a). P.931 In a laborator;, exp&ment conducted d;l a SO($2) lossless transmlss~onhne ternmated in an unknown load Ypedance, it is found that the standingwave ratio is 2.0. The successive voltage minima are 2$(cm) apatt and. the first minknum occurs at 5 (cm) from the load. Find (a) the load i m p d a n d , and (b) the reflection meffiEient of the load. (c) Where would the fint voltage minimum be Idtated'if the load were replaced by a shortcircuit?
jitrary
nofa n.
a load crlstic
.
mown Isa* L.
4
in
indtlng edance
P.932 Thc input imdduncc of a shortcircuited l o s e transmission line of length 1.5 (m) (ci/2) and characteristic impedance.100 (R) (approximately real) is 40  j280 (R). a) Find a: and /Iof the line. b) Determine the input impedance if the shortcircuit is replaced by a load impedance ZL= 50 + j50 (Q). c) Find the input impedance of the shortcircuifed line for a line leneth 0.15i..

P.933 A dipole antenna having an i n p ~ ~iili~diince t of 73 (0)is fed by i 200(MHz) source through a 3W(n) t w o 4 r e trsnsmission line. Desidn a quarterwave twowire air line with a 2(cm) spacing to matcg the antenna to the 300(Q) line. P.934 The singlestub melhod is used to match a load impedance 25 + j25 (Q) to a 50(R) transmission line. a) Find the required length and position of a ihortcircuited stub made of a section of the same 50(R) linc. b) Repeat part (a) asstming the shortcircuited stub is made of a section of a line that has a characteristic! impedance of 75 (Q). ' I'
'
)f char4
oltage
= Ro is The line (b) the )are the
=
P.935 A load impedance can be hatched to a transmission line also by using a single stub placed in series with the load I t an appropriate location, as shown in Fig. 927. Assuming ZL= 25 + j25 (Q), Ro = 50 (R), and KO= 35 (0). find d and I required for matching.
so (n)
(R) that
P Smm d open~g lnput
Fig. 927
Impedancg hatching by a series stub.
P.936 The doublestub method is used to match a load impedance 100 + j103 (Q) to a losrless transmission line of characteristic impedance 300 ($2). The spacing between the stubs is 3218, with
I
',
.
" ,
,,,one stub connected directly in parallel with the load. ~ e t e n n i n dthe lengths of ihe siub tuners if .. , (a) they are both shortcircuited, and (b) if they are both opensircuited. 6
"

,)l
L,
*
.
P937 If the load impedance in Problem P.936 is changed to 100 +j50 (Q), one discovers that a perfect match using the doublestub method with do = 3118 and one stub connected directly across the load is not possible. However, the modified arrangement shown in Fig. 926 can be used to match this load with the line. a) Find the minimum required additional line length dL. b) Find the required lengths of the shortcircuited stub tuners, using the minimum dL found in part (a). P938 The doublestub method shown in Fig. 924 cannot be used to match certain loads to a line with a given characteristic impedance. Determine the regions of load admittances on a Smith admittance chart for which the doublestub arrangement in Fig. 924 cannot lead to a match for do = A/16,1/4, 3118, and 71/16.
d, found ' I
lbads to :ces on a .ead to a
,,
! 1,
'.\
101

INTRODUCTION
I
'
In the preceding cha&er we studied the characteristic properties of transverse slec) guided by transmission lines. The TEM mode of guided tromagnetic ( T E ~ waves waves is one in which the tlectric and magnetic fields are perpendicular to each other and both are transvkrfe tc. the direction of propagation along the guiding line. One of the salient properties i ~ fTEM waves guided by conducting lines of negligible resistance is that the velocity of propagation of a wave of any frequency is the same . as that in an unbounded dielectric medium. This was pointed out in connection with Eq. ( 9 d 1 ) and wasreinforced by Eq. (955). TEM waves, however, are not the only mode of guided waves that can propagate on transmission lines;. nor are the three types .of transmission lines (parallelplatz, twowire, and coaxial) mentioned in Section 91 the only possible waveguiding structures. As a mattcr of hct, wc scc from Eqs. (945a) and (949a) that thc Litteriualion constant resulting iiom thc linite conductivity of the lines increases with R. the resistance per unit line length, that, in turn, is proportional to f i in accordance with Tables 91 and 92. Hence the attenuation of TEM waves tends to increase monotonically with frequzncy and would be prohibitively high in the microwave range. In this chapter we first present a general analysis of the characteristics of the waves propagating along uniform guiding structures. Waveguiding structures are called waveguides, of which ,the three types of ttansmission lines are special cases. The basic governini equations will be examined. We will see that, in addition to transverse electromagrfetic (TEM) waves, which have no field components in rhe direction of propagation, both tru'nscerse r~tnqrzitic(TM) waves with a longitudinal electricfield cornponefit and tmr~sverse electric (TE) \caws with a longitndir.~! magnetictield.eomponent crin also exist. Both TM and TE modes have characterisric cutofl fi.cyuerf~?c.s. Waves of frcquencics below thc cutolT frequency of a particillar lnode cannot propagate, atid power and signal transmission at that mode is possible only for frequencies highe;. than the cutoff frequency. Thus, waveguides operating in T M and T E modes are like highpass filters. Also in this chapter we will reexamine the. field and wave characteristics of parallelplate waveguides with emphasis on TM and TE modes and show that all
.,.
,
.,
&.>
...... $
f
.
,
, . ,
I . , . .
. . .  . . ..,.#> .., b
1
<
.
.  1 .
,
t.
, . I
' I
I'
I
I
can be expressed in teims of~E,'(z being thk diredkn of propagation) for TM waves, and in ternis 6f Hzfbr TE wave's. The attenuation ., constants resulting from imperfectly conddctihg prates will be detkmiined for TM and TE waves, and we will find that the attenuation cbnstant depends, in a comcomponents , transvers~~field
,a,
'?.
.
plicated way, on the mode of the propagating wave, as well as on frequency. For some modes the attenuation may decrease as the frequency increases; for other modes, the attenuation may reach a minimum as the frequency exceeds the cutoff frequency by a certain amount. Electromagnetic waves can propagate through hollow'metal pipes of an arbitrary cross section. Without electromagnetic theory it would not be possible to explain the properties of hollow waveguides. We will see that singleconductor waveguides cannot support TEM waves. We will examine in detail the fields. the current and charge distributions, and the propagation characteristics of rectangular waveguides. Both TM and TE modes will be discussed. An analysis of the propertie$ of circular waveguides requires a familiarity with Bessel functions as a consequence of manipulating Maxwell's equations III cylindrical coordinates. Circular w:~voguidcs will not be studied in this book. In millly applic:ltions wave psopagation in a rcctan,oular waveguide in the dominant (TE,,) mode is desirable because the electric field in the guide is polarized in a fixed direction. Electromagnetic waves can also be guided by an open dielectricslab waveguide. The fields are essentially confined within the dielectric region and decay rapidly away from the slab surface in the transverse plane. For this reason, the waves supported by a dielectricslab waveguide are called sudtrce waces. Both TM and TE modes are possible. We will examine the field characteristics and cutoff frequencies of those surface waves. At microwave frequencies, ordinary lumpedparameter elements (such as inductances and capacitances) connected by wires are no longer practical as resonant circuits because thc dimcnsions of thc clcrncnts would hove to bc cxtrcmcly small, because the resistance of the wire circuits becomes very high as a result of the skin effect, and because of radiation. All of these difficulties are alleviated if a hollow conducting box is used as a resonant device. Because the box is enclosed by conducting walls, electromagnetic fields are confined inside the box and no radiation can occur. Moreover, since the box walls provide large areas for current flow, losses are extremely small. Consequently, an enclosed conducting box can be a resonator of a very high Q. Such a box, which is essentially a segment of a waveguide with closed end faces, is called a cavity resonator. We will discuss the different mode patterns of the fields inside rectangular cavity resonators.
102 GENERAL WAVE BEHAVIORS ALONG. UNIFORM GUIDING STRUCTURES
In this section we examine some general chatactcristics for waves propagating along straight guiding structures with a uniform cross section. We will assume that the waves propagate in the +r direction with a propagation constant J = ci jB that
+
'
ti
firection muation for TM :acornx y . For or other ne cutoff
, 8 .
.
explain veguides rent and \
)
of circuveguides a rcctan
x i : field n
:ve: e. rapidly ives supand TE ~quencies :h as in
resonant :y small, :he skin .I ho!low nduc~ing In occur. :s tremely yhigh Q. ;faces, is :he fields
Ing along rhat the  j/3 that
,L
\,
.n*i: 1 .
I I
I
i
'
is yet to be aetermined. For harmonic time dependence with an angular frequency o, the depende&e on r ahd t for all field components can be described by the exponential factor i= e ( j ~t Y Z ) = eej(ut  BZ) (10la) AS an exa.mpl; for a. bosine reference we may write the instantaneous expression for the E field as E(x, y, z ; t ) = %[EO(x, y ) e ( j U '  Y ' ) ] , (10lb) a
where EO(x,y) is il twodimensional vector phasor that depends only on the crosssectional coordinated. The instantaneous expression for the H field can be written in a similar way. Hence. in using a phasor representation in cquations relating field quantities, we may replacc partial derivatives with respect to i and z simply by products witki ( j o )and (  7 )respectively; the common factor r ' j w '  " ' can be dropped: We consider a 9tra.ght waveguide in the form of a dielectricfilled metal tube having an arbitrary n o t s section and lying along the . axis, as shown in Fig 101. According to Eqs. (78.6) and (787), the electric and magnetic field intensities in the chargefree diclcktr:~region inside satisiy the following homogeneous vector llclmholtz's equations: V'E + k'E = 0 and V2H + k" = 0, (102b) where E and H are threedimensional vector phasors; and k is the wavenumber rk = Wd,LLE. (103) The threedimensiona! Laplncian operator V2 may be broken illto two parts: 2 FuIu2 for the crosssectional coordinates and Vf for the longitudinal coordinate. For waveguides with a rectangular cross section, we use Cartesian coordinates:
= V.
. (1018)
We note that Z, is the same as the intrinsic impedance of the'dielectric medium, as given in Eq. (825). Equations (1016) and (1018) assert that the phase uelocity and the wave impedance for TEA4 waves ore iirdependent of rl~efrequency of the ivnves. Letting EP = 0 in Eq. (107a) and H p = 0 in Eq.(108b), we obtain
Equations (1017) and (1019) can be combined to obtain the following formula for a TEM wave propagating in the + 2 direction:
which, again, reminds us of a similar relation for a uniform plane wave in an unbounded mcdium see Eq. (824). Singleconductor waveguides cannot support TEM waves. In Section 62 we pointed out that magnetic flux lines always close upon themselves. Hence. if a TEM wave were to exist in a waveguide, the field lines of B and H would form closed loops in a transverse plane. However, the generalized Ampere's circuital law, Eq. (738b), requires that the line integral of the magnetic'field (the magnetomotive force) around any closed loop in a transverse plane must equal the sum of the longitudinal conduction and displacement currents through the loop. Without an inner conductor, there is no longitudinal conduction current inside the waveguide. By definition, a TEM wave does not have an E, component; consequently, there is no longitudinal displacement current. The total absence of a longitudinal current inside a waveguide leads to the conclusion that there can be no closed loops of magnetic field lines in any transverse plane. Therefore,. we conclude that TEM waves cannot exist in a singlecondttcror hollow (or die1ectric;filled) waceguide of any shape. On the other hand, assuining pei$ect conductors, a coaxial transmission line having an inner conductor can support TEM waves: so can a twoconductor stripline and a twowire ,transmission linc. Wllcn (17c conductors havc losscs, waves along transmission iirics arc strictly n o lo~igcr'I'LM. 11o1cdi l l S c c t i o ~9 ~ 2. 102.2
Transverse Magnetic Waves
Transverse magnetic (TM) waves do not have a component of the magnetic field in the direction of propagation. H, = 0. The behavior of TM waves can be analyzed by solving Eq. ( I 0 .5) l i ) E, ~ subject to tlic hountl:lry conditions o r tlic guiclc :lnd using
Eqs. (109) ihrough (1012) to determine the other components. Writing Eq. (105) for E;, we have L ViyE: (y2 k?)~: = 0 (1021) ,.;  or
+ +
.
:diuni, ilocity waves.
..
, I
Equation (1022),is a secondorder partial diffdrential equation, which can be solved for E:. In thitsection we wish only to discuss,the general properties of the various wave types. he actual solution of Eq. (1022) will wait until subsequent sections when we examine patticu!ar wavcguidcs. For TM waves, we set Hz= 0 m Eqs. (109) through (1012) to obtain
ula for
2 we .TEM
loops 38b), :round mduc:. there . TEM .a1 diseguide lnes in ,r pl a 2 other :r con:owire In lines !
P
field in :zcd by ii using
It is convenient to combi1.e Eqs. (1023c) and (1023d) and write
L
where
denotes the grkdient of 1:: in the tmnsvcrse plane. Equation (1024) is a concise formula for finding E: an.i E,O from Ep. . The transverse components of magnetic field intensity, H, and H,, can be determined simply from Ex a r d E, on the introduction of the wave impedance for the TM mode. We have, from Eqs. (1023),
.
It is important to note thai Z.,. is not cquil? to jw,u/y, becausc y for TM waves, unlike YTEM, is not equal to jo The following relation between the electric and magnetic
,/a.
field intensities holds for TM waves: 1
J L Equation (1027) is seen to be of the same form as Eq. (1020) for TEM waves. When we undertake to solve the twodimensional homogeneous Helmholtz equation, Eq. (1022), subject to the boundary conditions of a given waveguide, we will discover that solutions are possible only for discrete ralues of h. There may be an infinity of these discretc valucs. hut solutions ilre not possible for all values of 11. The values of11 for which a solution of Eq. (1022) exists are called the c h i f r i r c t ~ r i ~ f i ~ of the boundaryvalue problem. Each ol the rigenvalues deteruolllrr or eioe~~crilues mines the characteristic properties of a particular TM mode of the glven waveguide. In thc following sections we will also discover that the ei~envaluesof the various wilveguidc problcms are real numhcrs. From Eq. (101:) we have 7 = ,/I,:  1;:  r? = ,jh  (?)/A€. (103) 7
Two distinct ranges of the values for the propagation constant are noted, the dividing point being y = 0, where w:,u E = 112 (1029)
The frequency,.j,, a t which 7 = O is called a c u t o l j ' j r c q ~ ~ ~ ~Ti" n c y .uuhe ($,I; for a partieuior mode in a wuueyuidr dcpeads or1 lie eige~lcaluruj this mode. Using Eq. (1030), we can write Eq. (1028) as (1031) The two distinct ranges of p can be defined in terms of the ratio (f/jd2 as compared to unity. )
(
>1
, > . i n this rrngr, w z y r ; /I' and 7 is imaginary. w e have. from
Eq. (102% = jp = j k
  .
dl .
 (Iencies below cutoK is purely reactive. indicating that tlicrc is no powcr llow associ:~tcdwit11cv;inusccnt wa\'cs.
102.3
Transverse Electric Waves
Transverse electric (TE) waves do not have a component of the electric field in the direction of propagation, Ez = 0. The behavior of TE waves can be analyzed by first solving Eq. (106) for H,:
Proper boundary conditions at the guide walls must be satisfied. The transverse field components can then be found by substituting H z into the reduced Eqs. (109) through (1012) with 4, set to zero. We have
SHP tr; = Lh 2 C),
_ ). : '. . . +
102 1 GENERAC
.
.i'
W A J BEHAVIORS ~ ALONG
. ,
c .
UW~FORM GUIDING STRUCTURES
453
I
7 e ' cutoff '
. . . , ;.. . ,.. !i1!$i' :
(1039)
Combining Eqs. (1042a) and (1042b). we obtain
., ,, ,
We note that Eq. (1043) is entirely similar to Eq. (1024) for TM modes. The transverse components of eiectric field intensity, EO and E;, are related to those of magnetic field intensity through the wave impedance. We have, from Eqs. (1042% b, c, and d),
ntuin the id is said highpass :he cutoff ' >.
:dance of
(1040)
Note that Z,, in Eq. (1044) is quite different from Z,, In Eq. (1036) bemuse ;) for TE waves. unlike ynln is ,lor c q l ~ a lto ,or fi.Equations ( 1 0 4 2 ~ ) (10 . 42di. ad ( I 0 44) c;in t w w bc c o ~ ~ l l ) i IO ~ lgive ~ d LIIC SOIIOWIIIC VCCLOI lil.~:iuIa:
3W c p f f ;meL ~ l t
Inasmuch as we hahc not changed the relation between 7 and 11. Eqs. (1028) through (1031) pertaining to TM waves also apply to TE waves. There are also two distinct ranges of 7, depending on whethdr the operating frequency is higher or lower than thc cutort'frcqu~nc~, givcn in Eq. (1030). JC,
Id in tlic Jyzed by
a)
(
> 1, or j' .
In this range 7 is imaginary. and we have a propagaring
mode. The expression for 7 is the same as that given in Eq. (1032): (1041)
(1046) xansverse . p . (109)
( 1042a)
(l0Pb)
*k)
( 10
( 1043d)
Consequently, the formulas for 1,)L,,u p , and u, in Eqs. (1033). (1034). (1036). and (1037), respectively. also hold for TE waves. Using Eq. (1046) in Eq. (1044), we obtain
I
b)
($)2
< 1, or / propagate with :I phase constant A given in Eq. (1033): and waves with / 5 J; are evanescent. Depending on the value of n. h e r e are different possible propa,.o ~ t i n gTM modes /I. Tllar. ~llcrc;lrc.lllc T k l , (e~genmodes)corresponding to l l ~ ciiilhrent cigcll~;~lucs mode (11 = I) with cutolYfrrquency ( j ; ) , = l/ll~,/~a. the TMl.mode (11 = 2) with = I/hdli+ and so on. Each mode has its own characteristic phase constant, guide wavelength, phase velocity. group velocity, and wave impedance; they can be determined from. respectively, Eqs. (1033), (1034). (1036), (103i), and (1038). When n = 0, E, = 0, and only the transverse components H , and E,, exist. Hence TIM, mode is the TEM mode, for which j, = 0. The mode having the lowest cutoff frequency is called the dornindnt mode of the waveguide. For parallelphte waveguides, the dominanr rnorie i s rlw TEM inode. iC
Example 103 (a) Write the instantaneous field expressions for T M , mode in a parallelplate waveguide. (b)Sketch the electric and magnetic field lines in the yzplane.
Solution
a) The instantaneous field expressions for the TM, mode are obtained by multiplying the phasor expressions in Eqs. (1054a), (1054b), and (1054c) with eJ("'a' and taking the real part of the product. We have, for 11 = 1,
where
,..
ncd Iionl
;.
i
. .b) In the
,
."
yz
E has both a y and a z component, the equation of the electric . ,:i.,$eId l lines ,  w a given t can be found from the relation: I.*
%.
,
, . .:
,
 .
d y = ddz E, EL
(1055)
*
(1059)
For example, at t = 0, Eq. (1059) can be Written as
d ~=' E,(Y, I; 0) 
[email protected] dz
(1056) h a phaswM . modes the T M , 1th (I;), = :nt, gutdc be @rcn M, jde qu$ncy is the domi
node in a I pzplane.
E,(y,z;O)
le
cot
(y)
tan f i z ,
which gives the dope of the electric field lincs. Equation (1060) can be integrated to give
which is the equhtion of the electric field line for a particular yo at ;= 0. DiLrent values of y,, givc dilfcrcnt loci. Severill i l c h electric licld lincs ;ire d r ~ w nin Fig. 106. The ficld lines rcpwt tl~cmaclvcsiur every cliangc o f 4 r by ?n rad. Since H ha3 only an s component, tNc magnetic field lincs are everywhere perpendicular to the y; pime. For thc TM, mode at t = 0, Eq. (1057c) becomes
The density of h, lines varier as cos ini/b) in the y direction and as sin 0 : in the z direction. This is also skctched in Fig. 106. At the conducting plates (y = 0 and y = h). there arr. surface currcnu because of a discontinuity in the tangential magnetic field and : urface charges because of the presence of a normal e1ec:ric field. (Problem 101).
5y multl 5 4 ~ )with
(1057a) (1057b)
r
I!
74

Electric t':c!d lines.
(1058)
a8 Fig. 106
Mngnctic field lincs (.\asis inld thc paper).
Field llnes I J r T M , mode in parallelplate waveguide.
.
& .
.
..
. .
460 . WAVEGUIDES AND CAVITY RESONATORS / 10 r
.
.
,
,.
!
9 ,
.!,
,,
Example 104 Show that the field solution of 9 propagating TM, wave in a parallel,piate waveguide can be interpreted as the superposition of two plane waves bouncing back and forth obliquely between the two conducting plates. Solution: This can be seen readily by writing the phasor expression of EP(y) from Eq. (1054a) for n = 1 and with the factor ejB' restored. We have
A1 [ e  j ( B :  n ~ l b ) =4
'.
2j

j(P:+n~lht
I.
(10G3)
From Chapter 8 we recognize that the first term on the right side of Eq. (1063) represents a plane wave propagating obliquely in the + z and y directions with phase constants3! , and n/b respectively. Similarly, the second term represents, a plane and + y directions with the same phase conwave propapting obliquely in the stants and d h as those of the first plane wave. Thus, a propagating T M , wave in a parallelplate waveguide can be regarded as the superposition of two planc waves. as dcpictcd in Fig. 107.
+:
In Subsection 86.2 on reflection of a ~arallellypolarized plane wave incident obliquely at a conducting boundary plane, we obtained an expression for the longitudinal component of the total E l field that is the sum of the longitudinal components of the incidcnt Ei and the retlectcd E,. To adapt the coordinate designations of Fig. 810 to those of Fig. 105, .c and z must be changed to z and  y respectively. Wc rcwritc E, of Eq. (886a)as
Comparing the exponents of the tcrms in this equation with those in Eq. (1063), we obtain two equations: (1064a) /3, sin Bi = p
.
X
p, cos Oi = b
Fig. 107 Propagating wave in parallelplate waveguide as superposition of two plane waves.
Y
.
.
/
.
/
p= which is the same as Eq. (1058), and COS
t.1063)
n /Z 0,= P,b  26'
(1065)
where I = 2n//3, is the wavelength in the unbounded dielectric medium. We observe thLt B solution o l Eq. (1065) for Oi exists only whcn L/Zb i I. At /./Zb = I. or /' = I I / ~= l / ? b , / ~ . whicll is the cutoff frcqucncy in Eq. (1056) for 1l = 1, cos Oi = I. and Oi = 0. This corresponds to the case when the w v e s bounce back and forth in the j9direction, normal to thc prailei plates. ilnd there is no propagation in the :direction ( b = 11, sin Oi = 0). Propagation of T M , mode is possibie only when /. c i, = 2b or f > ji.. Both cos Oi and sin Oican be expressed in terms of cutoff frequency /:.. FPorn Eqs. 11 0  65) and ( 1 064a) wc htvc
i 1063)
ns with "a plane se conaave in M'3 VCS, n
iClr
.L
he .:I cornxations ;~ively.
and
=5=2=/%.
sin Q i
/.g
(1066b)
I$
Equation (LO66bj is in agreement with Eqs. (1034) and (1036). 103.2
TE Waves between Parallel Plates I For transverse eiectnc waves. Ez = 0, we solve the following equation for HP()), which is a simplified versim of Eq. (1041) with no *dependence.
We note that H,(R ): = H (: y)e". The boundary conditions to be satisfied by H:[)I) are obtained from Eq. (1042c). Since Ex must vanish at the surfaces of the conducting plates, we require
H:(g) = B,, fos
(F) ,
.
i
a .
.
, '.
I .
In /k fI
\
L
'
:
1071)
TEM Modes The attenuat.on constant forTEM modes on a parallelplate transmission line has been discwed in Subsection 933. From Eq. (972) and Table 91
we have approximately
,
< . . ,
,
where E, ,u, and.a are, respectively, the permittivity, permeability, and conductivity is the intrinsic impedance of the of the dielectric medium. In Eq. (1073a) if = dielectric if the dielectric is lossless. Also from Eq. (972) and Table 91 we have
where a, is the conductivity of the metal plates. We note that. for TEM modes, r, is independent of frequency. and cc, is proportional io J?. We note further that a,, .0 as a + 0 and that r , + 0 as a, 4 cc. as expected.
TM Modes The attenuation constant due to losses in the dielectric at frequencies for 6. We have above ji can be found from Eq. (1055) by substituting E,, =e.+(aijw)
Only the first two terns in the binomial expansion for the second line in Eq. (1074) are retained in the third line under the assumption that
From Eq. (1056) wc scc t h a t
With this relation, Eq. (1074) becomes
from which we obtain
I
.

.... ..*::,*: dl2 and r: < d/2), the waves must decay exponentially We write
:. p, of space,
/?
s
zero.
2 s.
of the cutoff. iowing
where u is defined in Eq. (10126). Following the same procedure dr used for TLI waves, we consider the odd and even TE modes separately. Besides H;(J). the only other field components are H:iy) and E:(),). which can be obtained from Eqi. I 1042b) and f 1042c).
ii) In the upper free b > d. (b) n > d > b, and (c) a = b = d. where a. b, and d are the dimensions in the .u, y, and z directions respectively.
a:.
Solutiou: With the z axis chosen as .the reference "direction of propagation": First, for TM,,,,, modes. Eqs. (1014% b, c, d, c) show that ncirhcr 111 nor rr can bl; zero, but that p can be zero; second, for TE,,,,,, modes, Eqs. (10151~1, b, c, d. e) show that either nr or 11 (1x11 not both rn xnd 11) can hc zcro. hut that p cinnot be zcro. Thus. the modes of the lowest orders arc
I..
TM,,,,
TEoll. and TElo,.
The resonant frequency for both TM
and?^ modes is given by Eq. (10150).
a) For a > h > d: Thc lowest resonant Srcqucncy is
where c is the velocity of light in free space. Therefore TM, modc. b) For u > ti > b : The iowcst rcsonant frcqucncy is
,, is the dominant
1
i
and TE,,, is the dominant mode. c) For a = b = d, all three of the lowestorder modes (namely, TM,,,, TE,,,, and TE, ,,,) have the same ficld patterns. The resonant frequency of these degenerate modcs is
106.3
Quality Factor of Cavity Resonator
A cavity resonator stores energy ir? the electric and magnetic fields for any particular mode pattern. In any practical cavity the walls have a finite conductivity; that is, a nonzero surface resistance. and the resulting powcr loss causes a decay of the stored
i
i
i1 !
I
i
. The Gents
A
a+ytt,avity iable
; <
h, (b) n < h. and (c) a = h? R.1029 What is thc c u t o r wavelength of the T E , , mode in a rectangular wavcguiuc? R.1030 Which are the nonzero ficld componcnts for the TE,, mode in n rectangular waveguide?

R.1031 Discuss the general attenuation bchavior caused by wall losses as a function of frequency for the TE,, mode in a rectangular waveguide.
R.lO32 Discuss the factors that affect the choice of the linear dimensions n and b for the cross section of a rectangular paveguide. ....
..
R.1033 Why is it necessary that the permittivity d t h e dielectric slab in a dielectric waveguide be larger than that Of the surrounding medium? R.1034 What are dispersion relations? R.1035 Can a dielectricslab waveguide support an infinite number of discrete TM and TE modes? Explain. R.lO36 What kind of shrfxe can support a TM surfnce wave? A TE surhce wave? R.1037 What is the dominant,mode in a dielectricslab waveguide? What is its cutodfrequency '? R.1038 Does the attenuation of the waves outside a dielectric slab waveguide increase or decrease with slab thlckhess? R.1039 What are cavity resonators'? What are their most desirable properties'!
mode
rI"UIC
R.1040 Are the field patterns in a cavity resonator traveling waves or standing waves'? How do they differ from those in .I waveguide'!
, mode signify? The TE, <
R.10 11 In terms of field pmerns what does the TM,
2,
mode?
R.1042 What is the e%preision f a the resonmt frequency of TM,,,,,, modes in n rectanjulx cavity resonator of dimensicns a x h x d? Of TE,",,, modes'! . R.1043 Whar is meant by te~~clrel.c~te t~lodes?

k . 1 0  4 What are the mod:s of the lowest orders in a rectangular cavity resonator? R.104 Define the quality factor, Q, of a resonator. R.1046 Explain why the measured Q of a cavity resonator is lower ~ h a nrhe calculated vnius
PROBLEMS
P.101 Starting from the iwo timeharmonic Maxwell's curl equations in cylindrical zoordinates, Eqs. (78ja) and ( 7 4 5b), express the transverse field components E,, E,, H,, and H, in terms of the longitudinal corlponents E, and H,. What equations must E: and H,satlsfy'? P.102 In studying the wale behavior in a straight waveguide having a uniform bur arbitrary cross section, it is txpedient :o find general formulas expressing the transverse field components in terr~isoftheir longitudinal components. We write 1 .
.
,
.
*"
\
where the subscript excitation: a)
T denotes
i
.
#
.
.,
"transverse." Prove the followi& relatipns for timeharmonic
1
Er = p( y V T E ,  a , j q
(10163a)
x VTHz)
where it2 is that given in Eq. (1013).
P.lO3 For rectangular waveguides. . '*
a) plot the universal circle diagrams relating tr,/~r and P / k versus],jj,
b) plot the universal graphs of uh,. B/k, and ;.,,lib versus fjf,, C) find u,j11, u,/u, P / k , and i.,iiL' ~ 4 t = 1.251,.
P.10J Skctch thc c t ~   Ptliugr;~mof 3 p:~rallclpl:~tcwaveguide separated by 3 dielect~icslab of thickness 6 and constitutive p a ~ u n c t e r s( E . p ) for TbI Th4,. and Tb13 modcs. Discuss
,.
a) how b and the constitutive parameters affect the diagram.
b) whether the same curves apply to TE modes.

..
P.lO5 Obtain the expressions for the surface charge density and the surface current density tc Do the currcnts on the for TM, modes on tllc conducting piatcs o f ;I p a ~ ~ l i e i  p h w;~vcg~~iclc. two plates flow in the same direction or in opposite directions? P.lO6 Obtain the expressions for the surface current density for TE, modes on the conducting plates of a parallelplate waveguide. Do the currents on the two plates flow in the same direction or ii' opposite directions'?
P.lO7 Sketch the electric and magnetjc field !ines for (a) the T M 2modc and (b) the TE, modc in a parallelplate waveguide. P.108 A waveguide is formed by two parallel copper sheets0, = 5.80 x 10' (S/m)separated by a 54cm) thick lossy Jiclectric E, = 7.25, ii, = 1. a = 10 l o (S/nl). For :in operllting frequency of 10 (GHz), find /I, z,,,qC, up, u,, and i., for (a) thc TEM modc. (b) the T M , modc. and (c) the TM, mode.
P.109 Repeat problem P.108 for (a) the TE, mode and (b) the TE, mode. P.lO10 For a parallelplate waveguide, a) find he frequency (in tel'iils of the cutoff frequency 1;) at which the attenuation constant due to conductor losses for the TM, mode is a minimum, b) obrzin thc ror~nulilfor this mini~ri~~rn *~~t~:nu;~ constant. lion c) calculate this minimum 2, h r the T M , nwdc it' ~ l l cp:~r;dlcl plalcs arc niadc ol'coppcr and spaced 5 (cm) apart in air.
P.lO11 A parallelplate waveguide made of two perfectly conduciing infinite planes spaced 3 (cm) apart in air operates at ;1 frequency I0 (GHz). Find the maximum timeaverage power that can bc propagltcd pcr unit width of the p i t i c without o voltagc hrcnkdown for c) tllc 'l.ll, I I I C ~ C . !I) thc ' f M , nlotlc, a) the TEM rnodc.
PROBLEMS
497
P.lO12 Prove that the foflowing wavclcngth rclation holds for n uniform wavcguide:
.
; '%
&?
I
I
.
where i., = guide wavelength, i = wavelength in unbounded dielectric medium, and i, = u/f,= cutoff wavelength.
P.1013 For an (1 x b re~tas~gular waveguide
operating at the T M , , mode.
a) derive the exprcssicns for the surfacc current dcnsitics on the conducting walls, 11) skctch illc surfacc c ~ ~ r r c un n ~ Lhc s walls at s ;= 13 and at ) = 6. P.101.1 Calcuiare and list i 1 ascsndiny order the cutoff ticqucncics (in icmms ol' the cutolf frcqucncy of the dominant midz) of an u x b rcc~angularwaveguidt: for tht Ijl!owlng modes: TE,,, T E l , 7 T E l , , T E , 2 TE . c: T M , T M 1 2 .:mi TM,? [a) i f a 3h. anti ~ h i )T t ~= h

1'.1015 An airlillsd ' r x b I ! , < .I < 36) rec~angularwavsguidz
1s to b\: consrrucred to opcmrc ~ Srcqtlcncy 10 hc ;it 1e:lat 20'1; h~ghcr at 3 ICHz) irl the dornin;ill~i:~oclc.:I/c clcsirc t l opmrting Lhall l l ~ cculoif lrcqucncy ol't.ic c l u ~ n i n a nlodc ~ i ~ .1nd :dso at lens^ 20:'" below ~112cutotr rrquencq of the next higherordcr mod.
a) Give a typicril dtsiy ;for the dimensions u and b. b) Calculate for your dsign 1,I(,,,i.,, and .he ~c'.tveimpedance at thi opcratlng frequency. P.lO16 Calculate and cornrxs the values of /, u,, u,, i,,:lnd ZT,,,, for rcctangular wavcgu~tlcopcratng at 7.5 ICHz) P) ~f the Lvavegu~de 1s lioilow. h) if thc waveguide is t~lledwith
.I
.t
d~electncmcdiurn chnractcrizcd by
2.5 (cm) x 1.5 (cm)
E,
= 2. p, = 1 2nd
a = 0.
P.1017 An airfilled rcctangular wavcguide made oi copper and having transverse dimensions u = 7.20 (cm) and b = 3.40 (cm) operates at a frequency 3 (GHz) in the dominant mode. Find (a) (b) i.,, (c) a,, and (d) the distance over which the field intensities of the propagating wave
fc,
will be attenuated by 50%. P.1018 An avcrdgc ppowcr o, 1 ( k W ) at I0 (GHz) is to be delivered to an antenna a: the TE,, mode by an airfilled rectangular copper waveguide l (m) long and having sides o = 1.35 (cm) and h = I .OO (cm). Finti
ipsct..' power
P.1019 Find the maximum amount of 10(GHz)hverage power that can be transmitted through an airfilled rectangular wavegl~deu = 2.25 (cm), b = 1 .OO (cm)at the TE,, n o d e without .I breakdown.
,
P.1020 Determine the value,of (fl') a t which the attenuation constant due to conductor losses in an a x b rectangular waveguide for the TE;, mbdeis a minimum
".
P.lO21 Find the formula for the attenuation constant due to conductor losses in an a x b rectangular waveguide for the TM mode.
,,
P.lO22 Show that electromagnetic waves propagate along a dielectric waveguide with a velocity between that of planewave propagation in the dielectric medium and that in the medium outside. P.1023 Find the solutions of Eq. (10132) for k, by plottmg qd versus k,.tI for d = 1 (cm) and = 3.25 if (a) f = 200 (MHz) and (b) f = 500 (MHz). Determine /3 and a for the lowestorder odd TM modes at the two frequencies.
E,
i.
P.1021 Repeat problem P.lO23 using Eq. (10135) for the lowestorder even TM modes. P.lO25 For an intinite dielectricslab waveguide of thickness d situated in air, obtain the instantaneous expressions of all the nonzero field components for even TM modes in the slab, as well as in the upper and lower freespace regions. P.lO26 When the slab thickness of a dielectricslab waveguide is very small in terms of the operating wavelength, the ficld intensities decay very slowly away from theslab surface. and the propagation constant is nearly equal to that of the surrounding medium. a) Show that if k,d