Solutions Manual for

Applied

Electromagnetism

SECOND EDmON

•

Shen

Huang

,

solutions Manual

for

Shen and Kong ' s

APPLIED ELECTROMAGNETI SM

Second Edition

by

Liang C. Shen and Frank S . C. Huang

rlB 8C PWS-KENT Publishing Company Boston

.'

PWS-KENT Publishing Company

10 P"fk PI1.U Blmon. MUIKhowcu 021 16

Copyright c> 1987 by PWS Publishers.

All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transcribed, in any form or by any means - electronic, mechanical, photocopying J recording or otherwise - without the prior wri t ten permission of PWS-KENT Publishing Company.

PWS-KENT Publishing Company is a division of Wadsworth, Inc.

ISBN 0-534-07621-1

Printed in the United States of America

91 -- 10 9 8 7 6 5 4

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 16

CONTENTS

Complex Vectors

Maxwell ' s Equations

Uniform Plane Waves

Reflection and Transmission of Waves

Waveguides and Resonators

Transmission Lines

Antennas

Topics in Waves

Electrostatic Fields

Electric Force and Energy

Solution Techniques

Direct Currents

Magnetostatic Fields

Magneti c Materials and Magnetic Circuits

Electroquasistatic Fields

Magnetoquasistatic Fields

1

5

8

12

16

20

25

30

32

36

40

44

49

54

56

58

(eclors

"Rand Their

wi

11.19)

to half . B. B' . ler Ihe

(1 .20)

-domain ~' tJ have - f:!. and

r Problems 17

NOlicethal A · 8 * - -2pmrllhat (All). S (I) -11/ 2IReIA . S ' J - O. We should a lllO nollilhal A . B _ 0 and Ihal A ll] . 011]- 0 Thu$.lhe Iwo tlme-doma in Vf!CIOrll ate always perpeod icular 10 each OlhAr.

Example 1.20

Solution :

r.nnJlliriAt Ihe two vectON A _ i + iy and B - t t /y. IThey a re actually Ihe same vector.) We fi nd tha t A )( 8 - 0 and that A . B - O. Are Ihe Iwo veclOrll para llel 10 each other or perpenrlir:ular In eitch other?

Con ~ itl er ins lead A x B' and A • S' We fi nd tha i A x 8 ft - -t2;and Ihal A _ 8 ft

_ 2. Thus. (A ll) x 8 [1» - O. and tha i (A II)- BIt» - 1. Furthermore. A fl} x Blr) _ O. and AlII· 0 1'\ - 1. TllU~. Ihe time-domain veclor Is parallel to itself all the time.

Problems

1. 1 Lei a _ 8 + ;2 and h - - 3 + ;. Ca lculate fa) a t b, Ih) a - h, fl:) ab. lind (dl alb. Cive IhA an~wer in real and imaginary parts.

1.2 Repeal tc] and Id) in PmbhUIl 1.1 wilh Ihc answer given in phasot form .

1.3 Find the rea l ~rl, IhA imHginHry !Jitrl. and Ihc magnitude of e'" '. whArA '" andt are rAaI numberll.

· 1.4 lei c be a complpx number. Are the following sta temen ts alwaylllnlA?

(a) (c I c ' )isreal (b) (I: - 1: ' \ is ima,l(illury. (e) clc ' has a magn itude equal tn L

1 .5 Consider Ihp. P.<l llalion z· - 1 + ;. Find two values of 1. lhlll Mlidy this equulion.

1.15 Lei a be a rea l numhPr. IIml lei lui .. 1. Show that the square rool nf (t ... la] is IIpproximately equHllU 1. 11 + ja/ 2\

1.7 Lei 0 be 11 JXlSiTive tp.al nllmber. amI 11:\ a ... 1. Showlhallhe sqUMft rool nfft + ju) i~ apilroxinu:lldy t:4 uallo 1. (1 I iKa/2I" J·

.... 8 Obtain the pha.~or nolalion uf Ihe following time-ha rmonic fu nl:lions (if po~ilJ l e):

(a) Vltj_ G cos!"" + .. 14\ (b) 111 1- - 88illl",1) tc) AIII _ 3sinlwl) 2eos( ... 11 (d) t:(1j - 6 I:OS !l 20 .. 1 - .. / 21 ! e) 011) - 1 - eos (",'I ( I) Uitl - sin ("" ..... / 3)lIin Iwl + .. / 6)

1.8 Obtain GII\ in termsaf '" from Ihe following Ilha)l(jfll: [H)C - 1 + ;. (b) C -" exp 1/0.8 \. and tC) C - 3 exp! ; .. / 21 ... " exp I/O.tI).

-1.10 Show Ihal. if V - r + /X and U _ g I 1}'. lhen VIIlI1!1J *- Rft!VU e""I. Find the corrpct eX!Jrt:SSion for VII) L1 II) in tftrms of r. x. ~. y. arId ",t.

18 1 Complex Vectors

1.it Lt:IA- Ai+!ly iand l' -2x - 4y t 3t. FtnrtjajA ... B.(bJ" H, jr:IA. B,linu

fd)" )C 8. 1.12 Find the angle ~Iwccn A and 8 that life given In Problellll .11

1. t3 Show [hal for

VII I -V.cosl",11 tbl-ReIVe""l , al V(li- ""V

1.14 Find a VPf:torC that is perpenrlkuluf toA - 8t + Dr - t, has no t component. lind hall a IIIIIHrlitude equal 10 1.

1.15 "'Inri tht! vector C Ihat ill parallcl 10" - si - 8y f 21: Ann hltll II mognitudA Aqual to t

-1.11 Find II unit vectur II that points in the same dlrActiOIl as (Ill arrow drawn from point /\ to vaint B whAre the rectangular c.oordinates of A ,md B UN (1.0.2) and (- l,:t 21. respAr.lively.

1.17 Show thai thl! definition offhl! dot product v·u givtlll by (1 .10a) Is equivlilcnl to thlll givtlll by (1.10b). 'Ib simjJli£y the algAhrll. yuu moy choo!!fl thtl coordinates so tlnlt the x axill ill8100g Vand the 7.lIxis i8 pcrpondlr:u lar to Loth V and U. In other words, lat V • ux and U R b~ ... c9.

-1.1' Prove 11.11b) using the approach SUgg8SI00 in the texl.

1.1. Shuw that the definition of tbe eros!! product V x U giVAn by (1.12) is equivalent to thnt given by {1.14}. To simplify Ibe ii1gebra, do what ill sug8ested in Problem 1.17.

- t .20 Figure "1.20 shows Ihat II vector V is along the x llXi! and a vector U IS on the x-y plane formiog a 135- angle with the x axis. The magnitudes of those vectors are v pnd u, respeclivt:ly. Usc (1.14) to express V )I U In terml v oDd u. Now, we C8 n also say that the 8ngle betwfMln V 8nd U is 225°. UlW {1.U} and explain why the vector V x U is the same whether or not we choose the angle betW1;!cn them to he 135 - ur 225- . l lint: Follow the right-hand rule. In determining tho direction vuctor II. your ringers lIIust alwaYR point from V 10 U in the diroction you rnellSUf'C the angle be­Iwt.'en them.

1.21 Pr(Mt {1.15} usillgtho approar.h suggostod in the t!tXt.

_1,22 Find the phoser nota lions of tho following time-harmonic vectors: (ti) V{I} • 3 cos ((oil)! ... 4 sin (wtJ9 ... f: CUll (wi I ..-/2) (bl Etf) • [3 CUll (Wi) I 4 sin twillt ... 8[cos(wl) - sin (wIlli (e) H[I) • 0.5 cos I/o: - ",IIi

1.23 ~'jlld the phasor notatiun of the follnwinK vuctor:

C(z,I) . ta181) £ (Z.I)

where r. is !liven in Problem 1.22(h).

- t.24 From the following complex ver.lors. fiod C (t) in IArms u( '-iLia) C - i - IY. (h) C _ ,ti - /y),undlcIC-expl - jKz)R ! /exp(,KzlY·

Problems 19

1.2' LetA - J + /9+11 + /2j!,and let B - - R - 11 ;2)9 I /l, FlndlaIA .,. B, lbj A - B, Ic) A ' S, 811t1 fd) A )( S.

,

u

y 13!i ~

, Flgur. Pt .20

1.21 Find A . A' and Re I A x B ') fur the values uf A and H given In l'mhlAm 1.25.

1.27 Skelch Ihe trace or Ihe tip of Ihe vector All). where (01 A _ .t jf and where (bl A _ 4R + 139.

' .21 Ca lcu lale A . 8, given A _ i + j2yand B - 2i I jy AreA(I) and 8 (11 perpendicular I6 l allli llll~.!I?

,./ (a.) e,+ b • S-+j3 (IV 2.-£ • lI+j (e) 1l.·l.--2,+jZ ,i, gill. c-Z. Z-j'.4

J;l ~.: , .. jt· I.IS' I1" .pt' J lz.- -3"'j .J.I'II".S7~ !i'R - 2&.oBLI7'·"" lSI.! & :J." L-I ... 'J"

Ll Ne I tj.n J . (~~ (.oIe.J I", [ ej""t J • s·~ MIt' I J t ;wtJ ~ J

f;f (a) y.' '.J yes ") y~s I:L Z':I-tj - {feJ'(2ft"..,.~) .::;. z. Z~ ei(~"''P'.) . ..,,-" .. 1

:.z, • .2t$.~j~ . zc.:2.~ei(1r#~·)._24.eJ!Ji

!:l (Ifjo.)ll.. =['.fl/ .. )+.; .. ••··· J :::::t. I ,+j(~)J, if ''''«I !:J. 0.»1, ("j .. )i{ '" Ija.)J( &±/J(a)'I,.. :t{l+j)('Y.)J{

.'

J.:! (a)~::b~j!4 lb)I-j8 (c.).d.·j3-2 Cd) s.a-j6 (It.)i,"poJli.61c. fh/hlpot.S"6Ic.

l!l (o.) C(06l- R. {"fj)e;'" J • R. f,;r ",i ~e;""l & fl "S(AA! .. ~) ,h, C(~)-R'[4ej·" ~;,,,,}. -I~,(..n ... el

iC, CCt) II !?f 13ej1f{~.J·wt + -Iej···ei,.;t;} _ 3cpJ(,..;t.~) ... 4 "'Hut+".tJ)

1!! 'i:. r#jX "'* V(t)- rCDSuri: -)(s.~£.,)6; U -8+-jy ~ U('t)- g.C()iwt - Yl;l'Ll6lt

R( f:i y ~j",,}- R.{((rg-q J .jl IX'ry lJ t j .. , 1 - I r~ -XYJCoJ..,t -(9~+rf) s,,..,t B.t V(') U(~) c ra,os'we +XYS;n'",t - (gx +ry) s ... kIt c •• ..,t * R. [!til ej.,~ j

- ,.,.... ",....... -!J.! ,If, ) 1..8 =-6x,+SY<lZz (bJ A-o --IOx·n3y .. 4z <'J ,q·8c-,J,-31-3 ... S5

Cd) ~.e. 23x+211+14z 1.11 IAI'I-~x+qY-il'''146 , 1~1=12;-19'3il:(i;

~·O -S''' C4I5"(: ",,:r .. -o.S~S .', d...:r cD~"'(-".9 .. S)= 147. 7-JX a 00.;;;

. . jf f./~ It Vct)J: -fteV.(Ol (wt"'~»J. -v.c..J .s '~(wt". ¢) .: 1ft! [ j v.we)li" e JIoIt J _; V.w e.

13~t: Vf<)=V.C.'(wt.;)& R.fv.ej~ej ... tJ - R.fyejwtJ ~:t. v.ej~ ... ?t Vt-t) - j wv, tj';;s jw ~

" - '" ....... !:.!! lit CaX;+'IY CJ.A ->- C'A-O, lAJJ..ert. .A=-8)(oI-QY-2 -- 9 ~ 8 C·A·-8x.,q~.o :. ~:T~ Ah.,A/"" ,'+,'" =*~'=Trs" ".~-'¥' - '(". :'.-%,-,.: ~ C=:= 9~+8Y)

rt4S r 14S

,.,S Let C=";T~Y+};' eIIA-" AX c:t0 l IN"."" A.S;-B;+2~ I- c. (- e 3 -2,); +(zx- S ~) y + (S ~ 1 S ,) i -0

,', -Jt- 8 J-o } 2lI-")so '* 2'- i J """ ~'-4~ 5~+8"so

,,~t /E/ =/ ~ x"1"l"1 .. )'(.,.4 + 1'+)-/ - I ... ... ...

,'. C&~.=C5'·-8Y·2') ,9>

1

, .r,Ts

/ . J7 -

/. / F

/.19 -

/. 20 -

_ A "

=-1X+3!j-4J

-n =(-2X H)- - 4-})/ 14+r+16 / ~ " ") =-(-2X+3J-4J

(/. /0 .... ) J"'" V·U = Ilb

(/. / 0 ~) Jim jJ. (j = A. h'T;> (..s 8

~~ e..s8:; J. /1".,1

.' V·v = I\,J 10\" ""J8 :: .. b

- . " L~t A = A,,, + A~ J +

13 ~ 13,; + B,; -t - '" ... ,... C • C," -t '~::I + C, ~

'129 y -U

v

A' (If t c)" k[{ (3,+c,): + (13 .. + c, )~. + (IlJ + c) )lJ

'" A, B," c, A, + A, B, t A, C ... + A) 6,," ;\) S - - --A· B + A · c "" A, 8, + A, (3~ .. ,4J (3) + A, (, + A. c. + AJ CJ

Thtrt.j."t. A· (1ft c: ) :: A·a + A· C - - - " (I. (1) )'v<J Vx U = a. c. J

(j. Ilf - A

) ,;'{'IS V' u ., It. , ,,'&+,. S/" (} J

b...t so.B'" c Ih'.,'" - - ~ J.~,' $,',19 f' "

~ V xU = - "Co J -- - s.;'~J!'·) f r;: "" V ... U :: V-It '" i" V'''' J

2

x

/' 11.

f.l} -

- 1\ '" 1 - ""nAB" " Let A = A, x t A. , Tits} , (3::: P, X of ..... 'j'" J l/ ,... - A A C" ", C. ::: C,,Xt C, , .. J), "'1'\.

if ~ C = ; ( B, CJ - C. 8) J.r 9 ( 8J C, - c) 8,) t )" ( B, (~- CA)

:41 (B)«( ) = ) [ A. (B,C~ -c,B.)-A, (BjC,-C)B,)]

.. 9 [ AJ (B, c) - c, 8,J-A, ( /3, C. - a.e,J]

.. ; [A, (B/ (, - ~ (/) -A ... ( B .. S -aJc,)]

o.+~. "f~tr ~."',(,

'if (A' [ ) :. (f3, ; .. 8 .. i .. Br S ) ( A, c, + A .. c, ... AJ C) )

t (A , /3 ) = (c, ~ .. (, ; .. cd) ( .4,8, + A. L3 ... t ~) IJ )

,', 73 (A '[)-C (ft · S)

CA) ~ c 3; -j 4; .. j ;

(I) € - (3-j4); .,f/("jJ i ("B- o.s-e"iAJ-;

3

,

J.:.lj (A) ~U.). i' 'O)We ... ; S/~&.IJt

C• -') A. ~ t .) CL:ot --)( t.,,6o\It, .. ""0"11 (,) ~ (t). oX cosc",t -It) - Y S;~("'" +~J.)

_ _ . ... • A

1:ll Co.) t,'!}-(-I'J3)Y""'J)1 - - '" .... . .... (6) ~ - § • 2M +(I-;)Y+(U;) i

( c) .A.~ c. -'-<i+t.)"(i-2).-~

(J) l'~ c 4-r-CI.j)JY-(l-j3)1

1.2' A.A'uU .. jY"(i.j2JZJ-(,i-jY+(I-jz)iJ=I+I';-7 A .§. - U +j Y .(I.j2Jl.] x{-x-(l+jZ) 9-j ~] : (-2'jf)' - CI+ jJ 9 -(I+j) i :.Rt[d M§'1--2 t -Y-i

1.17 (~) e - x -j 9 .-. At.) - co • ..,t ;- .. • ;"wt y ...".f., ACt), c I y

(~)

,.~~It) / \. ,wt-o

." "c~:t

''Ii

.d: -4'·j 31 - A(t) c 4c.osloJ'i. y

A_ ;"'j2Y J Ee2;+jy .... A·i.2-1l:0 fjw.t AU) s co;wt ; - 2 $IAwt 9 4-nct. Bt-r)- 2 "',Swt X - ,,;"we; .:. At.e) ,.,.c, 8(-1) Ar4. ".1:: p.rp."dicu.U.,- ",t .....,y -I,.,.,e. .

4

~~-.---~--------............................................ ~

,

,-

I­n

d , .

n

)-

c

" !Y

Solution:

2.5 Poynting's ThftO rfim

Ell) - Rc!i E. e 10. c .... J- il::. COlI/WI - H1.1 k k

HIII _ Rely - E. e-,b 1'1""' - 5' - E. cos (wI .. hz) w~ w~

_ k ( Sfl] - E x H .. 1: ~C09' I.:.lI _ h J

"" 1 k (S ) .... Re (E)( ' ... 1_ I: - ~

2 ZI.,'I1 fl·;: •

U, - -cos (wi - hz) 2

kZE' 11H .. ----,-!cos1twt kz)

'"" (U,) .. ~ ~

k' , (Un)" -, Ei

""')1

35

Problems

2.1 Let A .. 5.2 + tsyzy + ~I:: fino 'Q' )( A and 'V • A.

• 2.2 Lei ~ .. xyz: find 't" 41 and v . 'iit/l.

2.' Let (I .. a,i -+ ali + oJ! lind b .. b,i + h" + bll:. Show that equations (2.91./2.11a), Knd /2.121 are true.

- 2 .• Show Ih,'11 'ii' ;It fa + bl -V x a ... V x b andV · fa t bl-V. a + v· b.

2.5 Show that V(<I>, "',1- <1>, 't""'J + "'I V til, and that 'Q' • (4)A) ... A· V<I> -+ tt>V · A .

.. 2.e ShowthatV )( 1"'''J-V<I> x A "f" <W x A.

2.7 In a source-free region. H .. zy + yi. Does D vary with time?

2.8 What is thtl dllll'Ke density in a region where 0 .. 2xx?

2.' Find the mllgnctic field S(y,I) ouucitttttd with thi! fl lfICotric nllid E(y.t) given as fol­lows:

£ (y.I) *' R 0.3 cos(tolt + ky)

where tol and It are cOII!Jh:lllts .

2.10 Express k in terms of the mll.llutllic permeability and dlllleotric permittivity of the medium whAn the eloclromagnetic Oelds oro givoll ill Prubhlm 2.9 In Ii ~oul'Ce- free region.

2.11 Lei £" 8"

H l ' lind D, SIItisfy Al'JUfltiOns (2 , lJ-{.!.4) with given " and P.,-Let also El • 8 t­HI' and D: satisfy equations (2.ll-IVI] with given 11 lind Pvl' What are the electromo8-IItltic fields due loa currAn t I, and charge P .. where I. - 1, t II and p", - PVl .. p.,? You must show that your proposed 1lOIulion SIIlillfiAJ Maxwell', equations. what is the appropriatA name for the theorem you have just proved?

36 2 Max~II '1 Equation.

_ 2.12 (8) II is known that the vtlClor 0 is equal 10 zero lit one point Does that Imply tha, V )( 0 _ 0 at Ihal point? Give a counter-example if your an!lwet is nu.

(b) Does ,.: - 0 on a line alwaYlI impl)' V x E - 0 on that lin c? Give a counler-t~lu:lrnple if the answer is no.

(c) II is found thallhe ,.; field i8 zero on a surrace. Dws it rollow thot aOla, - 0 on thallurface?

2.13 Show Ihal equations 12.221:) Bnd 12.2Zd) [;lin be derived (rom equations 12.22o). 12.22b1 and the I:UlIlItltvllt i on equation 12.231.

~ 2.14 To repreMlnllime.harmonic fiekls, most physics books usc the f3ctor e ""'nstead of u ..... which lIlost electrical engineering books use. For a time·harmonic ftlul functlull Ajx. y, Z, I) - o(x, y, 1,) COlI ,wI + 411. find the phallOT Ilototion Ihot corresponds to the phy,idlllll ' cunvention . Whal is Ihe corresponding conversion rule hy which ph880r8 can be transformed back 10 the tl!al·limA Axpre8!lion?

2.15 Rdcl' 10 Problem 2.14 about the nolalion 8'i00i adolJlw ill IIIOSt physics books. Wrilft IhA limit-harmonic Mltxwell'!J e4uotiolls using that notallon.

2.11 Whal is Ihe ronge of effective permittivity of the ionosphere al AM ltroltdcastinH frequencies" UMl lhe following dltla: N - IOU m-~ and f - 500 kHz to 1 M ll z.

2.17 Show thai the dimension of each term of equation [2.36) ill walts per cubic meter.

2.tl IndiCHls in watls. mclers. and joules the dimensions of the following quantiti($: (a) £ . D. (bl " . 8, and fc) S.

2.18 LetE _ (i t ;y)e II and H - tjl - It/e'". Find Sin lennsuf zand ",I and find .., S>.

2.20 Show thai S ,. Re IE x H c""'1. 2.21 Show that S "" Ke IE P."" )( H e,....j.

2.22 Compare the energy stored in a cubic: region onl! meier un a side which has a uniform": field of 10" Vi m 10 the energy stored in a similar region with a uniform B field of 10' C. (One C _ 10 • Wb/m:). The medium is air.

2.23 Repeat Problem 2.22 for the cllSe where the medium is water Instead of IIlr. U8e ~ - AO f g and ~ • I-'tl ror wator.

VL

5

-

2.e

.t . /.

v·o· v'( 2d) =2. =+ t;.. 2 ~/."

- ,'i, • Ii = • . J e ;:

v-E =-1 (o.JiiejA, ) = -j .. §

B = •. J ~ e ji.l l' -l3(y,t)~ .':.,'" e,.,r ... t+ -4tJ) ;

fl,( H == j ~ * 5, •• ~.f;kJ.,'J n - . J = 0, 1>" £. f == •. ] t Co, (wt ~ ~)) ~

'" .£,.., ~ .

V](~=·:t6, . v~~·J,+ho, ~ v.B,."; v·o,-I!, 'i/v.EJ·-fl!,. VI(J./z-J,+/Eo, , v·A;-c, 17'2\-""1 Q'1Ce;- v-( ~f"Ez)= v",~.,. V.~ • -lci,- /til- -It, (6, +4;)a -Jt ~ 'O'''Jif • 'l" eN; ~~)= VI. N, +VII ~ :r J," kP:".f, ~ It D, ae],.X)'" ~(5," ~) -1.'" ~ D't v· '8t r Q. ( 6,+ '1 ) '" fl· 8; + Q.F" r () ~ 0 &' 0

tI·~~wQ·(o;'Dz)W&V·5,"V.DL '" fV'''&l=-f ... ,.

.', il . £, f' ~ , H~. N, 1'#" • 8l- 8,,,E1 .... ct D~. s, +4 ,$.'1;,1'1 1'H...x1oJ-"; ~,~I,.() .. t s,,. ... ,.c;tJ J~~J',.,.j",....-, fl't-fVJf Iv:.

TA/$ ... ~ fAt. $~~"'e()S" 11'p"" .,I1tD,..,..,. .

) , '" ( tl.) NI), 4:.'r; ..... J.,.YS,'-;r+fs;,..,t.o .,It (D. tI,Dj. bl4..t v .. a: & £~D'1"'fUJ~f 4t(~ .• . ,,)

2.1l

2. ,,..

chI No. t s: £ s.:...) .. y I,:"'X co ." Y-AX"$ , h .... t V_E. yecS}. iUJ'~ ~ 0 crJ\. v-.••• cc.) No, Fe f J'--J ='0 -.,. )#0 I'~"e, I bJ v~F c VUJ} 4" 0 D'>\ I-"It.-.. ~ ~

0.",. +n.t p/~e

v·(~.!. )"0 ""'jwv·/J=0 .. v.ij.o v·(v'tl)"-9 Q'j+jw"'Q =D"*Jw(-tJ..,,.§)ao -+ "'ll:/. lI(r. l · ' . t)· a(Y, ' . lJ ",,,wi_;) _ a("'~. J) G<>S(-wt-<l-J

ej·e, A (q. l .t)~ K..[a.tr.l .nei·oiwtJ _ a.(x. l. J)ei~ eoi .. ': A (., t · l ,~)· Rf!fa.(%,t, J)i'f~ . i .. tJ - a.(" .I, )~-i. q·i=~"'5., flxitrl-iwe. v·g~o, V'e-!! £~[.(f-~) wp=!Nt.1 .[-/~""{U.~"'J1"J'If""(I'JJf 1C.S;'.x'D'1 I1,

w ".,. .I' ll I.- , ,

At SOOt#1 : E .. l [,_1'. ( .H'.,.", )') ~ -.13 3 f , (I ":ZIfr(~"'(J~ •

of I MN!: l. .. f. [1-( I::I'IIo/j" ?J = -80.43 E.

6

itt t. III <f j lr -.-L E· i4 ~ j NT £. Rc f ~ ej4Jt J • £~ CMtJt - Es s:,. .... t , il· R. £ lei ... tJ .. ~ c..SoJt - i1j $;/1p4AJt 5. i ,Iii =l~.i1~u~lwt+( E;x~ )s;...'4Jt - (£"''''iir ~ ~xN:t) S/~wt CAswt

8".'t H ej.Jt. (!i;u$we -liz ri".ve) ... j ( 17; JM../Alt +i/rc..'£Jt) ANi..

Rtf f ~j ei-.JtJ II:: ~1(#;e tDswt - ~ .",#/ t"~,.;t - Es-it s~t - Ez)( Nz tAh»t :f: s l.t l-ll(:#j fs .,..,.t H· H~ .jHz ! ejlJt =( ~ ~''-It _ £; , ;"pJ!) to j (;;.. j.:.....rt.; h (,IJ u~ ) E ~;~t # (~ c.swt - Hz s,'-u6)". j (;;; I~..,t: -I' N.r ~jwt) ~[i pj...,t,c i ej ..-t J =(~~~)US1~ '1-( Er)liijJ,~I ... t - (E",JC #j. + lrlt Z) s.~ t.oS6Jt

- (fs-x.)c.,'wt- (~~.H..),;.J"t - (E."lir ' Er_ ii,. ) ,.;""t"" ... e • ( f;q It ~ - E, 1/. iiI )( us'wt - ~:_"wt) -1,(r".i1z. .,.fs.,.iI~) rl~t ~Sf.Jt

,'. S ~ !?.{ ! .j",tx Be ;",t J !.!l (j,. ;t.E.e Sf; 1. J~.,,'IlIIJ&"_(l~"/ K' 4.4ZlI./ D&"(J,"''j,.,J)

U I r. - I I. - I (~'f)' S' ~C.TIl'M·HZ:T,t:t;o·OC ~.4"'J(Jtr' /0'''/0 =3,98JCIO .(J·""'o/m')

:. VI/V, ; 9 ",or

Ue • ..L£ E·g .. -L'.....!!-X/~-fJ((IOf)·_ 2. 2 Jb.,

UI " 3.98xlc' (J...4~,)

:. UYUt ~ / . / J 't.IO~

7

'o rrn Plane Wavall

If mankind rUr wing made be­:ure 3.111. Mod­; vllry grf:lfIIJy in dividual comet

comel's tail? II whelher or not his observation le SlIn Ihal lire the sun. Other­'resent explana­.1 pH rl ide ... II lid pressure of the m the sun, ilnd e ionized gllstJs e plasma forms now of protons til 8peeu lip 10 et's plasma and

1. Vul. I!.I!.I. October

4. April 1964. pp.

Problems

Prob lems

D' a 1ft iii !II 111

11 11 V 1<

1/

Figura 3.11 Various comet shapes drawn on silk fuund in Chinll. These figures were J.lainlcol bclwellll 246 10 177 H.C. Helow these figures arc Chincsc Humes for Ihese comets.

65

-It 1I

3.1 Eslimalp. the power density of electromagnfllic radiation from Ihl'! IIlln received on earth in the same frequency band as that of the VHF television channd 2 (54-60 MHz).

3.2 Consider the sun as an isotropic radialion source. Calc1 llale Ihe InIal power radiated by the sun in the television dumnlll-2 frequllncy banu (sile Problem 3.1). The distrlnce belween the sun and the earth is approximately 1.5 x 10' km.

3_3 Assume lhlll sulllr rauiatiull is i~ulJ'Opic. Estimale the total power radiated by the sun. The solar power density received on the earth is 1.4kW/m2

• See Problem 3.2 for nthp.r nata.

3 .4 Dilrive p.Sb) from (J .. 'ia), assuming that E .. Ex ~' and Ex is 8 functiull ofz uniy.

3 .5 The .'lIar a CAnlallri is approximalAly 4.33light-YAars rlistant from Earlh. A Ilght-yea.r is il unit of length thilt is the dis!ilncc a light wave covers in one year. How distant is a r.enlilllri in kilometers?

3.8 An electromagnetic pulse is sent from an earth station to the moon. and the reflected pulse is received 2.St! 9 later. Huw fllr is the moun from the Earth? fAn electrnmag. netic pulse consists of a wide spectrum of electromagnetic Wilves at different frAqllfmcies.)

3.7 Finu the 81 units uf the fulluwinx ltlllliltitills associatllu with a uniform t:!ectromIlS­netic wave: (a) "'. (b) k. (c) f. (d) T. and [e) >. .

... 3 .8 A heli um-neun laser emits light at a wavelength 6.326 x 10-7 m in air. Calculate its frequency. period. and wave number.

66 3 Uniform Plane WIVII

3.1 r-igure P3.9 ,;hnw~ iI dipolft aolAnnA II i~ \Iflry Rffllr:IIVA in receiving television ~i~ nilis willm its h!ll)!th is I!VVroXlIlIlllt:ly C4UIlI to ollc·llBl£ the signal wavelellKth. What arc approximate antenna lenglhs for roceiving signals for the following: (al Chllnnel2 (f ~ 57 MHzjslld (0) ChtlllucllJ (f - :n:i MHz)?

Two-wire lran5miMIon lim;

· 3 .1 0 The following .'lP.t of Illoctro magnAlir: fill Iii! slI.ti.!lfies the tirnll·hlirrnonic Ml:lxwell's

equatiuns in free 'poce:

E - E.e""'· x

"d H _ Il, e tV y

Expreli:I H, and k in terms of E. and ~ and I'e

3.11 00 Ihe fieldll in InA prfIViou!I problem represent II uniform p\ttne wave? In whllt dir~iun dOt:S the wave travel? Find its velocity and determine Ihe lime-average Poynting vector (5 )

3.12 nUl Ftlderal Commun ications Commission ar lhe United Slates requires 3 minimum of 25 mV/ m field intensity for AM Atations ('.ovflring thfl ('.ommflfdai area of a dty. What LS the lJO ..... er thmsity associated with this minimum field? What is the intensity of the minimum magnetic field II?

- 3.13 Study the following E field in a source·free region ;

E _ iE.t: ,h

Does it salisfy Maxwell's eqllatinns? If lVI, finellhA k anrt thA H fiflld , If nn!. explain why,

' 3.14 Show Ihal in [3.13J, if 1/1& - tI>!. - r/2 lind (l - U, Ihe wlive is ri,llht·iumd cin:ullirly IJOla rizt!tl .

. 3.11 "'ind the pnlaril'.JItion (linear. dn:ular, or ellipticlii llnd left·hllnd or riwht.iu:lnd) uf Ihe following field s:

I.) E .. Wi: + y} e-"" ( b) E-((l i i)y -+ (1 iJiJA Jl,.

(c ) £-112 + iii +13 _Il!je -flor

(d) E .. (j i: +j2y)e · ....

3.1' Show Ihat, if 0 _ b anel tP_ .. 4rt. .. r / 4. the wave is elliptically polarized {Refer 10 [l.131.} Do not try luoblain an analytical expression for 'hflioclIs . ju.~1 ohtain a pair of parametric equalions similar 10 [3.14J, r.alculale E. Hlld E,. allell poillts (",I .. O. 10· .

, 9Oo ~ and sketch the lucUli.

n Plane Wavell

televi.~ion sig­al "'iI~ltm8Ih, I following: {a}

1M: Maxwell's

Ive' In what time-aVCI1!.HR

sa minimum fell of II ciIY, the inltlOsity

not, explain

It! Circularly

hand} of the

d (Refer to lin a pair of 10'1- 0, to-,

Problems 67

3,17 Shuw that an ellirlirJllly pnlariMd wave can be dl.'CompuseO 1010 Iwo circularly polarized waves, one Idt-hanum.llluu Ihl! othllr righI-handed Hint: Let

E _ (Iii + urI e rho

whArA a and h are, in general, complex numuer.J. Then, lei

E -Ia'x I ;a'y) e !~, ... lUi - itfy) e '0,

and solve for a ' uou Il In termlt of a lind h

- 3.18 Show thai a linellrly polarized wave can lie uecornposoo inlo two ctrcularly polariZ4:d .... dves.

3.19 A dipole antenllu is in the x'y planA and makes a 45· angle to the x axis, A receiVflT allar-had to the antenna is calibraloo lu read tllI'Retly Ihfl component of the E field Ihul i~ plirallelto IhR riirnill . What are the readings when the fields aTR those given.· in (a)-(d) of Problem 3.151

• 3,20 An e lectromagnetic wave in vacuum has frequeru.:y r .. wavAlpngth >,., wave number k.. and velocity v.' Whfln It flntAr.olll dielectric medium chllrllcterized by iJ<,anr!, - 4~", what tire the f.;\. k. lind v of the wt!Ve in this m~rlilJm?

, 3 .21 Aluminum lUIS I - to. II - /I(Jo and rr - :\,54 "It" 10' mho/ m_ If on Dntennll fur UHF rf'CAplion is made of wood coaled with II lu)'er uf aluminum finn If its thickness oughl tu lie five lilnes grp.at~r than thfl $okin depth of the aluminum III that frC(illAnr.y. determine the thickness of tht:: IiluminulII IlIyp.r. 1~ ordinary aluminum foil thick enough for Ihllt purfJOSp·t Use f - I G llz Ordinary aluminulll fuil iliapproximatAly

3.22

3.23

3.24

1/1000 in thick

Cukulule the allp.nlllltion ratA ;lnd skin depth of earth for Il uniform vlane wavp. of 10 M liz Take the £ollowill)( dula for Iht! fUlflh: ,.. - ,..., I _ 4, ,,, and" _ 10 'mhO/in

Find the power uellsity 10 flarlh whel"9the field intensity is 1 VIm. Use the dRill in Problem 3_22 _

Suppostl that an airplane uses a mdar to m~ure Its 8ltilud8. t.Althe frequency of the radar be 3 CHz. SUpPOSfl furlhAr thai the ground Is covcnxl with a meter of hard-paclr::ed snow.

h

AIrplane

::::.-J I I I I , I I I I I I • I I i i All' I I I I I I ."nllw T I

"jm""""';"~"""",""","" C;round

Fillur. P3.24

(a) What is the ulrrecence betwften the apparent altitude mellsured by the ntdllr /Inri the lruealtltude1

68 3 Unifurm rhine WIVes

(bl How much attenuation in d B does the radAr !'ligna I !'Iuffflr hecause of the snow? Consider only the ;llIflnuation of the wave 11\ Iht: ~nu\\, antlncglcct the dfcct of snow on Ihl:! rent!Ctiuli at air-snow and at sn{)\'V-ground interfaces Refer 10 Fig­urcP3.24 . USCt. 1.5t-oand Ian 05 • 9 II( 10 ~rorhArd.packedsnowHt3GHz.

_ 3 .25 Thfl (0110\\'11111 uHta Hre Xiven for a u.ni£orlll plane wave in a dissipative medium;

(i) amplitude of E" at z· 0 is 1 VIm. (Ii) phasA of r:. ;I I 1, _ 0 is zero, (iii) k - 0.5 - i 0.5 (11m), (iv) d irection of propagation is in I , (v) intrinsic: imptldHnce or lhe metliulU is 1 + j uhms

faj find tht: phil sur expression for E. os a function ofz. (hI Find the phllsor expression for II as a function of 7..

[c) Skfllch E~ At 7. . 0 and al z .. 2 III, liS fum:tiulls uf w(.

(d) Skctch the time-domain H fields 011, _ 0 and 1,.2 m as functions of loll.

3 .28 Consider thaI a small space vehide with 100 kg of mass is iocHled in oute r space whArA the MraviiationHI field is Ilugligiblu and the fuel has been exhausted. A seo rchlight of 1 kW is turned on, with hopes that IhA veh ide will gain lIomA IIpood. I l ow mnch speed will it finally ga in if the seHn:hli"hl CII II last 41:1 hours? Hi nt: The li"ht wavc carries radiatioll pressu re, and thero Is a reaction force 00 the source of tho light.

3 .21 An ice pArtlde of radius a is r distance away frum the liun, The gravitational force HetlUM 011 the particle is givt:n by (3.46). The ice particle's mass can be obl;lin9d from its volume and its density which III A!IIIUmed to he 1 sram/cm3

, The iet! parti­cle III AIM) suhjecl to radiation pressure which is givcn in (3.47), The force acting 011 till:: icc particle due to radiation pressure is approximately f!CluAI to tha romss· sectional area of the parlicift timAs Ihe radiation pressure. Show that. when the particle's rHdiwi is les~ IhHIl a critical value, the radiation force will be grealer than the gravitational force, and this critical radill ll ill indflJl6ndAnt of r, the dislam:e from Iha IIUO. All a result, all particles with rllu ii siUolh:r thon this critical radius Itmu to be blown out of the sola r system , Find the 'Va lue orthls c.ri lir:al rAdh.II.

3.3 -

3's

li II II 3.' J. I.

3.Jl

CIiAPreR 3 pow." tlUi(i ly. c . 'rI/o·ll IN/MI.N,.

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pDWtr iVlSI/y ,:.... .;,Ie ~.tu~y ~e. S'4 -'.~IIJ • a., . ,,,-u W/",I"HS

e' , T.I .. l poto.).,. "UlJ i~ • ((,c. 5'4) _/""',, D.' 'tl/o"1;c 4..,r)(1.1.5){IOH)~. 10/ . ' MW . " f'= (,.I",O)'~"' (,.r.,.·') = ".,. Ii

(J. 5~) is tIt. ,~f~ .......

'/I./o1'/{u)( .".""tt,4ICl'S"Jt(J.J3. 4.of'/./o""". 4, 09-. 'O"l.'hI .

3)<10' ~.A"" (~ ' ~'/,) '''' a :J.BJ,x/>'m

(A) [W) E '''''I,,,, (bo(K.) - Vm) (C) [fJ' Vso<-, (<<) (rj· S«., «)fA):'"

f= ,"'0 1/" .329./0'" --I.7-1~/,,1.~ I T= '/1' 2.IIIIII·IS'SUI -k=.2-;!A':' 9.931'lo'(Ut.)

(D) 3"1.'112'~7'1") s :1.'-J m . Cb) 3.,.'/(2' ;1.13,,.')-0.7.9-""

V~ ~ : ·j,,·/I.il >9 Jj. rl s_..,...!... V'(E ej ., £) - _..L E.t jJS:: a /I eiltS, - /" - )~. ,"'1'-" - .... ,14. # 1/1 or

:. H.= -~ .. E. AilO Vl~.,.w~ .. 4L-D..;.-It.'."'~/IIf • • './e-UJIJl.t. D..-t, f.I.--IZ.£. - ~.

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r ri1At.-Ao,J. C''rt,J.I./'''''l~ P·(-0.~ tit ,.,t ...

3. 15 (L) EC(I·;+y)~.jI~ =fe-j(~J-f).;€J.) -; iil<).-£s; .. (w~-iJJ+;"'.(w/.-"1) /?,)At -A14,..J,. e'I"4I.IAr(, PDlv-tjQ#..

(bJ g<U<+j)Y+(I-j)iJ e~4> -+ E(~}~r(£C.S'wt-ft)<.~)+;r;;G3(t.Jt-Ax-I) /?' ;J...i -"".-vi c,',-e...I.." ':J P·t...rL.J&.d.. .

(e) ~. [.I--j)" +(3- j) ~ ] e -jk1 =7 E(t)- ; Ii us (wt-Al+M6) + 1:f" LDS(we -*1-0.") Loft - f,.~'" Lll<p(,'Jlj p.l .... ,j u.

(AI i. (jx+jlYJe j ;'1 ~ E(~)= -xS, ..... ( ... t+An-i:u;"'(..,f,+,AJ> i..if'le.rty P.lJJ." '3&4 ·

8

)

<)

a :;fa..b, ~.-~;':f; Ex(~J·4.Cl>;(wt-iJ"'I>A)' E, -a.c.:'(""--.tJ+I>.-;f)

a.t- "J ~, .. , E~'~). a.(.DJ"'{; ar.J. EI-,,-c...("t-~5·) wt o~ 10° 2.0. 30· ~. So· lJo' "70· 8eQ q,,11

Ei tt. .90S'... 84a. . gu.... .7"'" . '~Jo.. . SA . $42" • '1'44. 0

EJ . 7'~.. . a/fa. .9064. .9b'" . 996.. . ,.,... .ou... .94""- . a !fa. . 7.7a.

"---.-....

3.17 clliptica.ll!j p,/...,iJ'4. w.wt. "" 9"' ...... l. r.~, j-(zs.+?J.)e.JAJ ) "'''ere­~ d/ld i. are CDMI'/t~ nu.,..,6er.5 .

Let g ~ (; p,' .. yj~') e-J k , + ('!!' - 9 i.f) e -jA)

o(le/t-h4/>d ar=l.,-l, f'.IA~<j~Jt(e" At-~ ","!,ulAr'-:J p.lAriJU) rh,,,- (Rf!. 'YIl)t'iA1 = (;;e:+;j~')e·jn+ (H.'_Yji';e-JA1 .'. a.'+.k.'~ J!, ""do j(!!.'-.I!')-i q fiJ,'- f(g,.-j.k.) o.n4 .!'=; (g.+jb)

H,,,., "', <D~''''''- "'..., ... ... ,e., Ui:(4-iR)+9i(j!''''l!)]e-ilr~ (t.R-/. • .,Ju.)

""" [ % f( g')!) - H(jJ!, - k.)] (ri,~t-"-'u.) u''''''; pdo uj<4. wa.vt. ,;,.. J'""""" {,,"" it c(f"-+Y h )l!'jI,~ , wJ~ It. v.e/. '

6 fl.ft, rtAl. "u"'~uS. L.t [=(:4+Yb)e-j ·H = (~J!,'+Y;!!')c·ij)+(£ll'-YiR')e·jlo;, til..". CL~~.! '=0- al'lo( j(s,' ... j')=b => ~/= ~(a.-J6) A114 h l -T(AI)6)

:, 'Dw",/.J..I. """". "". [£¥a.:jb) +Y1:v ..... b)] e-j "1 (j,/t-II,.,h-l) ,..,.,I.. (Xt(a.·)b) - 9 ; 64 -b)]( r'1U -h,."tJ!..) , '

9

,/ Mlx/.> , L ~ ( 10" _)JS. _, crw£ Co ''''''/0' ~J'''tl.I. ~>I .', #~/~ c 71'".I()Jfrt"'''-.~.Jf..'(I'./ _:1.,..,X/P .....

S ~ .. 1 . .II-II /D·$'...., < 1.')tfIO·S",.., .', AI ... ...u,.~ i#,'f I~ fA,.'cJ::. IM#kflv .

rr/ ,,,-4- ""Ii'.' 1 ~"" ( • ~11''' ,,7 11 ~ - 4.S'll lp· «I I ~1 = ~ If - :-(If; c -f'llJ.DIT ~ ID-~)t 307Tc A otJ94-l)f..) ... tit- Tr - IC',I (1f"I)

7= I ~ _ I (]I; I I=- e.!...!J!i(I_j.,D4,)-YL_"!"q]j • torr-- YZ O-jrl/l.Il) - TIT. 1-}o-/41t '1 (7. :z.1 Y-o Ai h . O /<S>( _.!.... EA .... lj~ _..L~ = 2.(,S"'"W~L

-, (J i Z 11/ "6.rr .,,,

( a. ) L.t 11' .... ~ vc'.":Jy;~ IU;", lAc. -t • .2 [1.';.:,> {V'J..lm)l - :2 (- ~=) ... h - I + (/.i' = II.,.,. 09 1I'f'f' - It. • r;:; - leo. 72 S _ (It.,.,. - ..",..,..t oJ tz /uk. )

( bl !J. • "'/11. (I.sf.) ( /-j9'IO-')Y.. - w./)J,(1.5f.) (1-j+.~"ID-") -+ 1<.1 " "'/,1 ••• /1.5 a.~·I" t;J(~) /ltI-.n ...... ·I .. on .t.u. 10 S".W 1_ ..... : £"2<"') co ~""'3"',"{U"Z"I'~''''f= o.o'ff!'t., .... )

" ~K/(I

= ~ .• Lf" , .• ,/, - 0.6(18,; tA) E. c "f(-jlo"-jo" )J) ~ hrl- •. r}J"r(-j'.t)

(1) B =5 I~; .. ,(- • .r;)url-jo.t/):; ( •. J'-j'.!JtTfl-o.fJU'f(_j'.I";) (,) ~

"./J,t)~ <.,l-o.r;> CA,(4)t-'.Jj}

\0 H,II,t>= •. 7'7;'f'(-o'S"JJUJlo.Jt- •. t;_rr/.,. )

-- ---

-I

-2

10

f - ~s>A '" (oJ t:\.::..L = /01

" / 3. 2' - C. J 1 (Of i '" 1(/0'-/0' J>(IO'

3.n -

'V= 4t:: / ' Ilj./'XC. xl.:: t.f~I.-l""ll <'0

-II F, ;: '-'11/0

F, = F~ -. q J ~

'7'l, = TIT&.. . /0 g. ::. ). I

"', • /. 1 f • /0 ... ;,

1

1

I F. = ,." liD

11

Problems

92 4

,.leU,.4.23 Tn rf!Cf'i\'f~ II nc"rly poIOlri1.t.,t electromagnellc W.1V1!$, wi,,, grMing.~ may reo place metal plates for refleclor aOI*lOO&5.

Thus.

J. - y r~:8) cos (J e

Reflection a nd Transmission of Waves

(454)

Nole thalllu~ I:urr~nt flows in the y direclion Dnd Ihol no current flo\\'s in the i direction . In fact . if the ctJndllclill~ plHI~ is rf!I'IHC~1 by a grating of parallel conducting wi res orranged in the 9 direction , these wi res a1s(I serve as a fflfleclor thai is as effcctive as il solid conducting plale. Experiments have found that gra tes ~UH HffHf:ti ve when the spacing of the wi re in the grate is much smaller than the wavelenglh of the WHVtl. G " HIHfI (U' t! UStld instaad of cond ucting plates to refl ect linearly polarized eleclromagnelic wave~ hecfllIsH they rt'lduce weight. sove motorial. and decrease resista nce to wind . Based on these collsiciera1ions, some reflectors use wires to re place metal d ishes fortransmilling and receiving eleclrmnH8nelic waves. An f'lxHmple of such a structure is shown in Fi,Rure 4.23.

- 4.1 ThR E ARid measureu ill .. ir just above a glass plate Is cqual to 2 Vim in magnitude lind is direct at 45 ° 3way from the boundary.:ls shown in FtgurR P4 .1. The magnitude of the F. Aeld mp3SIJrP.O jllst I.lelow ttlt: boondllry is eqoal to /I Vim. Find the angle (} for the E fidd in the glass just below the boundary

•

- 4

4.

5'1

.he II~I

50

we ~ is or

.'es ,d. .,1 or

" i,

"

Problems

'. I i I

E, i

Perfect I I rMlOllCIQI /

• • "."". "4.1

1"l ... ,.I" •. 2

I

I v,

- •. 2 The H field in air JUS! above a perfect conductor is !!liven 0)'

H , _ 3x ! 4i amperes per meier

93

•

all .~ho",n in Fil!Ur9 N .2. Find the surface current J. on the surface of the perfect conductor. The conductor occupies the llpar:e y < O .

•. 3 Match IhA following dt::;t;ri~[ions with the figures shown in I'igure P4 .3. FiAirili a re lIell l' ,he interface but on opposite s tdM of Ihl! boundary.

(al mAdium 1 and medium 2 aredielcClric! wUh. , ;.. ' I

(b) medium 1 and medium 2 arA rlielflctrics with t, < t~ (e) imp"IMihle (d) impossible (e) Ihere is a poEillive lIurface chl:l/'Xt! un tht: boundary between two dielectria ( I) medium l is II J}t!rrecl conductor ~

r

,,1.uI.P".3

~ .::. k . -z-- r~ 1 ---~ E, 'P, ~< ~ -. , , , • E, - u , r

'-1'1 (II) ~ (III)

- ""-'-

~E' ~} 1°, , I ~ , , , , In, till) > 1·1

1"'1

94 4 Reflection lind 'l'r.n8mi!l1l1ion ofWav9s

.... Calr:lllale the critical <lnelA e~ uf an air-gJasslnlArflice lIimiiar to Ihe interface shown ill Figure 4 8 ThA dielectric conS'anl of glaSll ttl OpliCilJ frequendf!!I is 0! .25 timCSlhat of all'.

- " .5 A pearl is emheddecJ ullhe middle of II cllbic lu:avy-Icad glass If, - 3.6). Is it possible (0 cover II portloo oflhe surface of the cubcso that from outside the pearl will nnl be lleen at any viewing angle? If so. find Ihe shape and the minimum area uf the cover fin terms of the cubic surface areaJ. Hint: consider condillons of total reflection, and nflgleclllluhipic inteml'll reflections

... e In the IhMfHnecJia configuration lIhown in Figure P4 6. Ihtl wltve numbers are k, . k,. anr! "I' Find Ihe transmission angle in merl ium 3 in terms of I, ;md tilt: wave numbers. A5Sllrne all k's arc real.

'. ------> ,

k. k.

--- / I C laS!! rod L--_

•• 7 Solid -stale lasers (ruby or glass) arc orlen fahrkall!t.1 uf rods wiTh Thll entb bevelled aT the BreWSTer angll!. Lt!1 ( - ~ for the rOO. Skelch the proper arrangement of the exTernal mirrors and Their angles. Indicate the twlvellt!d angle of the glass rod. What III the polarizaTion of the uutpuT of Ihe IMllt i.lt:ulIl? (Sec Figure N .7.)

•• 8 A parallel-polarized wavll is incident from mtttlium 1 on rhe plane boundary tw.lwtlt:!o medium 1 and medium 2. Holh lIIt!dia arc dielectrk.jJ with ~ \ - 1'1 - J; and real permittivilitls t\ and {z. WA know that when thl! incidl;lnt angle II larger than Ihe criticK lllngle 9,. no lime-llverage power IlIlrllllldcrrcd to medium 2. Also. when the incidenl aoSIA is equal to the I-Irl!w~ter tlogle ~. thA ref1tlctcd power is 7.flro. Now Imaginl! It lIituation in which tlltl Brewster angle i~ greater than thll critical angle. A wl:lve incident III the Brl;!wster angle will nul be refiocled, lleClIUllt:! Ihe incident lingle is cqualln 9", nor will it be tranllmitted. because the incident angle Is grel:lh:f than 9,. Is thi~ situation poSllible? Why?

4.8 Consider an Illeclromagnelic wave of 1 MHz impinginH 11.1 60° on the inno~pht:!rl;!. ThilGase ill similar to Ihatshown in Figure" 13. Assume tho the plasma frL'quency of Ihe Ionosphere is 101,. '" 2r x II x Hf rad/ s. anti plot I E I Gsa function oh (like in Figure 4.14). Mark thAItCI:IIt! or z in meters. Solve only for IhA OIlIt! or parallcl polarization wtth Eo - 1 Vi m.

ofWlLves

cashown imes Ihal

I possible i ll nm be .he co\'er tion, and

Ire k" k~ he wave

elled al I of the '- Whll l

undal}' ~and

han the 'H:n the o. Now n,llie . A II angle Ihan {I,

5phere ency of

Figure iz.ation

Probl ems

x

~

Ivk..J,ulJI 2

-----~ ,

•

"_u,. P4, 10

l

3 2

"tu,. P4, 11

95

I E llvolts per IIIclcrl

2

/"', ~-

-'> ' lmelenl

• •• 10 A perpendicularly polarized electromagnetic wave impill)(es frOIll medium 1 (char. acterized by ~) - ~ and t, - 4(~ 1 10 medium 2 (characterized by ~ J _ ~ ond f l - IDI. This situation is shown in Fi)(urll N.IU.

( 8) Wh,lI islhecriliUlI angle" ~ '1

(b) Lellhe im;iuent all,l!le lre 00°; finu k. and k. in l11rlllll uf k" - !.J.fii7.. (c) Find k" in terms of Ko. (d) In the second medium. find Ihi! tlishlllCi!:to. al which the fii!l!! slrength is l i e of

thotolz-O ' 'e) Finrll R, l and the phase shifl <11')1 (Rrl.

- • • 11 A uniform plane \\'a\'e in air illlpill)(t:S lIoruliIlIy 011 a dieitJctric wall. The magnilude of the tulal E field measured in fronl of the wall is shown tn Figure P4 11

4.12

• • 13

•. 1.

( 8) Whllt is th fl [lflrmillivity of thp. diflifICtrir. wall? AssLlml! " 1 - I'fr

(b) What is the frequt:ncy of the wave '

A uniform plane WIIVt: in air impinw.'S on a lossless dielectric matt!rial at a 45 8 an.!!le, (IS sho ..... n in Figure P4 12 The Iransmilled W(lve propagates In a 30" direction ..... ith resf>ect to the normal. The frequeucy is 300 MHo:.

(_) I,'ind ' . in terms o f ~ ... (b) Find the reflection codficitJnt Ru. ( e) Obtain the mathematical expressions for

the incident E fi e ld . the refltlcltld E fitlld, ond the transmitted E field

( d) In Imth mp.dia. l'Ikfllr.h thfl l'Itonciin,!! WOVA potlt!rn of I E, _ r I os 0 function of z

~h!d i llm 1

k,

R'

Medium 2

k,

---~ ,

I"or t ..... o iMtropic media with~ ) ,. ~, 8nd f ) ". I" lind the Bre ..... ster8ngie for both tht: pc:rpt:ndicular pulari:tatiun and the parallel polarization

U 8 wire ilnlenna is attached l-'llrullel to the metililic surface of a vehiclt: and is insulated from the surface by (I thin layer of dielectric material with a thickness approxImately flqllalto 1 mm , would it fl'!CfliVft an AM ![snlll If _ 1 MHz)? Hint: Wire antenna interacts only ..... ith E field in the direction of the ..... ire

96 4 Rcn ec:: llon . lId n-lI nlmi ~lion of Wave.

- 4.15 It i,l known Ihllt the Iramlmilling anlp.nns of an PM slation lllioclited In the dirl;!Ction perpendicular 10 a metallic plate. as shown in Figure 4.15. The frequency of the signal is 94 MHz.

(a) Where should II rtlCCiving antenna ~ placed to t9r.eive m<lximum signal? The antenna is II dipole that interactll with the 10; neill .

(b) If the amplitude of the incident E field 1111 VI m, what is the amplitude of the £ ricld at Ihis optimum pusition't

• 4 .16 II is found that by placing 8 t:Ol1ducling plale 0.6 m lJehind II dipole antenna, the received sigll81 coming from Ihe normal direction is twice as strong as Ihe incident field . What is Ihe frequency of the signal? What would the strength of the lota\ E field be if the frequency of tht! wave is r:hlfllgcd to 98 MHz while the anten na is still placed 0.8 m from the plate·t

4 . 17 What would the E fiAid be if the rect:iving S)'MtJ!1Il in Prohlflln 4.1!:i werAlo delecl II W iW*! coming in al an anglA lOa off lilA normal? All/lllme that 311 nlher paramelel"!l relnain tilA same.

· 4 .18 I IArive (" .!:ill.

4 .19 For 11 parallcl-polarized uniform plane W!lVfl Impinging nn II perfect conductor at:1n An)!lc 9, find the eleclric. and magnetic: fields E ;1n(1 H for the incident and for the reflect'lft waves

4 .20 Consider a 90' "comflr rt:flcctor" shuwn in Fil!jun: P4 .20 11 is made of two conclllC:lin~ plateR placed perpflndicularly In 811Ch other. A uniform plane wllve with E - iE. cXPl/kx cos 8 t /ky sin 8J impinges on the structure a t an angle (1 Show that the tota l electric fielr! is E _ - 14E. sinfkx COlI 8) sin fky sin (1). Hint: Tht! field is the sum of four waVflS with four k·vt:Clors shown in Figm8 N .20

4.21 Ulle the fnrmula given in Problem 4.20 for the totlll electric field . Find the optimum poSition of II dipole antenna placed in fronl of the 90- comt:r reflector. The' angle of the mddent wllve is 30-, The frcquAnl.)' i5100 MH:£. Express tht! position in x - y coordintltes In meters What is the "gain" of this rccelvlnSll.ntenna? GHin Is defined as follow!!:

GHin - 20 lOS" l~ I (dB)

when! E. is the 1-; field at the Rntenna poIIiti!)n and f' .. Is the field strength of thl! incident wave,

y

> ,

!"i.yr. '4.20 Top villw of a 90G comer 1'8-

neclor aoll Ihe rour lI _veeloTI,

, ... lion the

The

the

the ltml 'ield still

~I a )Iers

II an -the

two v.ilh thot

i Ih~

"urn :Ie of -y

ined

f Ihe

lef re-

Problems

Medium I Medium 2

2 .U

o •

> ,

==~ ) ,

---4 0.67 --- - -o > ,

> ,

, (.,

Figur. P4.22

(b' Fll'!reP4.22

2.0

u

2.0

.. 0

o

t85

.J:<:

t­---1ll~

97

,

,

,

> ,

) ,

98 4 Rcneclion a nd Transmission ofwlilves

4.22 Malch tilt: following descri ptions In tht: stand ing ..... lIve palterns "howl! in Figure 1'4.z:l . The int:idelll wove in medi um 1 has an tuu phtudA Aquui to 1 VIm. Nole: There are thret: patterns Ihllt du not fit anv of the folJow inM utlscriplionll. Cross out the~fI patterns.

(i) Plot of E, ....... ,wilh mooium 1 hf>lnM Hir. medium 2 hUlling (l .. 4fD Hud #12 .. 1£0. t\urnu:al indrlem.:c.

(iiI Plot or 11-;, ..... ,with medium 1 being churaCleri7ffli by (, .. 4(0 Rnd p, Po-and mt:dium 2 MinK Hir. Normal incidence.

IIIL) Plot of I 1::, _II ' with mmliulU I being chlll1:lc lcrized hy I, .. "1 0 and 1', ~. and medium 2 being /lIT. The incicfllnce auglo Is grARler than the !"Tillettl anglo.

(iv) Piol of E,IIII&II. incidence anglA is tll.juoilo the Brewster angle. (vi r iot of I Euotell ' iucidencfI anMh: is equal to thu Brews'Ar aU8lc. tl is greater

than r~. (vi) Plot of I Ey1uIall. Meuium 1 is lIir 8ml modium 2 18 j.Jcrfoct conductor. {vii) Plot of ) II n",.1 I (Ih). MerllulII 1 is ai r and meuium 2 is perfet:lconductnr.

4.23 Cunsider the i.:ItSC of normal iucidellrA of a uniform plane W6\'C on 11 perftKt con­ductnr IHI ~hown in Figure 4. 15. 11 Cilil ue seen in [4.47) that an ollcilluli1l8 currAnt 18 IIU.lUCed on the surface of the conductor. Tlwfefore, the fullowing expre~~ion may be w ritten for the vtliudly of II chargc 011 the con d Ul.:!u l" :

v. tdqEncos(wl- h)

The abovtl aquahOIl is exactly the ~ltmo as f:f)u llliull (3.39). Con\IIlUO to wnrk alullg this line lind 1-l1"O\'0 thattha timo-averaSfI raJiation prellsure on the pflrfel.:! conduc­tor III twu limes that given in (3.45).

5.1 Pal

CIIAprEI( 4-4./ €,_ - E .. ;.... -+- e; co, B = Es (AJ'I-~'

- ,. - .... (...." ,. '" (-l/ ) 1:4 ~ = Y"'lI, c yx ~~"'4-I)" "'~-3* y",

4.3 U

<II C (iiJ f (;11) b (iV) et tV) d (vI) e S,. ~'"...-11;;.1~ - 41. SI'

si"" a:. r ~, , G()V~r PoI""I-/~ II" ~ I".r~e.. /.s A.. '.ire/t.. . • It .. _·~.A a.

Tirl..L C'JIV6.r Ar,ea" - 61<]rIC ("'£ft,.,.,p~i" -3..trJ(.z .M ... . ':;:! -1.812 (). I. '" /-s....,; " tA.-6,i:. .vk :;:,,)It a -= 'a"

:L '" ""vIr MIlt. • ..4~z. X/"',,!, _ 30.2 ~

s"e./!,' 1£"-1: /(,s; .... #,- ~,$.'".Ia c A,s,:"'1.1 ~ 19.1_ j'i",;I(~S~A!)

t. It, .)

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.c.:!-_--'.L-_J ~'. S;""6·/L$,;...(:-~)-rrc.s, •• J'/J-J..,,-1I-sf.7" ........ ..,. " Pi -1./ .. 8'~.7· ."J It. a I~'. '.'-S"4."1". ~s: , ..

i'"pI!Jssi&le.; /Jrt~Jl .. ,.. 4"J1e. (J'.) e-/wAf.s I~ss ~ cr':I,'W (U'J.3/e. (19,):

..Jw.O. z: (£I/E, S~~, .';11/1/ ~n;.kJn; ... 8c. > IJ. I, ~

Jl,~/', C/al .. J It.·,u • . f~=£Q[ :.tj ... ,O!.

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=

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rs J.)-

4.1J

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. -- ---:;--~. "'­~--"~---t----

1),7$8

0-' ,

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,1', - , -'#, ~ £, (I'l i _'pl·) 6 £'(/,," -",Il,' J

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c;.'''I-~ ", )', fa -I, ~I(Z,,"-I,')

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M /£1- 2/ •• {S;"("1> • ~ :'.8'" il: ,,,..V9ftL'. Z.t>' , JK /P .. ~

.: /£/-/£./. S,,,-( Z.OS ••• r) c /. 99> 11:.1

4.17 """1 /£/ • .2./£./5,' .... ( k,l). "J"It"'>!.· :. 1 E 1 % J E./~ ";"(/.f.9' •. & < Co>l.· > =1.999""./ •. il . , .

. ~..4~-"

13

+../

in

1

---;"*"'--J . ~.

Per/,e.1: C6~k>r: le, c I ,# 1i. 0

-k" rt: Ii s;""", -ti} 3 /({..Aj/1

R"• YII.e·J'lKI,;""-jJ.las/J

gi. (;''''SS- lS;..4XII.1)e-jAXr'..e -j A, ... e , r' /If-' il r _ ,.)/, '·jA'Xs;~")A}u'4 _ y.e . . •

r • -J~X$'..4'j"''''SI € =-(R"J8>2S,;"fJ)(lI.pe

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iff. t .... t"~ i" .. t' = ~ Eo[ ejAX"S~( e j l/ls,'J_ tiAl"~') -e -jixu,,( e;'1.s#-,_ e-.iJ;s:.,) • 'i t. (lj)S;~( R¥J,;"S) (:j)J, .... ( AxeD' (J) '" ; (-4/:.)s;...( lI."w,~) J';"(~IJ'~') -

f:. /()O;l)/) '9:t- 7"; --3'rr ,8PJ.·.~~,.l CIISIJ. ff '/IV" z. ' c

,'. ff. .l(-4E.)J.l,...('!"X)s;""(~d) ~~ ~ • /" -- f>' --+.::. ,. --{>'--, I

, ,

A-• ..... s

'! ./ : • • , • • .I, • ,

•• 7 I.'?J ••• .J, ••

(j) fl.r = ~.-~. ,., = -f i.+ (, --'+ ).

Iflt.t,,1 .. t J -0 : 1+ Rt = '/J

I (;ytoh., I ... A_j ..... "' .. I It/Ill. ~IJ I by M.iI "" ...... ....... I-I R.I '" >/)

14

1lAi~ k<...:J.,;,,,,, J"

1%-·17~ ¥ = I .$"M

-jitd r'"r,

<I.) f,t4/ t""''''~S''''', M ,.{k<h',,",. 1EJ,If;,.,tl - ... <.7><osf •• t ;"

•• /4 "'f.l.· .... ~ I c",.t .".,J,'w.-. 2.. ~ 'flu·lt .. 'jltt lo/lA,."'''J th.'rvJ "...r,

\0 +.+.1 frA"'''''So'.". • ..... re{Tt./" ... _ "V f',t.I iJ u ... t •• t ;. ,.., ... ~,

.. ,. is ",loCAl -Ip ~,..H.,,. '''~J t""A .. t ,'" ""I,'w"" ~. 7~ r,+io ~" .... I. "c..

tVi) N. E /.'.ltI ,'" t"IcJ.'", .. 1. J 'I);,I.J I" 0 ... 1 )., . 'Pia itt ..... : ffft C4/14"'~1 f..~s-t rooJ';

(vii) /I. H /.,.,14;" ,..,,/ ........ l.. 11-1" 1,#..//1 " ". .. ¥;tM"'_ c:1 J'. l'"tk" .. I I"f ul .......... ~ i-j'" r.V-.

- . II : 1

-F", = 1;;,.a" S (.", I/l~E: /1, ) .. ,'(111- ";>

l th.·~ l. 7"'" 1,wI ;.4 lwu.. /~ #'~4k*- (;.¥o ) J

F.llo"",,.'!' f4~ 4,,,,,'.,c.4-,_ Jb1t .. ,'" 'fL .. .J.o.t I bllf' OJ.~4""~

- .. (-P=c: .S)

15

" 15

:al

:bJ ,. Id ~t

loJ

Ibl

I a ,v. h. 'c.

h. ;ul

'c. ice ilh

00"5

lid lid h.

me

Example 5.8

Solution:

Problems 127

Using t5.45~ WI! QuIll in

- iJA • - fle Vx A -fl-' - . -- e ..

• , p

Cak:ulallt the IQlal time-averaije electromagnetic power Iransm iUed IIlon)l a coa~ial line when the fields ilft: ~ivcn by (54ft).

The time-average Poynting vector (5 ) is I!IVtlll by

(5 ) _ - Re[E )( WI - Rc - e I I IV' 2 2 P

''''P)(~ ·e' .... _ 1._'_ V I V ' 'IP 2'1p~

Therefore P _ (r V~Il))lJn{blaJ.

Problems

- 5.1

••• - 5 .3

Show thaI the complete fields of tilt: TE wave In II para lltd-plalc w3veguinf! aTe ,IIiven as follows

E,. - £.sin k.x e ,.....

- kE. H. _ --'-sin k.x e .... ., ;k.E. .... II, _ -- cos k.x e .,

E.-O: E.- O: HJ .. O

Find the r.omphlle fields of the TM wove In ;) paral14!['plate waveguide

What is the lowp.sl frt:qut:ocy of an eieclrulIIllHnclic wave thaI can be propagated in the '1'1': mode in the carth·inno91lIu:n: wavegUide? Mod~1 the latter as two perfectly r:onducting parallel plate~ ~eparatcd by AO km.

- s •• Find rhfl9udllcc·cha rge den9Hy p. on the upPflr lind the lower pllllps of Ii parallel. plate waveguide for lal the TE. mode. and Ib] the TM", mod A_

- 5.5 Finrl the IIIl1thcmatieal pxpre~ions for a Tt-:M wllve in a parallel.platt! waveguide thaI propagatl!l'I in the t dir€'r:lion I~e Figure 5. 11. Skelt;h the parallel.plate waveguide. and indicate Ihe rllTt!ctions of E. II . and J •.

-5 •• A rnicrostrip line has the dimeMions f,I - 0 Hi em and w _ 0.71 em. and the permittIvity uf the sullslralp. LS ( _ 2!i t ". I' - ,... II 0 Eslimlllp. the tUllc·averag€' powerlhat is Iran.~m ltted by the line whtm l E 1_ 10' VIm

128 5 Wavftgu lcl flfl and ResontllOrs

' 5 .7 The hrf'akdown vo]toge of Ihe dieler.lric .!luiJsll1lt~ used in Ihe Slriplinf! clAscribed in rrubll:!lTI 5 6 is 2)< 10' VIm. Ust: .. safety foctor of 1050 !hatlEI is ltlsS than 2 )( 10' Vi m eVf!rywhere in the linc. Find Ihe maximum tune-ilverage power Ihat the strilJlinl:! can transmit Neglp.cllhe ohmic loss

5.8 With the fields in a rcctangular wavJ>glllde, lind the surface current I. nn the top Iy - bl of the w;J\'egtlinll. If we v.anlio CUi a slot along 7., where should Ihe sial becul in order In mInim]',,\:! the disturbance it will r.aWle? Assume that only the TE,. mnde eXIsts ill the \\il\'cguide

5.9 Shnw that, If tht: .... avclcngth of an f'lectromll.,ljnetic wave In an unhnunrlP.rl medium chllrilclcrizcd by " and f is KreHIt!T than 20, then this Wa\"A cannot propagate in Ihe rectangular wiivtl.i\uide (shown in Figurf! 5.8) wllh the dielectric inside IhA wave-8U1de al50 Ch(lr<lcterized by" and I .

fl :;~, ml!tflfS

Exhausl 81r du~l

4 HI melp.T

U "­

Frc~h"lrdUCI

,..",."11.10

5.10 An AM rodio in an aUfomohile cannot rl..'Ccive any signal when the cor is inside a tunncl Why'f [.Ill m ilSlIUIIH: Ihollhe tunnel i8 thA LIncoln Tunnel. which was huilt ill unA nnder tlte Hudson River in New Yurko Figure PS 10 showA a cross sa;lIun of the Lincoln Tunnel·

5.11 "'incl Ihe fn:qucncy ranges for TE .. mooe optlr'l!lon for 'hMA rAClall,ljular waveguides listed in Table I.

5.12 DfI..!Iign an air·filled rectangular waveguide to be U8ed for Inmsmission of electro­magnetic power al 2.'15 CHz. This frequenr.y Ahnult.l he al the middle of 'hI! opP.rating frf!<Jllf!llcy band. The design should a],u allow maximum pnwflr iram:lfer without !ItIcrificing the operating frt:quency bandwidth. Find the maximum power thA waveguinA can transmit Use a safety {aelur of 10. Neglect ohmic loss. The hrflakt.luwn E in air is assumAd 10 ire 2 x 10' Vim

5. 13 Repe"t Problem 5.12. but assume thaI a t.lit:lcctric molerlal is ItAAd 10 fill the wlIvcguide. The malArial is charactcrized by , - 2.5Uto. " ... ,.... and" - O. The brcakdown ,.: fiAld in Ihe dielectric Is 10' Vi m.

5."4 Cnn:iiuer the size of a rectangular wavcguide to explain why il i:i nOI used 10 transmit clcclromagnAtir: wilves in the VH F rangfl. (Tllktl f - 100 MHz]

"G. t:. &tnOslrom. Turn"ll'lI.. New York: Holt. RinclwM.t WlnJllnn. 1963. p. 242.

•

ttnlltors

7iUetJ III :! w 10' hat the

the hJV

J( becu! moo,

nedium e in tht>

inside a \ huilt ill In orlhe

.. cguides

eleclro­p!'l<Itin,l! without

WI'r the JSS. The

fill (he O. The

transmit

Prohlem"

I I I I I , o

129

FI9""·",·16

) ,

5.15 The cleclromagnAtlc fiehJs ossociated with the TE ,o mode pmpaglltinx iii the t di­rection Me givt!lI by (5.23). Find Ibt! t!lcctromagnelir: frelch ossociated with the TEIO Illude propagating in the f direction, wilh maximum eiectrir: field equal 10 E,.

5.16 ConsinAr a reclunguinr wavegllinttllhuwlI in Figure 1·~ . 111. Fur the region z < O. the medium is nir ann for z > 0 tho modium II ciulrdclorizoo. by ,~ lind $/oz. A TElo mode with maximum Efield OCluIIlto En implngos on thA boundary from the iftft. Thfl resull is that somA power is reflected ann lIome is transmittod . Alisume tilal tho reilocted wave is tliso TE ID. with maximum E field aquallo EI, and the trans­mitted wa\1! is TEIQ mode with mllxiJlLum E-field ACl1l81 tu Ez' Find the ratio E,lEn in It:rlJlS of 0, W. fa. f'u.' l' alii.! $/oz·

5.17 Tilt! corncr refleclor sluLiied in Problem 4.2U rCi.juires Ihe solution

E - i:4E. sin{k)tcM8Il1l11ikysinOj

Show thai althou.IIh thc coordinates IIrt! different this solutlOlI is in fact the resonator mooe thai wc studied in Sedion 5.2. Placing condut:ling plates al )t - fI And y - b to furm a cavity ff'.sonulor as shown In ProlJlem 4.20, what are the restrictions on the incinpnl anKI!:! 8?

5.18 (a) Find (hI' real-time expression of tht: fiddsof the Tt';,... moot: in Ihe reclangular cavity shown in Figure 5.9 .

( b) Find Ihe tolal slun:d eleclrlc enArgy III the cavity as a funclion uf lime. Find the corrf'..spundinglotal storM lIlaHnclie energy

Ie) Show that enl'flI)' is stored alternMlngiy in !:!Ieclric and In magnetic fields. thai the maxUlium stored eleclrie IInt:f!~y is equal tn Ihe mllximum stored magnetic ellt!tgy. and 'hal Ihl! total siored elpclrolllll.llIH:tie energy In the cllvily is a constant inLit!jJt:ndenl of limA. Note thol these proPftrties IIrt: similar to those of the luw-frequency I.e rl!lfOnant circuits.

5.19 Find the lowelll resonant frf!C]Uflncy of the TE,., modI! in an air-filled rRClanHular OIvity III!:!Qsuring 2 )( 3 )( 5 crn l . Note thlll tht!re ore three different choices for designating thA Z axis ond thatthA.'If! result in three dlffenmt TEIOI modes.

1.20 ~:IACtrumagnelic waVAS in air with wavAlflnKlhs mnging from 1 1010 mm are called millimeter WHVI!S. Millimeter wavl!S may be gulden by dielectric slabs. r.onsidl!r a diAIACtric slab with" - Ilk, and I: _ f .. all !Inuwn in Figure 5.12. What should ils Ihickness he in orul!r that only Ihe TE. mude may be exdted fot' frequencies up 10 aoo Gil ... ?

-5.21 IJM direct substitulinn Inlo Maxwell's Aquatiuns to show thai the rll!lds given by (5.481 ore solutioos uf Moxwell's ACluations in cylindrical coonJinalcs

130 5 W.veKuidcs and Ruonaton

5.22 Use the formula!:! of divergence and curl in t;ylin drical coordin .. rell 10 provf;l thaI V. V x A _ OforanyveclnrA

5.23 Find the fP.Ctan)!uiar coordina tes of a p(lint P where the cyli ndrica l coordinules are p _ I.~ _ 30o.andz _2.

5.24 " lnd IhA cylindrical coord inate.-; of 8 point Q where the rectangula r coonJinales are 1(, y. and z.

5.25 Show thaI the di fferential voillme in the l.'ylindrical coordlnatM is p up dIP dz.

5.21 To convert It veclor expressed in cylindrk.al cOlnpullents inlo the same In tP.Ctan,!luil:lt oomrxmlwUI. or vice versa, il ill conven ient 10 prepare II ta hle fur dot products Ut: lwccn unil veclors in these coordinate s)'!!If'ms. For example. i . p _ (:0,0; 6, all IIhown in the fullowing lable_ Complete the table .

Dot Products aetween Cylindrical and

Rec tangular Unit Vectors

p ! • i cos .j

i'

• 0

,

5.27 Use the aLovtl table to find the roctongulor componpnls of the fullowing vector lucaled at II - Z, Ib - 30°, anu z _ 3:

A _ lIp+ 4~ 3 f

5,28 What is the maximum time-average power a coaxial line ca n transmit without call.'ling brtlokdown? Assnme that the coaxial line is a ir-filled and Ihal the break­down E of the air is 2 x lit Vi m Use a Aafety fac tOr of 10 50 thai the maximum E fie ld a nywhere in the line does not 8xct!f:!d2 x 10' Vi m Thfl rlimenllion of the line is 2(1 - 0.411 cm and 2h - 1.14:t cm. Neglect ohmic: loss,

-5.21 Cuns idc r the coaxial li ne shown in Figure P5,2A. HaH uf the line (z < 0) is filled wi th ai r, and half uf it (z > 0) is filled with a IUllterial charar.IArized by tl and ~I '

The 81ectrulllagnclic wave inc:ident frum tho loft hi" the folluwiu8 field s:

E' - V, - ,u... - p- c p

, • V, H _ q. - e i ....

'''' The fields uf Ihe reflected wave may Iut tlx lJ rcsscd as follows:

V' E' _ p~el"" p

atonl

. thai

'5 arc

)lular ducts 1/1. liS

:e<;tor

ithoul neak· .um E Jine is

filled nd ~1'

Problems 131

fa) Write down the fields of the transmitled wave In 1 '>' 0 What wave number k !lhould tw. used?

fb) Finu V~ and the amplitude of the transmitted fields in terms of V .. ",. and.,. by matching the boundary conditions at 7. _ 0 C'..ompare your r&sult with the reflection lind tnmsmission cot!fficitmls Obtoilloo in Chapler 4 ror waves reflected from dielectric boundaries.

I ',.1',

/ r =t= > Figur. P s.n ~ , ,I,

I , - 0

CN/iPTER S

£.1 V.,g-jwr.C ~ ryrJt 6--jwJlB willv _/ay.a

9 - If S1--i"'l'-tJ, i ~~J =~.ilJJ4~' .... (~"~.";~C)E',,-O OJ . ~).. .,:oJ -If '-______ _

fiUll - n Jl .. = J .... l IJ 0 A~~W""1l §_: Atr)e".,i- J ff.u. JJ1'l)+ w3M!-".' _0

"-:!ttl) 2 t 1 : • ..::l< Xl -+;7, -0'" A~) .. c,c.~lx'lC.+ (, Si",",.zX I 1.>J4"., It ... 1.},J.lt-Aj

/. §,- C,Cblllt;r.~-i~l!+CJ ~~2~·jl,i

ErJ'k.-o=br9 c,·o j '1/~ ... ·o .... It~.~ I .,..,,2JJ .. ··~ :. f, - E. s ...... ~~x e .j~/l (~UI.I..J.. II I; e.)

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• ", ti1l. .-!}A.Eos; .... ~lf%e ~jI4i? ; 1J. ... j :ft.-. c.>l~J:e·jl~i-)~ ,~ .. ~,-tt,:o H. AHu",e ~/a'S = 0 ~ tit· J.I(~)~-J.,I, I4tn.. (::1" 'i~: +w~t)t:!,. g;~(ff" (JJl£-R;)-O

H(a). A £DJD,.. ')!' .. 8 s, ..... ""x • IIJltart It.r I. ""~l-A.*,

S . ,

:. HI- A<u.,.e-j·.i.S.i"l,re-j ... • 4nL E-j.,f .~(:t){Au,.,r./l'~"'.)e-;;'l - • z (~)(-A"'I .. "8,",1,»~-j·.T-

~.I~.o= 0 + S cO :. I €. ~ ':z' H.GAJi,« -ji,l (","01.1.. ~ b, H.) E _ ';'Ix 11 $/"k )' e .j'.l _' 0.011' "

IJ; • 1-4. tb5.f"xe .. ji,f

t!:l, -Ji4 • {1 - 0

!. -= ---L- = JIt/o' _ / . 8'1S ~IIJ , 24~ ~1>i"'I(/OJ

Ca.) TE".: €,. E.fi ... (.s!1t)e·jl,f: ..,_ -t' f.,i x_#· f1/, .... 0" /J cO ",. ~oH. pt.'C$ • M A, (~. ) .jl,' I> L--_______ ~

( ) T ... : ~ It - iJl" H. CD) 7: X e (FroG'" p,..,p .. ,., l:.!)

d r.()~ t-~~·£EJ): •• • .y.",.t.-j'jl-

at %-4", jJl::'_;·t~/x ... ·-~H,(·,re-j ... i _(" )-'.!.LH.e·ji,, J 1t1.0 ... {.,.1'J' " _ W

r£M;" po.ro./I<l pl.f~s - 711. . f ·· J • -" J4~ = pnput ... "'j ."" -2 a4""t,,,,, : /1- Y H.e ~ /t. • ."Mr;.

E.:= ~"e • ,H. (-j4f) ejl~ _ -iN 1ej• l - JW[ Jwt ' at ):,;:01 .b.l: x,,8Iz,,0· 'iH,e ijl

at :r .. a.., ~ ··; ... BI, ...... iH.ei.J

P_-L/E/2.,!!!. v. f )". r.ITlIO·'II.I.S-rt/o·J

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fro,., ~j"~).;DI'I ($'.II,"), j6,JiU. m· 0,. .

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16

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O~)";. ~al'tlI"I." 1.91(/'" 41( .11.0 -). 3lI1W

~ :sIOD HHJ ~ 108 >(!c.),. OIl ~:':'D' .... a.>I.~"'. ph,liWt,.Jw,

r (E ,.-j-l." l + " ,.; 'ill ~) $:. ~ ,.. = t l' = • < "' • ..... J "'y.

fYJ = fy •• i i 00 ---+ IE. ~ €, = ELI J.i., :: fE, e-'J-I./l t • E, ejl\lf) -'ill 5 ' .lL!:

\.j l,.) A. . '" a... i

(E -E).:fu. • , ..... 0

s,I.~ Ii.~s .. &0 .. 1 .... 4/0 .. ,

.£. = ..... , {,l' ..... -1.11 Eo ..at '"'n + .M. ~u

",l..,< i'l = J w·",.f. - ~J'

17

I-j.!

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• +..,.9- ;t ; ,".'.I. '~'· · . ft..,,".' ..... So 18 (4)

(e)

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H E" ". r · ."A wi' t ;X • ;:;7.i4 7: "'~ L ,~s 71C S",. Q/

H • •• b:. CO$.!Lt S;,.,. ~t si",wt _"" CiI... Z

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VII - fE,,'lt£iJt['f'L i (-;b s;""(",:) CaS, t(!!J)S;"~*ti.t CD:.t~) ~;.,,1(¥) s;,.1~tl drel IIi ~ 'I I I I • l4tt s,· .... tt.Jt(AM)(7}.7f:)

u. - I. t d) .,Ito..z 1rJ. Z (--'- i Ii. -TE,, (4Jo ; v~ • ,,,,~E,,(.:L'«) jiT+JT)

At ,e~ • .1 fll-jinl-..t+;jr)Yz. ... ulc(21r/->t. ,;/ (-it" ,/.), tI.~nrore U~#. ,W:: ~11.(Ut1)(-ir+ir)x ~1(-ir""-;/rr/. : e,}'( ... ,,,,-). U~­

At asOI1Q.At.£.) lUI:. .M"/(-;!r+ir.) -.> UN, - +E...'( .. ~d)S,~2'''''t J

ihU"t/Dr8. Ul ,,.,. UN ~ £Iii.. {-~Ht( .. .t)[5;"'~l+Go$"w'tJ _fE4I' (Abd)_ ee-s.~r

(iiJ CJ.use 4." l tIIr1

18

.5" • .2 .1

- I w.e s:_ - p

w x.- COS 30"z; {J/ z; ~. Sin Jo". Vt ; E-. 2

S·'+ 1':: (7.'~'1' ; ¢ .,.;.".'(~/'lt ) ; !-..!

~ d,,= df·(Pdl» ·.i epa!,d?d'

Hi • ~ ~

• f' ~

• cos~ -S/"~ () • · 3/"~ C05¢ Y " I " () I

19

Problems

180 6 Tron,milll ion Lines

We can find thft rtlnecled wave by carrying out an analysis similar to the one for II transmission line with a capacitor. The result is as follows:

V _ VL

V, _ V.(e 2.0(. TVL - ~I [6.58)

Figure 6.39h shows the voltage VIz) on the line during the lime period T < ! < zT

1 .1. Wh", t81he voltage in the striplinc discussed in Example 6.1 whtlnlhe timtl-llverage power uein,l! transmitted is 10 kW?

e.l. ConllldfIT IhA coaxial line discussed in Example 6.3 . Calculate the maximum time-average power thai may lre tra nsmitted in the IinA. IlI1e Ihe hreakdown E - 2 )( 10' Vim and a safety faclor of 10.

' .3. Two coaxia l lines have equa l characteristic impedances: 50 fl Bolh ort! air-fill t:!ll. The firAl Itnll hall a power c.apadty of I MW. and Ihe ser:ond line's capacity Is I leW. Find the ratios a,loz and b,/ba. Consider only the breokdowll vollilKe.

11.4 . 115f116. Ibland the boundary condition (4.31 to obtllin the surfllce-currellt density J. on the lower plate of the parallel-plate waveguide Then calculale Ihe lolal current on the lowAr platA. Compare the curren! with the cif!finilion of Ilzlxiven by (6.3bl·

e.s. "'Ind the surface-current density J. on the inner conductor of a coaxial line. Then calculate the total current on it. CompBre the total curren! with I lzl defined for the coaxial line.

11.11. 1\ trunsmission line is short-circuited (ZL" 01 (_) FinclthA AxpMMion.( fo r j V(zl jand j l1zljll8 II function of kz. z.. II ncl V, . (b) SkctchjVlzlltlndjI(zll (e) Find VSWR nn the line.

II.? Rcpctlt Problem 6 6 for a transmission line with on open circuit at the IUlltlIZ~ - 001.

• •• • Ref}t!lit Pruuitlln 6.6 for II. transmission line with a mll!chAC! loarllZ/ - 2..1. -• • 8. 1\ tr<'lll8miAAion IlIlA is terminated with a normll1i7.ed load of n.R + /1.0 Calculate la)

the VSWR. fbi the posit ion of the voltage minimum. a nti IcJ thtl pt!rcentll..IIe of lhe incident power thaI is reHeeted by the load Sketch I Vlzljas a funCllo n of z/A .

• • 10. Solve the problem discussed in Example 6_6 by usin8 the Smith chart. Find the po!Iition of A Ahunt susceptance that can tune the line to have II perfect malch DelennllH~ the value fin mhos) of the shunt sUSCt!ptam;t! .

•. t 1. For an 0Jlfln-drr.uilArl 50 U transmission linft of lftnglh p. Ihft input ImpNtAnce Al lhe other end is ;33 0_ Find the lcngth 2 (in A)_

II. t 2. Repell.t PrulJlem 6.11 when the line is short-circuited lit ona find

• • 13. For Ihe first WAvegUIde in Tahle t of Section 5.2. design an iris Ihat will give a it_57 IiJulittanct: til f - 8 CHi:.

~ 8.1

0.1

0. 1

0 .1

0 . 1

o.~

0.2

.. 11.1:

. .2!

..21

·n Lines

he one

(6.58)

period

Ivenlge

ximum '.: - 2 x

--filled. ,1 kw_

,"sily I. ::urrenl 5.3bl.

'. Then for Ihe

L - <x».

• lale tal , of Ihe

!Od the match_

e at the

a j1_57

Problems 181

e .14. From the Smith chart. find r L for the following ZI,oo; fall + jZ. fbI "". fel O. And (dJ 0.55 iO.38.

~ e . 15. UI6 the Smith chart 10 find Zl.oI from the following f L (01 06 e 'll •. fbJ - 0.3. Cllld (clO

e. 1e. Fora load impedance of 0.4 ;0.5. find the location of the first voltage minimum and Ihe first voltage maximum allhA load end

.e.17. Prom Ihe Smilh cha rt . find Ihe ad miHances for tilt: fo llowi ng Iml>f!dancAlI: lal z. _ 0.3 - iO.6 amI (bl z... - 5 + i3.

8 . t8. A traMmiMlon line is terminated with a normalized impedance Z .... _ Z + ;2. as shown in Figure 6.19a. The incident V. - 1.0. linn Ihe characteriSlic Impedance of Ihe line is 1.0. Show Ihal V .... - 1-... - 1.62.1 vloll- 1.55.lvf - O.21!:l>.I1 - 0.76. V",,,, _ t.", _ 0.36,11(011 - 0.55. and I 11 - 0.219>.) 1_ 1.45.

e . Ut. A shunt IIdmillance of Vr ... - -i 1.57 is added to Ihe transmission line that ill terminated by o lood ZI.<o - 2 + j2. as shown in Figur8 6.19b. Thfl pMillon oflhe shunt ill 0.2IR>'. from Ihe load, so !hallhe line is perfectly motched. Let V. _ 1.0 and Zo _ 1.0 nnd show Ihal V .... - 2.lm.1 VIOII - 2.00. I ... , - 1.86.1 11011- 0.71. and I.", _ 0.49.

e .20. In Example6.B. find anolher set of solutIons of ~I and iJ (In cenlimcters).

e .21 . For the IIOlulion found In Example 6.8. how much lotallime·average power ca n be fed 10 Ihe array without causing breakdown in Ihfl difllftClrtc? Use the value 181.000 V/cm as Ihe breakdown sirength of Ihe dleJcclric. use 6 saft!ty faClor 10. and lei a - 2 mm. Hint: consider the standing waVe on Ihe stub tuners as well as on the transmission lines.

- 1 .2 2. For Ihe circuit llhown in Figure 6.25a. lei 2'. - SO II. Ht _ 70 n. It, _ 50 n. e _ 2 m, v-10' m/s..I1 - 10 's. and V. - 1. Plotthe vol!age and current at :t _ e/2 ItU function of lime.

e .2J. Ca lculate Ihe percentage of energy generaloo by the puilltl Jlenerator that is abllOrhect by the load in the circuit of Prohlem 6.22.

8.24. "or a four-digil code system. design a D-A converter sim ilar to Ihal discussed in Section 6.5 using Ihe transmi.ssion Iine8hown In "~18ure6 27a SpecIfy Ihe value of R. the location of the sampler. and the lime that a 1I1HIlpie shuuld he 18klln.

e .25. III plotlin,l! Figure 6.32. it ill implicitly as/lumAd that Hr > 7" and Ihal H, > z.. $0 that both r t Bnd f , ore positive numbers. Skelch a similar diaMram for Iht! clllle in which Rt - 0.52. Hnd R, - 0.52'. .

1.21. Draw Ihe voltage and the current refleclion dingrnms for Ihe trulUlinillSiun line which is sliort-circuiled as shown in Figure Pfi.26. Piol V and I (i5 funclions of time at z - t /2.

~o---===---, v. :

T z.

• ·1

_ 1.27.

• • 28.

182 6 Tra nsmission I.Inllll

~ J~ ~1.u,.".27

,- --1

Draw th~ vohoge ond thecurtenl reflection dj~KllIm5 for the transmission liRAlhal is perfectly matched. all shown in Figure P6.27. Plot V and I lUI fum.: tioll!J of time at 7._ 912.

Refer to Figure 6.31, and It!! R, _ 22'.0 and H~ - 057'\1' Draw the voltllMtl reflection dia,l!ram for 0 <. t <. 6T, and plot Val Z - 3i/4 tur 0 <; I <. 6T

• • 21. Refer In Figure 6.31, and let R. - 2 Z. and Al - O.S1... Draw the current reflection dill)lr8IU for 0 <. I < 6'1', and plntt al z - 3f / 4 fOf 0 .... I .... 61'

, .'o. Refer In Figure 6.388, and obtain an exprflAAion for I . Sketch I II} and liz) versus 1.

fur the time period 'j' < I < 21'. The SktllCh should be simtlar to Figure 6.3Sb and Figure 11.31k:.

8.31 . Derive (G 58)

F

-I ~

7.1 Vector ;

iI. -

.f---- 0') ~c Y""."'I'W#.~ ia. ~ICH· ~. 1(£·/0 e .. jAl • i(E.I'l.)e·j,r • " :.1. cIs W = W(E-It > e·,iAl ) S ..... L "'$ (,.3')

II ZL· 0 .;.. 11.·1

1')'iO) - 'i' ( .--j~}. r, e;·~) - l!' ( e-;')- ~ ;.) ) ~ -2h' s.;"(II}) ., yel)1 '" 2/~1I ~'~I'+~)I p!) = f y'{e-;"~-r. e;II): il£'re-jAI.ej·J) c f j:'·CD'(o!J)./ !Il)/ - f/v'u cos(41)/

• - • Iyt111· "tn'-IW -In ",,~

---L~-J .• --L-_~~~~o~--J

(C.) vswR.;: 1.,Cd _00

'·f£'d ' .7 21.=oc -=>!J .. cf .

I~l 'fl} ) ' y'( e-;tl. r,e;! I ) G ¥' (e oj',. e ,H) - 2'" <os( II l> ... 1>'(1l1 = ./ "'I/cosek)ll

PI)- f (e-jl) - r,oi",). f (e-j ·,- eJAJ ) - -2) f s,;"On ... I IIlJI = :.lltil f,;,.e~))1 • '~))I Il/.ll

,W I'r, ---- 21l1'l - 2. v' .

(0 V5w1!.- I+ICd I-Ifd =1

20

.. .ya. 0

1

1JI1l1 ------------1: ,.·r/ •• o }

'//0

g (~) [t :[( O·S< jl.O) -IJ/[r •. s+jl.O)+IJ • ()-D. ')/( j He) = O.49S'/'1~.'· "*/[,{= 0.49,

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21

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22

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24

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that find

.49)

j -

. 11 .. aH

Problems

Problems 221

to Puerto Rico. This mdio lele~r:oJle system is ca lled the Very Long Base­lin" Array. The angular resolution will be of the order of 10 • radian, which is smaller than the angle spanned by a dime located ill New York City when it is viewed from Los Angeles.

In ordinary arrays, individual antenna clements arc connected by transmission lines. For the very long baseline array. such physicHI inler­connection of antennas is DOt practical. IIISif!Hd, Ihe signal received by each 811huUlH in Ihe Hrray is recorded on magnetic tapes which arc later transported to a central facility where the tapes arc replayed simullfwl7 ously. The key to such processing is a ve l'Y Af!I~urHte lime standard for all rer:orch!ti dalH. AI present, lime synchronization of the recordings is provided by hydrogen masers which arc accurrate within 20 nanoseconds·.

- 7 . 1 Find Ihl! roclangular coordi nales of a poinl P where the spherical coonJinales are (r-l.8-60°,._300).

- 7.2 TI14~ reclonsular coorrlinates of a point Q are (1. 2. 4). Find ils llpherical coordinates.

7 .;) SllOw liIal V . V )( A - 0 in spherical coordinales for any vcctor A

-- 7.4 Show that the differential spherical surface element is tlqUllllo ds - r sin fJ dB dr;. Hill l: Rerer 10 Figure P7.4.

!lleur. P 7.4

- 7.5 To convert a voclorin spherical coordinates 10 the same in rcclangular coord inates. it is conven ient 10 prepart: Ii table fur dut prtxiucls belween unil vAC:lors in Ihese

'K. I Kellermann and A. K. Thompson. -rhe ..... ry lonll baseline.nay;" Science, Vol. 229, No. 4709. luly 1985. pp. 123-130 .

222 7 A ntenn ••

coordinate systems. Pnr example. i . t ... sin 8 cos lb. as indir.atM in the following tahle. f'.omplete this table.

, • t smScos,;

+--~-

i

- 1,8 I JM! thA I"ble prepared in the preceding prnhlp.m 10 express the fo llowing vector A located 01 It ... 1, 8 ... _ 60°, tP ... 45°} in rt:Ctongulor coordinates:

" ... 12t + 58 - ~ _ 7.7 Show thlll the distance function It ... r 'l llia\ appears In 17.7) and (7.8) can be

I!xpresscd in spherlc<ll coordinlltes as

Ir _ r ' ll_ r'+ r'l 2ft'COS )'

cos ')' ... COlI 8 COlI W + sin 8 sin 8' cos('; ... . '1 where l' is the angle helween the vectors r and " and (f, 8, tPl and If'. 8'. ,,' ) are spherical coordinates of r and r ', respectively.

7.' A vertical receiving dipole antenna at P Is 15 km away from \I capacitor.plaut antenna thai is abo placed vertica lly, 8uhown in Figure P7.8 The receivins antenna mellSUres an E field equal 10 10 mV/m. What Is Ihe vahut of E tlutlllle same receiving antenna will detect Ht a height 3 Icm ahovfl P? What must the orientation of the rACflivinX dipole be to obtain a maximum reading? (A maximum readinx is obtained if Ihe dipole is p3rallei1o Ihtl E field .)

--1 --_- I

I ---- I . __ --- I

- --~-- ---------H]= ~w"'p, .•

7.11 The power losl on a cylindrical conductor thllt is!).z lunx and that carries I amperes ut current Is given by

P _ _ YIl t H,!).z

where P _ Is the loss due to finite conductivity of the wirt!. It. is the surface rftllililanct!l(iven by 1/ (od,2TO). and d,ll! the skin dtlpth. The efficiency of the antenna

is given by

Power radiated 'I, - Power radiated + P_

AMumflthat a short antenna of length 0.7. hal an efficitlncy of ten percent. Is the efficiency improved by increasing the ienXlh 10 2 0.% while maintaining the same current and, if 110, by how much? Assume thai the anlfmna is still II short antenna after ils lenxth is increased to 2 6.7~

Antenn ...

2' following

18 Vf!Clor t\

'.8) can be

, 8', oil') are

lcifor-plale ngsntenna e receiving tion of the is obtained

I amperes

1e surface Ie antenna

:ent ls the the same

rt antenna

Problems

Power fl ivifler r

PhillltJ -I ,hifter r

90- [kIll)

223

y

, •

7.tO. Consirier the IIntenna system consisting of two short dipole/! arranged perJ.l8ndicu­larly to each other In space, aSllhown In Figure P7 .10. ThtlllC dipoles are driven by the same amOllnt of IlOwer from a common source, Ilowever. the current on Ihe i:-uriented dipole has a 90 v phase with rflSI>fK:t 10 thlll on the y-orientcd dipole because of a phase shirtHt ill~tlrted in the transmission line that leads to the former . Find thtl total radiated electric field on Ihe 7. axis. Verify Ihll t Ihis IInlenno system radiales a circularly polatizArl wllve ill t}u::! i direction. Is the wave left·hand or right-hand cin;ulllrly polarized?

'.11. Find the expression of thl! total nuJilltoo electric field on the x axis that is due to the antenna lI)'slem discussed in the preceding problem. What is its polarization on the x axis?

7. t2. A certain application rp.(luire~ that II fitlld strength of 1 V 1m be maintained al a poinl 1 km from all antenna located In free space. What power mtl.~1 be fed 10 Ihe IIntenua if il is (a) an Isotropic antenna. (hi It !lhurt diJ.lOle. and (e) a half-wave dipole? Neglcct ohmic loss. An Isotropic antenna radilltes an equal amOllnt of power in all dirtlClions.

7. t3. The curtfmt atlhe ctmterof on antenna Is 100 A ; what ill the E fieldl km away from il on the horizontal [0 _ goQ) planA al 10 MHz ir the IIntenna is (a) a dipole with hi _ h, - 0.5 m, (hi a eapllcitor-plule antenna with uZ - 1 m. and IcJ a half-wavfI clipole?

7.14. Show that if the radialion fil!ld pattern IIhown in Figure 7.4 for the infinitesimal dlpoll! or the CHllltcitur-plate antenna Is plolted in )(-7. plane in linear scale Ihtl vuttern is exactly formed by two circlfl.'l.

7.15. I,'jnrl the clirp.clivity of la) an isotropic antenna. lb) a C<lpacitnr-plaIA antenna. and lei It hilif-wovc dipole.

7. tS. Find thl! raciiuleu eltlt:lric field of a linear antenna IhM 1.'I:l m long r~ - 3 ml and Ihilt OJ.l8ratcs at 100 M Hz in ai r. Plot I!!! rarliation pallt!rll .

7.17. ( :nnslrler a uniform linesr array of two half·wave dlpolplI that are 1.5 wilvelenJojlhs apart. The currents on Ihe&fllwo dipolesllrtl ill ",lUIlit'. Sketch Ihe radiation pattern in the horl7.ontal [8 - 9O "llJlalle. Show clearly the number of lohel!;n this pattern. Also. estimuttl the beam width of each of IhA major 10UeIi. The beam width is the angle between Iwo rlirp.ctions in which the radiation intensity is one-half [ - 3 d B) the maximum valutl of the beam

224 Chapter 7 AnienD II

7.18. "'Igure 7.21(bl shows the array factor of a h~'O-element array separated by 20}... Find the beam width (ill terms of the ulIKh: between t\\'O adjacent nullA) of thill. array factor near 1ft- 90- and 0- 30e

•

(a) Use the approximate furmula Kivell by (7 .49). (bl Find the exact value starting from (7.45).

7.11. Find the dircctivity of the two-wire transmission line shown in Figure 7.27 with rarliation fiAlrlll. giwln by [7.42).

7.20. Find the field pattern of a two-element array with d - )' / 4 and", - O. Sketch the field paltern on the x-y plane.

7.2t . Find the field pallem of a four-element array wllh d _ ,\/4 and 1/--0. Sketch the field ptillern un the x·y plane. (a) Use 17.37) to oblltinlhe fie ld . ptiltern forlllu lH. !:tnll fh) USII

the result obtained in the preceding problem and In Figure 7.16 ond the pattcm­multipliclltion technique.

7 .22. Write a computer program to plot field pallerns of a len-element phased ilrray with d - '\/4 I!OJ varyins phases.

7.23. A uniform linear array consists of 6 short dipoles. Tho spacing botwoon ad jacent AhtmAntA ill. Al4, Itllllhnwn in f'iguffl 1'7.23.

la) What should the phase shift'" be. in order to point the maximum radiation in the 41 900 (thai ill.:91 direction?

(b) Suppose that the E·field due to the first clement (the dipole at far left) is givcn as follows:

1000 e-iit sin 8 , Calculate I f.:.1 of the entire array at point AIO.lOOO,O}, point R{IOOO,O,O}, point qo, - 1000.0), and point 1)( - 1000,0,01. separately. All positions arB given in rectangular coordina tes in meters. Use the phase shift found in {a}.

Ie) Sketch Ihe field pl:ltlt!rn uf tht! Iltfl:lY in the x·y plHIlt!. Irll Skell:h the fie ld pl:ltlern of the array intht! x-~ plane.

y

FJgure P7,23

,

8.1 Raleig

CHAPTER 7

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29

Problems

252 8 'lbpi£1 in Waves

Figure 8.17 shows 0 typical arrangement of H liquid-crystAl display.* II is operated in the so'(;HIlHd "dislortion-of-aligncd-phases" or DAP muue. Figure 8.1701 shows the normal stale ur Ilut crystal before activation. The light enlf'ring the crystal is polarized and then Irllnsmi llHd through the crystal with no aitentlilill in polarization. The second polaroid absorhs Hlllh61ighl, and no light IS IransmiU~1. In its activated Siale, the crystal changes the polarization of the transmitted light . which propagates through the second p(lt.lruitl and becomes visible.

8.1 The d£!rivalion of 18.5) only considers the electr ic field Why is Ihe magnetic fielti nf'gif!f:IP.d? Hint: Cum pare the magnitude of R, wilh I!H, nt!lll' Ihlt !:Ipllt!n:, or the stort,.'<i c!ectric·energy density 11/2~ 1 E 11 with tin: stored magnetic·energy dens!!}' [ 1 /2~I H I'.

8.2 Wh~' is the rising or setting sun rl!rI?

8.3 The smuke emiUoo from engines of boa's cnn lain .. fine purticles. Against a dark bilckground 'he smnkt;! Innh bill!! Imt lI)juinsl a bright oor.kground it looks yellow \Vhy?

8 ." Explain the :lpPp:lranr.A of shuTts of sunlight through bre:tksin a clOlld·r.nvAreri sky.

8 .5 Shuw that

I,-f' erdx_.;i

I-IInt:

I; - f ' e' dx . f ' e '" dy

Then. transform x.y r.oordinates inlo cylindrit,;al (:oordin,lles 10 perform the exact mtAgr.tllull .

e.e Show that

I, -f.· AXPi _p'.'(i+ q.'(IJ.'(- ;ex{4:') Hint:

pl,r ... qx ... _ px _ - +_. ( QJ' Q' 2p 4p-

ThUll. uS(: the result obtained in prohlem 8.5 after IrHllsforllling the inlegrillion varlahlp. from x In px - tj/2p.

8.7 Assume that on earth a microwavl! Imam of 10 GHz is rudiuleu by a 20·meter· diamPll!r rlisk antenna uillltlu til the moon Estimate the size of Ihe microwave heam on tht: moon

OSee R W Curlier ,lilt! C. Mare. "I.iqulo CoI"\·",,,1 Oispla)'s, . IU::t; Spectrum. Nu\'emt.t:r 1972. p. 25.

p,

~ / //

1.1 A

" d,

" I.' 0 h. r. Ih d

" 8 . 10 A

'I 'I

: . in Waveli

lisplay. * It AP mrxlB. !. The light ·he crystal Ithe light. langes the he second

wnetic field erc. or Ihe rgy density

insl a dark lkll Yl! lIow.

'\ered sky.

1 the exact

ntcgration

10-meler­.iI\·f! hPam

Problems 253

a.a J\ [)f!rson leaving his home by train mails a leiter home every utly. Suppose thaI the troin lrovt:ls ZOO miles per day and thai th~ ma ll moves at a speed of ZOO mile' per day. Ilow frequently do his leiters arrive home? Try 10 solvA thill rmh1em by simple relilJOning. nol by substituting numhers;n some formula

a .g On II foggyday.lhe driver of an automobile SIOPpt:J al a rtlilwllY crosl~ing hecause he heard u whillt le from II moving train . 'rhe sound of Ihe whistle came from hi' lelt. A few seconds/aler he heard Ihe ceho. ond Ihe lJilch of the fiNllllollnd was lower Ihan Ihut of the echo. AlISlImA Ihat thA echo was due 10 rencetion from 0 neurl,y mountain close 10 Ihe track. If you were the driver. wuuld ynu r. rf)llll the track- that Is. could you tAli whAlhAf the Irain was approaching Of leaving you? (See Fi)(Uft! PH.9.1

8.10 A Dopp/er radar sends a signolot 8.800 GHz. Hnu IhA TACAi vAr displays a frequency lIptlctnlm of fAlurnM ~ignals as shown in Figure P8.l0. Whot ClUJ you 9Ily aboutlhe speed of the ItlTKeqsj?

AmflHturie of rhe rclurn ... .,J sil'nuJ

10kHz 15 kHz - bkHJ. ,"!9ur. "'.10

8.800 r.Hz £requen

254

Randomly

l'g"1

Au.or l'lIon n. I

poolatiW I 1~;-----i!I7--I-----1I-

• .. i,wr. ".15

8

Passing "x ,~

Topic. in WIlVH

, Cfobserverl

B.1 t Fur th.: FM-C\\' 1lopplf>r rilJltf discussed In Sectinn 84. assulIlt: Ihat f .. the upper frequf'nr.y of Ihe l'IuJnr. is 8.8 G lb'! Suppose the fdunr is to measure targp.1 speeds rftn)(lUlI: from 0 to 3 "-lar.1I and IlJ~ l lini,;C5 from I km In 10 km. Fmu the system's opproxima lf> frfl11uenq hdnuwidth and ,hA lim p. inltH'VlI1 the system must be ahle 10

rf'lIOlvp..

8. 12 If d _ >.s/4. all shuwn ill Figure 8 16a, ann if rAflecliulIs III intcrfnc~ z - 0 ann z - tI art! ne,llli,!!iblc. ~, linearly rnlarized WilVC inciden t from IhA left will become II circu larly polarizHd \\ilVl:. as discussed in IhA Il!xl. Whal is the polarization of the pxitm.!l WIIVC if the rcncclion~ <It Ihf'S6 interfaces arc nol negligihll!?

8 . 13 If d _ X,./2 as sho\\11 III FiKUfe e 103 , whlll i~ Ih~ IXllaril.lltion of the exiting wllve if the lIlcid ent ""ave from the I tlfll~ cin:ulHrly pol.ulzcd"

8 .14 For II quartz cry,tll\' t. _ 2 11 t, IIno. ;!.JIk .. Find thc minimum Ihlc.kne.\l.ll uf a (IUllrll. 4uiHlcr·wave plall! for a h,llht having>. - 6500 A,

8 .15 In !llgnff> Pe.1S Ihl:! Polaroid film III A I.~ oriented ltut;h Ihat il passes lighl rm1arizal In

the x llirL'clion and IIh.'iorh.~ li,llhl IJOJilrizcd in Ih~ 5' dlrp.(:l\on The film al B passes r-polarb:f'i(! light Hnd .. 1)Surbs x-pobri7RfI lI ~ht A nllldurn l) pola rized light 8Oun:e. silch as a n.:lshlighl. shed~ hgh! from tilt: lefl illong z Clln an observer HI C sec Ihe lighl? Explain

8 .18 Cunsidcr Ihe arrangempn! ~hU\\1J in Figure pe 16 Thi'l figurl! dLffel'S from Figure P8.15 only in the 11Iat;t:Ulcnt of;) Ihlrd PolMoid IiIIII at D bclwCC!IJ .... and B. The ah~urptjuJi IIxisof thc IhirO film l.!1 ~5 ' frum eithcr Ihe If or Ihl! r aXL9, Nu\\, can Ihc ubserver al c.: AAH rhl' Ii,llhl? (If ~OU do not h(1l1l!vf> in yuur alls\\'cr. do an experiment with thrfle '1<111".::1 uf polarized slingla!l.M'!9l1ud stte for roursclfl

• ~r,. , ' , , , , , ,

, <":Iobservcrl

9.1 Ele

8.~

CHAPTER 8 r () ~ • .J. /~ I A.Vi.O /" I ", .. 8 Ir,l_ A"r~4 rrDnY S.'. "'~r""e. ~.~n:, r:::; AI 'lr~ I kr"" r' I ~,

:. 1/&/1 _ ~ r «I ~ U. _ 41lil' _j tH,' _(~r >'" I I!I u. flol' - T

Si .. ,e. Jv:.,J... Ir~Iu.Lncy .Ii.,J..t t M",-e.) ,"Ifers ~104''' s~ .. r"l'1j J.cJ~ e/.-. ~ {.''/,"~''1 j,;9"t (r04.l.

'Nil,1\; bAd;,N'u.."d,.. is dArk, GMt.. S~M ",,',4t; .sCJ..J'f~,L;(. " e'c .rht"lte. pa.rt4.du.

8/~ t.. L.;,,(t" ;~ $&A.1fl, U ,..,or~ sfr."",b:J ""'tIIa.". rU ~A.t.

Ajo.U, 5 t a. br,;gJ.t b-.d:. 31'O""M ~ c;v.Ic.. 'res ti.!J't P4lS$I',,-, +/ot'Ou'jt. -lite. ~17t"fft ..

.,.". hi •• LjA.< 5.t. ,c.J..-..;. ""'"- rd ....,( :I"'~ J:,IJ ..... /6.,. ~ , .... ~:tt;,n.,.J .

~M .. Sl.dl~·", of AJJu- t:V'Id.. d"st Inc/LCAA./u

(-r- ('.~') f.-r'· t' [-' I -1;1.,. ) ... L. e- ~ d.t:ti# -= (I '0 e - ftlPJ..; - 211 0 pe- f d.f ... Z,7T' Z f, l-Yd.t:: c 7T

, /' Z· If- -u' "'" 7P _e7?·, .• e ..t1J..pe

.', w,oI~ c: WZ/Zf := IOK3.B'I4JoBjf.0I'1 ..... /0 .. ~ J .GoJ K JO~ _ .J /J 3 RI11

M e.vv1 (JM~'" ~

6,'

9 . 10 -

-

.f, - f .of fl.- f +a.t.f

I I I I I I I I I I

I, ~ 1.-.. 1 ...... " o~ ("HE ~,,,,c -St,,_, ... " ~ - frt/wf1CY II¥' ..... c. ~c.h.:J

ZDOIfIO "or lorI'S 8 . .,1C./0'" Z • '4-1

30

9.11 A/of, (2ujC) _ 8.8J(ID'lIlJ< q93 • _I ' JI, • 3"ID' ~o $""" J

Z"/C - 2J1/~J/311IDa". h , "J(IO-~S • " . 7,Na

,", a......L~ ... "J.... 59 ~"'.J r..,." r"'ot....t.'_ ~ l,flr! - • . 7,kS .

L!!.. £llt'pt,'c-.({, f"t'L.r .. ;<Ul MUle

.... Ci"cwt.~ f1te'LD~'J-'- hu.t cppoJde. ~ • ....I. .

% o/'l. *-= ' .' ,soo A-L- i&- i •• &;y,;:o(ff.-IT) - ;;(;./1'). 4([iir-/i'fi)·

!!£ NO .

8." ye's .

T

31

288 9 F. lflCtmll talic Fieldi

Problems

e . t Considt:r tht: dipole arrangement shown in FiHUI'C !I2a Ull Q _ 1.6 X 10 - 19 C. find 4> at:

(a' x - O.l .y- O.l .Z-01 (b) x_l , y _ l,z _ l

U~ the exact ronnula 19_ 14) first . TIltm use Ihe approximate formula [!l. 1 7~ and find the accuracy of the laller. The medium is air.

' .2 Thret: point charges are lor.ated 011 tht: x axis with q, - q al ,,< - 0, 4 , _ 2Q at x _ 1. and q, _ 3q 1'11)( - 2. Find the positionls) on rhl! x axis wlltm: 4> is equal to zero.

11.3 Four point charges are located on Iht: corners of a reclanglf!, fill shown in Figure P9.3 Find Ihe pll1np~'I on which the potential is eqnal to ztlro. Skt:lch these planes

y

• 201 "I q ---tlin

I"'e"r. PI,I

I 'm I

- Q

y

" > • , q - 2q

' .4 Two ('IOint CltHr'MtlS arc scparated by 0 meteNl III air. as shown in Figme P9.4.

(a) "'ind the poltmlial funclion <l>(x, y. zl. ( b) Calculate 4> at x _ 1000. Y _ lOOa. Z _ 0 (c) Show Ihat, for distances much gre"lpr IhlUl U Imay from these charges, the

potential is approximately given by

- q 1 . -- ;-;,---;'---"",", 01 .. ~ Ix' + 'I + z' f~

whAfAlx! + y' + ZI)11I » o. UM Ihill approximate formula to r.alr:tltll le <l>1]()O<l, 1000. OJ. and compare it with the n:9ult obtained in I h I.

e.s Find thA E field ill air due to .. point chArgA of to'q .. (q. - 1.6 x 10 I. q . Sketch a diuHrHIIl similar to Figure 9.4.

t .1 For Iher.harge distribution given in Prohlem 9.2. cah.:ulate E~ at (a) x - - 1.th) x - 0.5. Ic) x _ 2.5. and (d) x - 3.

1 .7 Apply,.; - - V<I> to 19.14) to find the": fiflkl\O(:aleJ ulthe origin and produced hy two chuf)lfls + q and q locatAd allO. 0.11) ElIld (0. o. h). respectively. 1I1IlihoWll in Figure 9.23. where h - 1 Clll. Show that E _ !tq/2 .. ~hf).

e .B Solve Ute same prohlf!ffi all in 9.7. Lut usc (9.19c).

e.1 SkAlch the dirt:Ction of the fo: fialrllocaled ul tile center of a square shoVl.'Il in Figure P'J.9. The E field 18 producetl by four charges al four comerll uf the square Ot\BC. Ihe56 four charges corry q, q. q. anrl {j coulombs. respectively ,

Field.

. Find 4>

:es. the

1. 100o,

ketch II.

JC _ 0.5,

bytwu Figure

Figure OARC.

Problem!!

y

Q

I 2h

C 8

I-

-----. -q , ,/1 , / I 'x/ I /' I // ',I

-Q

A

2h---l

287

,..,.".. H .•

•

g.10 A line chnrgc 2h mllter.!! long ill located along tnc 2: axis II.lI:ihuwn in Figure 9.6a. The chargA ctAnsity is p, coulombs pllr meter.

(a) Calculate thl! electric fif!lct at p - O.lh, 4J - 0, and z _ 0 Wling thA exacf formula IY,21).

(b) Calculate fhe ell!ctri(.: fillld ol lhf! IIRme poin t using fhe 8:iSUllllJliun thilt the li ne is infinilA ly Inng.

(c) Find Ihe I}f!rr.f!nfagc error of the value outllined in /bl iI~ r.ompa red wilh thl! I!Xl:lct value

•. 11 "or the same line-chaf)!1! dll.!ICribed in Vrohlem 910,

(e) Cakulatl! the elf!r.lrir. field al p - 20h, tP _ O,:t _ 0 U.!!1Og thf! Axact formula fb) 1>0 the same usinlil thl! assumption Ihat lhe line is a point chaf}(e I:Itlhf! origin. (c) Find the pAN:f!nlage error of va lue obtil in et.l in /b}.

g.12 A plan(' charlile of p. coulomb:. ~r SflIJaM mPler 15 located on Ihe )( _ 0 planA, and another plilne of - p. coulombs per SQuare mt:lter I.!! localAri on Ihe )( _ 1 plane. Find the lotal electric-field in the Mgion Ill) x '". I. /b) 1 ~ JC :.. O. ond I(.:I}( ..: 0

'.13 COllsidt:lr the pmhlpm discussed in Example 9.11. Axsumlng thai everyth ing is the same except for the far.:1 Ihl:lt the tnlal charge on the conductinH .!!hllllls nnw equal to l.ero. r:alr.ulalp. Yo everywhere. and skl!tdl E, Vef"5I1A r Aimllar to the sketch shown in Figure !U5.

g.14 A dlill")(e dislrihllfinn of the following form is set III) in Illr I~pherical coordinates):

( a) Fill£! Ihp D fiAld forO < r< 0

fb) Find O foru·, ,.' U. (c) Finri J) forb < r

8.15 A charge distributiun uf the followIng fnrm i~ SPI up in air

p, 1U ' . e r:nlllnmh..~ per cubic meter

IJ!p Gil 11M' law 10 find the £ field cvel)wherl!. Hint: Tn finrl thA Intal charge in a Gaussian :.urfilce. you mllst rln thp integration becau$C Ihe charge is not unifonnly fh~lrlhllted Ilowever. symmetry still exists \\ Ilh rf!lI(lfM:t In ~ and 1/

'.11 Electric chargl!:' aredistnLuled IInHonnl), in the region 0.1 .... JC <. +0 1 with density fl. 10·' :/m' Elsewhere. Ihe density is ~u"llo zero. FlRrilhA "; field everywhere.

2BB 9 Electrostatic Field,

Plot E, \tArtlUS)t, Find the potential difference V. V. for iI point x with fp.specllo

the origin.

1 . 17 Fmd the [)(llp.nlia! rliffp.rp.nce V ... - V. for two points A and B IOClitOO 81 r _ 0 and r _ 1 in the E field obtainoo in Problem 9.1S.

1 . 18 Tilt! solution for the electric fielrl of an oscillaling Hcrttion dipole with anxular frequency w is given in 17.14) a&"follows:

E - ~/hl.l.z t: '~'!'1 . ,1 I ~12cos8 .,. ell +.~ + . ,~ ,Isinsl V 7 <llI'f I ftT f /I\fl /f{f IIl1t)

Derive the solution (9.20) for a Sialic di ~le by !JtIUin!l W - O. Nnlir:elhal k _ wf,utl,n

lind 1 ~z - npliH - , ... p.

,UI In the electric field E - 3x + 4y - 5i. find V" - V. if A islocatAl'l aliI. 1,2) and B isa! til l! ori~ln. Dop.~ the difference depend on the p3th of !he integration?

9.20 (:onsider the spherical-shell problem shown in Fixure 9.1 4. Find thfl pntenlial <t>( r) at (") r t:: (hI h .:; r < c; (r.) b :> r :> 0, lind (d) r - o. Assume '" - 0 at in finity. Plot .pfrl versus r,

• • 21 Rcpt!" t the preceding problem for thl! r.Ase In which the total charge on the conducting shell is equal to zero whill;l all otil!;!r cuntiiliunll remain unchanged . You may v.ant to use the re.'Iul! ohtained in Problem 9, 13

• • 22 Consider the coaxial line shown in figure P9.22. The inner conductor is a solid conrluctlng cylinder with a radius equalla U.l m. The outer conductor has an in­ner nuJiulI equal to 0.4 m and an ouler radius equal to 0.5 III. Tht! morlium bet"''fIen the Inner and the outer conductor is air. The inner conductor carries a net charge of - 3fo CJm Rnnlhe outer conductor carries ollet charj(e of - 1S10 CJrn. The sym· bol to used hero represt:uts 8 constant Aquai 10 8.854)( 10- 1l.

(II) Pinn E. In Ihe region 0.1 m < p < 0.4 III. (bl Find £~ in the reMian 0.4 m < p < 0.5 m. Ic) lo'ind E .. in the rcgionp > U.5 III. (ti l Fil1ti ~ at p - 0.2 m, knowing that'" _ 0 at p - 1 III. (e) Skotch £, 05 functio n uf p foro <p< I m. Mark thesca!e ror E, ondp.

Ftgur. P.i .22

"OItath:: Fields

\lith n:spect 10

:falr-Oand

with angular

I, Zland B ill at

)Iential '11 rJ at inity Plot o$(r)

harge on the changed . You

:tor is 8 solid tor has an In· hum between ; a net chargc 1m. The lIym·

I p.

Problems 289

'.23 Modellhe dome of II Van tit: Grllarf )ltmerator as a conducting sphere. The dome is charged to hold the maximum amount of electric charge Q", before the ai r SU t ·

rounding the dome breaks down. Ulle the following data :

radius or the dome - 0.11 m , breakdown E of air _ 3 II 10' VIm.

(8) Calculate the maximum Q", accumulated on the dome just before the break­down .

(h) Calculllt" th" 1IOlta~e of til" dome in reference to the potential at infinity just before breakdown occu rs.

rcl When the dum" is c1l1irjoloo with the U1uxiUluUl charge Q",. a person uses a con­duc ting rod to discharge the electricity. Assume that the discharge takes 0.01 seconds to complete, how IllronglllthA dlAChatging current Inn thl! avtlrage)?

CHAPTeR q J.:.L !"~A,r. (a.) It AI, o.~ 0.1) • J.1."";:"iTltlq4 {[(w!.(#.,I+taDfj jr_u".,)'+(o.l)·T(D.II)'rl)

¥ ,-" ., --t--"1 1 - , I I - '.L • -i--'- . - , !

~.,'" ~ 'J -I. s- ... ,l-iL

1 ---+-":--x i - r ,

v. f (o., PIX. ~ .11 < .t. [cx"Y'+ J'j(Z[C<'A)" ,'. J'r t J (h ) f<, ... ,/O<JA,.)' k [[(I_>'+(IO<JIl/'J+-z[cqq",+{'O<J4.)'Ji 1 -4;;1 .. (O.""7/~) ('J S;'_<<. (xJ."1'·J 1)Yz.)I>A,. I ((KPA.)1+ 11+ J'JYa_(1:' .. ~I+)')1i. I

F.- ("), if(x. , . J) - -Jr f (x'.,'+ J ']'C z[.··"+1 'Jf} --:JrrX'+"+l')-r f(10c:>A1,CCA./O}- ~ll4.~· -4;£I2.(II.CO'T01)

i1. 'E- - 1·':~"r~·'ttO·I<J'7TJ(/O' ~ XV("il.,4-.4- ~WI 0" 0:1= I,m .- 0«'," F r . ..

r X~X

32

(ij ~ -1.434

,

o 8 I -------"-1 . , ' . " '··~i , , , , , " o •. ':..' ___ '",'"A,-_

I -I

/1,10 (41) f. Dol ~. cJ.o c 1.,,"( ~/tJ.I"') • 84.S-

e-· p'" '" S ' , ...... ). f PL (f,.') ltrl.(~. IIt) .... %~t." ~

(h) F'",.. ;,,·h,...I.ly UIlJ 1.4",,, J ~., ir/ll.ili) - P .a:1~ (10)

(~) Ewor = IC~.::S "'IO~" c o. S'D' ~

P-w. , ... ~~-'(i,/ZO.).Z,U· (.) g- p I:/,.{:,Ol,) 5,;'(1.8'-)- f40!\J1 (tJ.o.I99)

(bl po;"t cll .. ,..a£ L!JAfJ~X""'A~i .... , ii .. 11 r.p2~ _ ~ fL(o oS) '" -r r f,"" { !OII)I r "4iii1I .

(C) .,,.,. .. ":'!;::fff)t/Ot'% • 0.% r-(a.) ~ itO /0,. ~~I (h, E" 2 ({f-! : -;({) {;~ D<Jr< (

(C) ~.O .f.r- a"'<o ~,.

£=Ol'orr<o.. - - I Eaf7iilfJ ~r 4frcb

E ~() f-r bfr<c

E-f 4n-i r c ,., r-."

\ ,

9.1" to, Ixl(D. I~ lf~t(£).)_"ltl{_O.)~ ZTTJ.I(D.).Pv"J.t(I~) ,', p)CKfvx 4> E-f~x J """are (v·,O"

~, ')()tJ.1 1 'trAt (D.)·fv '1T1i1.(D.l)

... 'tI,.. 0 . / Pv .., E·: -f;(tJ,/) I~t f.,.a,D4

~ "<-D,' , lnh'(-o.> - p. "~'(A')

-/I

... , ~

':/; ,'. o~ ~ .. D.' p., .. i ... 9';; (tJ.I)~ ,.J,ut f" -1<"· 'v· /O

•• I

33

• .,

.0.1 J, /

---~T-~~~----1 !/ ... oj

'/.17

4Uicr)

0 0 ---+-_0 I : o 0 ,

• .. • < r

~~--f--+-----r

34

35

Problems

332 10 .. :lItc:l ric Furee and Enerl)"

This electrostatic adhesive surface is widely used in desk -top calcu­lator-driven curve tracers. A typical voltage used to charge the embedded conductors is 300 volts. and typit:1I1 spacings between them are approxi· rrmtely 2 mm "

10. 1 A point charge of 4 cuulom bs is located at the uri)Cin (0.0,01, ami (I. !lecaml point cha rSA of lJ' cuulombs Is a t II , 0, OJ. A ~Ula ll test-charge il placed at 13. O. 0). and it is ruumJ 11'101 the 101111 force 011 Ihtl test charge ill ~uallu .:era. Find Q' in IfOrms of (j.

10. 2 Two irienticill small balls are altllcht:d 10 weightless IIITin!!!! 15 em long. Each hall carries 10 • C of c.harge, lind cach has a ma.'IS of I )! The), achieve an Aquilibrium state lIndp.r the influence of electrostatic force and gravilalinnai fun;t:. as shown In Figure PlO.2. Find the angle cr. Hint: n is small

10.3 Consider a long \intl·charge with fI1 - 1U-~ C/m . Find thl! fon:tl acting on 11 dust parllde carrying 10' C, I m away from the liM CharM!!.

10.4 A line charge wUh p, _ 10 6 C/m isl(lf:aled in tlil" ot )( - 1, Y - 0 A plane charge with P • .. 10 • e l m is located al x - O. A positive point chArge of 10 I C is at (1{z, 0, 0) in rt!chUlMulaf coordinates. WI,al is the 'olal force actinM on .his point charge?

10,5 Cha'"8e is uniformlydilii,ributed in the spherical volume r.$ 0 with p~ - 2 x to 'C/m' andp. - 0 forr ~' u.

tal u~ Gauss· law 10 find E for r .:. n (b) Find Ihe fon;tl acting on .1 1 1'~t churgtl uf 10 I. C al r - 11/7.. (e) Is the force ob'aiMd in [hila be changed if 'he dntrgc distribution extendslu r-

20 instead of heinM limited to r ~., o? 10.1 In a .'IPOO sorting machlnp-, undesirable seens IIftl deposited with an tlh:ctrostalie

r.hargt! while they p<I!IS lUi automatir. r.olur-!ltlnsitive or sizf!-sensitive monitor The good seeds art! passed unchal)!t:d. All seeds arf! druvpcd belween a high-voltage pardlltll-plate reginn 10 surt out the undtlsirable seeds. Lei Iht! charge on lhe undcsirahle.~ btl q. its ma.1\S be Ill. the voltage Mtwetlll Ihe parallel plates be V. anrlthe plate separatinn bt: d Assume Ihal the !M!t!ds enler the [Iaralltl\-plate region at vdocity v,. and finu the dlsplacpment y of the bad 1Ifl~ "S8 function of x. Figure P10 6 ;l1llstrat~ this situation. Consider on Iy Ih~ tnllt!(;IO!), inside thf! paralld-plate

'P. Lorram 1111(.1 0 R r.OT!IOn. E:1t.-cltomogM.tism ISan FnlllcitOO W. II. Freeman lind Co.

1!171!11. P 189

I [ner~y

• ca lcu­hedded pproxl-

j tu r-

oslatic r The ·oltll.lje 10 the be V,

region Figure plille.

,

Problems 333

VlvQhsl

"" 5

[milllsecondsl

At room temperature f:lU"CI and standllrd Atmosphere, whfll should be the S17.e of the cnrona wire if b - 3 cm. Vo - 10 kV. and the fOughM" factur uf the wire is equal to 0.8? (RefAr to Figure lOA..I

What should the loweS! vollage on A Van de Grad! generator be in order to htiVA It product! corona on its surface? Assume that ,.;< _ 4 x 10" V 1m and Ihlll the radius of thA melal sphere ill equal to O.H m.

Refer 10 Figure 10.1. If Ihe voltag9 applied to the parallel plall!llsihe sawtooth signal IIhnwn in figure Plo.g. find the loclls of tht! electron on tlll~ fluorescent !K:reen located at x - 20cm.

for the cathode-roy luhe shown in f igure to.8. whlllllhnuid Ihe voltagM V~and V, be in order 10 make the electron beam trace 11 circular palh on the x:reen at 60 TflVnlulions per second ? AssulOt! Ihat the vcl1iclliand the horizon tal deflection plates are identic.a!.

~n Aleclron is IIcceleraTed by a difference In poten tial of 1 kV between Ihe anode and Ihe r.alhode. It enlers the pnrallel -plate region with Ihls kinetic energy. Its velocity makes a 5" angle wilh thA plane of thtl fllu-allel plate lit thA entrance end, as shown in Figure P10.11.

(a) Find V,. v .. ' und v"" al I _ O. (b) OLtain Iwo equatiuns for the coordinatAII of Ihe eltlCtron (II. z) as [unclions of I.

NOlc thut x - Oand z _ 0111 1_ O. (e) find the posil ion of the electron lit the exil end uf IhM parallel jJllltA.

- 3cm_1 T--F======~ ~ 100 V

10m -+-~"'"' __ --,.,,.-_ > 1 -1.0' ,

L--F====~'V m._gll x 10 " kll q.- 1.60xl0 " e

...... r. "10.11

334 10 Electr ic force Bnti Energy

10. 12 Conyider the ink-jel printer shown in FigurA to.l0. Define

q~ _ c:hargt! on the ink drop /lid - mass of the drop V. _ deflection-plate voltage d - deflt:CIion-plale spacinH v~ _ velocity of the ink drop al entry 10 the deflection plaIA

9dp - deflection-plale length ~ _ distance from the deflection-platA Antry 10 the print plant!

Show Ihl:lllile vertical dillpla!:flmeni of the Ink rlrop is Kiven by

'. - --'-I "p - - t1dp L1~ V. Rdp [ ') mdUVoj 2

10.1' Find the capacitance orlhe IIpherical capacitor 8hown in Figure 10.13 by HllinH (10.42J

and (lO.50). Start from

Q £ ___ , ' for b > r > o 4 ... u

and show that your rellull agrees with (10 47).

10. 14 Find the cllp8citance of thA cylindrical capilc!lor shuwn in Figure 10.14 by using (lO.42) and (10.50). Shnl from

E -~ jJ for h :;:.p> u 2lffP

and show thai your result agre4!S with (10.49).

10.15 Consider the parallel.plote capacl10r shown in Figure 10.12. What is the maximum capacitance on~ can obtain by Ilsingmica as the Insulalor1 L~t the area of the plate be 10 em' and the voltage rating of the capacitor be 2 kV, with 0 safety factor of 10 Use Table 10.1 for the value of ( for mica.

10. 10 Consider the cylindrical capacitor shown in figure 10.14. What is the maximum capacitance one can obta in by lIsingoil es the insulator? Tak~ 0 - 1 cm, h - 2 cm and the volta)!!! rating _ 2 kV, with a safety factornf 5. UlW Table 10.1 for the value of t for

oil. 10,17 A pmallel·plale capacitor Is filled with two dielectric: mB!t:!riais In a configurBtion

shown in figure PIO.17. The lulal area ofthA plBte is A (a) Find the cBpacitance Gin lerms of A. d, f" Bnd (I ' (bl Suppose Ihal the poslllvA plale carries Q coulombs of charge. and find QI and Q1 in lerms of Q, whArA QI and Q, are charges un the left· and on the right.hand sides of the plate. respectively. Neglect fringing fields.

10.18 Consider the capacitor shown in Figure PIO.17. Letl , - 3 ••. (, - 5fp. d - 0.6 mm. and A _ 20 cml. The potential between thA plat~ is 300 V. Find the 10lal slored eleclric energy In Ihis capacitor.

t{( E:Fn w12 wlZ 1

fora: Bnd Enerxy

13 hy using r 10.42J

ra HlI4 hy using

is the maximum "ea of the plate be factor of 10. Use

15 the maximum cm,h-2cmand . the value of f for

1 II configuration clt~citam:e C in s Q coulombs of uges on the left­ing fields

d - 0.6 mm. and .:II stored electric

Problems 335

10.1' Find the capacitanr::e per unil length of Ii c.;ultxla] capacitor with two layers of insulating materials. asshuwn in FigllTe 10 15c. ExpressCIl1 in tennsof n. h. c. ',. and

'" 10.20 Find the capacitance C of a parallel-plale capacitor witla two lavers of insulating

materials. al'! .~hown in Figure P10.20. Expre.u C In tt>rms of .A {the orea of the plalel, d,. dz. 'I' and ( J.

10.21 Refer 10 the capacitor shown in Figure PlO.:W. Let (, - 3fo, f , _ 5(0' d, _ 0.3 mm. dl

_

0.3 mm. anu A - 20 r::m z. The vohage across the capacitor is 300 V. Finrlthe total stored eltXtric energy in this clll-lticHur.

10.22 Derive {lO.tI:ll.

10.23 A pltrallel·plate capacitor ctll"l' ics ... Q un onR plaIA lind - Q on the other plate. The orca of ellch plale is 1\ and the separation between the platfll'l is S. The medium is air.

(0) Finu the tolal stnred energy UII; In this capaci tur ill lerms nfQ, 1\, Sand (0.

(b) What is the c1cctrosllltic fUTCe acting nn the platos? Is it attructive or repulsive·' Hint: finn the change in VB with fCSpt!t;1 tu S.

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368

r.onnl1l::lOr Conductor

., % ~

<to, _ II +, _ 0

Case II

Problems

Cuu,ju"lor

<t,·O

Cue JIJ

11 Solutiun Thch niques

"''lure "'1.1

11 .1 Consider the thrAll LlOunuary-volue problem!! !lhowllm Figure P11 I The Milltion or UlSf'l I is <1> ,. lind the solution /'If Case II is "'1< In Case III , IhA r.hllr,!jI!.!H/,ltnd Q2 l:HC the sumll charges Ihal appellr in cases I and II, and thAy IIfJVt!ur in exactly correspondinll positionH. ExprllSll <S>, in terms of $, and "':.

11.2 Consider the IhTRA buunuary,villuc prohJAm.~ shown ill Figure P112. ThA Milltion uf CaSI! I is "',. and Ihc solution of C:ase IT is "'I' In Case Ill, IhA .. harges Q, lind Q2 orc Ihe SUint: chargcs thaI appear in C8St:S I and II, anI'! !hAy IIppt!I:Ir in exact ly corresponding positions. Nole Iht! differences in the boundtlrY conditions for Ihe three ClilIeS. Can.J

hfl Axprt:SSed in terms of "'1 and 4>21 U so. obtain the exprp..Mioli Ir not. explain why_

11 .' The radius IIf the innl;!r conductor of a r.oaxill.llint! is 0 and that of the ouler conductor ill o. The potential of the illllerconductor is Vand that of the outer conductor is zero. There is no volume chltf)!t: density between band u. Slur! from the Laplar.e f!quation to ohlain the potential in the coaxial line.

11.4 Two concenlric conduclingspheres have rarlii ultnd b. respcctivelyth;;. 01· The ouler IIphere ill CIt zero potential. anrlthe inner sphere Is main lainp.{1 at V volts. There is no space charge helwAen the conductors. Start 'rum Ihe Loploce cquallon til ohtain the potential <I>f rl for b ;>0 r ). o.

CunJuClnr

\ • _ 0

Ca.II@I

Conductor

4> _l 00V

C3s.>11

ConJuctor

Cas.' III

l'e(;hniques

1.1

;ululion of Ql 3re the

!sponning

:olution of Q1 3re the ~sponding

~s. Can 4>a l::Iin why.

:onductur or is zero. t!(jualion

fhe uutAr lere is no 'utllin the

11.5

11 .6

11 .7

11 .8

11.0

11. 10

11 .11

11 .1.2

Probl ems 369

InSlilarillg8up ,~ - 0

In Figure P11.5 a {;unuucling conp. i.~::It a potential Va. and a small gap separates its verlex from II conducting plane. The axis of the cone is per~enrliclJlllr to the conducting plane, which is m<lintained at zero pultmtial. The angle of the cone is 0,. Becausfl of the symmetry of this ~rublem ::Ind the fact that the buunn::IT)' conditions on the potelltilll 4> involve (/ only. '*' is independent of rand 41 when spherir:al coorninMes are used . Find the pulen tial <l>{O) in the region 8

1 .:!O 8 ~ 90°. Hint: f (l/sin

0) dO ~ In (I!HI 012J. Find Ihe surface charge density on Ihe cone.

The uppflr pl<lle of the parallel.plale r.::Ipacltor discussed in Example J1.1 is maintained III 100 V, and the lower plate is III 110 V. All other conditions remain Ilnr.h::lnged . Find '*'. Model a de vilcuum · tllhfl rectifier as two parilllel platfls with a Spilce dlllrge in Iw.tween, as shown in Fi.l!ure 11.3 . Let Ihe separtltion be 1 cm. Find the voltilge needed lu prmlllr.fl I Alml current.

Find the surface chnrge distribution on the vertiC31 and the horizont::ll conducting Willis for lhe case rlisr.l1ssed in Example 11.8. Plol p. for 7. '> 0 and x _ y _ U. Lei Q _

10 ~ C,andQ - b _ l.

Find the inHl)o!es uf a point charge netlr n corner uf a conductor similar to the one shown in Vigure 11 .11 excejJllhlll ¢.O _ 45°.

Find the c!ectru~tlltic forcfl Ih::ll acls on the point ch<lrge Q al to. 0, d) and is due lu imluGf!n s11rface ct13rges 3t z _ 0, <IS shown in Figure 11.5.

Calculate the capacitcull;e IJflr mp.tflr of a 12-inch (0.3048 mJ-diamfltflf steel pipe locllled 6 ft II.A.1 m) above and pOflllJel tu the grollnrl.

Example 11.10 slote~ llml the milximtlm electric field on lhe surf::lr.e of the conduct. in)o! cylindAr is located ot the point ll1mrest the ground. Show the villidily of this st3tement by jJlulling Ollt E. nn Ihe surf3ce as <I functiun of ¢.. I Jse the following dat<l; Vo - 100 V. h - 2 m. 3nd a _ 1m.

11.13 For the point char)o!e 4lucaterl rI meters from a grounded cunducting sphere shown in Figurfl 11. 14, find the surface chal".I!e uislriblltion as a function of 8.

11.14 Repeat the precedinH jJfUulem for an iso lated conductill.l! sphere r.ilrrying no net dl<ll')(e.

370 11 Solution Techniques

Cundll~tor I'" 01

Flgur. Pltt,tll

11. 15 Equation 1I1.42J )lives the polenti:!1 dUll In II lJUint charge in the prf'_wnce uf II ~ rounded wnducling sphere Equatiun jll.ol41 gives the pOtfmlia! ,Iue Iu II point chorge in the presence of an isol. tt'd sphere carrying no opt elia!')!!:. From these resultli., find IhA polfmlial JUt: to Q poinl charsI' q. (/ IIlCIt:I'S from an isolated r.oncluclin)o! sphere tAm y ing 0. net charge of fill_

11.16 A lint'! r.har}!!: PI is insidc 1\ conrh lCo lill,l( tUllnu l of radiuli. 0, a.~ shuwn in Figure PII .t6 Nutice that the liM char,llt: is b meters off r.AnIAr. Firu.1lhc potential fuor:tinn in tht: tunnel. Ilinl: This i~ I) compiemAn rary pl'ul,lcm of the one !lhnwl\ in Figure 11.1 2.

11.17 Calculate Ihe forCA [lpr mettlf aCling on the linA char,!!!! in lht: lunllclshown in FigurE' PI Llfi .

11.18 A point (;harMt: Q is inside a J;pherit:a.! cuvity 01 a connur.tor. Ii~ ~ilown in Figure Pll.16. The radius ofthftclivily is 0 and Ihe cav ity ill rllltKI with air.

(a) Jfind the VOlential" in thA cllvity wilen b. O. [h) Find Ihe surface chuge dellsity of the cavity Willi when b - o. (c) Find the potenlilli 4> in tho cavity when h - ul'J. . (d) Finn the sur£ace charge denf;ily or the cav ity wall when b 011..

11.te Calculate the elP.r.trolltlitit; force acting on the VOinl charge in thp cavil)' shown in Figure PI1.HI.

11 .20 Sketch the E linp.ll due to 0 point chargp nfMr the intc rfac(' of Iwo dielectric media The ,llilualiull is similar 10 thA nne lIhuwn in figure 11 .17. except that ( 7 - 0 ·5f1·

11 .21 A rectangular conducting trough of widlh u and height h j,ll maintained at zero pnlRntial. as shown in Figllrf! 1'11 .21. The potenlial nn Ibft IU[J plate. which covf!l'!Ilhe trough. is known to ue 4>{K. bl - ZOO lIin(2u/ul volts Find the (lOlential 4> in the trough. There is no volume chargfl in the trough.

11.22 Three sides of a recll:lll)(utllr conducting pilH:! 111'1: grounded . whllfl the fuurlh side is maintflined allOO V. OS shown tn FiKure Pll .22. Find Ihfl pohmliul in the pipe. ThArA

i9 no volume charge in the pil)C.

Pie"'. "11 .22

'" '0./" ~I----I /oo\->

• • ---"

,u ..

of , ..... fintl 1ere

1.16. 'he

1 in

:Jia

em 'he 'he

flis

"'"

Problems

,~

A~,·" ) fu. OJ •

1"19U'.P1'.2~

'1 I

.-,oovo E

I"leur. P1' .114

0-0

f • I . - 0

371 .

-> •

11.23 The lJoumJulj' potentials of II rRr.langular conducti ng pipe arc shown in Figure PI L23 Find the potential in the pipe. There i,l no vol ume chaTgfl in Ihe plpi'! .

11.24 CUlilliuer the boundary valu!! probl/lm lihnwn tn Ftgtlre Pli 24 The upper Qnd the lower conducting plates are mainlQintld at :':I:I\J ~lentilll. Thl! IllatA III the left is IIHllnlf1inflfl fit 100 V. Two gilps insulate the side plBte from the grounu . Tht:re is no volume charge in the reKion and <z> Il~prollche~ ZAro HlI x apprnaches infinity. (a) 11M rhe method of separation of variables lu uutu ill two oNli na ry differential

tl4UUlioll1l.

(b) Solve Ihe differential equations. (The rUllet ion involving y must be Il sine (unction.]

(e) Mateh the boundary condit ions. and find thft fiMI Mllnion

11 .25 A sphftrical capacitor is filled with a dlelectrie material of (1 in half or the space and with another moleriul of f l in the remaining IpIlCIt, I'll shown In Figure Pll .2S.

{a! Find thtl pollmtial function <l>(rJ in the region o<r<b. The potelltial al r .. a ill Vn and it is zero at r .. b. Hint: Thti pottinlilll IllllillfiAII Ihe r.aplace equation t I 1.21 and it may be assumed that it is II functio n of r only.

(bl Find the electric field in the r8gion 0 < r < b. {el Find the D fie ld in the reKion 0 < r < b. Hint: 'l'hfl D filtld in medium 1 Is differ­

ent from that in medium 2. Note: the bounda ry conditiolls 011 the tangential Ii: IJm.l oillhe no rmal D field s arA AAtiSriAri ulllngihe suggested approach.

(d) Find the total charge 011 till: inner conductor and Ihlt CApACitAnce of this ca­IHlr.ilnr.

Figur. P11 .25

" b

• -CHAPTER /I

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:. ¢(fJ- VJ.f/J..(t)-VI.-,/J-(~) - VJ..(-f)/J..{~) v'~M-}.:r(rltt-).o .. ~(r)= C.-Ci;,. l(A)-VOc,-t./a. ~ 1(6)'0~C,-('./o :. £(,)- y.~y~(1/6-llrJ ' v'~(')· r's';,.,+,(Si'l.9 :[) -0 *' :1. ~f.', -> ~(8). c,£..(f.tt.,..1)"'c~ ~(9.).v. .c,J.,.{Ic.t)+c, .,.{. ~(qD')-O- C. ~ c,~v.I'£"(""{<) """- cz·o :. 2(~)z v.J..U ... f-)/.L.(~:) aNt if'-vf(l)o-;·Hf.~ -; I'o/,.I-.(J..,,~)r:..e :. (,.£E·§c-£V./rL.( .... ~)$NJ, (1;. ... ""'f." f,>0 iJV,>O) v'I.,;ll/~~'&'o ... I(IJ-',,+cz ~(D).'O.C, .""~(().I)&'a.IC,l>eo~/oo*c, •.

:. flO° -Zoot- 80 _ 4-f K .I-_"itrrlll-' 2",,'WID V~ .... VCI:J..2, V JJ.:L J - ,;r{2q," V .,.. I 9"(10"')' q.II~/(r~1

~ 1= f,;tf v~, -YR,' Y.f, - Yft. J, ",~ve Il, _~·~('x:._-,T)l-.-:.,.·.-(l-_-,-;>' , R,. )r(-•• -,~>.-.-~~l.-(I-_-,).,.·, I J • (X~I)'~ 11+ (;'+1)' ~J. I?". J(#t-d.,. '1'+(-1#,)1 10 ....

Ea- ." .. -Lfx.Ot·,)I-;,t1('-t'_ ~C1Ctl) .;'f'i(J~) of' ~{a:t,)+rl+1(1"") _ ;'cr.,).,.;/.#(J.t)j Vr 1rf,1 1<,' RI ~"l R.!

0'1,11e.. " .... J.ofo.t "''''', 2· 0 """ E _ .!. [[(,x.,f. 6,.,]:il- [(~"'" ~'" J~ 1 ; .'. r. = Uf·' • • ~[[I.;J)·.~·.,J~-[(.-,l·~··,r'~l

01'\ tJ.e. v~·cd 1,oJ4iL 1 ),-0 4#'tt:/. E - k{(,. ,,'.('·')'1""'- - [""'1' +(1.')1.].* J;

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11.'1 t! .. ~ ... ~_~Z ~~u) _, . ' ··11 '.·t , .

: .~ ~ -~--.-. -.~"'--""" I / , ~ ,,' . "

+\ \./ ,/, , '... , .,--"'\

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1!J.! F~ (.II./D). c" 3."£../UJIJ,·'C!> - :".. Ji".1t/D-/~sIt-I("'::') - I70S ~~f/n,

40

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zoo / -

, I • ,

t', , " ,--

lC)

::

~,

" ·k ,

"-

-J~

•

.. .",. ,,.

~ t= (rl.rd.'_arJ.U){PI'a.

A~ -cr'#r.'-u·r;uu)"i

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a:t I.b,~~o """- jo -I'.. ",;!Iv ... -",'/6

41

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11. 20

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/ I t.-T ----7'"-- t ---';:F , ,

. A

J'

--.-~\ .. -~ ---{~ \ I . ' ,

(-,1.) \ I ; '-

\r :,~t

v'1 - 4 ""'- iI-o .. :!..:~~ -() .. 1t .. ,J- (AC'S~%'B.;.ix)(CU.!t..l·Ds;.J,'}) Ito,O·o. A-D. a"",>-O. C-O :.I("-')-65;....~ .. ;..;..I.J 1(4//>-0. * -~, "'. ' .. z, 3, .,. .", #, .. ,> .lA_ ~;""(~J:) S;....f(TIJ ." ~(;t:,i)- zO() S;,." (Rn/ .. ) .. t.,A.... s.;"'(=Ek)S;'..,Jd~') ... Ma2 .... ~ .,fa • .200/S~(~'J .', ~(·,1) - ~(iE4) S':'(.~.t)~.:..I.(~!i)

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J.',DD$, .... (T'I)dl. ~(I- us_r) _ [ 0, 1ft. Il ••• ' •••• • "." .. ..,y_rr , '" ." J, e; ,.,

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11 f

(~"'''':''''':''''')~~ ""~~".:.;...-.,. -. Ca.') T C·,,)

Cu. L W. 2-

C4.Sfj A., III"" tot.. ...... M. !',._II , 11.11. ("4.1(.2. C-A-II be s.lAJ .... 'J cM"g;,., iJ.c.. c-Nf.#Iai.

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42

11, 2+ (4) -

(tl 1>r' ::; Vil (, -L • ... ""w" I (I_I) r> 'n ~ b

1:>n = V. £ ... / ttA~J;IA.'" 1. (I_...L) y;: • ,,,

!) ~ b

(d ) It, == lIT ... • D., (,-... ) = V. (/

lIT (if -tl

' .. (:: '").((((, +t,)

ri - t) v.

43

lrnmtl

12. 11 '1 and

,dary h.811 exact IIff'nl

"\' 1001

dlf· boun· t reo ~ 1i ve

Q

Solution:

Pruulcmll 387

The resistivity read hy the !IOnrlJ! willlWl infhLf'nr:P.rl fly mooillm 2. Thus, IhA Mmde Will nol feMcJ p, to \1'11\. IIlthouxh it is IOCillt:u cOlinll), in medium 1. To find the cxpt,.'Clcd read ing. we must first calculate Ihe potential detected at /I We solved Ihe polf'n liaillroh]pm In F.xampif! 12.S. In the presen t Clist!. \\e Imve"l _ 0.1,)( _ U, Y _ O. I. " 16 in. x 12.5-1i HlOl m/ in _ 0406 m, Z f h - 32 )( 25<1/ 100 m - 0813 m. and

l' _ 1[° 1 0.01) _ 0.8181 01 + 0.01

'l'hPfPfoff!. according to r 12.2211},

-, I [I O .81~ I -;--'--;" -- + -- - - 134.11 4. x 0.1 0.406 0.B13 4-w

Substituting the above value 101('1)1 2 26), WA ohtain

2 . 5~ 1 p, _ 411' x 16 x - x - x 34 .7 _ 14 I n·m

100 4 ..

Problems

12.1 A p(lf<llIAl plaIA b~ lillflrl with two IIIl1terilll:; jill! conf'i1<uration shown in figurt:: P12.1. The lul(ll area or the plote is A The dielectric constan t and the conductivity nf nne materia 1 are. , ann q h rf!RpoctiVf~ ly , ThCllle uf Ihe Olht!r lIIult!rilllllre ~ 2 unu 0'2' Finu Ibe f+(llIlvu!tml dn:uil fur this parallel pl(1tc. and express the circuit paramete rs in terms

of A . d, ' " 0" ' 2 ' and fJ l

12.2 A pilr'dlld vlule is filled with t ..... o materinls In Q configuration shown in Figure "" 2.2. Find its equivalent circuil , and exprp..M IhA r:ircllit paramfllfu's in terms of A tbe .. reu nf thf! 1l1a1f!. and d,. <12, t ,. t 2' II ,. 1llU11l2, which ore defined in the figure .

12.' A cooxialline haslwo layer.'l nf in.'lllla tinn . FigurA Pt2.3 shuws the Malmetry. Find

( . ) Ihe IJUlt:ntial 4> , fo r a < p ... b (b) the potential 4> , fnr h ..: p '" c (c ) the resistuJ\<:e uf a section of such a line'! meters long

~ltI .. · . P12.1 ~Igur. P 12.:1

I . ,. ", '., ". j] ,,.c--r5 -,,:; C2>' h) ~

1 1 .-wl2_~ -1 " ./

wI'! , I I I • I I I I "'tI",. Pt2.2 -1 I I

"

[ ~ '-- ' I I I "

I jd. \ ," __ J I

' " 0, :--........ ..... _- - --"'

I .. '"

388 12 Direct Currenls

y

I , .• ,

/ ,

,

',20 ern , , *'_IOA ' IOcm .......... I ' / , , I

Fleur. "12 .•

/ ... ~ R r 2{lcm

... :IUcm I fI-O.Glmho/ m

Perfect conductur ,

y

-~-T I I

• I p'. I I

0 PArfocr conductor

,

.,

12.4 A lIphericctl conuuctor of radius a is inside a spherical conducting shel l of radius c. Two malerlals arc used 10 fill the space between these cnnnll!':l nn!. The dielectric con51an~ and the conductivities of Ihelle malerial.\! a rt! (,. 0'1' f l . It:_ respectively. FiMura P1 2.4 shows the configuration. Find the equiva lent circuli of Ihis system, and express Ihe ci rcuit pa ramelp. fs in fArms of Q, b, c, t o. fl ' <I" ant! crt.

12.S Two oil wells ilrc 1 km apart. The resislllncc between IWO steel pipes in these wells I, measured li t lAIIi. What is Ihe conductivity of the ~roUIIJ I1Imr Iht:lltl well.\l? Use the fulluwinH d(l\(l : the length of both pipes - 1 km. and the diameter of bolh pipes - 10 em.

12.' A currenl e l~lrode is ncar a pcrf~t1y conducting plate that Is bent to form a 90-corner, as 5hnwn in FigurA Pt 2.6. ThA output from the electrode is I UIJl~r~~, und the material fillinJij th~ spact: has a conductivity cqual to a. Find the potential function "'( x, y, 1.1.

12.7 A currellt electroUe is nt:ar a perfectly conducting plate that is bent to form a 60· corner, as shown in Figure PI 2.7. The electrode prodllcftS 10 A nf r.urrAnl. anrl lhe mate rial filling the region defined lJy 0 .~ •• ~ tiO° ili wuter wit h conductivity equa l to 0.01 mho/ m. find the potential at point B shown in the figure

u .s A point electrode puis oull amptlrtlS of current above a conducting piane, as shown in Figure P1 2.8.

(_) Find "'Ix, y, zl for z > O. (b) Find Ihe current density 1.lx, yJ al the surface of the cnndllr:lor. (e) SlcAtch the paths of Ihe current flow.

" f, _ eo

ius c. celric IVAly. t. and

ells is ,p.lhA

. - 10

• 90' d Ihe clion

ill SO" d the nallo

4

12.8

Pmbl ems

f C I I

3m I

I i 0a.1 SlIIl

'%\"\\\'@. '{ {~·;\ .... :,fu"\'\~\~ C7b 2 S/m

FJgure P12.10

389

For thtl Coltse shown in "'igure 12.9, find InA pfHGAnlagA of Inti curren t emitted from Ihc elect rode crosses the boundary ond cnters iUlllcdium <!.

12.10 A source 4 meters below an interface of two conducting media emits 2 A of diroct t:urrent, u show" in FI,I(ure P12. 10.

(al Calculate Ihe potential at point B. (b) Calculate the polentillial point C.

12.11 A wAII.logging rAsislivHy lnol similar 10 Ihe one shown in Figu re 12. 1215 near a buumlury between two beu!!, as shown ill FiKure PIl.11. Tht! boundary is making 060" angle with the well . Find the ilpJXlrcnt resistivi ty measured by this tooJ al the po'!it!on "hown.

12.12 Rflfflf to Examplfl 12.6. Ohtain Pe (the apparent resislivity mtl8surtKi by the tool) as o function of tool position for Zo" j 160 in. to Zo" 160 in .. where Zo is the posi­tion of the center of the tool (the midpoint between electrodes A And UJ relative to lile uuulluliry. Clilcullilc Po for Ilt ICllst 21 PoiUls. IInu!)lut Pit V(1J'~US lon.

12.13 RCPCOf Problem 12.12 fortne situation shown in Figure PI2. tt .

".~: 1"" -"'",'" • I I

I'" P. - 1(11)l1-1n

",' I I

-! .. '

Figur. P12_ t 1

390 12 Direct Curre:nll

12.14 A point electrode is located at (0. YI' nJ. lind 8 perfuctiy conducting sphere of ra· dius 0 is located at (-t, 0,0) as showl! ill Figure P12.14. The electrode gives I am­peres of cUfmnl. The cunductivity of the medium Is (7. I"ind the potentilll oil on the y axis. Hinl: usc (11 .44).

12.15 Consider a woIl-logging resistivity 1001 simillif to the one shown in f'igure 12.10. Lei the spacing bel'oWtlll the current eloctrode A and the potential electrode B be 6 m. The tool measures the conductivity oflhe earth furmation as it trawls in a well. Assume thai the well passes near II mineral deposit modeled by a perfectly alii­dueling lIphertl. 8S shown in Figure PI 2.t5. Find the Apparent resistivity measured by the tool as a function of y. Use the follOWing data : 11 • 0.0 1 mhnlm for Ih. grou nd ; the radius o f the mineral deposit. 50 m; And the dillhmctl between the Ctmttlf of the sphere and the well • 70 m. Plot 0. __ versus y for - 70 < y < 70. Hint: use the result ohlAined in the proceding problem.

,

1-1'-'- - 1 y

• 6m

• - 0.01 mholrn

•

Flgu ... P12.1S

,. 1 ~

•

13.1 Magn

2. ~

CHA PTiiR 12

a,c ....... +-;t.ii.J.. E.. f.·.Id.

"~J t. W """t,·"MO\,U.. F, = E1. .

CO.,,/" ,"II). V= 6," = 8,01

........t we C.An c. .... s:Dle,.. i.f 4lJ

I3tc ...... ,t fA. G4.V'r"f" t h ••

t. 4t (..o"+i", k..' '" I J r = , X.

".~ ...... ""'" c,,,, Si .of.,. il-

v T

'< +"" i'"f~rf~J- c·rc.tjft,I"J /" Se',., ·~J.

:r

"" f' oQ iii. II. '" (~far.+Pi . , :0/,. {; .. ~,k oS i.

'1--1£, ~, = Ifnl"'j

(I " X_I , -I.. ... -b < ~ C-

r, UI, q, = ~lT~ = .......... .l.._..L..

" " ... •

44

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C .. I"I r. ' ..

c,

to--;; ... ~ ..

" . 7

II, T " 4, £.,tr. f II, t:,.'i " (. ~.

r ,I ;/1. 8 (4)

,"

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C, Ii,

4

,'.1

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7r.t, r.c/o' "'7:Ti': 7·2 f~/P ""/,,,

Use. ,;......'c ""c.H-.d..:

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(~"f;,0)

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Jj (".1)- a; EJ h·. - -~ (;Z,.:C. J:)W

(c) SaJO"Ic, A' p" " '''''L 11.1

45

-for D4~~r'.r i" IIt#J;w," ~ # -tJ..,.

S:I"'ATi,," looks ','ke .,1..... "'" " I So""t'rt.x. ., ... t.1l _t J • .( J

,"',. " A.'".), .. c .... J ,..,J ........ of U; . n,Uf{.r-t. i .f I II .,,.,, (.r,'S +~4.

x-A.iS I.t. t"- /.",1£ A.lf ~ tI.. S,."" . . ~ ~_ O'i

c.rl1lf'IfJ "_ .. ,,., = ~I - a;-.,.~ I

c-..t'( ... f eNi,:, )1- ... !t's :: [ 0-; / (o;.r. ) ] Jr I" ,. v· ·f I'

l> )

I I :0 -:Ir.::'c.-_r.:::.. r cr.., ... r 4

i ct. ::.

• '" -J

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l= f 11 o-.. rJ,.

I"; 2. CI"'" 't = aa. .. 0;

i ... /J

"t ( ::. ~lr' I· 7

... J

-= IS. 2. .. V

46

I I I

", I --/-1- -

t( I , r; \1

'I I

1 •

I I r. I I •

iI-I/ ~ I"

I.

I~.II

'i

48

, , " z4 , ,

J ~.

• g;(f!T ...... S", I I ... I I I

.. SiJ I!: ; v , ... " I

" , " , i __ .J

s'

I

L

I !

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'ID' ,.~

------ -- • H' ••

•• •• fo'

, .. ,., d ... ...

·s .~ . . , , I • • • 12.15 F 1'0,... p,.., 1.3.:..!J:. I

1.(1,). 4trfN,/!IlIJ/r: I!:-ld (,/,.1- -,.,;; . .-:-:,-:-:.~a.::-, ~[~ .... ;-=~+ r;;""'~'J ... I)i I "'Ji(~,~.tt .. ) +~ '(',I .. .l'rA'':U ,.. A · ,

f..rl,). 'DD{-f- 5'9 Ji/!!!. ('I,l;Zf«»1. .. ~i .. !IH l U."'X .. c.;4'.J-as-,j l

:. d;(,.,,- ?.~,.) ~ t:r...(I-'>- ~(V.) I!II P (n.~)

A--'e...c -------------~--- ---------

'-..

! , • II !

47

Problems

422 13 MIl8ntl lOILIa l ic Field,

Oecause magnetic fi e ld is present in the wflxial li ne. we know that magnetic energy is stored there. The mngnclic field is giVtm hy (13.7):

{

I - b > p > o

1-1. _ Z/rp

o e lStlwhere

SubsWuling the ithove ex pression in (13.34). we obtain

Utt .. ; JJ l l o dq, 1" p dp I: l _ ,1011 % In (~)

2 0 g 4,.. P 4". U

T his result is the stored magnetic energy pe r unit length o f rhe cooxinl line. Conseq mll1t ly. we Cun calcula te lht! inductance per un it length IIf tht: line from [ IJAl j:

L .. ~ In (~) {lJ.4Y} 2'11" I I

Th is inducta nce per un it length also appea rs in Ih H tra nsmission · line repreStlrl tHtion of the coaxia 1 lim~ in (6.191 of Chn pie r 6

13.1 "'inri the IIIll,l!nctic fi eld " althA cflIlttlr of a square loop carrying It current I. The side uf the square loop i.~ b meltlrs long_

13.2 A r:in:ulllr loop that has ratiiull u amI thai carries a current I IJrOOuces the some Ul lIgnetic·field strpngth III its center as thot al thA cAnler or u square loop that hall lIine band Ihat cll rries the some current I. Find tht! ratio of b 10 0 ,

13.3 Consider a la rge concl uct i n~ plate of thickMAA II locutt:d ilt d/2 .s y . d/Z. all shown in Figure PI3.3. U niform current of de nsity I is flo\\ing in Ihe f: ni rection . Find H in ulll'egions

13.4 The eH rth 's magnetic fie ld al the norlh mognetic pole IA approxilllHtcly 062 G 11 C _ 10 • Wh/ m:j Allsume thot this magnetic fill id ht vrwuccd by a loop of CIIftllnt fl owing II long tht! t:qua tor. £Slimal/! the IIIH)!ui tude of this Climmi. The rutl ius of the mirth is approxi mAte ly 11,500 kill ,

y

l-~·-/ d . ___ . Y- Ftlur. ~ U ,:I

l,

,. i.

" '" 11

13.5 An Infinitely long tubulul' (,:undut:lur uf inner radius 0 and outer radius b carrit!9 II direct curTent of I amperes. as shown in Figure P13.5. Find the H Rflld at fI: where (a) p :;, a, (bl a .so p .so b. IlIllI (\:1 b ~ p.

13.8 An infillltely long tuhular conductor has outer radius b and inner radius a offset by a distance c from the axis of the Olllflr cylinder. as shown in Figure P13.6. This eccentrir: tuhular conductor carries a direct current of 1 !impArl',&, Find the " field at point A shown ill the lisurfl. Hint: Consider the tube to be a superposition or IWOllOlid r:ylinders that have radii b and a amI thlll !;1m), IIniform current density 1 in opposite diret:tiolls.

13.7 An infinitely long wire is bent to form II 90- \:ornflr, al shown in Figure P13.7. A di rect (':UlTt:llt I nnW!! in tllfl wire. At point A find 11'10 H field due 10 this current. Follow the steps given below.

11 (a) Usc tilt: Biot-Stwllrt law 10 express the " field at A due 10 It typical segment of Itt wire dyon the wire axis. Express til t! field in rectAngular coordinates

(b) InteNl""d tt! the rp.!!uit ohlained in (al to find the H field dUt! to the semi-infinite wire OC. Note. to faci litate inteHration, lilt y - fltan fI, 110 thai dy _ a SCCZ 8 d6.

( c) Finrllhe H RAid Al A due 10 thc ~mi-infini le wire 80. (d) Add the results olJtuined in fh l and (c] to yield the lotal field at A due to the

current in the wire DOC.

13.8 Follow a similar procedure 10 the one rlltllCrihed tn Problem 13.7 to find the H Relrl at ~inll\' . 8.'1 .~hnwn in Figure P13.7.

13.8 Consider a circular loop \:lIrrying a currenl I counterclockwise. as shown in Fi8urt! 13.11. Plolthe mAgnetic field 11, on the z axis for - 0/2 -::: :t -::: u/z. Find the value Z. in lerms of a. such thllt. ;rl:.d ..: 1.., then H. ill uniform within 10% of the valut! uf H.!:It Ihe (".f!ntf'r of the loop_

424 13 Magnetos' . I\(; Fields

,

PltU,. "13.10 Helmholtz t'OI!t. ,

13. 10 '1'0 ImprovA the uniformity of Ihc magnetic field along Ihc Axil! of II circullif loop (see Prublem 13.9). onc may use two iden lir:alioopil S8!)l:1rl:ll~ by tl distollcC cqu31 !alheiT radII, as shown in Figure P13.10. Such II pair of current-carrying 1()('1pt; i.. r.illlild Helmholtz coils. Find H, as 3. funClion of z on Ihe axis of the Helmhult:e coils. Piot H. for 0 < Z <: o. Find, in terml! of c, the valut:!:t,. such thot. within the range I zl < 7. H. il'luniform within lO"k of the magnetic field at the middle of IhA two coil" CompiHe your result with thai obtained in ProhlAm 13.9 for Ii sin)!]e loop.

13.11 A 5tjuArA conductor loop 20 IUcters long on each side carries a (Hreet cu rMol1 u shown in Figure P13.11 .

laJ CaJc:uilltf! Ihe magnetic fit:ld B al (b.O.O). Express the magnetic Reid in terms of 4 integrals, where each represents the contribution from thtl currellt all each side of the squaNl. UIWt the Biol-Sllvilrl law. Do nOI try to integrnte those Inte­

gtlll9. (b) Assume thnt b is much greater than o. Now, INllluulll thc inlcgtnis approxi­

mately to ohtain an approxinllilll value of B at (b,O,O).

y

c

T .

I " lb. u. uJ

,

Flgut.P13.11

A B

13.12 A surface chAtge of P, C/m1 is uniformly distributod on a t9Cord disk. The inner radius uf thtl disk is a and the outer radius Is b. Thfl rACord disk is turllinlol at II constant angular velocity w radII in the duckwiStl diruction. Find the mngnetic Reid at the r..anter of the disk dUll 10 the surface charge on the turning disk. Ignore the prestlllctl of the metal post on the turntable.

< Fields Prublflms 425

J (see Their liliAn 0 1 H .

".h II, l~Dre

t I AS

'01> of cach int&-

.roxi·

no. :tt Ii

clic 0 ..

13.t3 The tlHrlh's magnetic field at the equtllor illllpproximlltcly B _ to 4 Wb/m7, Calcu­late the cyclotnm freqllfmcy of the electron in the ionosphere.

1:1.14 lIecause natural uranium contains a slight Illlluuni of Uranium 234, llit! electro­magnetic: Isotope separator I:tlil also yield U. If tho radiuli of the circular path for I)lU particles (st!8 Figure 13.14) is cquollu 10 m, where should one place collec­torI for U~U and 2:HU partidell? Expreu spacings In melers.

13.1e Rcrcr lu Figure 13. 17. The maglletic fittlri ill r::hanged from 5 x 10 ~ In 10· J Wblm1, All other parameters remain unchanged. Find the following:

(0) tbe position of the electron 1I1lhe exilltide of tho magnet ic·field fegion (h) tho exit anNie (the lingle between tho truitlctury lind tho K axis after the elecl ron

hA~ passed through tilt: magnetic field)

t3. tO Consider an electron having initial kinetic onorgy III. ~2 IIml enlering B region uf uniform magnetic field. as uelJicted in Figure P13.16. This ~ilulttiC)n il> similar to that SIIUWII in Figure 13 .17, except thot tlJtl!:!leclron In Ihe prescnt COstl is inclinmi AI an a anglc with n:specl to Ihe x axis.

(a) Shnw Ihal v. and v, of tbe electron after il enters the UUI,I(nelir: field are given by

v, - V. cosj"',r , uj

v. - v,sin(\o.'<I'" a1

wheT!:! 101< - Cl,.B./m~ And I - 0 corrcsporuJ:! to the mnmenllhe electron !:!lItel'lllhA magnetic field

(b) Fint.1lhe cooroinares x and zofthc !:!1t!Clron i!.llime t NOle that x _ 0 and z _ 0 al ,- 0

(e) Finn Ihe poinl where Ihe electron Ip.awt' Ihe magnetic field. Alisump. v, ... 2 x 10" mis, a _ ao. "', - 8.77 )( 10" rad/s. and d _ " cm

(d) Find the ongle beh-\cen the x axis and the Iraj .... 'Clory uf the e l~lron after il hall left the mAgnplic field . Sketch 1111:: entire Iri!.j9CIOry, and compare il with the one shown in Fi)!ure 13.17.

, J----j

------, • • • . , Figur. P13.1' , • , , .,

.l::lecrron " ,

• • • , , t )

• • ~--~ • • • K K I x )( X 1< I --------'

t3. t7 'I\\-,., flarallel wires arc carrying 100 A of current in 0pPQ:!lte diroclions. On each wire £ind Ihe force pAr unit length due tu the mllgnetlc field produced by the other wire. Is the force repulsive or attractho-e? Assume that thf!linf!S IITf! 1.5 m apart.

426 13 Mll8nelollltic: Fifle.

13.18 " wo idtmticul circular loops of radii n ATflllepartltoo by a distance d, where d «(l. Qne of tho coils carries I amperes or current clockwise. and the other carriM I amperes coun terclockwise. Find the fort:e betW8fln Ihft8f! coils. Hint: Becaux these coils IlrC close togethor. you can appmxlmate lhe magnetic field that Is I' ODe

coil and is produced by the current 011 the othor as III .. Izl'f2Trl). the field due to an Inn nileiy lonx win:. Let 0 .. 1 m and d .. 0.05 m. How much current is needed tu produce a force of 9.8 N?

13.1' A circular loop of radius 0.5 m 1:II ld l00 lurns is excited by II 2 A direct curreni. This loop is placed in the Earth's magnetic fiflld, which b IIpproximatcly equal 10 5 )( 10 - ~ WbfmJ pointing nnrth . How do you orient this loop to produce II max;· mum torque? What is the value of this torque? Find thAt oriAnttllion uf the loop in which it t:Xpt!ricnccs no torque.

13.20 'fhc »quare conducting loop AIICD ~hown in FiJ\urc P13.20 cn rries 2 A of direct current. I~ach sine of the iuutJ is 0.1 III long. The loop 15 pillced in II. uniform mag. netic fitlld B. Find the force on each side of IhA lootJ II.nd the torquo on the entim luop if:

(II) H - ~ 0.2 Wblm2

(bJ B · t 0.2 Wblm2

, D

-T U.lm

-.

C

f .. 2 A

•

Flguf. P13. 2. 8

13.21 An Infinitely lung conductor of radius n cllrries II. uiroclion current I as shown in Figure P13 .21 .

(A) Finel the H field in the region 0 <Il < o. (oj Calculate the stored magntltic II11CfMY jJCr unit longth In the region 0 < p < u. (eJ Find thA inductanctl per unit length of the connu r.tor. CUlilIidtlf only the mag·

nctil: tlllcrgy in the region 0 <II < o.

13.22 'fhree infini tely long parallAI wil'tts tlll.cb cu rry 10 A of current In the t direction. liS shown in FigurA P13.22. Find the fOfce per unit IAngth ActinM UII tbe.3 wire due to the mHMlletic fields produced by fhA other two wire9. Cive tho nurnerieRl value of the force. Its direction, and its unil.

Itatic f'jelds

111M d « o. ler curies I ~t: Because halls alone field due to It is nettd9d

}Ct current. lly cqutll to lell II maxi­the loop in

" of direct form mag­the culil'9

hownin

<no he mag-

:tion, 0.5

e duelO Jalue of

Problem.

y

t

'i I

T " 1m

I ,

" !---1m-...j

Figur. P13.22

FlgvrePt3.2t

13.23 The mo.gnt:tic field in a cOIlxiallille is giVfln by

1'.- fOllP forO.lm <p< O.2 m ~ elsewhere

427

" ,

Tho mcdium i8 Air. What is the totiil stored magnotic ellcrMY pltr unit length ill the line" Give tho numerical value and indictl te its unll .

13.24 (a) CHku lale lhe stored U1tlgnelir: energy per uuille nglh of Ihe parallel·plate mn­ductors shown in Figure 13.5.

(hJ If the parallcl plate is u~ed as a capacitur to II lore elect ric CllerNY, finrlthe volt· axe Vo for which the stored electric energy is cqutll to thfl ilored magnetic en· ergy found ill (a). Let , - l A, W _ 10 cm, And (] - 1 c m. ExpteSll Vo in volts. The medium is air.

13.25 CalculAlfi the inductance pllr unill fl ngth of the coaxia l line shown in Figure 13.3a.

13.21 Calcuillte the inductance per unit length of the parallel·plate conductors shown in Figure 13.S.

• ·r 'f.--+-~

" IJ.! Fo..- 11/" d/~; ~}lz - ~1j

:. ii·(-'):r~. 1~/f"-h

F.r I~/'''-;'; 2/-1. = 2 (¥Z)7,p 1-1,,-1("-1,)

... H_{(:')I(d/.) , ~''';' xU"!.), ".~

d 30

CoS'(' .t, 1)+(6/2)\

,

LIse .!/,.4A..J;Ol'l. (/3.!J()) ,.;iH..-} .... ' $"" k,.o. .,.$./C'," J ~~, _ f A..ra.2 ~r Co ~4 / ' . '11110 rt~1i l( (. ~1t./O 9

fa 1- o. ,UIO- • 2( .'<t-d.t )ill - <#Ii... ... I· o.'UID )rt -I-II JI 4/,u. <: .,rJ< ,,,_Op , I.SIt!1> J

(a.) fl4j Hzo (7W) c~l'Te .. t ."doJcJ.) r A I ,,:t:.a.:I

,bJ a..ff$b; j- "'bt::4.1J ""'Ii J.lTI'H,E'j·"(f7-a!) 9 ll-; ~n;a' bi_iA..' ,/ 4,(f!},

(C) p'p; .nfllf-I ... P-i';f . hlf

c..J.t I

:::rr~f---_A

t·c , - -

/' "

" ~\ , I , , 1 .. I \ ,Cc.seZ '- , -.-

/3.1 ( 4)

49

.. I~1fI(F I"~;.(-'H""-S;"" ... 9J A r x.4Ll ~ (4) d.JI"," +rr~ - "'l .. J#>{I-~.)iJ - Z" (&iH1-1.'fjJJi

- f. - A n,r dl ( " ', ............ ) (.) H~. d.~ - * "'#f). (.x:.I+C I·,.)1JJfl.&. 41- .r..s...\t. .... A E> I rJ{ A r "

Z Z H 'x! ).J4 CAS.(..t_ - i:;:;rz: (''''r)

X. a a. uHS·" ': 4 • H,;.. Z <tf;'a. · (~?)11j. _ i ~(~6:). jU'2f: i te) ~ t. b s,-_Iry J H, ~ ii.t. ().19!J. i • (4) H.~"'iJ;,- a.lSl-ii-

~

I'! Fr,,"" .,"4+t" .... (IJ.20>, III - .2(a..i"rl)'1& r I (f,Iott,. I

• , ',': ••

"- ]1. , : '~ ,,' ---.........-

"

ar i. 2., HI./N;""t.,,_ [1+(J./t1iJ-o/J.o., .. -1'12)}..(I">'/,,)'] =.1.. D.,

-"34:" ' (,+(i'/G.),-]¥.i - H, . l,.,t~~)J,{J. ( N:~ H,/, .•• .f ) ::"-'1'-.-,_

••• .. ).,..!.I .,'-/#>l J c. ().D7 .",. (~/().:/r O.()7;e .. 2._ r ().'Z7Cl.

/3 . ID t;uperp.,"·liM I H~. ~tL' [(a·"(I-~lJ-~+(a.'+(I.tlJ-~ J -~ ([I"I : -'.~fTI; .. [1'll.AS)')""';

ai i_O; H: c. ·k::J.·(I+". t.~;-J{ c (~"/~S) ! ct ~·i..1 Jl1 -kfcU'(£-AS)'r41a. [J-t<.f."dsly'4j

• 1 .(~!ISS) I(I·(~-o.(tylh+[I.(.:c+~'l J- ¥J 1 ZJ" I..'JI [(I'*(l!-o.S-lrJ;, .... [I.t{: .... s1f'nJ .o.t'/

:./t,.> .. (I'fc:z.- o.n 1r'Jt+[I+c .. +O- r rr .... -1.2'S-D ,'""'Cf" z.~ ::;. x.-::-:t o.,,,~ (r_,_ of ~.;$ ~.,. ~ p"'~'.M. p""''-')

13.11 lo.) -

-" -.'

. , ••

H' I

•• •• •• .,

.... 3 () . J.

- ,I" .... , ' , ,I" - /I • • r, .. = -. of" 6>-x)

50

l~-"'; J ( )'. 1> .... ,. J'"

f .

I

I

" ~.

I

fl./2. .6J = f's ('tI.~ tl.f)

.;;: = fw (-il; r., =(-rf + IjJ/JrV

Iii ~ ~1lr = J', c,#'rJI",I-fJ . [" ~ dH = 1',"''''fdvc-lI x -ff+HJ

4711 f' + J ') ". -fslJpJ'r'~

I

/2.13 r,a: 18 • ,4./0"'"'0''' ~.Ti. .. ""..., ~~,,~,.II.I. ·)I & 2.B "'HI

!!1! PS/Il, ~ (hlr/""',)JS.. (,JS'/,,8)'i = O.Q9J7 .... ~s" "l9.17_ I!.;/ K, • (1I¥/1If.,Y.. (Uf./ue)5Sc • f),q'fS'.... R.,.. 9 .• 14..,

~ R. ,.,V; ,,".1 •. '1/. 2 .. ,," • IS. - I, (,~JO··'. I,.·j • C.I/~ ~

(4' loS-. V; t.I.s~C.; IfJ • - V, S·;., .... t J wJ...,,. W. Vi/Ro :. ,: _jt\l"rt't • ~S~'""'/t- .. J!i~tw,"

• fN ...,

J _l'v:,Jt • ...!!! ,.,..,t I~ • ~ ("'Jl.fl!-/) , "" .... Ai: ~ ~j,t;; :ttt.) .. ~.()# = -t!s,.;.....,f; • .. ~t •• 't _ D. Jr J.li.) • ..:e iii· a..", - 0 .. IU ( 1-(#.1$)* -I) - - (1. "D7.2,."

(. ) tL~ ~c. .... is.t,, -t .. i .... 1/. $i",wl." f).'S'"

~u .. ) = v. CASwI. c v. f ,-s.· .. ·fJ't. .. 11; ,rl-_(~'-. '-rjl~ w (). 9.J7 Y;

'itl.)--V;SI:"""~. -=-I,ls"'O

.... ~,,",,-'( "j (4)/",(4>] & +--'(-", "V .. 9J7) • - 'o.S" ,

.. ... or I V. ,

Lt ""', m45"";"'c.. /It.U. r .. ,i."", ~,.)+ ~I(,I)~ v.l. .. 'V",.tI>-v.'AS("4t+6) ~ "JW-\$f'~("'lt~f t.~,c. ~(,,)c V; uSoI.- J l.tJ '.).c:'-~f/~ 1$ -v. le)c '" ~J{~t40 fII.) ~ 'Ii ((): .. V. S';"("4:tt,.( )

(.) I,. tilt.. '""'j"'/'" /.0c,t,l "~j;DI\ ~ :t'(4)-;: tI;,(t)Jt' ..,..£ JftJ-1.4"5 ct)"t: tor. • • v.;; •

:. )'l<6)- ~s.;,.(44tt")4C, AIttL Ju:)--i/t ~sc~t4>"')+'l.

r;'"" %l~).J(").oJ c,=--..J:,;,...{ ~ C".-~c..os.t.. :~ 'l'(.e): ~[s. ... (~t+,L)-I,:..J. J ..,w(. 'k).*[GOJl~t"''''J- '.ht.J

(C) J. TJ. • • "';I~~ -i r~. Wi" %l.,.)_ tL -;r[''''(IoJ,,~.''''')-S~''J . • silt (1J,t,,+.t ). ~" + j,~ .. -= 1."";;~:;f=If"'·' .,.s;,..U·). O~Z,ZI, :. h's (c.r.l.:~. +.(,) .y 1_ s;.",.'(1IJc4+tJ.) ~ '" I. (~.1,,{,)t .. D • • ,r

•

:. at -Mc ~,'Ii~ xt,.)- t>.c4-,., 4..IttI. "{4>- ~"/_" ,(".,s--u,(s..,J - -D.DC'7li/o", , ' .'7./0

ll.> Itt /1" 4.J( " J~~ ~ .... ). "" COJ( ... ,i-.+.-4) - ''.II'S'V''. ~ 1)(01.). - ".2''-' 'Jr.

cll isi .,.,It. 8. ~"" al (Vj(to/~t4.)J • ~ ·'r -1>.1Itl'l ,.,,~) _ .. IS.U·

51

.. ,

/J. 11 ~ - I, 7, ~ /4I.l.

~ITZ

/3 .'0

1l2..!

/J. n

# i "*, ~ W-~'" , E . oS ! .5 " :

£.0, sl..tl/t/... 'I.. pt..(ui.. A.,.i.J""'IwJ(~ 0'­

IIui,"wi.r ,,",-..H.c. £_11'1/ ~,. .. ,f,"O" +0 oh-!aJ",. -;IlL. ~i-"" .,.,., .... .

T ..... )<-msc "(".SJ1."'IDOK.%M5'KIO·r c: f.SS'X/D- J "'-m NO ..J.rfu~ ~~/."'i~«J ...,;.'" b.p ;1 piNeL VW,'c...J/!j l~ Hoc N- S d.·,.~t.h ·O'"

(ct) Fltu=-O, Fa,,= ,t,illfij. '/1I0 ./A.j·X ~ . l; rrf.#41- ~ i-i) . F~. 11: 0 ,

~P ... = 0· ·'''1.)(1); T=(o. /)I.If'2jx , .,.;:: ""1110"" J/.,.. (.,"')

l~ )

(A •

II'

0 . /62. -X X (. O. '2; ) =

• . Olf "';; T: 0 .

'~r fi+ = r.r-:f.!) .... lJ.· JJJ t .... H; 4v a

I ,'" LlI.

1-1+ '" If ;'ro<1' ( ... 1.1f"''"

II. 1.. 1" f." l ".~... Jf t' (1'11.1) = Alr ,J., "1f

I' I UH ~ iLl" L= 1 ~\. .A. .,,,.-1 H/~ -: Y. II" ;-- /0'" /l = , ~r{t

;; - ,.~ y' "',- -."

• • "Y

I' Ir. fi

) F, = /.;,,[ /~;' (i;y+y)]: ':;"(f_1;J:,o-rrf_J:J

tJ~: -t ... t;( trur (-if / = "M..I.. t = 1.7~'" -I J/,., • L

52

/.1/.

/3. ,"

53

Problems

452 14 Mil3uellc M. leriaJ. and Masnetic Circuit,

From 8'01, we find the correspondin)! I-t0) (Figure 14.1 7b):

/l ilt _ 2~OO Ai m

The iterative method calls for substituting the above value Inll) (14.23) to obtain the "first-order ' approximation of B, which IS denoted lUi BI":

/1111_ 11000 - 24(0)( 0 1211.u.. _ 0.176 Wblm1 0.005

The corresponding I Jl I' mllY be read fmm Fi"ure 14. 17h;

H I' I _ 2100 Ai m

WA ubta in the "second-order" approximation of B by substitutin8 the above H" va lue for H in (14.23):

B'" _ cll~OOO=_2~'~OOo::;x~O~.~12~I~"''' ' / • - O.167Wu III 0.005

This procedure (".an IJe repealed to find the n_th itljl'lltivc result of BI~I. When a rli8illli computer is available, the mlignc ti7.<llion curve can he HIlProximated hy Ii !itllndard polynumial-curve fi lling and SIOTfld in the computer memory. A simlJlt: program may be written 10 cllrry out thA iti!rativc proc:eduft!. which rACluirt!9 very lillie computer time l!lei! Problem 14.71.

Tht! problem ill hand can also 00 9011U!. by a gmphicaJ method. Nutc that l14.221 or. equivalently, {14.23} is an equtttion of II AlraiMh! line on tht! B-H plane. As shown in Figure 14.17b this lint! intersects tht! B oxls al 0.251 Wblm1 and tht! H axis at R264 Aim. II 81110 intersecls thlt nonlinea r magnetizotion corve lit B _ 0.19 Wb/m1

•

This Ttlsuit a)!rees fairly \\..,11 with the result obtained by tht! iterative methtxl.

14.1 RefAr to the magnetization CUrvA shown in "'igure 14.3. The malt!riol is a nnnlint!8r nH:dium bec-.8wt: J' depends 011 the magnitude of If. For mognelostalic fidds. u is equal 10 the slope of the line joining the ori!jill to lhe tHo BI poin t on the mllKlletizaHon curve. In this WilY, Figure 14.3b is obtolned frum Figure 14.30. Now. if the matltrial is placed in Ii time-harmonic field . thA Afft!ctive J' will btl different from thA Il for the magnetostatic fi e lds. Consider a field H - 110 + H, cos (101' + t/lJ. where H. is the nias magnelolltatic field and H, is tho amplitude of the timt!-hormonle eompont!llt of the lotal field. Lei H, « 1I. ; IhAn the cffecllve Vt:rmcobility of a malt!riol is thA slope of the tanKtmt of thA ffittMncUzatlon t;urvc at f-I •. Skt!teh the effective I' Vt!I'SUS H. for the curve !lhuwn In Figure 14.3a. Compare it with the magnlltoslatic JJ "hown io FigurA 14.3b. and IIhow that the JJ 's in these two Cl:ISt:S are t!Qualto eilch other al " ,.

14.2 Point out the differt!nccs between the foliowillK pairs of IArms: (o) diamagnetic \IS.

paromagnAtic,(bl remanen(;e vs retenlivity. and leI cDflrdvc force vs. cUllrdvily.

14.3 What ore approximate valuAS of the retentivity and thA (;UIlrcivity of Iht! ferrite shown in Fiwure 14.91

ic Circuita

ohtain the

,hove H ilt

, a di~itul standard ~ram may :::omputer

hat (14.22) As shown H axis at g Wb/m1

•

n<i.

on linear ~lds. JJ. is

on the . Now. if enl from I. where .. rmonlc lity of 1:1

-etch the .... itb the ases are

letic vs. vily.

! ferrite

•

Problems 453

14.4 Consider Iht! r.arbon steel. alnico V. ond cunico malerials listed in Table 14.2. Which has the highest IJf:!nnanen l magnellc.Reld strength? Which has Ihe most difficu lty in losing its permanent magnetism once it i~ magnetlzed '~

14.5 A permanent ma~net of fUJIUS 1.5 cm and thickness 0 3 cm is put in It magnetic fie ld Ihal is parallel tn the disk. as in the siluatioll depicted in Figure 14 .7. The torque on Ihe disk is cquallu 1.2 )( 10 ' N m. and the magnetic field is equal 10 10- 1 Whim' Whal is the remanence orthe permanent magnet?

14.8 To writc "onc" in the memory corf! X' YJ shown In Figure 14..11, how should the current pulM'.'I he sen t along the wires? S~ify the pola rity of these puLses.

14.7 Consider the magnetic·core melllory sketchflnln I"'gure 14.11 and the corresponding hysteresis curve for the cores shown in FiHurc 101.9. Now suppose that. because of malfunt:liun in the CiN':lIitry, a positive pulse of amplitude I, which alone is capable of producing the switchinH ml:lHlletic fielri strAn8th HI' is scnt down the line YI lind Ihlll slmulianRolIs\y an identical pulse is scnt down lhaline x •. Assume that all cores arc initially ill thtl 'zero~ sta tR, which corresponds to havinK the magnelir. flux cir· culation point ing either tOWtlro thtl uppflr Iflh or the lower left (usinH Ihtl risht. hUlld fule). What afe the states of aU of thu l.:ures after Ihlts", pu ls8S have passed through?

14.8 Compare tbe hysteresis lonp.~ of two ferritcs shown III Fiijufe PH .6. The curve labeled til is "thinner" than lilul lullelecl 1/2. Which fcrrite core requires leM switchin,ll current? Which ferrite has a better ability 1o Withstand magnetic Interier· pnccs'l

14.$ Consider the lOal!Cnetic r:irr:uil shown in Figure PH.g. The material is steel. and Figure 14.1 7 shows its magnetization curve. The flux density in the air gap is 0.5 Wb/m'. Find the curren t ' needed 10 produce this flux .

14,10 The magnetic circuil shown in FiHure P14.10 i~ made of a materia l with Jl - 600,"".

"'

Find the flux dellIllues HI and B:. and indicate thdr directlOn~ .

"' > fI

"1." r. P1 • •• Hy~teresilloope ror two ~ ferril

P , P, /' , r- TDi I I

100lums i 0 , i , , I ' I ... ___ ='.l .. ------B,

r. p.

lOOOr\1rns

Ipngth~.

P,P,-lIcm P,P, - PIP, _ 10 em P,",", - Ucm /',/',-9cm

J. f O.S cm

cross' N!CtioMlllrella:

":/'. - 12 em' all other branches _ 9 em'

454 14 Magnetic Mllterillis and Magnetic Circuit,

12cm

14.1 t To produce a magnetic flux of 0.5 Wh/ m1 in I h~ air gap of Ihf! ma)(nt:l icci rcuit sho ..... n in 1-"gIlfA PI4.11 . what s hOllln be lhtl magnitude of the current in the coil? Take.u • 200.uo. The cross-seclionaillrt:a of all brandlfls is t:!(juollo 4 emt,

14. 12 Write II cornpuh:r program 10 f'.H rry uut the Iteralinn proceuufc outlined in Example 14.5. First. approximate lin: nonlinear CUr\lft in Fi)!uro H .17b by a polynomial of fiflh order. Than carry out the Iteration five timos to obtAin the rourth-order ap­proximatiull for B.

14. 13 Find the approximate value of Ii in tin: magnetic r:ircuit ~hown in Figure 14.178 for excitatiUIl currenl J _ 15 A instead of 10 A. All othor conditions givtlll remain unchanged. Ca rry out the iteration a liufrident number of timt:1I 10 obtain an ar.cu· racy to thfllhiro digit.

15. QU;

CIIAPTE I< 14.

-------::,;----...

o SL

o

~ f d.:_~,,,I''c. : ,rt..,,, • .uc t..J«s. p~uL 61 or/H'/_t ~4. .p/,.",'"S ..Ie r.~, clMcd J co"pl,,/,'; .

1'4.r~"J..·c : "'A9"e.~iC .,o-'J~ .e -h .l.Cc.. o,..j,".J.&/...,,4. &";It"'~ ~/"cA-ro ... b ~.~ ca,.,d ~w..( C#.,pl"d"

[ rc_et'lc£: MJ.e,.. H· I:> I -He. -~,.:ItuIe, "'.-fe. "4!':~ l3 ~ C4.JLuL 1Y1tIAII..,.,&.

rc/~ ·fivi'l-y: .,.,,~ Tt,...,.".,CIVC /$ ;s wlu/. ~;u..J...",,-Iy MIA..,.. boN. ent4 "I -M .. I'U.J,.~:t'J.J,.o,. Curl/I! (V~ Ae s~1"o"",

{

COerG/Ilc.. /Drce: Tl-c. .-Lue 0,1 /I i" ~c. ".,t'.,e dircc.Jjt>tt .10 hwe.fy +lie. 6 1),.£:1. .

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1+.3 r~ju.J.;II;tr - 0,22. w~/I"'Ia. i eOCrciv/+y. 100 014/....,

14 .4 AA.9AL:.e pC"'''''(VJl"nt ",,,,"c./-" I,'~ 6J,.c"fHo.. MAlc"".t;~ tt.l..nico V

'h/)J~ at.;/f.Ut..ti. +0 .t.IL perft'll}J."'6'1t "'-:'J,.c.lis," I"'IAlp,A/. i& (;u.nic"

I+. '! •

HI

14.'

¥. ,-J

X,

ill ~

Yo ==#==~==!!;

T=I.Z)(/o·l a""", "",)(10·' .. m-/.2,

m_Nrc."a1) ... HI- ".::1 - 11"(;::'$)'-'&''17. ~~ u .J J'."'~ ~ ,.,. "I z: "'I/IL -. 3",/0.j - ~." K 10

.-- B ""iloH • 41'1'>110''')1. ~."7I.ID r:; 0."7/ Wj4"

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u~r ·~'/e .,. 4,.,.". _ L .. /t dt',.,I-1'o ..

... x,Ya-, ,X.Y, .. / j }(,,>t-o; X ... Y,_I }(,Y,-/; x"y.ao; f,y .. _,; ~Yf,e)

111-.8 #/ nlu.irc • .Lt.~. sw;-IJ,,'''J ~"H".I'I~ .

#z. """-s A. bcHc~ UI'Iily ¥o lIIi-H.slw.Nl "''''*''eW(, k1.Ju-/'cr.,., •• ,

~, ' rl"'p'" FiS"" I .... I,.,~ IT.... ,;,"'. PI+. f,

~iNw 8 .. ,. D. ~ W6/~a ~ 1\- J~_o AI,.., I~ -I4c. ./nt. (B I~ -.;~ ... B,~ slid,'

JODe r. ~.1t(5X/t;J) + H, 11'( ~""'/o-'- SKID-')

• , ~

1+.11

_

9·$-SIl./0·' I 4'''/0-' .Z'Url'M.I.·I_~rllti) - ~JJ3.4--

.'. I= 4Z33.4/1_ ~ 4.23 A

I ''''j II = ?) H'~"' I~" ~

, .,u·j,eJt1A

/.O()pI t 5rlIDO_ H.".o.2e.,.H,XAI ••• -. - - _ .... -. (1)

i. DIlf'll: 0 -J.l,~'.Z'-liJat).l .. H,.Z.'II, ____ .@

tP# • ~ ~ ~ • ,uJ4ll('XJo~). ,uH,'rl( IS¥l.I.-4)"'JAN~ /IC ('IC ID"')

... 1'(,- 4-H,*H,-(T/U".,)H, -4.·H7H,- .. ~ ,"".U . .I."~ ® 4.,.( (]) I.", <P ~

5Do-(,.28"'4.4'7.,..D,'.6·")"'1 qHI.-!"Si -33/ H,.~.t.HI.. 7.''''.1''. SfI,D.S-

-'. 8,.:!'-I-I, ... 'DOlC47frlul"" 86A~" a.USS wll/h" (}J'tI/.. S~ -,u.JI,= ~D¥l.PJlID"'J(;JI ~tJ.~'I9' tv""­r,,~(a.ir!J~, a-o .. - 0.$' ..... 1/,... ... -vL Ha-s.../p..-O.S/u..

¢,., ... ¢, ~ ¢ .. * 8. k (,JIVO-4) I;: (8,+11. 'If ~",,-# q. 6.- ~"'6~ 81.u.u" "I Hot.,. ~y~ 8,-81. ~ 8.-_8, S;"" 8;".;,. - l3 i .... ~~ q "0 -IS .. - o.S'~"' .. +H .C.S/~~ 6 , • /-B, - T - -i=. D. t.S W'Y_" .... H,- o.25"/.u.

Fro-~ Z" 3()CI - J.I .. 'tIS,,",I;J~ 1I.r«(I"'-S'UD·J)"'H,.(/J."",./1+~.D') L!:!::~~~~~=~!J

:. 3OoI" ~;.S'tC";'+ ;~!.]lCO.IIS"+ ~.:ttCD. 28 = .!t.)(3.Ii'1S,qo-~

~ 10

:. I - 3 .'3'TS-"/O-'/(3N)~f.""~·") .. 8.32. A

H-O 20 30 40 60 70 eo 90 100

I11AGE 7A.OD.7A,D.DDD,17A~ODDDD.Dt6A FOR 1-1 TO 5 J=l-I B=(1000-.121*H)/3978.9 H=(25815.47-(129360.27-(354706.29-(447912.425-214034.26*B)*B>*B)#.B ) tB PRINT USING 20;" ORDER-",J," B-",B," (Wb/m**2) Hz .. ,H," (AIm)" NEXT I END

ORDER­ORDER"" ORDER­ORDER­ORDER-

o S- .251 (Wb/m**Z) H- Z37~.6 (AIm) 1 B= .179 (Wb/m**Z) H- 2090.4 (AIm) 2 B= .188 (Wb/m**2) H= 2127.8 (AIm) 3 s- .187 (~lb/m**2) H- 2123.0 (AIm) 4 8= .187 (Wb/m**2) H= 2123.6 (AIm)

ftE.. ~ " ().DDS + 1-1 .. )( D.III-/S_IOO -ISDO ... B _ C!~DD. 0./1.1 N'l/IC,u. "'. c.",.,s e/·)= ISDDIt,IT rl-IO·"/o. ct1S - (). 377 .,.. H'·J;'300D (':;"'_J:..~...".e.I4.n)

st,'= (ISCO-C./2/'t1..3000)KIITII./O·Yl1,t,OS = 0.%8' .. H M • ~,oo C. Fff_ "fIVe 1 • • 1'1)

/j (j) ~(!S'"o "4./21 'It ~"'o) I( 4N/O·'V •• IoS- - D.l'S .... 1-1 fa), '-'OD (F,...... J;;gtJ.f'& JI./.,)

G,.,.pAic..L ~,f4.6J.: H -0, B - O. '17" J B - D. 2 ~ H. (1$0()- !t!-- 0.1>11$")/4../21 - S820

... 8'* 0.' ""y,.,a

55

Problems

468

I,

d.

CirCtJilS

v

15 EIcc:troquasistaUc Field.

1f18ur. 15 •• Poynting', theorem fOf quasi­Italic flflkiL

p - - Ii ria . IE x H I - Ii do . ((V<I') x HI

-1IJ, rlV" . II"") x H) - 1IJ, rlV,,· I" x I .. Hl .. " x HI

- - .ff£ dV V . (<I!V x H) - - f1 do· IIlV x II

- -#do "(1 + ~~) Ignoring the dispIHct!lUt!nl current iJDli) t. we find

N ,

p - - #do. "1-S V.[-1l.. do .1)-~ V.I. where I. represents the currenl nowing into the volume through the surface An. whose surface normal is pointing outwArd. Thus, we have established thai the circuit theory's concept of power input is va lid only when the displacement current is negligible.

15.1 Considllr lilt! short-circuited parallcl plate shown in FigurA PIS.I, wilh wiuth w. length t and separation o. Find the zeroth-, fil'llh secondo, and third-order electric lind magnfllic fie lds. Show tll&llhe sum or the quaslslatic solution is equal 10 Ihe ruil wave solution as presented in Chapler 6 for II IIhort-circuit Il'I:Insmission line. Assume that the current at 1. _ 0 is 1fl) _ I. t:os(loIl) and thot a ll fields are funcllons of I and z only.

15.2 Calculate the tolal 7.f!roth-ortlt:r stored e lectric energy In the parallel-platA rAgion shown in Fi)!ure 15.1. What is the zeroth-ordAr IIIOfflci ma)!netit: enef1lY in the some region ?

15.3 Calculate the total first-order slored magMlic eneTtO' in the parallel-plate region shown in t-·igurfl 15.1. What is tilt: first-order stored electric energy in Ihl! same volume? Ot!note Um. the maximum 10la1 fil':lll-nml!r lItored ma)!IIt:lic energy in the

c Fields

. rface ished , the

th w. eclrlC e lull slime lod z

_'8 iun "'me

Iglon oma I the

Problema 469

•

, , --! F======~" 1-1.--- .- -

, 11"1.",. P1S.1 A PIU'llllel plate whh ,horl. r.ll'f':ull CUfrenT , - '. CUll 1",11 at z _ O.

region nnd V~ the maximum lotal zeroth·order stored electric energy in tht: sume vol1lme. Show Ihat

V'I' H"_V,lk.r

Vt!~ J

15.4 Calculate the 1011:\1 second-order Siored eleclric encflO' in tha parallel.plate rCl!lion shown In Figure 15.1. Comparc il with the 7,ernth-order stored electric Anergy. If\? is O.V, 10111!1. whllt clln you say aboul the relolivtl mll,ljl1ltlldA~ of the zeroth-ordel' !>Iored fllecl rlc energy. the firsl-onler slorM mllsnetlc energy. and the second ·order slored electric enATS}'?

15.5 Find Ihe total zeroth·onltlr slured alfl(;lric lind magnetic energies in II parallel plale with 1:1 !>hort-circuil current I - I, cos(OOI'llit ;t - 0 [rAfer 10 Problem 15.1) .

t5.a Find the higher-order stored flleclric and magnetic energies in the parallel.plale reMion shown in Figure P151 up to the thim omltr. If P _ O.IA. compare Ihtl relativA magniludcs of thtlse stored energies Use the 10lal U!roth-ordar stored magnetic energy fOllnd in Problem 15.5 for comparison.

15.7 A coaxial line 'melers lanK is filled WIth II mlllerial char.1cteri;ted by f I:Ind 11. The radii of the innAr lind the outer conductors artl U I:Ind b. rp_'1pectively. The voltage between Ihe coaxial conductors is V I r.0I11"," . !'Ind the zerolh-orderel~troqllasjstatic field £101, IhA r.urrentl lOl, the charge Qt-I, ano thtl fint -order currenll' lI. Express thtls~ in terms or tht: pl:lra melers VI ' fl, n, ", t, 6, and lol l

15.8 Two cunr.flnt ric spherical e lcctrodtls or radii fl and h, respcct ively, urA filled with a material choracltlrized by f and 6 The voltage iJt!tween Ihe e lectrodes is V, cos IOOI'J Find Ihe 7.erOlh·order clectrut.jul:lslstatir. fie ld ,.; ... , the currell t I t-1,lhe charge Q IOI. ilno the first-ord er current/ III Express IheSt! in Iflrms or V" a. b. t. 11. tlud wi.

t5.a Show Ihatlhc time nt:eded 10 r.harge a Van de Craurr Meneralor shown in FigUl"t! 9.241:1 with radius It to a maximuill VOIiH8f1 of V .... by applying a dm'1!inl!l r.llrrent I is equal to bl. RV ... .I1. Calculate Ihe chal'gin)! tillle I if H _ I m, V, .... _ 10~ V lind I - lO 'A.

IS. 4-

Itt): I.cOJwt. ~CI""O.H.,. ()r4~f'I:

CIIAPTeI! I~

... J(-IJ- ~ fJ~""'t' SI·J. O I,..,.,., -Me 'o_d.~y

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2",1 tJrdcr: i i-' • (') U'I) 3

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:. E - -~!1 wp.s/nwr(1 -.;; w~, .. .... ) --i-!T-r SM~ [,U~ J~ ("J~"' •. J ~ -;/!1S/"~ si,,(.oU)

H" -Y!:'-.5Wf: (1- fw!u( ....... ) - -9fJ cos~ '~J'.f) L. pIt .. ~.r np"l..I-, • .,s J &. - ;. ~ 'l (e· j ,,- e i• i ) IJNt il • .. y z...·t (e·jl,+ e i4" ) ; "lbU&

~w _ .... tJrL <Hoe. 4// W&i'& 60'-'"" 1-D 4. ~~t. - clr",~f ~"'iIJ"lJ" LI .. ( .

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~'" -0 H(". -y" CJJ~!Y! s~""t '" ..1.1. ut) =0 ~ U!". tfU.H'!JRMtt.,.. ':(':·J'ilJls;,,1we.(a..W)~'lr.Ji _ :t (~).i2V.2S;,,\;t

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I(¥)L'V.2

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I II: n-. - 2'1fw,L£ V.S;"wt/ ~,./ .. )

56

IS

57

510 16. MagnllloqualiltaUc Field.

Solution: from 116.721. Wt! find

2 .0: 4. X 10 1 )( 2 X to} x 2 x 101 X 1.0 ,.' ... 1 - - 64 X 101 N ... )( 0.5

- tl.5 1Ilt!lric tons

Problems

18. 1 A 911lull circular loop of 5 mm radiull ill placed I m IIWd)' from il 50-Hz power line The voltage induced on this loop is measured al 0 fj mir:rnvolt WhallS the currt:nt on IIII~ power linc?

16.2 Assume thai the Cllrrent on the infinitely long line shown in FiS1ITfl 16.1 IS the Iria ngull1r pu]:re shown in Figure PI6.2. "'I ncllhe Induced vohil~t: un the rectangula r loop. Usc the fo llowing rl::llll : U - Z em. b _ 4 em. and d _ 1 em

18.3 Consider the network shuwn in Figure P16,3. The magneTic flux is increa.~ing al II nllil of 0.5 Wb/s in Ihc direction pnlnling into the IHlf)tlr. Find Ihe readings of thc voltmeters shown.

1S.4 Find the readings of the vullmtl tt:rs shown in Figure VI 6A. Thll mAgnet ic nux is increlillilll( al a ratc of 0.5 Wb/s In the directlnn poinlinl( into the paper.

, ,

PI.",. "16.2

=

d.

oe 00

he I"

I , he

;,

v,

Problem I 511

~----~ kQ ---....... I p;! PI

I I 0' -+ I , lO

/ \ I (000\ I \ 00 I llO

I \ 0 In " / \ J'>, \ ././ "-\ ./ "-

./ "-,,-

P3 HO p.

Flgur. P11.5

U5.5 Four resistors form R circuit 8S shown In I,"gure P16.5. The total magnetir. flux linking thA circuit is increasing at R rale of U.5 Wbls. In the direction pointing out of thtl papor.

(a) Find the dirt:(;lion and magnttmle of the induced current in the circuit. 11>1 Find the readings orlhe \IO]lmclcrs VI and Vl .

18.& Two resistors are conlloctoo by wires to form 8 circuil as shown In Figure P16.6a. The magm:tic flux linking the circuit vorios with lime. Figure P16.6b shows the time variation of the 1l1ltKilCtic flW(, The positi", value of the flux curresponds to the flux dirt:clOO into the pHper. The mognitude of the flux is for a lingle turn of 8 circuit loop Ihal flncircltlll the magnetic flux.

"0

(a) Plut the current/tl) vtlrSUlJ time. Bo sure to mRrk tht: scale orthe curNlnt. (b) Piol the voltage V(I) vcrsus time. Mark the scale.

r----------, I 1111 I

I I

HOO VIII 0.'

I I

I I L-______ _ __ -l

{-I

Figur. Pte.'

,t{WulJcrs'

,

"I

512 16 Magoe luqullli.tat ic Field.

18.7 What illlhe EMF induced on a propeller hlade 111111 is 1.& m long and is rotating al 10,000 rlmin in the earlh's magnetic field (0.5 x 10 I Wb/ m' f'

18.' Find the volta)!!: induced in the rectangular loop shown in Figure 16.1 If it 15 rotating "wu! thc axis parallel tn Ihe z Itxis located at x _ d .,. ~_ AAAlIrnfl thai the angular frequency of the rotation is", and Ihallhe infinitely 100M wire carries a direct currenl of I IiIllj)efCS Show that IhA Induced EMF is not a pure sinusoidal voltase. It is approximately !'inusoidal when d ,. 0

18.8 A magnt:!tic core is made of a mllleriAl whuse hysteresis loop III "hown in Figure P16.9. Noto that thill hysteresis curve is not a MsqU8r8 loop.~ To read the content of the core, two pubes arc applied 10 the wirfUI. The currents gonerate an J I equal 10 200 AJm. Thccore hasan liMa of3 x lU 'm2.

(a ) What is the voltage induced in tht: sensing wire if the corA i.~ originll.l1y III the "zero" .o;la!f! (at point C)? Assume that swtlchlng from C to A is linear with time IIml !hat it is completed In a micro.!JflCOIUl.

(b) What is the voltage imJucoo in the sensing wire If thA t:Ort' is originally 01 Ihe "one" stllle (at point AJ? Assume that .o;wlh.: hinl( {"om A 10 A' is linear wi th lime and thnt it is comptAtoo in U.5I/-s. This voltage is Ihe "noi~A" voltll.~tllrecause it would ideally be zero if the hysteresis loop were tI IJt' I'fecl square.

S#lTlslng wire

Flgur. P11.t t'erntll t;ore OIemory Ilnd III hy~llIrll5llloop.

- 200

,., 'z;.:;.-. 0.20 .......

" 100

- 01

- OJ

H (WIlOOI'» ""r lIqU<I

A' ------A

100 200 H

C

rll meter)

(omperes per mf'ler)

18.10 Find the 10Ial expanllion fOfCtl acting on the surface of lin IIlr-core solt'noid that has 100 lurn.o; of coil and radius 0 - 1 em, length 1/ - 10 t.:ln , lind currenl l _ 10 A.

t • • t1 HAptlll.t Prublem 16.10 for the CMIA in which 100 turns of coil are .... ,ound OVt'f II.

(ueromagnetic core with II- - 100011-0' The current hi 10 lilA, with 0 _ 1 em and f - 10cm.

t8.12 A r:npper IJipt: of radius 0 - 2 cm and Ihkkness d ... 0.1 em Is placed In a IIOIf!noid that has 200 turns per mAtflr and is t'xcitcd by n 1000 I ll. 10·A r:mrenL Thtl conductivity of the cOPllPr is 5.92 x 107 mhol m Ca1culatA the POWt'f per meter dissipaled in Ihe copper pipe.

t S. 13 A tran .• forlllt'r similar to Ihe one shown in Fil!lure 16.11 is made of a steel with relative IJCrmeability equal to 1100. The effective length of the core I! 40 em. and Ihe flU)( density III B - 0.3 WhI m'. N l - 100, N, - 1,000, I , - 60 A,

(a) Find I ., assuming that the transformer 15 an Ideallrnnsfofmer. ( b) Find I ,. using 116.37}. (e) ComPQre the two an.w/ers.

ltie Fields

-olating at

is rotating e angular -ct current tage. II is

in Figure ;ontcnt of , equal 10

Ill)' at the with time

Illy 1:11 tht! with time JeC8USll it

Ur<! meier)

jltmpere! IJ'l'r rn..,tO!r]

1 !hlt! hCll>

\

lei over a ) cm and

noid Ihal Ictillity uf ~ in the

~ relative [ the nux

-Problems 513

11.1. Con!\ider II magnelic circuit lIilllilar to the one shown in Figura 10.11. The efree­tivtlltHlMlh of the core is 0.4 m and its permeability is 2000 I'U' The cross-sectional area of the core is4 x 1O - ~mz.L.elll '" lOA, 12 - 24A,N\ ... 50.and Nz _ 20.

{al Caicu18te the B field in the coro. Give both the direction and the mllgnllu.nA. (h) Jfllisa~ wilhf - 60Hz, whlltIl.I'fl I V ll llmI I V21 ?A9sumelhalthcmo8nctic

flux always staYI iUlhe core without any leakage.

t8.1s The primary coil of II trAnsformer has 150 turus lind the seco ndary coil has 450 turns. The errectiV6 lCIlMlh of the core is 0.5 m ond the flux density in the core is 0.25 Wb/m 2. The transformer is similar to the nnA shown in Figure 16.11 . As~ullle that J 1 ... 60 A ann there is no flux lettkllMe.

(0) Find 12, Ilssuming ideal transformer r:nmli t ion. (bJ Find V 2, assuming ideal lransformtlr condition and V I - cos(120wtJ. (el Find 12 , taking into consideration that the core material has a finite permftahit.

ity equal to 1000 ~. (dJ The hysleT6!'1i!'l loop of thtl core material has Illl area equal to 90 Wb-A/ml.

Willit i!l the power loss due to the hysteresis in the tran!'lformer1 A!'Isume that the core has a cross-sectionAl Are/! Aqua\ to 4 cm~.

18.18 Figure P16. 16 showlI/! magnetizl:ltiull curlle of a core used in a transformer. Notice thttllhtl hystere!lis is negligible in this case and that the curllA Is linear in the ranse o ~ I H I ~ 150 AIm but salural!Ki when H is incrtll:lyed beyund this nmge. Lei us now review Example 16.10. BllCllUse I VI I - wNt "+, we want to tll\e mllximum "+ in order to minimize the number of coils in the Irllnsformtlr. UyiUK Figure P16.16, explain whll l will happen to the shtlpe of the '1'(1) and consequently to the shape of VI(tl if 'I' is too high - for example, if "+ is 50 high liS 10 cOITespond to B _

1.2 Wbfm2.

18.17 Estimate the approximllili power loss attributed to hysteresis in the ferrite core shown ill Figure P16.9 if the core is switched back and forlh between W"lftmM and "oueM states 1000 times in a second . Assume that Ihe core has I:In IIl1erege redius of 6)( 10- 4 m and Ihllt itscross-sectiOIlHI area is 3 x 10 -7 m'.

1.0

05

- zoo - 100

B Iwebers per ~uart: meier)

100 ZOO /ilamperes per meter)

- 0.5

-1.0 Flgur. P11.11

18.18 Show that the mecha nica l lorque required to drive an ac gener::Hor is nol constant with time or, 10 he eXllct, that il consists of II (;onstant limn lind a term Ihat varies sinu:wit.il:llly with timt;! with an angular fr!XIucncy lw. What is the time average of the torque? Express the torque in te rms of the Mila of the winding A, tht! currt!nt I , the magnlilic nllx density B, and tltt! phastl angle" between Ihe voltage and the current. Plot T as a function of t for tt - O.

514 16 Mtlgnetoquasistatlc Fields

11.18 Fi)lure 16.15 depicts an <Ie gtmeralor with (I single oolll>fling rulilleu in a constant mtlgnellc fiflld . It illustrates Ihc operating principle or Ii single-phase ae generator. Let us nuw consider <I three-phalW! flC generator How would you physically (mooge three sets of coilll in oruer to generate Ihree-phillWl p.Jeclricily? To illustrate your design. lIKelch Ii diagram similar 10 Figure 16.15.

tl.20 WhAI is the lulBlmcchanical tOrqu8 naeded lu dri\IQ tho three-phRSfi generatur that you hove designed for Problem 16.19? Expross this torqllA liS 8 function of time in terms of the appropriele parameters. Pial'/' AI II ruuctiuu of timo, and compatfl it wilh that obtained in Problem 16.18. 15 tho \nslantRnooU8 1Ilt!Clllmical torque "smoother" [riDe!! its time-averago value Aucluale ll:!ssj compared witn that for R

"lngle-phllsl:! Htmerator?

11.21 Design a coil configuration similar to the one shown In Figure 16.17. Desig" it [n such a WllY that it will produce a rotating magnAtic field in the armature-slator air gap and that the field will haVA an anguillf speed equal to w/ 2 when fflf.l with the 3-phase current "hown in Fixure 16.18. Prove Inal your cl ~i)ln is correct by draWing inHtlllltlilleous-currenl diagrams .'Iimilar 10 those shown in Figure \6.17.

18.22 Show qllalilativtlly thai the torque genera led by an induction motor may be varied by chlm)ling the resistance ('If IhA windlllKS of the rotor Figure 111.2 \ .~hows lorque cllrves versnll v,lv, wilh three different values of rotnr conductivities. What are the fflillt ivtl lllllgnitudes of "1' <J r. a nd (1.?

11.23 Re fer 10 InA If}'nchrunollS molar shown In FigurA 16.22. What happens when the tOfllUt: angle is negative-that i.e;, when the position of Ihe rOlor magnAtic-moment is ahead of Ihe magnetic field?

11.24 Consider Ihc colis of Ihe magnepllllltl 's track shown In VlgurA 16.25. For the magneplane 10 travel lit a speed of 250 kmlh. what IIhould btl thl:! distance 2, in melAl'II? AssulIle that the power Is providAd by a thrl:!l:!-phose SO·Hz power line,

App.

Sym

A

A II

~.

C

o 01'. , d.

E

• r F18. ( I I,

e H I

/ /. i k k k,

I C. , c· ) -

i..Jw4Jl VPLt...JL ~ l«>p A,I,A1 iJ V. ().~ v (,.~~ t...-... t "".uc '_~p A,A,A, i.

.. •

!iV, ~P.. P, ,

?, r ... t; v. p~ P, If 109f \I" P.,. P, V'" :

EMF .-% .. ~ lit

llt): EMF/ loiJl.

I "1).SIc JH.f'-"1:r) • to,.,A in .,.A& 'pC.l.ldb"c/~UJise.

... V,.:r"7,~'A- o.,,.v V, c - r .. 4,s'¥A.-·a/S' V

al,.,V- v,-V, ..".. V.-v,-V--O.JSV

u.lftltcul. vaJ;t.t ... 4-tJ 4-,. ,I,A,I, ;. V~/, O'" l~u.(.. ~"# .. t M. ~L 1..'1' A,"~I1, ,', I_ •• ~.: • .,..r -rr ... A

V,-ZM'1.:f'JCn-O. v,.,:_'?$'_ e.G ...

~~ -.u..." ..(....., U& h.p A,".A~~JI, ;J V .... D.S Y

... V' - r.I.$'~· Va..~ ... V, .. D·$'-iT"~, s -0.2 Y' /NiMIU wa.., ... ""'-j 1-,- A,4,A+"s i, v"'. D.r ... :. v"'-r-.,.srn-vl • V.J.o.~-;J"«S"·OV

o :: 1/, + o. /l-r + o. or .... V, :. - 0, )./' II

- o. ,t , 1/1. r o,,\.r ~ v,. = _ D . h,t \I

olt ( • . J X ( .... ",

I.f a (/. 1-I.r < t , l. t .t

•

-_. hi'" V

I (.

&,/Vff'· EkF v( Y- .. 4tf'" .. ..J

EMF'_///t)+ 7(/0) ZitI ....

···r

",t) (~.,.) , tl$r(O<&"f )

0

I 1

-) 58

....

•

11:.1 V'~ rw i dl .. tfStl.r - ~8 "d~ .. £ .. iwer'. iX.1f)(IO .. t;ItD. .. ""o"'~("SJ'- ~. 9 "'v ~ At (,_~ -t , -tAt. rtJt.u~ tlVI.!Jlc,. ,. tfJ-t ~ IUV/. J. I ~

d.'~ d ... -f -+,",S..-t: d'-a/. tL .. ~ ... tCOJ.wt

:. 1'. •• £ p .. )' / ' '''4jc.. ~J,.. -u, .u.p d "to .... .,. "l!-

i. l" - -i!b). ("'~'l')-:: b.t.(~:t:tC:;;:) v=-;: . :Of bI.;...s ...... t)[(d·t~~< ... ,.j~(.I· t·;"..,tJ'J

I.4r h QtJJ(d+ U . - "iir (d'-f).. (II .... )' ~/ ... Wt

.', I-t/Mlul ",../ i, ~I 4. ~ ' i" .... ,iAAL ",u-.,c. • 4,t i4 AfFW~v-J.':! J"I't .. ,.~...t".,ltJo I/»c..

/6 'I ( 4) F"f)I't, C.,. A JAB: a". M/...,1 . v.-~" .. ... .dM __ 61"" u tJ-f' ....... ,JV " 4T" ~t. / "/0" V·

(bJ Ft· __ At-A; 458 . D . .,2 ""./ ..... • ~ - .1&1 , •• /e Je/!'" V .. V ·-~t: -.......-r.-- ~. ~_/.. __ 0.0'2-

r

• • • • • .15 • • • •

~ l W",. II 'T4' 1.; F- ~I~. -f~P9tI17'al -z~"I" .. l • (,Df~'".~o-"J( (J;;l .. 7T~ 40"'1( ~ ·I" (), ,.fN

!.i:.l1 P .3-~",""A-t .(I(J.#,.6J~"I<J<»rt.""/O-". (if!-) .... ,,'" 11.1"'11 •• /- 0_ 3fGII.ID-J ItI

.!!!:..B Frp,.. (I{,.I~), < PJ>/i .. (" .. " .. ) 11t:"~::J1:ii};i I ..,.4.." No· ""I .. 2CD./~ -Z4f)O "''''''

Tr'fIJI ~ _ ,r- tJ." Z 1( f) .DDI" S. tlll./oT. ~7/f,' • Wtl.I.I _ H. /'Z. - 1.7rI"O~. tt.Dl. "- ~1f't1.,o··. 2()()t)/Z. • (J,I.?', 1AJJ,l. Ir":" / : .. 2"JL'D~1t4",Ut> ' '''"" , .'0·" II. DZ_ " , ""IIz. • 4 . ~'T~ .

. ', (f"'>1l - ,-'" ,).. {o. ,,?}'. 4,oSil ""/,... - u , ,,,{tI,,?#-)£' I~

(4.) N,I,-"z-I,,'" .z;.~I,-~_,C)_ ,,.. (b) ~f"OWI (l1. •. s..,) .. N,I,-AJz,I, -1P.L"uA)" 1t#II..'D~/fX}D.I •• ~~.J.. ':>" . tt... ~, 81 I" _ "'r.ttl·"~/~tI·-cI'

.-. II Riii!;;l'OO'llllD-8"'t}- S..,A

(C') Mr.-Hz-I, " .. S(lOtI:( ~..,...~~ ~I' (1' , ~'),4.. ric. ,.,.,..,JA,.,;..,i...,~ ~c J.4.

lD.) II,~, - IJ, ~. = HI

S.o -{Jlo:: ,.., •• ,V -') k.,ro ..... 0 = lO"O •• '~';'.(.ro = •. p.' wl./ttt 1.

\ll J "01 '" iJ, w'"i- ( c/"K ",,'J' )

= l-ItJ.~"!i.'.~o,n.'J'f~'D-'I-~ o· ,r II'

" = />.1/ ~ W, •• V h -.·,r= .. JI

59

,,, ) (I )

(t)

V:a.lv, =-- KIlt, -+ v:a, :(9h16.)e.J(,u~fJ::

/J,l, -/J.t. =HI.(O/A).l

+h:I. :lro:-6o- , . It''x ,.r!r / ..... ",.,/0.7) . r, ; ".7' 1\

(J) l' = 1.'["".-~' •. r]·to ; , . • 1'" (,,...,_.« J flr.,w,,,,,,

,,~~) •• •• • • S

7.r Cd (lhU)

~~-~--;l~""":=-~=~-, _-'-_-1-_-'-__ t - Hcll-l

."

+----t-H

• ~ (.)U (I'fl). wr· (68H)I(~"")

"TIt • . slope. oj '1m, "'" v,ttJ, ",;u. 6. ""Io,toA. .1 .. i. -I- "'14..

Fret- "",lOre. Y'",7 J .tJ.BH" D'%"/~O"4- _ 1<#-4- jill'll

,", We 1+4.." 27r-''-''' leJ· <I-". 3_,o-7c /.'Z.'1IC IC-"j , ." ., •• 1 $,;",e ,,.,,.'~" 1000 r, ..... $ pt-r S"""''''. P_/~D"/.~a. '10 .. C.I.3-vv

!£!L Frc,.. (1'-, $() , r c It BX CDSb ... t.)GD~'I.tJ't~ rJ.) - ~[ C .. ,_ ... e.1 (J'-It • .t>j <r> = A:r coS.c. ~ ,4... .. _0 I T _ A:r [''''CPs(2wt)j

~ •• O) ~

;;r~~\.::. "-----''';1''----;:-----'----'''t

,-\ 7

5 N

60

I

~ r. t!:r [!CtlJ.t. 'H~J{ ,,.,4,,. tL) + CDJ (36J't +ti 010 Z4D·). CDse Iltllt .. wI.+IU- J

• A:I (3e.$-'.+ "CO$ (11D-)(AS(Jwt+.(0I0/U·) +UJC Ju4 ~O(.IJG·JJ

• A:r [3~J.c.+ ~5(2wi ",.c.~/b-J(~"'J/b·""») • j4:r (IUJStA.J UJ~CJtt..; we. Tl.e __ o) ~

,

c a.' •

1". 22 Fro_Oj.,.,), (1'"(1-11';/."..) - Ad./t ,J,t. V, > r., ,nI>)':_'''" w,..,,, ... . Fir l.rJv r; (1- fI;/V;) slo~ .... U J" s .... &.11.,.", tn" v,../JIi ,J. • .JJ.. S~ dose,.. t. I .

FI'PAo I-rJHrl. I'. ~I J ..,A,,.. v;./II; cloa -h I, Ii 1>1& II Ij .

,b.ll r-i b4.~ 4. Se,,1.rca.t:;,...·

~ 11" - z50 ""'/lIr - Z~DJ(ID'IJ'OD • ,,9. U- ""h it· ""'I"': v/; • '''ff/'D. /.Is7_

61

Appendix E Answers to Odd-Numbered Problems

Chapter 1

Chapter 2

Chapter 3

Chapter 4

l ,l(a) 5t;3 (b)ll+ j (C) -26+;2 (d) 2.2 j t.41.3 r.olwl;sin wt;1

1 .5 !: 21140. ''') 1 .7Proof t.9 (a) .Jicos (WI !~) (b)4 cos(wl + 0.6)

(e) 3 cos (WI ... ~) ... 4 cos{wL ... 0.6) 1.11 lal - GI ... 59 ... 21

(b) - lOR -t 139 - 4.t (e) -55 {dl Zlt -t 229 I 141 1.13 Proof

1 1.15 ~ (5t 89'" zt) 1.17 Proof ollliltild 1.19 Proofomttl&d

v9' 1.21 Proof omitted 1.23 iwllJ -14) t ... 8(1 ... Jlt] 1.25 lal( - 1 ... j:l)9 ... (1 ... j3) t (b)2:t+/1-119+(1+J1t (e) 5 (d)4t - Il +j3)1.( - 1+j3)t 1.27 Sketch umittod

2.1 - 6yt 3x2,.6z 2.3 Proof 2.5 Proof 2.1 No 2.9 B{y,tj .. O.3(Ie/w) cus (loll t icy) I: 2.11 £ 1 ... Ez• H I + Hz 8 1 t 8 2 and OJ ... Dl I IUptlrposilion theorem 2.13 Proofomiued 2.159 )( E . 1...,8 , V)( H _ r - ;",0. V' B .. OandV' 0 .. P. 2.11 Proof omitted 2.191, f: 2.11 prooromittcd 2.23 UIIIV~ .. 1.13 x 10'

3.1 3.6)( 10 · II" Wlm 2 3.34 x l~W 3.54 .1 )(1013km 3.7(a)rtu.llloc (b)m - ' (el ll8c - ' (d) soc (e) III 3 .9 (al 2.63 m (bl 0.704 m 3.11 Yos, f direction.

~. - .1 ~ E~f 3.13 No, MRxwell's equations not satisfied v /Ag!, _ 2 .uo

&.s~.) Rixhl-hand c ircular polarl7.11f1on (b) Rixhl-ha nd circular polarl7.11lion (e) Loft-ba nd elliptical polarization (d) Linear poiafi7.Atlon fi ,!.pruof omitted 3.19 (61) 1 (hI 1 (c) 1.58 (d) 2. 12 3.211 .34 )( 10 - ' m , aluminum foil is aoout9.5thick 3.23 2.65mW/m2 3.25(aIE._c- o.h e -I'u. (bJII .. f(o .5 - ;o.5) 0 - 0-,. e - /I.I·51 Ie) Skfttch omitteu (d) Sketch omitted 3.270.6)( lU e In

4.1 61.9 0 .4.3 (II c (iI) r (iii) b (Iv) a (v)d (vile " .5 Yes. circulaf shape, 0.30' on each surface [0 il{ the !ensth of t=ach side of Ihe cube,) " .7 hevelled tin,lj!e - 35 -; , 1 . mirror making 70- with i: axis; i polarized 41 ·. 1 Eh 1- '21 f' .. lIl - t= .,. , c ... ·I.

11::11 1- ~1,.:. lIl + e-pu·cil;,' I.IEh l_OS II-:' llt - e-.a'· le -·",..,.".I Ebl_ ;'! 1,.;. 11 1 + e -rlu' lc-'-"'~' ... 11(a) 9t, (hI 75M Hz

520

ChapterS

Chapter.

Chapter '

Chapter

+ J3l ",7

• em ).

: 107

I)m - I

'.

,n

I foil is - jO.5)

Ingle

r, 60

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Answlln to Odd-Numbered Problems

t , l~ztJ

-Pllt: ttl 4.13 cos-' ...,t"", .. - p , f,l cos I

' Ilp! - p! ) • 4.15 lal 80 cm in front of the plate (b) 2V/m 4.171 ·9'.)61r..1 4.19 H...:JI/~""-·~ E' _ Ii c058 - .f: lIinS) H,.qe ,.- . H' - ;-1 I~ I~- .. "-

E' _ Ii cos/J + f sinS) H.,,~ ".-., • ..-. where" - ~~ 4.Z1 x .. 0.87 m. y .. 1.5 m, 12.04t!1I 4.Z3 Proof omitted

5.1 proof 5.3 1.875 kHz 5.5 E - i: En~. H .. - 9 ~ c'". J .. -E.,

-f.-alb:

E • on lower platt!, J ... i ~ c IU on upper plAte 5.7 89.33 kW • 5.9 proof

•

521

5.11 5.26- 10.52 GHz for 2.85 x 1.2ti2 (em) wavogulde. 21.1-42.2 GHz for 0.711 )( 0.355 (crn] waVAguide 5.131 .318 MW 5.15 Ey _ E\ sin (·uJ(JJflk.. 5.170 .. tAn - 'lnnlmbl 5.195.83 CHz

H~ ~ (P.,k.tw~) sin (rxlo) elk,. H ... U E ,rlw~a) cos (rxlaJe'k.. 5.21 Proof omitted 5.Z3 (0.866, 0.5. 2)

whcrok. _lw2 III _1 ... 10)21112 V

5.25 proof omitted 5.Z7 A _ 4.93:i of 7.469 - :It 5.Z9 (II) I I - P ....!e-ik ,., ,

~ Va jI."a 2rVo _Jh 8.12000V 6.3 (allol ) .. (u,lbJ ...... 1000 6.5 I.- t - C ,I - -- e •• • 8.7(8) IV{zll - 2 I V.llcoskz l,l l{zl l" 21~.IISillb l (bl Sketch omitted (c) QI:I 6.9 t_}2 .96 (b) z .. - 0.35).. (c) 24.5% 8.110.342)" 6.13d _ 0.2Scm 8.15(_)1 .26 t ;\.61 {bI0.54 (cll 8.17 (a) 0.67 + ;1.33 (h) 0.15 jO.09 6.19 I'roar omitted 8.2114 .2 kW 6.Z348.6'lb 8.25 Sketch omitted 6.27 Sktltch omitted 6.29 Sketch omitted 8,31 Proof omiH9d

7.1 (0.75, 0.433. 0.5) 7,3 Proof omitted 7.5.t · 6 .. cusO coS<P. i . ~ .. - 8i n~, t . r .. sinS sin¢>. y . 6 • coshin¢>,9 . ;, - COSIP, t . r • cosO, f. . B - - sinS, i . ~ .. 0

7.7 Proof omitted 7.9 Yell, imlJrovcd to 18% 7.11 I! _ (_ P) jkJaz'le-Ib. ." lineAr 7.13 (8)0.314 Vim (b) 0.028 Vim (cIa VIlli 7.15(8) 1 (b) 1.5 (c) 1.ti4 1.17 Six lobes: beAm width . 19.2° olong41 .. 0; beam width .. 2ti.4 ° 010ng¢>_4·1.8° 7.190=4 7.Z1Sketchomilled 7.23(a)-!:IO- (b) 6: 1.414:0: 1.4 14 Vim (cl Sketch omitted (d) Skelr:h nmitl~d

8.1 Uw/UE - {krf « 1 8.3 When bockground is dl:lrk, one sees light sr:attered by the smoke particles. Blue light Is scattered \IIure strongly than red light. Against a bright background. one Sflelllight IJl:Issing through tha arnow. The blue Iighlgelll !lCaucrod. and r&d and yellow lights suffer llt!!1 scattering. 8.5 proof omittoo 8.7360 km in radius 8.9 Train lA movinK toword the inler'lection. 8.11 Band· width _ 59 kHz, 6.7 jd for t kUl resolution 8.13 Circuitlr but opposite hand

8.15 No

Chapter 9

Chapter 10

Chapter 11

Chapter 12

522 An.wen to Odd·Numbered Problems

9.1 Exact: la1 5.5302 x 10· l 0y (bI 5.54244 x 10 Uy Approxlm~IB : laI5.5426" 10· 111y (bl 5.54256 ,,10 ' u V 9.3 x _ 1m plnne, y .. 1.5m pillllc 9.5 ( - PI

1.44 II 10 '1 ' I Vim 9.1 Prooromitted 9.9 SKtllCh omitted 9. 11 Cal (J L

40'll'th (eI 0.2% 9.13 E .. 0 for r < (0.0499) Y/m (bl (J .-!!.- (0.05) VIm

40nh oandh " r < c. E .. '~ ruru '" r < h~ndr ;;J c 9.15 10"

4rf~ ,r' (2 _ e '(? .. 2r t 21J 9.11 1~ -1 (30- 1 1) lIint: I drt! '(1 -t- r )J? .. - e"lr

9.19 3 V. ind!!pondonl of path 9.11 (a) -.!L (bl ~ IC1 .!L(1 t ! 1b)

4'11'te 4T,r: 4n r e

Idl .!L( l -t- ! 1) 9.23 lal 4.03 x 10- 8 G (b) 330 kV Ie) 0.4 rnA 4 '11'l a e b

10.1 _ .! fj 10.3 1.8" 10 -~ N (allraclivt:ll 10.5 (a) 112.4'11' x 10') Vim

• (hl n)( 1.2'11',,10 · aN(repulsivtl) (e) No 10.70.36mm 10.9z .. '" 3. 14 cm 10.11 (a) Vo .. I.R74 X 10' m/s. vo..- 1.867)( lO' m/s. vu• - _ 0 .163)( lO' m/s (hI X(I) .. !S.78 )( 1014)11 - (0.163 x 10')1 Ill, :.:(1) .. ( I .Sfl7 x 10' )1 III (e) x .. - 3.52

x 10 4 m. z .. 3" lO 'l m A ( tl -t- l2) 10.13 I' ruoromilted 10.15 5.3 x 10 10 farad 10.17 (a) d - ,-

<,Q .,Q 1[' , ] (b) QI .. - - , Q, • - - 10.19 2'11' -In !clu) -t- In (hIe) t \ '(2 f)-t-f l f l t 2

10.214.97 x 10 A I 10.28 (a) ciSl2Afo (hI - (j/2Aco (allnclive)

11.1 . 3 .. . ) t <1>2 11.3'" - Yin (blp)lln (binI 11.541(6) .. Y, In ( 'lin i )1 In (lan~). p . .. Yot:4r'n (taU~) Sin9] 11.7 12.3V 11 .9 Sketch omitted

1l .1l 17.5(,IJ. F/ m 11.13 - Q{d' - a2}1(4Talol + rl ' - 2ud coser"] 11 .15 ~ t _ -d' ) t q(o/d) I CJo. where fl, _ II'" + d ' _ 2rd c0s9 )'IJ.

4Tt HI Rl 4l1'tr

( " ",' )'" H, _ r l I d ' - 2 d case 11 .17 IPrJ'/ 12rtfd - bll where d - cr/h, "I\ractiv~

11.19 (d/ olq l/14Tfjd - bfl. where d - cri b, aUrar.live 11.21 200 slnI2'11'x/ u) sin hI2Ty/ o)/ 8inh(2Tb/ ol

/ 2::- 1 [Sin\ nTx/ oj sinh(n'll'y/oj sin[nry/b1 !linhf nn/ b)l

]1.23 (400 '11') - + . n sinh(nrb/ o) Innh(nTo/ bl .-..

11 .2.5

V, (' (-) (; ;) ;

_.!V!!:"'''-' _ (e) 0 " • ~

(~ ~) ~ V""

(d) Q

12.1 C .. A\,) -t- tJ/2d. C _ A{o) + cJl )l2d; C in parallel wilh G 120.3 (_) lin {olp)l

I I'n (bIn) In (clb)] (2To\t) Ibl lill (aJb)l(2'1fcJ1t) + lIn (blp)l(21rotl) (c) - , 2"" 0 \ (11

•

Chal

ChI

ChI

ChI

•

Chapter 13

Chapter 14

Chapter 15

<,)

Chapter 16

Anlwers tn Odd·NumbeNNl Problems 523

tz.$O.92 x 10 ' mho/m 12.7100V

12.9 (~\ )C 100% 12.11 12.3 O·m 12.13 Sketch omiltAd jJl + oJ

12 15 p .. 600 C! _ 50 t 50 ) . y lts cll(y'" 6 - roY1J'f I pO - 7OrJ d)lJIIl d[(y ... fil~ + {7otl'!2j'

where d .. 1t70)2 ... V11I2, fO .. (501 ~1f{70)1 t YII12

13 .1 ! 2 .J2 fI{ .... b) 13.3 H .. (- Illy fur I y I < ~, II .. ( - .tlJ(d/2) for Y ;::. d/2. 2

H ... J{d/2J fur y < (- dl21 13.5 (a) 0 (bl I(l - (l2)/I2rP(b2

- (II)] (cjIlZ,.-p 13.7(1I){ - t)fo{dyJ/14';uJ t y}lI'l] (b)( -I)U(4TO) (clQ (d)( tjll{4Tnl

13.9 Zo" T 0 .270

13.11 H All .. t 2!. 1m

J dx 12 4..- -G [0 + Ih - xft

H ili.:" t I b - a

4.

IICD .. H"D

arlllI

HIM .. t Jib ... 1,1) I' <Ix 4..- __ II ' (b + 0)1)112

13,132.8MHz 13.15(a) x .. O.04rn,z. - O.OO725rn (b) - 20.S' 13. 17 1.33:0: 10" N/m Impulsive) 18.19 I.} Loop should be placed horizontally or vertically in the Mst·\WSt direction (b1 7.ISS)( 10·) N-m (cl Vertically in nurth-south direction

I' 13.21 {II} H • .Je.... (hI UH

• E- (eI 5)( 10 . 8 Him 13.28 2.14)( 10 • ~ 2TlIt 16w

'" [1 c4

In (db) c! c! + b1 J

Joulelmeter 13.za - - + In (blu) ~ ~ z t -~ + t 1 2 ... 4 (c-h) c- b 4fc- h)

14.1 Sketch 14.30 21 weber/ fll3• 100 AIm 14.5 0.11 wel~r/mJ 14.1 x,y, - 1. X'Yl _ O. x,y, _ I. x.y. - O. x,)') · 1. xlY) . 1. x.)'t - 1 14.!14.233A 140 .11

B.32A '4.13 0.29 weber/ mt

15 .1 [I" _ 0, 1-I1t) _ - Yl,cos(wl}/w. [1'1. il..,/oI7.lIin(wIVw. ~1I. 0: ~ - O.

Hill _ Y I fI-J'f/oi Hl t CQII(wIVw: £III - tl~V( Hl Hl r sin(wIVw; K II - 0

15.3 uW _ ~ ... l f l\1 (~) sln1lwl) (i) owi·. U~l - 0 15.5 U':' - o. • I ' uw - -.....; cos1lwtXwm!1 2w

15.7 £~'I • v . cO!l(wtVIP In (b/ ull. l it) _ Z..-V",,, cos(wlj/ln [b/o~ ott)_ 2 ... Vaf i cos(""tj/ln (bIoi. 11'1 - 2 ... V tI""" s in[wlj/ln (bIoi 15.9 11.1 sec

16.1 101 A 16.3 VI 0.25 V. Vt _ -0.15 V. V, - -0.25 V 16.5 (.}O.12~ rnA (clockwise) (hi V,. 0.25 V; VJ - 0.625 V It1.7 58.9 mV 16.9(.) -O.llV (b) -0.012 V 16.11 0.395)( 10 · l N le.13t_1 6A (b) 5.91 A 16.15(1I)20A (b)33U cos {120..-tl (c)19.78A (d)1 .08W 16.110.18mW 18.19 Sketch umitted 16.21 Sketch omitted 18.23 It b8C1l1l11:19 a generator.