Haus Melcher solutions manual

MIT OpenCourseWare http://ocw.mit.edu Haus, Hermann A., and James R. Melcher. Solutions Manual for Electromagnetic Fields and Energy . (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-NonCommercial-Share Alike. Also available from Prentice-Hall: Englewood Cliffs, NJ, 1990. ISBN: 9780132489805. For more information about citing these materials or our Term s of Use, visit: http://ocw.mit.edu/terms. Solutions Manual , Electromagnetic |“Fiekis andEnergy _ .me a oe bane Solutions Manual Electromagnetic Fields and Energy Hermann A. Haus James R. Melcher Massachusetts Institute of Technology ill PRENTICE HALL, Englewood Chffs, New jmey 07632 PREFACE TO SOLUTION MANUAL We are fortunate that electromagnetic aspects of engineering systems are ac­ curately described by remarkably concise and general laws. Yet, a price paid for the generality of Maxwell's equations is the effort required to make these laws of practical use to the engineer who is not only analyzing, but synthesizing and invent­ ing. Key to the maturation of an engineer who hopes to use a basic background in electromagnetic fields for effectively dealing with complex problems is working out examples that strike the right balance among a number of interrelated objectives. First, even in the beginning, the examples should couch the development of skill in using the mathematical language of field theory in physical terms. Second, while be­ ing no more mathematically involved then required to make the point, they should collectively give insight into the key phenomena implied by the general laws. This means that they have to be sufficiently realistic to at least be physically demon­ strable and at best of practical interest. Third, as the student works out a series of examples, they should form the basis for having an overview of electromagnetics, hopefully helping to achieve an early maturity in applying the general laws. In teaching this subject at MIT, we have placed a heavy emphasis on working out examples, basing as much as 40 percent of a student's grade on homework solu­ tions. Because new problems must then be generated each term, this emphasis has mandated a continual search and development, stimulated by faculty and gradu­ ate student teaching assistant colleagues. Some of these problems have become the "examples," worked out in the text. These have in turn determined the develop­ ment of the demonstrations, also described in the text (and available on video tape through the authors). The problems given at the ends of chapters in the text and worked out in this manual do not include still other combinations of geometries, models and physical phonemena. These combinations become apparent when the examples and problems from one chapter are compared with those from another. A review of the example summaries given in Chap. 15 will make evident some of these opportunities for problem creation. After about two decades, the number of faculty and teaching assistants who have made contributions, at least by preparing the official solutions during a given term, probably exceeds 100, so individual recognition is not appropriate. Prelim­ inary versions of solutions for several chapters were prepared by Rayomond H. Kotwal while he was a teaching assistant. However, finally, the authors shared re­ sponsibility for writing up the solutions. Corrections to the inevitable errors would be appreciated. Our view that an apprenticeship of problem solving is essential to learning field theory is reflected in the care which has been taken in preparing this solution manual. This was only possible because Ms. Cindy Kopf not only "Tex't" the manual (as she did the text itself) while taking major responsibility for the art-work, but organized and produced the camera-ready copy as well. The "Tex macros" were written by Ms. Amy Hendrickson. iii SOLUTIONS TOCHAPTER 1 1.1THELORENTZ LAWINFREE·SPACE ·1.1.1 ForVi=0,(7)gives andfrom(8) V=J-2~Ez(1) (2) yso v=2(1X10-2)(1.602X10-19)(10-2) 31 (9.106X10-31) =5.9x10m/s z(3) -t-o.:::::~--------X xy Figure91.1.3 1.1.2 (a)Intwo-dimensions, (4)gives mcPez=-eEdt2 z md2ey__E dt2-ey so,because Vz(0)=Vi,whilevy(0)=0, dez e-=--Et+v"dtmz•Figure91.1.3 (1) (2) (3) 1 1-2 Solutions to Chapter 1 de" = -!....E t dt m" (4) To make es(O) = 0 and e,,(O) = 0 (5) (6) (b) From (5), es = 0 when and at this time 1.1.3 The force is so, f =0 if Eo = viPoH o. Thus, (7) e = -.!...-E (vi 2m)2 " 2m" eEs (8) (1) dvs =0 dv" =0 dv" =0 (2)dt dt ' dtI and vs , vII and v" are constants. Because initial velocities in :z: and 'Y directions are zero, Vs = v" = 0 and v = vii•. 1.1.4 The force is so (2) and mdvs dvs ~ =ev"poH o ~ dt =WeV" (3) md~ d~ ~ = -evspoH o ~ ""dt = -WeV s (4) where We = epoHo/m. Substitution of (3) into (4) gives (5) Solutions toChapter1 1-3 Solutions aresinwetandcoswet.Tosatisfytheinitialconditions onthevelocity, inwhichcase(3)gives:(6) (7) Furtherintegration andtheinitialconditions onegives (8) (9) xz ---H-Eo 0-- yz / x Figure91.1.4y 1.2CHARGE ANDCURRENT DENSITIES 1.2.1 Thetotalchargeis (1) 1.2.2 1-4 Solutions to Chapter 1 Integration of the density over the given volume gives the total charge (1) Two further integrations give (2) 1.2.8 The normal to the surface is ix, so (1) 1.2.4 The net current is (1) 1.2.5 (a) From Newton's second law where (b) On multiplying (1) by Vr , v r dEr -­-dt (1) (2) and using (2), we obtain (3) Solutions toChapter1 (c)Integrating (3)withrespecttotgives 122"mvr+eEoblner=Cl Whent=0,Vr=0,er=bsoCl=eEoblnb and ~mv:+eEobln; =° Thus,1-5 (4) (5) (6) (d)Thecurrentdensityis Jr=p(r)vr(r)=*p(r)=.l(r) Vrr Thetotalcurrent,i,mustbeindependent ofr,so .1=-'-r211Tl anditfollowsfrom(6)and(7)that , m p(r)=21frl2eEobln(b/r) 1.3GAUSS' INTEGRAL LAWOFELECTRIC FIELD INTENSITY(7) (8) (9) 1.3.1 (a)Theunitvectorsperpendicular tothe5surfaces areasshowninFig.81.3.1. Thegivenareaelements followfromthesameconstruction. (b)FromFig.81.3.1, (1) r=vz2+y2 Thus,theconversion frompolartoCartesian coordinates gives(2) (3) 1-6 z ySolutions toChapter1 -iy FlpreSl.3.1 (c)Onthegivensurface,thenormalvectorisixandsotheintegralisofthez component of(3)evaluated atz=a. ffaA,11faaEoE·dal,,,=a=-2- 2y2dydz 'll"Eo0-aa+ =~tan-1 (~)r=~(~+~)=A, 2'11" a-a2'11"444(4) Integration overthesurfaceatz=-areversesboththesignofE",andofthe normalandsoisalsogivenby(4).Integrations overthesurfacesat11=aand y=-aarerespectively thesameasgivenby(4),withtherolesofzandy reversed. Integrations overthetopandbottomsurfaces makenocontribution becausethereisnonormalcomponent ofEonthesesurfaces. Thus,thetotal surfaceintegration isfourtimesthatgivenby(4),whichisindeedthecharge enclosed, A,. 1.8.2 Ontherespective surfaces, {l/a2 E·da=-q- 0 4'11"fo1/62(1) Onthetwosurfaces wheretheseintegrands arefinite,theyarealsoconstant, so integration amounts tomultiplication bytherespective areas. (2) 1-7 Solutions to Chapter 1 r Figure 91.3.2 1.3.3 (a) Because of the axial symmetry, the electric field must be radial. Thus, inte­ gration of Er over the surface at r= r amounts to a multiplication by the area. For r< b, Gauss' integral law therefore gives (1) 2nl€oE r = 10(' 10r" 10r pdrrdl/Jdz = 21rl 10r p;2,.s dr Po,.s Er = 4€ob2ir < b For b < r < a, the integral on the right stops at r = b. b<r<a (2) (b) From (17) (3) (c) Because it is uniform there, integration of the surface charge density given by '(3) over the surface r = a amounts to a multiplication by the surface area. (4) /. That this is the negative of the net charge within is confirmed by integrating over the enclosed charge density. ( pdV = (' r" r po(-)2rdl/Jdz = -n{~ob2 (5)1v 1010 10 b 2 1-8 Solutions to Chapter 1 (d) As shown in the solution to Prob. 1.3.1, I.. = (zlx +yl"¥ )h/ z2 +y2j (6) and substitution into E=.~ {(r:/b2)I.. j r<b (7)4Eo (b /r)l.. j b<r<a indeed results in the given field distribution. (e) For the surfaces at z = ±c, da=±ixdydzj E .n = E,.,(z = ±c) (8) while for those at y = ±c, da = ±I"¥dzdzj E· n = EII(y = ±c) (9) The four terms in the given surface integral are the integrations over the respective surfaces using the field given by (d) evaluated in accordance with (8) and (9). According to (I), this integral must give the same answer as found by integrating the charge density over the enclosed volume. This has already been done and is given by (5). . 1.3.4 (a) For r < b, (1) gives (1) Thus, _ por.Er-3Eo, r<b (2) (3) Er = -3 [b3Pb b3 j b<r<a (4)1 -2-+(r- 2")Pa] Eo r r (b) At r= a, (17) can be evaluated with n = i..,Ea =0 and E b given by (4) 1 [b3Pb b3 ]u. = -3 ---;;F +(a -a2 )Pa (5) (c) For r < b, Er is still given by (2), while for b < r < a, (3) has an additional term on the right 471'"b2uo • Thus, 3 3 2 1 [bPb b] buoEr = ---+(r --)Pa +-j b<r<a (6) 3Eo r2 r2 Eor 2 Then, instead of (5) we have 1 [b3Pb b3 ] b2 uou. =---+(a --)Pa -- (7)3 a2 a2 a2 Solutions toChapter 1 Pa Figure81.3.4a1-9 1.3.6 Usingthevolumedescribed inExample 1.3.2,withtheuppersurfacebetween thesheets,thereisacontribution tothechargeenclosed fromboththelowersheet andthevolumebetween thatsheetandtheposition, z,oftheuppersurface. Thus, from(1) (1) andthesolution forEzgives (2) Notethatthechargedensityisanoddfunction ofz.Thus,thereisnonetcharge between thesheets.Withthesurfaceabovetheuppersheet,thefieldgivenby(1) withtheintegration terminated atz=8/2isjustwhatitwasbelowthelower sheet,Eo. 1.3.6 Withtheunderstanding thatthechargedistribution extendstoinfinityinthe yandzdirections, itfollowsfromarguments alreadygiventhattheelectricfieldis independent ofyandzandthatthatpartofitduetothechargesheetscanresult onlyinazdirected electricfield.Itthenfollowsfrom(1)thatiftheregionsabove andbelowthechargesustainnoelectricfieldintensity, thenthenetchargefrom thethreelayersmustbezero.Thus,notonlyis (1) butalso, (2) Fromtheserelations, itfollowsthat (3) 1-10 Solutions toChapter1 1.3.'1 Thegravitational forcehasacomponent intheedirection, -MgsinQ.Thus, thesumoftheforcesactingontheupperparticleintheedirection is Itfollowsthat,fortheparticletobeinstaticequilibrium,(1) e=41rfoMgsinQ(2) 1.4AMPERE'S INTEGRAL LAW 1.4.1 Evaluation of(1)iscarriedoutforacontourhavingtheconstant radius,r, onwhichsymmetry requires thatthemagnetic fieldintensity beconstant andin the¢direction. Because thefieldsarestatic,thelasttermontherightmakesno contribution. Thus, (1) Solvingthisexpression forH<f>andcarrying outtheintegration thengives (2) 1.4.2 (a)Thenetcurrentcarriedbythewireinthe+zdirection mustbereturned in the-zdirection onthesurfaceatr=a.Thus, (1) (b)Foracontourattheconstant radius,r,(1)isevaluated (withthelastterm ontherightzerobecause thefieldsarestatic),firstforr<bandthenfor b<r<a. r<b (2) b<r<a (3) Solutions to Chapter 1 1-11 (c) From (1.4.16)' H~ -H: = K z :::;. H~ = Kz + H: (4) This expression can be evaluated using (1) and (3). 2 2 Ha =_b Jo + Job = 0 (5) '" 2a 2a (d) In Cartesian coordinates, Thus, with r = ..jx2 + y2, evaluation of this expression using (2) and (3) gives (7) (e) On x = ±c,H· ds = ±H . i y while on y = ±c,H· ds = =r=H . i x so evaluation of (1) on the square contour gives (8) The result of carrying out this integration must be equal to what is obtained by carrying out the surface integral on the right in (1). (9) 1.4.3 (a) The total current in the +z direction through the shell between r = a and r = b must equal that in the -z direction through the wire at the center. Because the current density is uniform, it is then simply the total current divided by the cross-sectional area of the shell. (1) (b) Ampere's integral law is written for a contour that circulates around the z axis at the constant radius r. The fields are constant, so the last term in (1.4.1) 1-12 Solutions toChapter1 iszero.Symmetry arguments canbeusedtoarguethatHisq,directed and uniform onthiscontour, thus 21f'rH.;=-I=>H.;=-1/21rr; 0<r<b (2) I 2[1 (r2-b2 )1]21rrH.;=-I+1f'(a2_b2)1f'(r2-b)=>H.;=I-21f'r+a2_b221f'r(3) (c)Analysis oftheq,directed H-fieldintoCartesian coordinates gives Hz=-H.;sinq,=-H.;y/Yx 2+y2 Hy=-H~cosq,=H.;x/Yx2+y2 wherer=yx2+y2.Thus,from(2)and(3), l(yix-xi:r){1; 0<Yx2+y2<b H=21f'(x2+y2)1-(z3~~;.t); b<r<a(4) (5) (6)(d)Inevaluating thelineintegralonthefoursegments ofthesquarecontour, on x=±c,dB=±i:rdyandH·dB=±Hy(±c, y)dywhileony=±c,dB=Tixdx andH·dB=THz(x,Tc)dx.Thus, faH.dB=[CcHy(c,y)dy+f:c-Hz(x,-c)dx +[Cc-Hy(-c,y)dy+ [CcHz(x,c)dx Thisintegralmustbeequaltotherighthandsideof(1.4.1),whichcanbe evaluated inaccordance withwhether thecontourstayswithintheregion r<borisclosedwithintheshell.Inthelattercase,theintegration overthe areaoftheshellenclosed bythecontourisaccomplished bysimplymultiplying thecurrentdensitybytheareaofthesquareminusthatofregioninsidethe radiusr=b. c<b/V2 b/V2<c<b b<c<a/V2 (7) wherea=cos-1(c/b). Therangeb/V2<c<biscomplicated bythefact thatthesquarecontouroverlaps thecircler=b.Thus,theareaoverwhich thereturncurrentintheshellpassesthrough thesquarecontouristhearea ofthesquare(2c)2,minustheareaoftheregioninsidetheradiusb(asinthe lastcasewherethereisnooverlapofthesquarecontourandthesurfaceat r=b)plustheareawherethecircler=bextendsbeyondthesquare,which shouldnothavebeensubtracted away. Solutions to Chapter 1 1-13 1.4.4 (a) The net current passing through any plane of constant z must be zero. Thus, (1) and we are given that K za = 2Kzb (2) Solution of these expressions gives the desired surface current densities Kza I K _ I (3)= 1I"(2a + b); zb -21I"(2a + b) (b) For r < b, Ampere's integral law, (1.4.1), applied to the region r < b where the only current enclosed by the contour is due to that on the z axis, gives -I 211"rH", = -I~ H", =-j r < b (4) 211"r In the region b <r< a, the contour encloses the inner of the two surface current densities as well. Because it is in the z direction, its contribution is of opposite sign to that of I. 2a 211"rH", = -I + 211"bK zb = -(--b)I (5)2a+ Thus, H -_~(~). b<r<a (6)'" - 211"r 2a + b' Note that if Ampere's law is applied where a < r, the net current enclosed is zero and hence the magnetic field intensity is zero. 1.4.5 Symmetry arguments can be used to show that H depends only on z. Ampere's integral law is used with a contour that is in a plane of constant y, so that it encloses the given surface and volume currents. With z taken to be in the vertical direction, the area enclosed by this contour has unit length in the x direction, its lower edge in the field free region x < -8/2 and its upper edge at the location z. Then, (1.4.1) becomes i H.dS=Hx(Z)=-Ko+!Z Jydz (1) G -0/2 and for -8/2 < z < 8/2, z 2Joz Jo [2 2]Hx = -K o + --dz = -K o + -z -(8/2) (2)! -0/2 8 8 while for 8/2 < z, (3) Solutions to Chapter 1 1-14 1.5 CHARGE CONSERVATION IN INTEGRAL FORM 1.5.1 Because of the radial symmetry, a spherical volume having its center at the origin and a radius r is used to evaluate 1.5.2. Because the charge density is uniform, the volume integral is evaluated by simply multiplying the volume by the charge density. Thus, 2 d[4 s ()] rdpo47rr J. +--7rr p t =0 => J. =--- (1)r dt3 r 3dt 0 1.5.2 Equation 1.5.2 is evaluated for a volume enclosed by surfaces having area A in the planes z = z and z -= O. Because the the current density is z directed. contributions to the surface integral over the other surfaces, which have normals that are perpendicular to the z axis. are zero. Thus, (1.5.2) becomes (1) 1.5.3 From (12), a~. = -n . (JG -Jb) = -(0) + J:(z = 0) = Jo(z, y) cos(wt) (1) Integration of this expression on time gives Jo(z,y) . 0'. = Slnwt (2)w where the integration function of (z. y) is zero because, at every point on the surface, the surface charge density is initially zero. 1.5.4 The charge conservation continuity condition is applied to the surface at r = R, where Jb =0 and n =il" Thus, Jo(tIJ, z) sinwt + a~. = 0 (1) and it follows that 0'. = -it Jo(tIJ,z)sinwtdt= Jo(tIJ,z) coswt (2) o W 1.6 FARADAY'S INTEGRAL LAW 1-15 Solutions to Chapter 1 1.6.1 (a) On the contour y = sx/g, · d· d(. dy.) d (. s.)ds = dXIx + yIy = X Ix + -dIy = X Ix + -Iy (1) X 9 (b) On this contour, while the line integral from (x,y) = (g,s) [from b ---+ c] to (O,s) along y = s is zero because E . ds = O. The integral over the third segment, [c ---+ a]' is (3) so that fE . ds = Eos -Eos = 0 (4) and the circulation is indeed zero. 1.6.2 (a) The solution is as in Prob. 1.6.1 except that dyjdx = 2sxj g2. Thus, the first line integral gives the same answer. (1) Because the other contours are the same as in Prob. 1.6.1, their contributions are also the same and the net circulation is again found to be zero. (b) The first integral is as in (b) of Prob. 1.6.2 except that the differential line element is described as in (1) and the field has the given dependence on x. (Note that we would now get a different answer, Eosj2, if we carried out this integral using this field but the straight-line contour of Prob. 1.6.1.) From b ---+ c there is again no contribution because E . ds = 0 while from c ---+ a, the integral is {" -Eo~l_ dy=_EoxYI_ =0 (3)J 9 x-a 9 x-ao which makes no contribution because the contour is at X = o. Thus, the net contribution to the closed integral, the circulation, is given by (2). 1-16 Solutions to Chapter 1 1.6.3 (a) The conversion to cylindrical coordinates of (1.3.13) follows from the argu­ ments given with the solution to Prob. 1.3.1. (1) (b) Evaluation of the line integral amounts to recognizing that on the four seg­ ments, (2) respectively. Note that care is taken to take the endpoint of the integrals as being in the direction of an increasing coordinate. This avoids taking double account of the sign implied by the dot product E . dB. (3) These integrals become (4) and it follows that the sum of these contributions is indeed zero. 1.6.4 Starting at (z, y) = (s,O), the line integral is £E . dB = ldEz(z, O)dz +1d EII(d, Y)d y-1d Ez(z, d)dz -ldEll (0, y)dy +l'Ez(z, s)dz -1'EII(s, y)dy (1) This expression is evaluated using E as given by (a) of Prob. 1.6.3 and becomes i [ld 1d 1d AI dz Y Z E .dB=-- -+ dy - dz o 211"£0' Z 0 rJ.2+y2 0 z2+ rJ.2 dy z y ] --+ dz- d-0 l, d y l' 0 z2+ s2 l' 0 S2 + y2 Y -(3) Solutions toChapter1 y cos<piy 181 II r--:..:coildaI,. :J:=d Fleure81.6.51-17 (1)1.6.5 (a)InviewofFig.S1.6.5,themagnetic fieldgivenby(1.4.10) B=I",(J...-) 211"r isconverted toCartesian coordinates byrecognizing that • •A.. A.. -'11I+z. -/22(2)I",=-sm'f'lx+cos'f'1;y= x.J l;yir=Vz+'11Vz2+'112Vz2+'112 sothat(1)becomes B=i.-[-'11Ix+zI] (3) 211"z2+'112z2+'112;y (b)ThesurfaceofFig.1.7.2a,shownintermsofthez-'11coordinates byFig. S1.6.5,canbeusedtoevaluate thenetfluxasfollows. r rVR~-d.~ >'1=Js"'oB.da=lJo -PoH",(d,y)dy l·lVR~-d.~() I" (4) =_Po' -ydy=Po'In(R/d) 211"0 d2+'112211" Thisresultagreeswith(1.7.5),wherethefluxisevaluated usingadifferent surface.Justwhythefluxisthesame,regardless ofsurface,isthepointof Sec.1.7. (c)Thecirculation followsfromFaraday's law,(1.6.1), 1E.ds=_d>'1=_po',n(R/d) di (5)Ja dt 211" dt (d)ThisfluxwillbelinkedNtimesbyanNturncoil.Thus,theEMFatthe terminals ofthecoilfollowsfrom(8)as tab=P;':In(R/d) ~~ (6) 1-18 Solutions to Chapter 1 1.6.6 The left hand side of (1.6.1) is the desired circulation of E, found by deter­ mining the right hand side, where ds = i,.dzdz. 1 E. ds = -~ r/SoB.. ds 0' dt 1s d J'/21V1 = --d /SoH,Az, 0, z)dzdz (1) t -1/2 0 dHo = -/Sowl""dt 1.6.7 From (12), the tangential component of E must be continuous, so nx (EG -Eb) =0 => 1:; -E1 =0 => E; = E1 (1) From (1.3.17), foE; -foE2 = 0'0 => ~ = 0'0 + E2 (2) f o These are components of the given electric field just above the 11 = 0 surface. 1.6.8 In polar coordinates, (1) The tangential component follows from (1.6.12) (2) while the normal is given by using (1.3.17) Er(r = R+) = 0'0 cosq, + Eosinq, (3) f o 1.1 GAUSS' INTEGRAL LAW OF MAGNETIC FLUX 1.7.1 (a) In analyzing the z directed field, note that it is perpendicular to the q, axis and, for 0 < fJ < 11"/2, in the negative fJ direction. B = Ho(C08fJi .. -sinfJio) (1) (b) Faraday's law, (1.6.1), gives the required circulation in terms of the surface integral on the right. This integral is carried out for the given surface by simply multiplying the z component of B by the area. The result is as given. Solutions to Chapter 1 1-19 (c) For the hemispherical surface with its edge the same as in part (b), the normal is in the radial direction and it follows from (1) that PoH . ds = (PoH ocos O)r sin OdOrdtP (2) Thus, the surface integral becomes (3) so that Faraday's law again gives (4) 1.7.2 The first only has contributions on the right and left surfaces, where it is of the same magnitude. Because the normals are oppositely directed on these surfaces, these integrals cancel. Thus, (a) satisfies (1.7.1). The contributions of (b) are to the top and bottom surfaces. Because H differs on these two surfaces (:.c = :.c on the upper surface while :.c = 0 on the lower one), this H has a net flux. 1 H.ds= AHo:.c (1)Is d As for (b), the top and bottom surfaces are where the only contributions can be made. This time, however, there is no net contribution because H does not depend on :.c. Thus, at each location y on the upper surface where there is a positive contribution, there is one at the same location y on the lower surface that makes a contribution of the opposite sign. 1.7.S Continuity of the normal flux density,(1.7.6), requires that IJoH: - IJoH l = 0 => H: = Hl (1) while Ampere's continuity condition, (1.4.16) requires that the jump in tangential H be equal to the given current density. Using the right hand rule, H; -H2 = K o => H; = K o + H2 (2) These are the components of the given H just above the surface. 1.7.4 Given that the tangential component of H is zero inside the cylinder, it follows from Ampere's continuity condition, (1.4.16), that H",(r = R+) = Ko (1) According to (1.7.6), the normal component of PoH is continuous. Thus, poHr(r = R+) = poHr(r = R_) = Hl (2) SOLUTIONS TO CHAPTER 2 2.1 THE DIVERGENCE OPERATOR 2.1.1 From (2.1.5) DivA = 8(A z ) + 8(A,,) + 8(A z ) 8z 8y 8z Ao [8(2) 8(2) 8( 2) =--z +-y +-z (1) tJ.2 8z 8y 8z 2AO( )= tJ.2 Z+y+Z (2) 2.1.2 (a) From (2.1.5), operating on each vector V.A= Ao[~(y)+~(z)] =0 (1)d 8z 8y V· A = Ao [~(z) -~(y)] = 0 (2)d 8z 8y V· A = Ao [~(e-1c" cos kz) -~(e-1c" sin kz)]8z 8y ~) = Ao[-ke-1c" sin kz +ke-1c" sin kz] = 0 (b) All vectors having only one Cartesian component, a (non-constant) function of the coordinate corresonding to that component. For example, A = ixf(z) or A = iyg(y) where f(z) and g(y) are not constants. The example of Prob. 2.1.1 is a superposition of these possibilities. 2.1.3 From Table I 18 18A", 8AzV·A= --(rA r ) +--+- (1) . r8r Thus, for (a) Ao [18(2V·A =--- rd r 8r = ~O[2coS24J for (b) 18 r 84J 8z ) 8. ] cos24J --(sm24J)84J (2) -2cos24J1 = 0 18 . V·A = Ao[--rcos4J ---sm4J] = 0 (3)r8r r 84J while for (c) Ao 18 AoV·A=---r3 =-3r (4)tJ.2 r 8r tJ.2 1 2-2 Solutions to Chapter 2 2.1.4 From (2), DivA = lim _1_ 1 A. ds 4V-+O~V ls (1) Following steps like (2.1.3)-(2.1.5) t A.da~~~~z[(r+ 6;)Ar (r+ 6;,~,z)] _ ~~az[(r -~r)Ar(r _ ~r, ~,z)] (2) a~ a~ , +~raz[A<t>(r,~+ 2'z) -A<t>(r,~- 2'z)] az az +r~~ar[Az(r,~,z+ 2) -Az(r,~,z- 2)] Thus, the limit DivA= lim r.o.<t>.o.z-+O { ra~az[(r+ar)Ar(r+ ~,~,z) -(r-~)Ar(r- ~,~,z)] ra~azar [A<t>(r, ~ + ¥,z) -A<t>(r,~ - ¥,z)] (3) + ra~ [Az(r,~,z+ ¥) -Az(r,~,z -¥)]} + az gives the result summarized in Table I. 2.1.5 From Table I, 18 2 18. 1 8A<t> V· A = 2"-8 (r-Ar) +-.-(J 8(J (AB sm(J) +-.-(J 8'" (1)r r rSln rSln Y' For (a) Ao [1 8 (5)] Ao (2 V.A =---r =- 5r) (2)d3 r28r d3 for (b) Ao 1 8(2V·A=---- r )=0 (3)d2 rsin(J 8~ and for (e) (4) Solutions to Chapter 2 2-3 2.1.6 Starting with (2) and using the volume element shown in Fig. S2.1.6, (r + ~r)u8 (r -~r)u8 Flcure 82.1.8 Thus, 2-4 Solutions toChapter2 Inthelimit 1a2 1a. 1aA",v·.A=2"-a(rAr)+-.-./Ia./l(smOAo)+-.-./Ia'" (3)rr rSlnl7 17 rSlnl7 'I' 2.2GAUSS' INTEGRAL THEOREM ~y iydxdz-iydxdx..---- ixdydz,/2.2.1 Figure83.3.1 (a)Thevectorsurfaceelements areshowninFig.82.2.1. (b)Thereisnozcontribution, sothereareonlyx=±dsurfaces, A",=(Ao/d)(±d) andn=±ixdydz. Hence,thefirsttwointegrals. Thesecondandthirdare similar. (c)From(2.1.5) V.A=Ao[~x+ ~y]=2Ao daxayd Thus,because V.Aisconstant overthevolume [V.AdV=2~o(2d)3=16Aod2(1) (2) 2.2.2 Thesurfaceintegration is IA.da=~:[jdjddy2dydz_jdjd(-d)y2dydz18 -d-d -d-11- +jdjddx2dxdz_jdjd(-d)x2dxdz -d-d -d-d(1) Solutions to Chapter 2 2-5 From the first integral = ::(2cP) (~d3) The others give the same contribution, so 4Ao 4d5 16Aod2 = d3 3= 3 (2) (3) To evaluate the right hand side of (2.2.4) V .A = Ao [!""' Zy2 + !.....z2y] = Ao (y2 + z2) d3 az ay d3 So, indeed (4) (5) 2.3 GAUSS' LAW, MAGNETIC FLUX CONTINUITY AND CHARGE CONSERVATION 2.3.1 (a) From Prob. 1.3.1 E A [ z. y'J= - lx+ 1211'Eo z2 +y2 z2 +y2 ~ From (2.1.5) A [a ( :& ) a( y )] V·E-- ­+­-211'Eo az Z2 +y2 ay z2 +y2A[1 2:&2 1 2y2 ] = 211'Eo z2 + y2 -(z2 + y2)2 + z2 + y2 -(Z2 + y2)2 A [y2_ Z2 z2_y2] = 211'Eo (z2 + y2)2 + (z2 + y2)2 = 0 except where z2 +y2 =0 (on the z-axis). (b) In cylindrical coordinates (1) (2) (3) Thus, from Table I, 2-6 Solutions to Chapter 2 2.3.2 Feom Table I in cylindrical coordinates with a( )/at/> and a( )/az = 0, v .foE = -f o -a (rEr ) r ar so r<b b<r<a r < b b<r<a 2.3.3 Using B = Ho(i x + i)') in (2.1.5), a(l) a(l)V . /LoB = /LoHo[- +-] = 0ax ay 2.3.4 In cylindrical coordinates (Table I): 1 a 1 aH~ 8H", 1 a (--i ) (1) (2) (3) (1) V·B=--(rH r )+--+-=-- =0r 8r r at/> az r at/> 2'll"r (1) 2.3.5 If V ./LoB =0 everywhere then the integral of its normal over an arbitrary dosed surface in that region will be zero and (a) V/LoB = 0 (b) (c) HoayV . /LoB =--= 0 a ax Thus, only (b) will not satisfy (1.7.1) 2.3.6 Evaluation using (2.1.5) gives aE", 2poP= V 'foE= fo-=-z az 8 which is the given charge density. Solutions to Chapter 2 2-7 2.3.'1 Using V· F in spherical coordinates from Table I with %f) and 0/0; = 0, V oJ = _.!.~(r2Jr) = _.!.~(,.s dpo) =_ dpo r2 or r2 or 3 dt dt which, since Po is independent of r, checks with (2.3.3). 2.4 THE CURL OPERATOR 2.4.1 All cases have only z and y components, independent of z. V X A = [i; iX i~] 8s 811 Az All 0 =i.[oA II Bz _ oAz ]oy Thus (a) V X A = Ao /1- IJ = d 0 (1) (b) V X A = Ao 10 ­d oj = 0 (2) (c) = Aol-e-kll cos kz + ke-kllV X A coskzl =0 (3) To make a finite curl make a single component having any dependence on a coordinate perpendicular to the vector. All = I(z), Az =0, A. =0 (4) Say, (5) 2.4.2 In all cases A. = 0 and B/Bs = 0, so from Table I, . [1 0 () 1 oAr]V X A = I. --rA~ --- (1)r or r 0; (a) Thus (a) ~ V XA = i. Ao[!.~(_r2 sin 2;) - !.~(rcos2;)] d r or r 0; (2) = i. ~o [-2sin2;+28in 2;1 =0 2-8 Solutions to Chapter 2 (b) ==>VxA = i.Ao[~ :r(-rsinq,) -~ :q, cosq,] (3) = i.Ao[ _ sinq, + sinq,] = 0 r r • 18 (AO r 3 ) • (3Ao r) () ()c ==> V x A= 1.;:-8r 7 =1. 7 4 (b) Possible vector functions having a curl make A = A<f>i<f> where rA<f> = j(r) is not a constant. For example f(r) = r, r2 ,r3 , in which case (5) 2.4.3 From (2) (curlA)n = lim } 1 A· ds (1) .o.a--+O ua fa Using contour of Fig. P2.4.3a, (VxA)r= lim {[6.ZA z(r,q,+¥,z)-6. ZAz(r,q,-¥,z)] r.o.<f>.o.z--+O r6.q,6.z _ [r6.q,A<f>(r, q" z + ¥) -r6.q,A",(r, q" z -¥)] } (2) r6.q,6.z 18A z 8 A", = ;:-8q, -8z Using the contour of Fig. P2.4.3b (VxA)", = lim {[6.rA r (r,q" z +¥) -6.rA r (r, q" z -¥)] .0. r.o. z--+O t::.rt::.z _ [t::.zAz(r + ¥, q"z) -6.zAz(r -¥, q" z)] } (3) 6.r6.z 8Ar 8Az = 8z -a;­ (V XA)z = lim .0. rr.o. "'--+0 [(r+ ¥)t::.q,A",(r+ ¥,q"z) -(r-¥)6.q,At/>(r-¥,q"z)] { 6.rrt::.q, _ [6.rA r (r, q, + ¥, z) -6.rA r (r, q, -¥, z)] } (4) 6.rr6.q, 1 8(rAt/» 1 8Ar =;:- 8r -;: 8q, 2.4.4Solutions toChapter 2 /'\d.8 Nrsin8t::..dJ From(2)(r~~r)sin8t::..dJ Flsure83.4.42-9 ('l"7A) ,{[raOAB(r,o,tP+ ¥)-raOAB(r,O,tP- ¥)]vXr= hm - .rt:.BrsinBt:.",-O raOrsmoatP [rsin(0+¥)atPA",(r,O +¥,tP)-rsin(0-¥)atPAfjI(r, 0-¥,tP)]} + raOrsinOatP =__1__8A_B+_1__8(.>.-si_n_8A---,fjI~) rain08tPrsin080 (1) (VXA)B= lim.{[arAr(r,8,tP+¥),-arAr(r,o,tP-¥)] t:.rsinBrt:.fjI-O arsmOratP _[atPsin8(r+~)A",(r+~,O,tP)-atPsintP(r-~)A",(r-~,O,tP)]} arsin8ratP 18Ar18(rAfjI)=-:r(-sin-O-=-)8tP-;8r (2) (VXA)",=lim{[ao(r+~)AB(r+~,O,tP)-aO(r-~)AB(r-~,O,tP)] rt:.9t:.r-O raOar [arAr(r,O +¥,tP)-arAr(r,0-¥,tP)]} raOar 18·18Ar=--(rA9)---r8r r80 (3) 2-10 Solutions toChapter 2 2.4.5 (a)Stokes'integraltheorem, (2.4.1)is£A.ds=1VXA.da (1) WithSaclosedsurface,C-+0,sotVXA.da=0=1V·(VXA)dV (2) BecauseVisarbitrary, theintegrand ofthisvolumeintegralmustbezero. (b)Carrying outtheoperations gives V.(VxA) =~[aAz_aA,,]+~ [aAz_aA"]+~[aA,, _aAz]=0(3) axayazayazaxazaxay 2.5STOKES' INTEGRAL THEOREM 2.6.1 y h-----,...--......----, Z(9----""---~----~x g FigureSJ.5.1 (a)UsingFig.S2.5.1toconstruct A·ds,£A.ds=19+~Az(x,O)dx+lhA"(g+~,y)dy -lg+~Az(x,h)dx-lhA,,(g,y)dy =19+~(O)dx+lh~(g+~)2dy (1) -lg+~(O)dx-rA;g2dy g10d =~;[(g+~)2h-g2hl Solutions to Chapter 2 2-11 (b) The integrand of the surface integral is [ ix i~ is] aA 2A x V X A = a/;x 1" ~ = is ax" = is ,; Thus r r ru+6. 2A x A1 s VxA·da= 1 0 1 u ,; dxdy= ,p[(g+d)2_ g2jh (2) 2.5.2 (a) Using the contour shown in Fig. 82.5.1, faA . ds = ~o [ iU +6. (O)dx + lh(g + d)dy -iU +6. (-h)dx -lh gdy] Ao[() j 2Aohd = d g +d h +hd ­gh = d (b) To get the same result carrying out the surface integral, [ ix I)' is] aA aA V X A = a/ax a/ ay 0 = is [T -T] All: A" 0 x Y = Ao [1 +1] = 2Ao d d and hence l(vX A)· da = 2:0 (dh) (1) (2) 2.6 DIFFERENTIAL LAWS OF AMPERE AND FARADAY 2.6.1 r<b b<r<a (1) r<b b<r<a (2) 2-12 Solutions to Chapter 2 2.6.2 Ampere's differential law is written in cylindrical coordinates using the ex­ pression for V x B from Table I with ajat/J and ajaz = 0 and Hr = 0, Hz = O. Thus VXB=i).aa (rH</»=i • .!:.aa {Joa2 [1-e-r/a(1+.!:.)]} = Joe-r/ai. (1) rr rr a 2.7 VISUALIZATION OF FIELDS AND THE DIVERGENCE AND CURL 2.1.1 (a) For p and E given by 2po z p=­ B Ez = ~[z2 _ (~)2] (1) foB 2 the sketch is shown in Fig. 82.7.1 Figure 83.1.1 (b) X[ii.]VxE= 0 iy0 ajaz =0 (2) o 0 Ez (c) The density of field lines does not vary in the direction perpendicular to lines. 2.1.2 (a) From Prob. 1.4.1, -J e-r/a• Joa2 [ -r/a( r)]Jz - 0 , H</>=--1-e 1+- (1)r a and the field and current plot is as shown in cross-section by Fig. 82.7.2. (b) From Prob. 1.4.4, the currents are a line current at the origin returned as two surface currents. K _ {I/,rr(2a+b); r=a ,,- ~Ij7l"(2a+b); r=b (2) Solutions toChapter2 Intheannularregions, H__.!-{l/r; 0<r<b 4>-211"2a/r(2a+b);b<r<a2-13 (3) Thisdistribution ofcurrentdensityandmagnetic fieldintensity isshownin crosB-section byFig.S2.7.2. (a) (b) Figure 83.7'.3 (c)Because HhasnotPdependence withitsonlycomponent inthetPdirection, it mustbesolenoidal. Tocheckthatthisisso,notethata/atP=0anda/az=0 andthat(fromTableI) (d)See(c).1aV.H=--a (rHr)=0rr(4) 2.1.3 (a)Theonlyirrotational fieldis(b),wherethelinesareuniform inthedirection perpendicular totheirdirection. In(a),thelineintegral ofthefieldarounda contoursuchasthatshowninFig.S2.7.3amustbefinite.Similarly, because thefieldintensity isindependent ofradius incase(c),thelineintegralshown inFig.S2.7.3bmustbefinite. 2.7'.42-14 Cr--'4--, "I I., I I I IL__+f--J (a) Figure92.".3 Therespective fieldsareSolutions toChapter 2 (b) (1) (2) andthefieldplotisasshowninFig.82.7.4.Notethatthespacingbetween linesis lesserabovetoreflectthegreaterintensity ofthefieldtl;J.ere. LII!I! l~//y, 2.7'.5Figure92.7'.4 Therespective fieldsareFigure92.7'.5 (1) (2) 2.7'.6andthefieldplotisasshowninFig.82.7.5.Notethat,becausethefieldissolenoidal, thenumberoffieldlinesaboveandbelowcanbethesamewhilehavingtheirspa.cing reflectthefieldintensity. (a)Thetangential Emustbecontinuous, asshowninFig.82.7.6a,sothenormal Eontopmustbelarger.Because thereisthananetfluxofEoutofthe interlace, itfollowsfromGauss'integrallaw[continuity condition (1.3.17)] thatthesurlacechargedensityispositive. 2-15 Solutions to Chapter 2 rL L 0-----...., ..• z L L (a) (b) Figure 82.7'.8 (b) The normal component of the flux density 1"011 is continuous, as shown in Fig. S2.7.6b, so the tangential component on the bottom is largest. From Ampere's integral law [the continuity condition (1.4.16)1 it follows that K", > o. SOLUTIONS TO CHAPTER 3 3.1 TEMPORAL EVOLUATION OF WORLD GOVERNED BY LAWS OF MAXWELL, LORENTZ, AND NEWTON 3.1.1 (a) Replace z by z -ct. Thus E-- E·olxe-(z-ct)' /2a'., (1) (b) Because 8( )/8z = 8( )/8y = 0 and there are only single components of each field, Maxwell's equations reduce to (2) Note that we could pick these expressions out of the six components of the laws of Faraday and Ampere by first writing the left hand sides of 3.1.1-2. Thus, these are respectively the y and z components of these laws. In Cartesian coordinates, the divergence equations are automatically satisfied by any vector that only depends on a coordinate perpendicular to its direction. Substitution of (1) into (2a) and into (2b) gives 1 c=-- (3).j#lof o which is the velocity of light, in agreement with (3.1.16). (c) For an observer having the location z = ct+ constant, whose position increases linearly with time at the rate c m/s and who therefore has the constant velocity c,z -ct = constant. Thus, the fields given by (1) are constant. 3.1.2 With the given substitution in (3.1.1-4), (with J = 0 and p = 0) 8E 1--=--VxH (1)8t fo 8H 1-=--VxE (2)8t #lo 0= V· #loH (3) 0=-V .foE (4) Although reordered, the expressions are the same as the original relations. 1 3-2 Solutions toChapter 3 3.1.3 Notethatthedirection ofwavepropagation isobtained bycrossing Einto B.Because itwouldreversethedirection ofthiscrossproduct, agoodguessisto reversethesignofoneortheotherofthefields.Inthatcase,thestepsfollowed inProb.3.1.1leadtotherequirement thatc=-1/';~ofo' Wedefinecasbeing positiveandsowritethesolutions withz-ctreplaced byz-(-c)t=z+ct.Following thesamearguments asinpart(c)ofProb.3.1.1,thissolution istherefore traveling inthe-zdirection. x }'---'" Hy ~--------1~ Y Figure83.1.4"f---- E. "----+-----z -HIJ 3.1.4. Theroleplayedbyzisnowtakenby:z:,asshowninFig.S3.1.4.Withthe understanding thatthezdependence isnowreplaced bythegiven:z:dependence, themagnetic andelectricfieldsarewrittensothattheyhavethesameratioasin (1)ofProb.3.1.1.Further, inordertopreserve thevectorrelation between E,H andthedirection ofpropagation, thesignofHisreversed. Thus, E=Eoi.cosP(:z:-ct)j 3.2QUASISTATIC LAWSH=--~Eoiy cosP(:z:-ct)V~o(1) 3.2.1 (a)Thesefieldsaretransverse tothecoordinate, :z:,uponwhichtheydepend. Therefore, thedivergence conditions areautomatically satisfied. Fromthe direction ofthevectors, weknowthatthe:z:andycomponents respectively ofthelawsofAmpereandFaraday willapply. 8H"8foEz-8z=at (1) 8Ez 8~oH"8z=---at"" (2) Theotherfourcomponents oftheseequations areautomatically satisfied be­ cause8()/8y=8()/8z=O.Substitution of(a)and(b)thengives wP=W';~ofo ==- (3)c 3-3 Solutions to Chapter 3 in each case. (b) The appropriate identities are 1 w w )cos fjz coswt = 2"[cosfj(z- pt) +cosfj(z+ pt] (4) sinfjzsinwt= i[cosfj(z- ~t) -cosfj(z+ ~t)] (5) Thus, in view of (3), the fields indeed take the form of the sum of waves traveling in the +z and -z directions with the speed c. (c) In view of (a), this condition can be written as fjl = wy'IJoEol = wllc <:1 (6) Thus, the condition is equivalent to having the electromagnetic delay time Tem =llc short compared to the time l/w required for 1/21r of a cycle. (d) In the limit of (c), cosfjz -+ 1 and sinfjz -+ fjz and (a) and (b) become the given fields. (e) The electric field of (c) is irrotational and hence satisfies (3.2.1a) but not (3.2.1b) while the magnetic field has curl and indeed satisfies (3.2.2a) but not (3.2.2b). Therefore, in the limit of having the frequency low enough to satisfy (6), the system is EQS. 3.2.2 (a) See part (a) of solution to Prob. 3.2.1. (b) The appropriate identities are sin(fjz) sin(wt) = i [cos fj(z - ~t) + cos fj(z + ~t)] (1) cos(fjz) cos(wt) = 2"1[ cosfj(z -wpt) -cosfj(z + wpt)] (2) Thus, because wlfj = c, the fields indeed take the form of the sum of waves traveling in the +z and -z directions with the speed c. (c) See (c) of solution to Prob. 3.2.1. (d) In the limit where Ifjll <: I, the given fields become E ~ wIJoHozsinwti x (3) H ~ Hocoswti~ (4) Thus, the magnetic field is uniform while the electric field varies linearly between the source and the "short" at z = 0, where it is zero. (e) The magnetic field of (4) is irrotational and hence satisfies (3.2.2b) with J =0 but not (3.2.2a). The electric field of (3) does have a curl and hence does not satisfy (3.2.1a) but does satisfy (3.2.1h). Thus, the system is magnetoqua,... sistatic. 3-4 Solutions to Chapter 3 3.3 CONDITIONS FOR FIELDS TO BE QUASISTATIC 3.3.1 (a) Except that it is in the z direction rather than the z direction, the quasistatic electric field between the plates is, as in Example 3.3.1, uniform. To satisfy the requirement of (a), this field is E = Iv(t)/d]i x (1) The surface charge density on the plates follows from Gauss' integral law applied to the plates, much as in (3.3.7). cr -{-EoEz(z = d) = -Eov/a; Z = d (2) • - EoEz(z =0) = Eov/d; z =0 Thus, the quasistatic surface charge density on the interior surfaces of each plate is uniform. K.(z) 17.(z) K.(z) y c (a) (b) Fisure S3.3.1 (b) The integral form of charge conservation is applied to the lower and upper electrodes using the volume shown in Fig. S3.3.1a. Thus, using symmetry to argue that K z =0 at z =0, for the lower plate ocr.zw ZEo dv wIK.(z) -Kz(O)] + --ar:-= 0 ~ Kz(z) = -7 dt (3) and we conclude that the surface current density increases linearly from the center toward the edges. At any location z, it is that current required to change the charge on the fraction of "capacitor" at a lesser value of z. (c) The magnetic field is found using Ampere's integral law, (3.3.9), with the surface da = ixda having edges at z = 0 and z = z. By symmetry, Hy =0 at z =0, so (4) Solutions to Chapter 3 3-5 Note that, with this field and the surface current density of (3)' Ampere's continuity condition, 1.4.16, is satisfied on the upper and lower plates. We could just as well think of the magnetic field as being induced by the surface current of (3) as by the displacement current of (3.3.9). (d) To determine the correction electric field, use Faraday's integral law with the surface and contour shown in Fig. 83.3.1b, assuming that E is independent of x. (5) Because of (a), it follows that the corrected field is 2 E ( ) = ~ JoLo€o (z2_ 2) d v (6)xZ d + 2d Zdt2 (e) With the second term in (6) called the "correction field," it follows that for the given sinusoidally varying voltage, the ratio of the correction field to the quasistatic field at at most (7) Thus, because c = 1/VJoLo€o, the error is negligible if 1 l -[-w] ~ 1 (8)2c 3.3.2 (a) With the understanding that the magnetic field outside the structure is zero, Amper'es continuity condition, (1.4.16), requires that 0-H y = K y = K top plate H y -0 = K y =-K bottom plate (1) where it is recognized that if the current is essentially steady, the surface current densities must be of equal magnitude K(t) and opposite directions in the top and bottom plates. These boundary conditions also require that H= -iyK(t) (2) at the surface current density sources at the left and right as well. Thus, provided K(t) is essentially steady, (2) is taken as holding everywhere between the plates. Note that this uniform distribution of field not only satisfies the boundary conditions, but also has no curl and hence satisfies the steady form of Ampere's law, (3.2.2b), in the region between the plates where J = O. 3-6 Solutions to Chapter 3 (b) The integral form of Faraday's law is used to compute the electric field caused by the time variation of K(t). 1 E·ds=-~ 1lo'oH . da (3)fa ats (a) (b) Figure SS.S.Z SO that it links the magnetic flux, the sudace is chosen to be in the :z: -z plane, as shown in Fig. S3.3.2a. The upper and lower edges are adjacent to the perfect conductor and therefore do not contribute to the line integral of E. The left edge is at z = 0 while the right edge is at some arbitrary position z. Thus, with the assumption that EI/ is independent of :z:, (4) Thus the electric field is Ez (0) plus an odd function of z. Symmetry requires that Ez (0) = 0 so that the desired electric field induced through Faraday's law by the time varying magnetic field is (5) Note that the fields given by (2) and (5) satisfy the MQS field laws in the region between the plates. (c) To compute the correction to H that results because of the displacement current, we use the integral form of Ampere's law with the sudace shown in Fig. S3.3.2. The right edge is at the sudace of the current source, where Ampere's continuity condition requires that HI/{l) = -K(t), and the left edge is at the arbitrary location z. Thus, (6) Solutions to Chapter 3 3-7 and so, from this first order correction, we have found that the field is H = -K( ) WfoJJo (12 -Z2) cPK (7)1/ t+ W 2 dt2 (d) The second term in (7) is the correction field, so, at worst where z = 0, IHcorrected I = f o/Jol2 ...!....I cPK I (8)IKI 2 IKI dt2 and, for the sinusoidal excitation, we have a negligible correction if (9) Thus, the correction can be ignored (and hence the MQS approximation is justified) if the electromagnetic transit time 1/c is short compared to the typical time 1/w. 3.4 QUASISTATIC SYSTEMS 3.4.1 (a) Using Ampere's integral law, (3.4.2), with the contour and surface shown in Fig. 3.4.2c gives (1) (b) For essentially steady currents, the net current in the z direction through the inner distributed surface current source must equal that radially outward at any radius r in the upper surface, must equal that in the -z direction in the outer wall and must equal that in the -r direction at any radius r in the lower wall. Thus, 21l"bK o = 21l"rK,.(z = h) = -21l"aK .. (r = a) = -21l"rK,.(z = 0) b b b (2) => K,.(z = h) = -Koi K..(r = a) = -Koi Kr(z =0) = -Ko r a r Note that these surface current densities are what is called for in Ampere's continuity condition, (1.4.16), if the magnetic field given by (1) is to be con­ fined to the annular region. (c) Faraday's integral law 1E .dB= - ~ { /JoB· da (3)'e atls 3-8 Solutions to Chapter 3 applied to the surface S of Fig. P3.4.2 gives (4) Because E.(r = a) = 0, the magnetoquasistatic electric field that goes with (2) in the annular region is therefore E. = -J.&obln(a/r) d~o (5) (d) Again, using Ampere's integral law with the contour of Fig. 3.4.2, but this time including the displacement current associated with the time varying electric field of (5), gives (6) Note that the first contribution on the right is due to the integral of Jasso­ ciated with the distributed surface current source while the second is due to the displacement current density. Solving (6) for the magnetic field with E. given by (5) now gives Htf> = !Ko(t)+ EoJ.&oba2{(:')2[!ln(:')_!] _(!)2[!zn(!)_!]} f1JKo (7)r ra 2 a 4 a 2 a 4 dt2 The last term is the correction to the magnetoquasistatic approximation. Thus, the MQS approximation is appropriate provided that at r = a (8) (e) In the sinusoidal steady state, (8) becomes The term in IIis of the order of unity or smaller. Thus, the MQS approxi­ mation holds if the electromagnetic delay time a/e is short compared to the reciprocal typical time l/w. SOLUTIONS TOCHAPTER 4 4.1IRROTATIONAL FIELDREPRESENTED BYSCALAR POTENTIAL: THEGRADIENT OPERATOR AND GRADIENT INTEGRAL THEOREM 4..1.1 (a)Forthepotential (1) (2) (b)Theunitnormalis 4..1.2 For~=~zy,wehave y (a,a) -----¥----- ...xo(3) (1) Figure94.1.2 Integration onthepathshowninFig.84.1.2canbeaccomplished usingtasa parameter, whereforthiscurvez=tandy=dsothatin ds=ixdz+iJ'dy wecanreplacedz=dt,dy=dt.Thus, l(a,a)lav. E·ds= -;(ix+iJ').(ix+iJ')dt=-V a (0,0) t::::::oa Alternatively, ~(O,0)=0and~(a,a)=Vaandso~(O,0)-~(a,a)=-Va'(2) (3) 1 4-2 Solutions toChapter4 4.1.3 (a)Thethreeelectricfieldsarerespectively, E=-V~, E=-(Vo/a)ix (1) E=-(Vo/a)i), (2) 2Vo(• •)E=--2XIx-YI),a(3) (b)Therespective equipotentials andlinesofelectricfieldintensity aresketched intheX-YplaneinFigs.S4.1.3a-c. .... --4- -.......--fII (e) (f)(d) E _._--..of> ~-- ------- (b) (r) Figure84.1.8 (c)Alternatively, theverticalaxisofathreedimensional plotisusedtorepresent thepotential asshowninFigs.S4.1.3d-f. Solutions toChapter4 4-3 4.1.4 (a)InCartesian coordinates, thegradoperator isgivenby(4.1.12). With (J>de- finedby(a),thedesiredfieldis (b)Evaluation ofthecurlgives ixi)' VxE=:s:y EsEy 11'2 1I'Z 1I'y~ 1I'Z1I'y]=[-cos-cos---cos-cos-abababab =0 sothatthefieldisindeedirrotational. LL_-I==:=Jt:::==L_...L--.J-.. z(1) (2) Figure84.1.4 (c)FromGauss'law,thechargedensityisgivenbytakingthedivergence of(1). (3) (d)Evalvuation ofthetantential component from(1)oneachboundary givesjat z=O,Ey=OJ y=O,E s=OJz=a,Ey.=0 y=a,Es=0(4) (e)Asketchofthepotential, thechargedensityandhenceofEisshowninFig. 84.1.5. 4-4 Solutions to Chapter 4 Figure 94.1.5 (f) The integration of E between points (a) and (b) in FIg. P4.1.5 should be the same as the difference between the potentials evaluated at these end points because of the gradient integral theorem, (16). In this particular case, let x = t,Y = (bla)t so that dx = dt and dy = (bla)dt. b -P fa 1r 1rt 1rt f E·ds= [( 1 )2 0( Ib)2] [-cos-sin-dt a f o 1ra + 1r a/2 a a a 1r • 1rt 1rt]+ -sm-cos- dt a a a (5) -Po fa 1r • 21rt d =f o[( 1rla)2 + (1r/b)2] a/2 ~ sm -;- t _ Po -f o[(1rla)2 + (1rlb)2] The same result is obtained by taking the difference between the potentials. (6) (g) The net charge follows by integrating the charge density given by (c) over the given volume. Q = ( pdv = r r r posin(1I"xla) sin(1I"ylb)dxdydz = 4Po:bd (7)1v 101010 11" From Gauss' integral law, it also follows by integrating the flux density foE· n over the surface enclosing this volume. Solutions to Chapter 4 4-5 (h) The surface charge density on the electrode follows from using the normal electric field as given by (1). (9) Thus, the net charge on this electrode is (10) (i) The current i(t) then follows from conservation of charge for a surface S that encloses the electrode. (11) Thus, from (10), (12) 4.1.5 (a) In Cartesian coordinates, the grad operator is given by (4.1.12). With ~ de­ fined by (a), the desired field is a~. a~.]E =-[ -Ix+-Iaz ay ., Po [1r • 1r 1r. 1r 1r. 1r • ] (1) = f [(1r/a)2 + (1r/b}2J ~ sm ~zcos ;;Y1x + ;; cos ~zsm bY1.,o (b) Evaluation of the curl gives so that the field is indeed irrotational. (e) From Gauss'law, the charge density is given by taking the divergence of (1). (3) 4-6 Solutions to Chapter 4 (d) The electric field E is tangential to the boundaries only if it has no normal component there. Ez(O,y) = 0, Ez(a,y) = 0 (4) Ey(:Z:, O) = 0, Ey(:Z:, b) =0 (e) A sketch of the potential, the charge density and hence of E is shown in Fig. 84.1.4. (f) The integration of E between points (a) and (b) in Fig. P4.1.4 should be the same as the difference between the potentials evaluated at these end points because of the gradient integral theorem, (16). In this particular case, where y = (b/a):z: on C and hence dy =(b/a)d:z: r(b) E. de = fa {Ez(:Z:, ~:Z:)d:Z:+ EII(:z:, ~:Z:)(b/a)d:Z:} ita) a/2 a a Po fa 21/" • 1/" 1/" = [( /)2 (/b)2] -sm -:z:cos -:z:d:z:Eo 1/" a + 1/" a/2aa a -Po = -E-:-::[('-1/"/7"a~)2~+':""";-( 1/""""7/b:'7)2=:"]o The same result is obtained by taking the difference between the potentials. (6) (g) The net charge follows by integra.ting the charge density over the given vol­ ume. However, we can see from the function itself that the positive charge is balanced by the negative charge, so (7) From Gauss' integral law, the net charge also follows by integra.ting the fiux density foE· n over the surface enclosing this volume. From (d) this normal flux is zero, so that the net integral is certainly also zero. Q =tfoE· nda =0 (8) The surface charge density on the electrode follows from integrating foE .n over the "electrode" surface. Thus, the net charge on the "electrode" is q = tfoE· nda = 0 (9) Solutions to Chapter 4 4-7 4.1.6 (a) From (4.1.2) E ( a~. a~.) =-az Ix + ay I)' = -A[mcosh mzsin klly sin kzzix (1) + sinh mzkll cos kllysin kzzi)' + kz sinh mz sin klly cos kzzi.1 sin wt (b) Evaluation using (1) gives (2) =-Asinwt{ix(kllk z sinh mz cos kllycos kzz -kllkz sinh mzcos kllycos kzz) +l)'(mk z cosh mzsin kllycos kzz -kzmcosh mzsin kllycos kzz) + i.(mkll cosh mzcos kllysin kzz -mkll cosh mzcos kllysin kzz) =0 (3) (c) From Gauss' law, (4.0.2) p = V· foE = -EoA(m2 -k~ -k~)sinhmzsinkllysinkzzsinwt (5) (d) No. The gradient of vector or divergence of scalar are not defined. (e) For p = 0 everywhere, make the coefficient in (5) be zero. (6) 4.1.'1 (a) The wall in the first quadrant is on the surface defined by y=a-z (1) Substitution of this value of y into the given potential shows that on this surface, the potential is a linear function of z and hence the desired linear function of distance along the surface ~ = Aa(2z -a) (2) 4-8 Solutions toChapter4 (3)V ~=_(z2-!l) a2 Ontheremaining surfaces, respectively inthesecond,thirdandfourthquad­ rants y=z+ajy=-a-ZjY=Z-a (4) Substitution ofthesefunctions into(3)alsogiveslinearfunctions ofzwhich respectively satisfytheconditions onthepotentials attheendpoints.Tomakethispotential assumethecorrectvaluesattheendpoints,where z=0and~mustbe-Vandwherez=aand~mustbeV,makeA=V/a2 andhence (b)Using(4.1.12), E (a~. a~l) V(••)=--Ix+-=--2ZIx-2ylazay¥a2 ¥ FromGauss'law,(4.0.2),thechargedensityis(5) (6) Figure84.1.7' (c)Theequipotentials andlinesofEareshowninFig.S4.1.7. 4.1.8 (a)ForthegivenE, ixi¥i.aavxE=a/aza/ay0=i.[-(-Cy) --(Cz)]=0(1)Cz-Cy0az ay soEisirrotationaL ToevaluateC,remember thatthevectordifferential distance ds=ixdz+i¥dy. Forthscontour, ds=i¥dy.Tolettheintegraltake 4-9 Solutions to Chapter 4 account of the sign naturally, the integration is carried out from the origin to (a) (rather than the reverse) and set equal to q>(0, 0) -q>(0, h) = -V. 1-V = lh -Cydy = --Ch2 (2) o 2 Thus, C = 2V/h2 • (b) To find the potential, observe from E = -yrq> that aq>-=-Cx· (3)ax ' Integration of (3a) with respect to x gives q> = -"21Cx2 + f(y) (4) Differentiation of this expression with respect to y and comparison to (3b) then shows that aq> df 1 -=-= Cy '* f = _y2 + D (5)ay dy 2 Because q>(0, 0) =0, D = °so that 1 (2 2) q> = -"2C x -y (6) and, because q>(0, h) = V, it follows that q> = _~C(02 _ h2) (7)2 so that once again, C = 2V/ h2 • (c) The potential and E are sketched in Fig. S4.1.8a. 1...-, I :,...­ '" ---1----.:; II II I --''----7---.L----~ .. X X =-d x=d 1 ~--------l~X w (a) z (b) Figure 84.1.8 4-10 Solutions to Chapter 4 (d) Gauss' integral law is used to compute the charge on the electrode using the surface shown in Fig. S4.1.8b to enclose the electrode. There are six surfaces possibly contributing to the surface integration. t EoE ·nda= q (8) On the two having normals in the z direction, EoE.n = O. In the region above the electrode the field is zero, so there is no contribution there either. On the two side surfaces and the bottom surface, the integrals are W Jd2+h2 q =Eo rr E(d, y) . ixdydzJo Jh1 W Jd2+h2 + Eo rr E(-d, y) . (-ix)dydz (9)Jo Jh1 w d + Eo rr E(z, hI) . (-i), )dzdzJo J-d Completion of the integrals gives (10) 4.1.9 By definition, ~~ = grad (~) . ~r (1) In cylindrical coordinates, (2) and ~ifJ = ~(r + ~r, ~ + ~~, z + ~z) -~(r, ifJ, z) a~ a~ a~ (3) = -~r + -~ifJ + -~z ar aifJ az Thus, a~ a~ a~ ar ~r + aifJ ~ifJ + az ~z = grad ~ . (~ril' + r~ifJi", + ~zi.) (4) and it follows that the gradient operation in cylindrical coordinates is, (5) Solutions to Chapter 4 4-11 4.1.10 By definition, Aw=grad (W) .Ar (1) In spherical coordinates, Ar = Arir + rA9i8 + rsin9At/>i<f» (2) and Aw = W(r+ Ar, 9 + A9,t/> + At/» -W(r,9, t/» aw aw aw (3) = arAr+ aiA9 + at/> At/> Thus, and it follows that the gradient operation in spherical coordinates is, (5) 4.2 POISSON'S EQUATION 4.2.1 In Cartesian coordinates, Poisson's equation requires that (1) Substitution of the potential (2) then gives the charge density (3) 4-12 Solutions to Chapter 4 4.2.2 In Cartesian coordinates, Poisson's equation requires that o2~ o2~ P= -fo( oz2 + oy2) (1) Substitution of the potential Po ~ ~ ~ =f o [(1l"/a)2 + (1f/b)2] cos ~zcos bY (2) then gives the charge density 1f ~ P = Po cos ~zcos bY (3) 4.2.3 In cylindrical coordinates, the divergence and gradient are given in Table I as V.A = !~(rAr) +!oA~ + oA. (1) r ar r aq, oz au. 1 au. ou.Vu = -1_ + --1... + -1 (2)ar· r aq,'" az • By definition, V2u= V. Vu= !~(r ou) + !~(! ou) + ~(ou) (3)r or or r oq, r oq, oz oz which becomes the expression also summarized in Table I. 21 0 (au) 1 02U 02U (4)V U = ;:-or r or + r2 oq,2 + oz2 4.2.4 In spherical coordinates, the divergence and gradient are given in Table I as (1) (2) By definition, V2u=V. (Vu)= ..!..~(r2aU) + _~_(! ou sinO) r2 or or rsmO roO 1 0 1 ou (3) + rsinO oq, (ninO oq,) which becomes the expression also summarized in Table I. V2u = ..!..~(~ou) + _1_~(sinOou) + 1 o2u (4) r2 ar or r2 sin 000 00 r2 sin2 0oq,2 4-13 Solutions to Chapter 4 4.3 SUPERPOSITION PRINCIPLE 4.3.1 The circuit is shown in Fig. 84.3.1. Alternative solutions Va and Vb must each = fa = satisfy the respective equations I(t) dVa VaC---;jj" + R dVb VbC-+­dt R v ( ) t; (1) h(t) (2) R Figure S4.3.1 Addition of these two expressions gives which, by dint of the linear nature of the derivative operator, becomes Thus, if fa => Va and h => Vb then fa + fb => Va + Vb. (3) (4) 4.4 FIELDS ASSOCIATED WITH CHARGE SINGULARITIES 4.4.1 (a) The electric field intensity for a line charge having linear density AI is Integration gives (1) (2) where ro is the position at which the potential is defined to be zero. 4-14 Solutions to Chapter 4 (b) In terms of the distances defined in Fig. 84.4.1, the potential for the pair of line charges is A, (r+) A, (r-) A, (r_) ~= ---In -+ --In -= --In - (3) 211"lO o ro 211"lO o ro 211"lO o r+ where Thus, A [1 + (d/2r)2 + !! cos 4J] (4) ~= --In r 411"lO o 1 + (d/2r)2 -~ cos 4J For d <: r, this is expanded in a Taylor series 1 +:C)In(-- = In(1 + :c) -In(l + 1/) SI:$ :c-1/ (5)1+1/ to obtain the standard form of a two-dimensional dipole potential. (6) 4.4.2 Feom the solution to Prob. 4.4.1, the potential of the pair of line charges is ~ = -A-In [1 + (2r/d)2 + ~ cos 4J] (1) 411"lO o 1+(2r/d)2-~cos4J For a spacing that goes to infinity, r/ d <: 1 and it is appropriate to use the first term of a Taylor's expansion l+:cIn(--) ~ :c-1/ (2)1+1/ Thus, (1) becomes 2A ~ = --rcos4J (3) 1I"lOod In Cartesian coordinates, :c = rcos4J, and (3) becomes (4) which is the potential of a uniform electric field. (5) Solutions to Chapter 4 4-15 4.4.3 The potential due to a line charge is CI> = -.A-Inr o 21r€o r where ro is some reference. For the quadrapole, (1) (2) where, from Fig. P4.4.3, r~ = r2[1 + (d/2r)2 + (d/r) sin </ll With terms in (d/2r)2 neglected, (2) therefore becomes (3) for d ~ r. Now In(l + x) ~ x for small x so In[(l + x)/(l + y)] ~ approximately CI> = _.A_ [ ­(d/r)2 cos2 </l + (d/r)2 sin2 </ll 41r€o -.Ad2 = --2[cos2 </l -sin2 </ll x ­y. Thus, (3) is (4) 41r€or -.Ad2 = --2cos2</l 41r€or This is of the form A cos 2</l/ ,..r with -.AdA=--, n=2 (5) 41r€o Solutions to Chapter 4 4-16 4.4.4 (a) For,. <: d, we rewrite the distance functions as (la) ,.~ = (d/2)2 [(d2")2 + 1+ d4,. cos 1/>] (lb) ,.~ = (d/2)2[e;)2 + 1 + ~ sin 1/>] (Ie) ,.~ = (d/2)2[(2;)2 + 1-~ cos 1/>] (la) With the terms (2,./d)2 neglected, at follows that (2) Because In(1 + z) !:::! z for z <: 1,ln[(1 + z)/(1 + y)1 ~ z-yand (2) is approximately >. (4")2[ 2 .2 I 4>.,.2~ =-- - cos I/> -sm I/> =--- cos 21/> (3) 411"f o d 1I"fo d2 This potential is seen again in Sec. 5.7. With the objective of writing it in Cartesian coordinates, (3) is written as (4) (b) Rotate the quadrapole by 45°. 4.5 SOLUTION OF POISSON'S EQUATION FOR SPECIFIED CHARGE DISTRIBUTIONS 4.5.1 (a) With Ir-r'l = .vZ'2 + yl2 + Zl2, (4.5.5) becomes (1) 4-17 Solutions to Chapter 4 (b) For the particular charge distribution, ~ Uo fa fa z'y'dz'dtj = a211"fo 11/'=01""=0 Vz,2 + y,2 + Z2 (2)a = ~U l[Va2 + y,2 + z2 y' -Vy,2 + z2 y']dy' a 1I"fo 1/'=0 To complete this second integration, let u2= tj2 + z2, 2udu = 2tjdy' so that Similarly, (4) so that (c) At the origin, (6) (d) For z > a, (5) becomes approximately ~~ uoz3 {1 + ea2 + 1)3/2 _ 2(a2 + 1)3/2} 3~11"~ ~ ~ 3 2 2 2 2 (7) = 2uo z {1+ (1+ 2a )(1 + 2a )1/2 _ 2(1 + a )(1 + a )1/2}3a211"f z2 z2 z2 z2o For a2 /z2 <: 1, we use (1+zP/2 ~ 1+!z and (8) 4-18 Solutions to Chapter 4 Thus, 2w= 20"0a (9) 31rE o Z For a point charge Q at the origin, the potential along the z-axis is given by Qw=-­ (10) 41rE o Z which is the same as the potential given by (9) if 2Q = 80"0a(11)3 (e) From (5), E=-VW = -8W i• = ~lz(2a2 + z2)1/2 + Z2 -2z(a2 + z2)1/2]i. (12)8z 1ra2Eo 4.5.2 (a) Evaluation of (4.5.5) gives W_ r fr (fr 0"0 cos ()'R 2sin ()'d4J'd()' -1</>'=0 1(}.=0 41rE o1R2 + z2 -4Rz cos ()'j1/2 0"0R2 (fr sin2()'d()' (1) = 4Eo 1(}.=0 v'R2 + z2 -2Rzcos()' To integrate, let u2 = R2 + z2 -2Rz cos ()' so that 2udu = 2Rz sin ()' d()' and note that cos()' = (R2 + z2 -u2)/2Rz. Thus, (1) becomes 0" l(R+ll)w= ~ (R2 +z2 -u2)du 4EoZ ll-R = ~[(R2 + z2)(R + z) _ (R + z)3 4EoZ2 3 (2) - (R2 + z2)(Z -R) + (z -R)3 ]3 0"0R3 = 3EZ2 o (b) Inside the shell, the lower limit of (2) becomes (R -z). Then W= O"oZ (3)3Eo (c) From (2) and (3) E= _t7w= _8wi = { ~~:~; i. z > R (4) 8Z • -3<01. Z< R .!!.J:I..' (d) Far away, the dipole potential on the z-axis would be pj41rE oZ2 for the point charge dipole. By comparison of (2) to this expression the dipole moment is 41r0"0R3 p= W3 Solutions toChapter 4 4-19 4.5.3 (a)TofindQ)(O,O,z) weuse(4.5.4).Forr=(O,O,z)andr'=apointonthe cylinder ofcharge,Ir-r'l=v(z-z')2+W.Thisdistance isvalidforan entire"ring"ofcharge.Theincremental chargeelement isthenO'21rRdz so that(4.5.4)becomes .....()l'O'021rRdz' fO-O'021rRdz' ":l"O,O,Z= +o41rEoV(Z-Z')2+W _I41rEoV(Z-Z')2+R2 Tointegrate, let<I=z-z',d<l=-dz'andtransform thelimits Q)=O'oR[-1%-' d<l+1% dq'] 2Eo %Vq'2+W %+1Vq,2+R2 R[ 1%-1 1%] =0'0-Inq'+VR2+q'2+lnlq'+VR2+q,2 2Eo % %+1 Thus,(1) (2) Q)_O'oRI[ (z+..,!R2+z2)(z+..,!R2+z2) ] -2Eon(z-I+VR2+(z-1)2)(z+1+VR2+(z+1)2) =uoR[21n(z+VR2+z2)-In(z-1+VR2+(z-1)2) (3) 2Eo -In(z+1+VR2+(z+1)2)] z r'=(x',,,',z') r=(O,O,z) x FigureS4.5.S Solutions to Chapter 4 4-20 (b) Due to cylindrical geometry, there is no ix or i)' field on the z axis. -2(1+ I) 1+ (I-I! E __ aCb i_ i.uoR [ yR'J+I'J + ( YR'J+(I_I)'J) -az· -2Eo z + yR2 + z2 Z-1+ vW + (z-1)2 1 + 1+/ R'J+(I+I)'J)] (4) + (z+ I + VR2 + (z+ 1)2 • uoR [-2 1 1]=1 - + + ~-----,.--= • 2Eo yW + z2 V R2 + (z -1)2 W+ (z+ 1)2 (c) First normalize all terms in Cb to z uoR [ (1+J1+~:)(1+J1+~:) ]Cb= -In 2Eo (1-~ + J(Rlz}2 + (1-~)2) (1 + ~ + J(Rlz)2 + (1 + ~)2 (5) Then, for z :> I and z :> R, uoR , [ (1+1)(1+1) ] ~-n / / / /2Eo (1-:;+1-i)(1+ :;+1+:;) _ UoR,n[ 4 ] -2Eo ,.(1 -(IIz)2) (6) = ~:~ln[1_(~/z)2] ~ ~~ln[1+(llz)2] uoR l2 ~-­2Eo z2 The potential of a dipole with dipole moment p is ''1 p cos(J ()dipole = -4---2­ (7)1rEo ,. In our case, cos (JI,.2 = 1/z2, so P = 21rR12 (note the p = qd, q = 21rRluo, deJ/ = I). 4.5.4 From (4.5.12), ld/ 2 >.d1/' Cb(:z:, 1/, z) = -:--r:===<=:::====;==~=~ (1) ,/,=-d/2 41rEoV(:I: -a)2 + (1/ -1/')2 + z2 To integrate, let u = 1/' -1/ so that (1) becomes >. j-Y+d/2 du ()-­-41rEo -y-d/2 vu2 + (:I: -a)2 + z2 (2) = 4:E In[u + vu 2 + (:I: -a)2 + z2] =:~:~: O which is the given expression. 4-21 Solutions to Chapter 4 4.5.5 From (4.5.12), A{1' z'dz' Z'dZ'}~ (0, 0, z) = --O ----;==:;;===;==~ 4'11'fo1 :z'=o vz'2 + (a -z)2 v z'2 + (a +z)2 = ~{2z + vl2 + (a -z)2 -vl2 +(a1 +z2)}4'11'fo1 4.5.6 From (4.5.12), A z'dz' A z~(O,O,z) =la 0, = _0_ la (-1 + --,)dz'z'=-a 4'11'foa(z -z ) 4'11'foa z'=-a z -Z (1) = ~[-a -zln(z -a) -z+zln(z +a)]4'11'foa Thus, -AO[ (z-a)]~(O,O,z) = -4-2a+z1n -- (2) '1I'fo , z+ a Because of the symmetry about the z axis, the only component of E is in the z direction a~. Ao [(z-a) {1 1}]. E=--I.=- 1n --+z ----- I. az 4'11'fo z +az -az +a (3) Ao [1 (z-a) 2az]. =-n--+ I.4'11'fo z +a z2 -a2 4.5.1 Using (4.5.20) 1b (J' (d-b)la ~ =- 0 Inld -x'ldz'dy' 1/'=0 :z'=-b ~2'11'fo(d -z') =_ (J'o(d -b) r 1n(d -z') dz' 2'11'fo J:z'=-b (d -z') =_ (J'o(d -b) {_ ~[ln(d _ z')]2\b } 2'11'fo 2 -b = (J'o(d -b) {[In(d _ b)]2 _ [In(d +b)]2} 4'11'fo 4-22 Solutions toChapter4 4.5.8 Feom(4.5.20), 12dalnld-z'l10alnld-z'ldz'~(d,0)=_° dz'+ _0=----"'---_...:.-_ :z'=o 211"Eo :z'=-2d 211"Eo Tointegrate letu=d-z'anddu=-dz'. Thus,setting ~(d,0)=Vgives 211"EoV ao=3dln3 y -2d 2d Figure84.5.8(1) (2) (3) 4.5.9 (a)(Thisproblem mightbestbegivenwhilecovering Sec.8.2,whereastick modelisdeveloped forMQSsystems.) Atthelowerendofthecharge,ecis theprojection ofcona.Thisisgivenby Similarly,(1) (2) (b)Feom(4.5.20), (3) 4-23 Solutions to Chapter 4 where With (J defined as the angle between a and b, Idl = Iblsin(J (4) But in terms of a and b, . laxbl sm(J = lallbl (5) so that d= laxbl lal (6) and (7) (c) Integration of (3) using (6) and (7) gives (8) and hence the given result. (d) For a line charge Ao between (z, y, z) = (0,0, d) and (z, y, z) = (d, d, d), a = dix + di)' b = (d -z)ix + (d -y)i)' + (d -z)i. c = -zix -yi)' + (d -z)i. b. a= d(d -x) + d(d -y) c·a = -xd- yd ix iyi. I (9)axb= dd 0Id-x d-y d-z = d(d -z)ix -d(d -z)i)' + d(x -y)i. la x bl2 =~[2(d -z)2 + (z _ y)2] (b· a)2 = d2[(d -z) + (d -yW (c .a)2 =~(x + y)2 and evaluation of (c) of the problem statement gives (d). Solutions to Chapter 4 4-24 4.5.10 This problem could be given in connection with covering Sec. 8.2. It illus­ trates the steps followed between (8.2.1) and (8.2.7), where the distinction between source and observer coordinates is also essential. Given that the potential has been found using the superposition integral, the required electric field is found by taking the gradient with respect to the observer coordinates, r, not r'. Thus, the gradi­ ent operator can be taken inside the integral, where it operates as though r' is a constant. E = -V~ =-r V[ p(r') ]dv' =_ r p(r') V[_1_]dv' (1)lv 4'11"£0Ir -r'l lv' 4'11"£0 Ir -r'l The arguments leading to (8.2.6) apply equally well here 1 1 V[--] = - ir'r (2)lr- r'l Ir-r'12 The result given with the problem statement follows. Note that we could just as well have derived this result by superimposing the electric fields due to point charges p(r')dv'. Especially if coordinates other than Cartesian are used, care must be taken to recognize how the unit vector ir'r takes into account the vector addition. 4.5.11 (a) Substitution of the given charge density into Poisson's equation results in the given expression for the potential. (b) If the given solution is indeed the response to a singular source at the origin, it must (i) satisfy the differential equation, (a), at every point except the origin and (ii) it must satisfy (c). With the objective of showing that (i) is true, note that in spherical coordinates with no 6 or q, dependence, (b) becomes (1) Substitution of (e) into this expression gives zero for the left hand side at every point, r, except the origin. The algebra is as follows. First, (2) Then, 1 d (Alt -lCr e-lCr ) 2 Ae-lCr Ak2 -lCr Ak2 -lCr --- -e +-- -It-- = -e +-e (3)r2 dr r r2 r r2 r = OJ r",0 To establish the coefficient, A, integrate Poisson's equation over a spherical volume having radius r centered on the origin. By virtue of its being singular 4-25 Solutions to Chapter 4 there, what is being integrated has value only at the origin. Thus, we take the limit where the radius of the volume goes to zero. lim { (V.V~dV-1I:2 (~dv}=lim{--!.. (sdv} (4) r-O Jv Jv r-O fa Jv Gauss' theorem shows that the first integral can be converted to a surface integral. Thus, lim { 1 V~· da -11:2( ~dv} = lim{--!.. ( sdv} (5) r-O Is Jv r-O fa Jv H the potential does indeed have the r dependence of (e), then it follows that (6) so that in the limit, the second integral on the left in (5) makes no contribution and (5) reduces to . ( All: -lCr Ae-lCr ) 2 Q11m --e ---- 4'11"r = -4'11"A = -- (7)r-O r r2 fa and it follows that A = Q/ 4'11"f o ' (c) We have found that a point source, Q, at the origin gives rise to the potential (8) Arguments similar to those given in Sec. 4.3 show that (b) is linear. Thus, given that we have shown that the response to a point source p(r')dv atr =r' is p(r')dve-1C1r-r'\ p(r')dv ~~ = 4'11"foI r -r'I (9) 1it follows by superposition that the response to an arbitrary source distribu­ tion is p(r')e-lClr-r'l ~(r) = dv (10) V 4'11"fo(r-r'l 4.5.12 (a) A cross-section of the dipole layer is shown in Fig. 84.5.12a. Because the field inside the layer is much more intense than that outside and because the layer is very thin compared to distances over which the surface charge density varies with position in the plane of the layer, the fields inside are as though the surface charge density resided on the surfaces of plane parallel planes. Thus, Gauss' continuity condition applied to either of the surface charge densities 4-26 Solutions toChapter4 showsthatthefieldinsidehasthegivenmagnitude andthedirection mustbe thatofthenormalvector. (a)11l (oJ ~~!::+t:~:J:d I z (b)z+6z (b)(1) Figure94.5.13 (b)Itfollowsfrom(4.1.1)andthecontour showninFig.S4.5.12b havingincre­ mentallength I::1.xinthexdirection that Divided byI::1.x,thisexpression becomes _EaEbdaE"=0 ",+",+ax(2) (3) Thegivenexpression thenfollowsbyusing(1)toreplaceE"with-~land recognizing that1/".==u.d. \ho 4.6ELECTROQUASISTATIC FIELDS INTHEPRESENCE OFPERFECT CONDUCTORS 4.6.1 Inviewof(4.5.12), lbA(a-*') ~(O,0,a)=4t-c ')dz'c1/"Eoa-z(1) Thezdependence oftheintegrand cancelsoutsothattheintegration amounts to amultiplication. Thenetchargeis~(O,O,a)=4~o)(b-c) 1/"Eoa-C 1a-bQ=-[Ao(-)+Ao](b-c)2a-c(2) (3) Solutions toChapter4 4-27 Proviedthattheequipotential surfacepassingthrough (0,0,a)encloses allofthe segment, thecapacitance ofanelectrode havingtheshapeofthissurfaceisthen givenby Qc=~(O,O,a)=211"Eo(2a-b-c) (4) 4.6.2 (a)Thepotential isthesumofthepotentials duetothechargeproducing the uniformfieldandthepointcharges.Withr±definedasshowninFig.84.6.2a, where z=rcos(Jq(1) dr±=r2+(d/2)2T2r2"cos(J Towrite(1)intermsofthenormalized variables, dividebyEodandmultiply anddivider±byd.Thegivenexpression, (b),thenfollows. z 5 1 (a) (b)o 1-r.2 (2)Flsure94.8.2 (b)Animplicitexpression fortheintersection pointd/2<ronthezaxisisgiven byevaluating (b)with~=aand(J=O. r=i_q -(r.-~)(r.+~) Thegraphical solution ofthisexpression ford/2<r(I/2<r.)isshownin Fig.84.6.2b. Therequired intersection pointisr.=1.33.Because theright handsideof(2)hasanasymptote atr.=0.5,theremustbeanintersection between thestraightlinerepresenting theleftsideintherange0.5<r.. 4-28 Solutions toChapter 4 (c)Theplotofthe~=0surfacefor0<(J<1r/2isshowninFig.S4.6.2c. z 1 (c) (3)1 Flpre94.8.3 (d)Atthenorthpoleoftheobject,theelectricfieldisz-directed. Ittherefore followsfrom(b)as(0.5<d E.=-a~=-Eoa~=-Eo!....(-r+i1-~)ar ar. ar r.-2"r.+2 =Eo[1+q2=q2] (r-!)(r+!) Evauation ofthisexpression atr=1.33andi=2givesE.=3.33Eo• (e)Gauss)integrallaw,appliedtoasurfacecomprised oftheequipotential and theplanez=0,showsthatthenetchargeonthenorthern halfoftheobject isq.Forthegivenequipotential, 9.=2.Itfollowsfromthedefinition of9.that 4.6.3 ForthediskofchargeinFig.4.5.3,thepotential isgivenby(4.5.7) ~=0'0(VW+212-1211)2Eo At(0)0,d), ~(O,O)d)=0'0(VW+d2-d)2Eo(4) (1) (2) Solutions toChapter 4 and Thus4-29 (3) (4) 4.6.4 (a)Duetothetopsphere, andsimilarly,(1) (2) Atthebottomofthetopsphere whileatthetopofthebottomsphere(3) (4) Thepotential difference between thetwospherical conductors istherefore (3) Themaximum fieldoccursatz=0ontheaxisofsymmetry wherethe magnitude isthesumofthatduetopointcharges. (4) (b)Replace pointchargeQatz=h/2byQl=Q~atz=~-1J.2andQo= Q[l- ~latz=h/2.Thepotential onthesurfaceofthe-topsphereisnow Q(5) Thepotential onthesurfaceofthebottomsphereis () Qo Ql Q bottom=411"€o(h-R)+411"€o(h_R_~2)--411"-€-oR-(6) 4-30 Solutions to Chapter 4 The potential difference is then, For four charges Ql = QR/h at z = h/2 -R2/hj Qo = Q(1- ~) at z= h/2j Q2 = -QR/h at z = -h/2 + R2/hj Q3 = -Q(1- ~) at z = -h/2 and Cbtop = ~+ Q(l R) + (Q2 R2)41rfoR 41rfoR 1-h 41rfo h -R -h (7) + Q3 41rfo(h -R) which becomes (8) Similarly, Cb Q(R/h) Q(R2/h2) bottom = 41rfo R + 41rf o R(1-~ _ *) (9)QR/h Q(1-R/h) 41rfo R(1-f) 41rfo so that (10) v Q 2R R/h (R/h)2} {1= 21rfoR -h + 1-R/h -1-~ -(R/h)2 (11) (12) Solutions toChapter4 4-31 (1)4.6.5 (a)Thepotential isthesumofthatgivenby(a)inProb.4.5.4andapotential due toasimilarly distributed negative linechargeonthelineatz=-abetween y=-d/2andy=d/2. ~=~ln{ [~-y+./(z-a)2+(~-y)2+z2]4~ 2V 2 [-~-y+J(x+a)2+(~+y)2+Z2]/ [-~-y+J(x-a)2+(~+y)2+z2] [~-Y+J(x+a)2+(~-y)2+z2]} (b)Theequipotential passingthrough(x,11,z)=(a/2,0,0)isgivenbyevaluating (1)atthatpoint (2) 2 ~i 1 o 1 2 Figure84.8.5 (c)Innormalized form,(2)becomes (3) 4-32 Solutions to Chapter 4 where ~ = ~/~(~, 0, 0), e= :&/a,,, = y/a and d = 4a. Thus, ~ = 1 for the equipotential passing through (~,O,O). This equipotential can be found by writing it in the form f(e, '7) = 0, setting '7 and having a programmable calculator determine e. In the first quadrant, the result is as shown in Fig. S4.6.5. (d) The lines of electric field intensity are sketched in Fig. S4.6.5. (e) The charge on the surface of the electrode is the same as the charge enclosed by the equipotential in part (c), Q = Ald. Thus, c = Aid = 41rE d/ln{ [d +va2 +d2][-d +V9a2 +d2]} (4) V o[-d +va2+d2][d +v9a2+d2] 4.7 METHOD OF IMAGES 4.7.1 (a) The potential is due to Q and its image, -Q, located at z = -d on the z axis. (b) The equipotential having potential V and passing through the point z =a < d, :& = 0, Y =0 is given by evaluating this expression and taking care in taking the square root to recognize that d > a. (1) In general, the equipotential surface having potential V is v--.!L[ _ 1 ] () 1 -41rE o V:&2 +y2 +(z -d)2 V:&2 +y2 +(z +d)2 2 The given expression results from equating these last two expressions. (c) The potential is infinite at the point charge and goes to zero at infinity and in the plane z = O. Thus, there must be an equipotential contour that encloses the point charge. The charge on the electrode having the shape given by (2) must be equal to Q so the capacitance follows from (1) as Q (~- a2 )C =-= 21rE o ":"""---<- (3)V a 4.7.2 (a) The line charge and associated square boundaries are shown at the center of Fig. S4.7.2. In the absence of image charges, the equipotentials would be circular. However, with images that alternate in sign to infinity in each di­ rection, as shown, a grid of square equipotentials is established and hence the boundary conditions on the central square are met. At each point on the Solutions to Chapter 4 4-33 boundary, there is an equal distance to both a positive and a negative line charge. Hence, the potential on the boundary is zero. -------L "i I-~-------i-------~.I I I I I ' I I I I I I I I I I + I - : + I , I I I 'T-------­I I I I I I I I ~---------- +-< '"--------!~L------ -~ I I I I I I I I , I + I I + I I I I I I I I I I I I ~t-------.± --------1--------,[-1 Figure 94.7'.3 (b) The equipotentials close to the line charge are circular. As the other boundary is approached, they approach the square shape of the boundary. The lines of electric field intensity are as shown, tenninating on negative surface charges on the surface of the boundary. 4.7.3 (a) The bird acquires the same potential as the line, hence has charges induced on it and conserves charge when it flies away. (b) The fields are those of a charge Q at y = h, z= Ut and an image at y = -h and z= Ut. (c) The potential is the sum of that due to Q and its image -Q. ~_ Q[1 1] () -411"E o y!(z -Ut)2 + (y -h)2 +z2 -y!(z _ Ut)2 + (y + h)2 +z2 1 (d) From this potential E a~ Q{ y-h "=-ay = 41l'E o (z -Ut)2 + (y -h)2 + z213/ 2 (2) y+h } -[(z -Ut)2 + (y + h)2 +z213/2 Thus, the surface charge density is 4-34 Solutions toChapter4 U-EEl-QEo[ -h •-01/1/=0-411'Eo[(x-Ut)2+h2+z213/2 -[(x-Ut)2:h2+z213/2] -Qh =211'[(:Z:-Ut)2+h2+z2]3/2 (e)Thenetchargeqonelectrode atanygiveninstantis lw{' -Qhd:z:dz q=.=0},,,=o211'[(:Z:-Ut)2+h2+z213/2 ITw<:h, {' -Qhwdx q=1z=0211'[(x-Ut)2+h2]3/2 Fortheremaining integration, x'=(x-Ut),d:z:'=dxand j'-Ut-Qhwdx' q=-Ut211'[x'2+h2]3/2(3) (4) (5) (6) Thus ---"'---,/ (2) "Qw[l-Ut Ut] q=-211'hV(l-Ut)2+h2+V(Ut)2+h2 (7) (f)Thedabsedcurves(1)and(2)inFig.84.7.3arethefirst·andsecondtermsin (7),respectively. Theysumtogive(3) q ----r-- ... (\)"", " --=~--+:--7"T---==--. Ut (a) II (h) Figure 84..f.S 4-35 Solutions to Chapter 4 (g) The current follows from (7) as . dq Qw [-Uh2 Uh2 ] (8),= dt =-21rh [(l --Ut)2 + h2]3/2 + [(Ut)2 + h2]3/2 and the voltage is then tJ = -iR = -Rdq/dt. A sketch is shown in Fig. S4.7.3b. 4..7'.4. For no normal E, we want image charges of the same sign; +.A at (-a, 0) and -.A at (-b, 0). The potential in the z = 0 plane is then, 2.A 2.A /~ = --In(a2 + !l)1/2 + -In(b2 + y2)1 221rf o 21rf o .A a2 + y2 (1) = -211"f In( b2 + y2 ) o 4..7'.5 (a) The image to make the z = 0 plane an equipotential is a line charge -.A at (z, y) = (:-d, d). The image of these two line charges that makes the plane y = 0 an equipotential is a pair of line charges, +.A at (-d, -d) and -.A at (d, -d). Thus ~ = -_.A-1n[(z -d}2 + (y-d)2] -~'n[(z + d)2 + (y+ d}2]41rf o 41rf o + ~ln[(z --d)2 + (y+ d)2] + ~ln[(z + d)2 + (y-d)2] (1)41rf o 41rf o __.A_{ [(z -d)2 + (y+ d)2][(z + d)2 + (y-d)2] }ln -41rf o [(z -d)2 + (y-d)2][(z + d)2 + (y+ d)2] (b) The surface of the electrode has the potential ~ aa = _.A_ ln { [(a -d)2 + (a + d)2][(a + d)2 + (a -d)2] }= V (2)( ,) 41rfo [(a -d)2 + (a -d)2][(a + d)2 + (a + d)2] Then (3) 4..7'.6 (a) The potential of a disk at z = s is given by 4.5.7 with z -z -s ~(z>s)= {70 [VR2+(z-s)2--lz-sl] (1)2fo The ground plane is represented by an image disk at z = -s; (4.5.7) with z -z + s. Thus, the total potential is 4-36 (b)Thepotential atz=d<sisSolutions toChapter4 w(z=d<s)=!!!!..[y'R2+(d-s)2-Id-sl-y'R2+(d+S)2+\d+s\]2Eo =(To[y'R2+(d-s)2_(s-d)-v'R2+(d+s)2+s+d]2Eo =(To[y'R2+(d-s)2+2d-v'R2+(d+s)2] =v2Eo(3) Thus, 4..'1.'1 From(4.5.4), 12'11'lR !!D.rdrdq,12 '11'lR-!!iJI..rdrdq,W(O,0,a)= R + _----;~R?=::::====:;::;: "'=0r=O41rEoy'r2+(h-a)2 "'=0r=O41rEov'r2+(h+a)2 (To[lRr2drlRr2dr ]=2EoRr=Oy'r2+(h-a)2-r=Oy'r2+(h+a)2 =...!!.2-[R(y'R2+(h_a)2 4EoR2 h-a-yR2+(h+a)2)+(h-a)21n( )y'R2+(h-a)2 +(h+a)21n(R+yr=~::-2+-+-;'a(h:--+~a)=2)] Thetotalchargeinthediskis Thus, 0=~={21rR3Eo}/{!J[y'R2+(h-a)2 -y'R2+(h+a)2] h-a+(h-a)21n( )v'R2+(h-a)2 +(h+a)21n(R2+~~+:}h+a)2)}(1) Solutions toChapter4 4-37 (1)4.7.8 Because thereisperfectly conducting material atz=°thereisthegivenline chargeandanimagefrom(O,O,-d) to(d,d,-d). Thus,fortheserespective line charges a=dix+di)' f=(d-xlix+(d-y)i),+(±d-z)i. c=-xix-yi)'+(±d-z)i. b·a=d[(d-x)+(d-y)] c·a=-xd-yd axb=d(±d-z)ix-i)'d(±d-z)+i.d[(d-y)-(d-x)] laxbl=d2(±d-z)2+d2(±d-z)2+~[(d-y)-(d-X)]2 Thepotential duetothelinechargeanditsimagethenfollows(c)ofProb.4.5.9. A{2d-x-y+V2[(d-x)2+(d-y)2+(d-z)2] Cb=--In 41/"Eo -x-Y+V2[x2+y2+(d-z)2] -x-Y+V2[x2+y2+(d+z)2] } .2d-x-y+V2[(d-x)2+(d-y)2+(d+z)2] 4.8CHARGE SIMULATION APPROACH TOBOUNDARY VALUE PROBLEMS 4.8.1 Forthesix-segment system,thefirsttwoof(4.8.5)are(2) Because ofthesymmetry, (3) andsothesetwoexpressions reducetotwoequations intwounknowns. (Theother fourexpressions areidentical to(4).) (4) 4-38 Solutions to Chapter 4 Thus, V 0'1 = 2D [(822 -825) -(812 -815)] (5) V 0'2 = 2D[(8 11 +813 -814 -816) -(821 +823 -824 -826)] (6) where D = (811 +813 -814 -816)(822 -825) -(821 +823 -824 -826)(812 -815) and from (4.8.3) (7) SOLUTIONS TO CHAPTER 5 5.1 PARTICULAR AND HOMOGENEOUS SOLUTIONS TO POISSON'S AND LAPLACE'S EQUATIONS 5.1.1 The particular solution must satisfy Poisson's equation in the region of in­ terest. Thus, it is the first term in the potential, associated with the charge in the upper half plane. What remains satisfies Laplace's equation everywhere in the region of interest, so it can be called the homogeneous solution. It might also be made part of the particular solution. 5.1.2 (a) The charge density follows from Poisson's equation. V2~ = _.!!... => P = Pocos{3z (1) Eo (b) The first term does not satisfy Laplace's equation and indeed was responsible for the charge density, (1). Thus, it can be taken as the particular solution and the remainder as the homogeneous solution. In that case, ~ _ Po cos {3z. ~h =_Po cos {3z cosh {3y p- Eo{32 ' Eo{32 cosh {3a (2) and the homogeneous solution must satisfy the boundary conditions Po cos (3z ~h(Y = -a) = ~h(Y = a) = -:.....:...----::-=-­ (3) Eo{32 (c) We could just have well taken the total solution as the particular solution. ~p =~; ~h =0 (4) in which case the homogeneous solution must be zero on the boundaries. 5.1.3 (a) Because the second derivatives with respect to y and z are zero, the Laplacian reduces to the term on the left. The right side is the negative of the charge density divided by the permittivity, as required by Poisson's equation. (b) With 0 1 and O2 integration coefficients, two integrations of (b) give ~ 4po (x -d)4 0C (1)=-d2E 12 + 1X+ 2 o Evaluation of this expression at each of the boundaries then serves to deter­ mine the coefficients (2) 1 Solutions to Chapter 5 5-2 and hence the given potential. (c) From the derivation it is clear that the Laplacian of the first term accounts for all of the charge density while that of the remaining terms is zero. (d) On the boundaries, the homogeneous solution, which must cancel the potential of the particular solution on the boundaries, must be (d). 5.1.4 (a) The derivatives with respect to y and z are by definition zero, so Poisson's equation reduces to tP. = _Po sin ('II"z) (1)dz2 Eo d (b) Two integrations of (1) give PotP . ('II"z) .= --2 sm -d +01Z+02 (2) Eo 1/" and evaluation at the boundaries determines the integration coefficients. (3) It follows that the required potential is .... PotP. ('II"Z) Vz"I/!=--sm -+- (4) Eo 1r2 d d (c) From the derivation, the first term in (4) accounts for the charge density while the remaining terms have no second derivative and hence no Laplacian. Thus, the first term must be included in the particular solution while the remaining term can be defined as the homogeneous solution. Vz.h=­ (5)d (d) In the case of (c), it follows that the boundary conditions satisfied by the homogeneous solution are (6) 5.1.5 (a) There is no charge density, so the potential must satisfy Laplace's equation. E= (-v/d)i. = -8./8z v2• = ~(8.) =0 (1)8s 8s (b) The surface charge density on the lower surface of the upper electrode follows from applying Gauss' continuity condition to the interface between the highly Solutions to Chapter 5 5-3 conducting metal and the free space just below. Because the field is zero in the metal, u. = folO -E~I = f~tJ (2) (c) The capacitance follows from the integration of the surface charge density over the surface of the electrode having the potential tJ. That amounts to multiplying (2) by the area A of the electrode. foA q = Au. = -tJ = ev (3)d (d) Enclose the upper electrode by the surace S having the volume V and the integral form of the charge conservation law is 1J. nda + ~ rpdV = 0 (4)J8 dt lv Contributions to the first term are confined to where the wire carrying the total current i into the volume passes through S. By definition, the second term is the total charge, q, on the electrode. Thus, (4) becomes (5) Introduction of (3) into this expression then gives the current dtJi = e (6)dt 5.1.6 (a) Well away from the edges, the fields between the plates are the potential difference divided by the spacings. Thus, they are as given. (b) The surface charge densities on the lower surface of the upper electrode and on the upper plus lower surfaces of the middle electrode are, respectively (1) (2) Thus, the total charge on these electrodes is these quantities multiplied by the respective plate areas (3) q2 = folwu m (4) These are the expressions summarized in matrix notation by (a). 5-4 Solutions to Chapter 5 5.2 UNIQUENESS OF SOLUTIONS OF POISSON'S EQUATION 5.3 CONTINmTY CONDITIONS 5.3.1 (a) In the plane y = 0, the respective potentials are (1) and are therefore equal. (b) The tangential fields follow from the given potentials. (2) Evaluated at y = 0, these are also equal. That is, if the potential is continuous in a given plan, then so also is its slope in any direction within that plane. (c) Feom Gauss' continuity condition applied to the plane y = 0, (3) and this is the given surface charge density. 5.3.2 (a) The y dependence is not given. Thus, given that E = -V~, only the :z; and z derivatives and hence :z; and z components of E can be found. These are the components of E tangential to the surface y = 0. If these components are to be continuous, then to within a constant so must be the potential in the plane y=O. (b) For this particular potential, Es = -f3V cos f3:z;sin PZj Ez = -pV sin p:z;cos pz (1) Ifthese are to be the tangential components ofE on both sides of the interface, then the :z; -z dependence of the potential from which they were derived must also be continuous (within a constant that must be zero if the electric field normal to the interface is to remain finite). Solutions to Chapter 5 5-5 5.4 SOLUTIONS TO LAPLACE'S EQUATION IN CARTESIAN COORDINATES 5.4.1 (a) The given potential satisfies Laplace's equation. Evaluated at either :r; = 0 or y =0 it is zero, as required by the boundary conditions on these boundaries. At :r; = a, it has the required potential, as it does at y = a as well. Thus, it is the required potential. (b) The plot of equipotentials and lines of electric field intensity is obtained from Fig. 4.1.3 by cutting away that part of the plot that is outside the boundaries at :r; = a, y = a,:r; = 0 and y = O. Note that the distance between the equipotentials along the line y = a is constant, as it must be if the potential is to have a linear distribution along this surface. Also, note that except for the special point at the origin (where the field intensity is zero anyway), the lines of electric field intensity are perpendicular to the zero potential surfaces. This is as it must be because there is no component of the field tangential to an equipotential. 5.4.2 (a) The pote~tials on the four boundaries are ~(a, y) =V(y +a)/2a; ~(-a, y) =V(y -a)/2a ~(:r;, a) =V(:r; + a)/2a; ~(:r;,-a) =V(:r; -a)/2a (1) (b) Evaluation of the given potential on each of the four boundaries gives the conditions on the coefficients vV ~(±a,y) = 2aY ±"2 = ±Aa+By+C+D:r;y VV ~(:r;, ±a) = -2:r;± -= A:r; ± Ba + C + D:r;y (2)a 2 Thus, A = B = V /2a, C = 0 and D = 0 and the equipotentials are straight lines having slope -1. V ~ = -(:r;+y) (3)2a (c) The electric field intensity follows as being uniform and having :r; and y com­ ponents of equal magnitude. E= -V~ = -!.(i x +i)') (4)2a (d) The sketches ofthe potential, (3), and field intensity, (4), are as shown in Fig. 85.4.2. Solutions to Chapter 5 5-6 y x Figure 85.4.3 (e) To make the potential zero at the origin, C =O. Evaluation at (x, y) = (0, a) where the potential must also be zero shows that B =O. Similarly, evaluation at (x,y) = (a,O) shows that A = O. Evaluation at (z,y) = (a, a) gives D = V 12a2 and hence the potential v C)= -zy (5)2a2 Of course, we are not guaranteed that the postulated combination of solu­ tions to Laplace's equation will satisfy the boundary conditions everywhere. However, evaluation of (5) on each of the boundaries shows that it does. The associated electric field intensity is (6) The equipotentials and lines of field intensity are as shown by Fig. 4.1.3 inside the boundaries z = ±aand y = ±a. 5.4.3 (a) The given potential, which has the form of the first term in the second column of Table 5.4.1, satisfies Laplace's equation. It also meets the given boundary conditions on the boundaries enclosing the region of interest. Therefore, it is the required potential. (b) In identifying the equipotential and field lines of Fig. 5.4.1 with this configu­ ration, note that k = 1rIa and that the extent of the plot that is within the region of interest is between the zero potentials at z = -1r12k and z = 1r12k. The plot is then adapted to representing our potential distribution by multi­ plying each of the equipotentials by Vo divided by the potential given on the plot at (x, y) =(0, b). Note that the field lines are perpendicular to the walls at x =±a/2. Solutions to Chapter 5 5-7 5.4.4 (a) Write the solution as the sum of two, each meeting zero potential conditions on three of the boundaries and the required sinusoidal distribution on the fourth. .... _ T' • (1rZ) sinh(1ry/a) TF' 1ry sinh[;-(a -z)] (1) .., -YoSln . h() + Yosm .h() a sm1r a Sln1r (b) The associated electric field is E =-as::~1r) {[cos(1rz/a) sinh(1ry/a) -sin(1ry/a) cosh [;(a -z)]]ix +[sin(1rz/a) cosh(1ry/a) + cos(1ry/a) sinh [;(a -z)]] iy } y (2) Figure 85.4.4 (c) A sketch of the equipotentials and field lines is shown in Fig. 85.4.4. 5.4.5 (a) The given potential, which has the form of the second term in the second column of Table 5.4.1, satisfies Laplace's equation. The electrodes have been shaped and constrained in potential to match the potential. For example, between y = -b and y= b, we obtain the y coordinate of the boundary '7(z) as given by (a) by setting (b) equal to the potential v of the electrode, y= '7 and solving for '7. (b) The electric field follows from (b) as E = -VCb. (c) The potential given by (b) and field given by (c) have the same (z, y) depen­ dence as that represented by Fig. 5.4.2. To adjust the numbers given on the plot for the potentials, note that the potential at the location (3:, y) = (0, a) on the upper electrode is v. Thus, to make the plot fit this situation, multiply 5-8 Solutions to Chapter 5 each of the given potentials by tI divided by the potential given on the plot at the location (x, y) = (0, a). (d) The charge on the electrode is found by enclosing it by a surface S and using Gauss' integral law. To make the integration over the surface enclosing the electrode convenient, the surface is selected as enclosing the electrode in an arbitrary way in the field free region above the electrode, passing through the slits in the planes x = ±l to the y equal zero plane and closing in the y = 0 plane. Thus, with Yl defined as the height of the electrode at its left and right extremities, the net charge is Y1 q = dfo -Ex(x = -l)dy + dfo lYl Ex (x = l)dy l ~o ~o + dfo l~-, -Ey(Y = O)dx [lY1tld1l"f o . 1I"l .h 1I"Yd=- -sm-sm -y (2) 2b sinh(;:) 0 2b 2b lY1 1I"l 1I"y+ -sin -sinh -dy o 2b 2b 1I"x]-cosbdx+j_// 2 Note that .h k sinh ka sm Yl = --kl- j -sinh2 ky + cosh2 ky = 1 (3)cos and (2) becomes the given result. (e) Conservation of charge for a surface enclosing the electrode through which the wire carrying the current i passes requires that i= dq/dt. Thus, given the result of (d) and the voltage dependence, (e) follows. 5.4.6 (a) Reversing the potentials on the lower electrodes turns the potential from an even to an odd function of y. Thus, the potential takes the form of the first term in the second column of Table 5.4.1. 1I"Y) 11" X ~ = Acosh (-b cos- (1)2 2b To make the potential be tI at (x, y) = (0, a)' the coefficient is adjusted so that coshky k =_ ~ ~ = tI cos kx cos h ka j 2b (2) The shape of the upper electrode in the range between x = -b and x = b is then obtained by solving (2) with ~ = tI and y = '1 for '1. '1 -_-!k cosh-1 [COShkka] (3)cos x Solutions toChapter5 5-9 (b)Theelectricfieldintensity followsfrom(2)as E=-tJ:kl-sin(kz)cosh(ky)lx+coskzsinhkyly] (4)cosa (c)Theequipotentials andfieldlinesareasshownbyFig.5.4.2.Toadjustthe givenpotentials, multiply eachbytJdividedbythepotential givenfromthe plotatthelocation (z,y)=(0,a). (d)Thechargeontheelectrode segment isobtained byusingGauss'integrallaw withasurfacethatenclosestheelectrode. Thissurfaceisarbitrary inthefield freeregionabovetheelectrode. Forconvenience, itpassesthrough theslits tothey=0planeintheplanesz=±landclosesinthey=0plane.Note thatthereisnoelectricfieldperpendicular tothislattersurface,sotheonly contributions tothesurfaceintegration comefromthesurfacesatz=±l. q=2dEo1"[co::kasin(kl)COSh(kY)]dy 2dEotJ.kl.hk=hksmsmYIcosa Withtheuseoftheidentities coshka cosh(kYI) =kljcos(5) (6) (5)becomes 2dEotJ• q=etJ=hksmklcosa (e)Fromconservation ofcharge,[cosh(ka)] 2_1 coskl(7) .edtJev..t=-=-cJJJsmwtdt 5.5MODAL EXPANSIONS TOSATISFY BOUNDARY CONDITIONS 5.5.1 (a)Thesolutions superimposed bytheinfiniteseriesof(a)arechosentobezero intheplanesz=0andz=bandtobethelinearcombination ofexponentials intheydirection thatarezeroaty=b.Toevaluate thecoefficients, multiply bothsidesbysin(m1rz/a) andintegrate fromz=0toz=a Solutions to Chapter 5 5-10 The integral on the right is zero except for m = n, in which case the integral of sin2 (n1r:r:/a) over the interval :r: = 0 to :r: = a gives the average value of 1/2 multiplied by the length a, a/2. Thus, (1) can be solved for the coefficient Am, to obtain (b) as given (if m -+ n). (b) In the specific case where the distribution is as given, the integration of (b) gives 32 1 0./' n1r:r:An =. (Rfrb) V1sin (-)d:r: a~mh -G 0./' a (2) 2V1 [ n1r:r: ] 30./'= ----=:,--:7" cos (--) n1rsinh (n:b) a 0./' which becomes (c) as given. 5.5.2 (a) This problem illustrates how the modal approach can be applied to finding the solutions in a rectangular region for arbitrary boundary conditions on all four of the boundaries. In general, four infinite series would be used, each with zero potential on three of the walls and with coefficients to match the potential boundary condition on the fourth wall. Here, the potential is zero on two of the walls, so only two infinite series are used. The first is zero in the planes y = 0, 'II = band :r: = a and, because the potential is constant in the plane :r: =0, has coefficients that are as given by (5.5.8). (The roles of a and b are reversed relative to those in the section for this first term and the minus sign results because the potential is being matched at :r: =O. Note that the argument of the sinh function is negative within the region of interest.) The coefficients of the second series are similarly determined. (This time, the roles of :z: and 11 and ofa and b are as in the section discussion, but the surface where the uniform potential is imposed is at 'II = 0 rather than 'II =b.) (b) The surface charged density on the wall at :J: = a is 8~ a. = fo[-Es(:r: = a)1 = -fo 8:r: (:r: = a) (1) Evaluation using (a) results in (b). 5.5.3 (a) For arbitrary distributions of potential in the plane 'II =0 and :r: =0, the potential is taken as the superposition of series that are zero on all but these planes, respectively. (1) +L00 Bn sin (n;'II) sinh [n1r (:r: -a))b'1=1 The first of these series must satisfy the boundary condition in the plane '11=0, ~(:J: =0) =f: An sinh ( -mrb) sin (n1r :r:) (2) '1=1 a a 5-11 Solutions to Chapter 5 where .(:z: 0) _ { 2Vo.:z:/a; 0 < :z: < a/2 (3),- 2Vo.(1 -:z:/a); a/2 < :z: < a Multiplication of both sides of (2) by sin(mll':z:/a) and integration from :z: = 0 to:z: = a gives 2V. 10./2 mll':Z: 10. mll':Z:----!!. nin (-)d:z: + 2Vo. sin (-)d:z: a 0 a 0./2 a 2V0.10. . (mll':Z:) (4) -- :Z:SIn -- d:z: a 0./2 a a . mll'b = Am-sinh (--)2 a Integration, solution for Am -+ An then gives An = 0, n even and for n odd 8Vo.sin (T) n211'2 sinh (n:b) (5) Evalution on the boundary at :z: = 0 leads to a similar term with the roles of Vo. and a replaced by those of Vb and b, respectively. Thus , Bn = 0 for n even and for n odd 8Vi sin (!!!t) B __ b 0. (5)n - n211'2 sinh (n~o. ) (b) The surface charge density in the plane y = b is a. 0'. = fo[-EI/(Y = b)1 =f o 8y (y = b) ~ [ (nll'). (nll':Z:) (nll'). [(nll') ] (6)= L..J An - Sin - -Bn -b Sinh -b (:z: -a) ..=1 a a odd where An and Bn are given by (5) and (6). 5.5.4 (a) Far to the left, the system appears as a parallel plate capacitor. A uniform field satisfies both Laplace's equation and the boundary conditions. E =-V i)' =>.0. = Vy (1)d d (b) Because the uniform field part of this solution I .0., satisfies the conditions far to the left, the aditional part must go to zero there. However, the first term produces a field tangential to the right boundary which must be cancelled by the second term. Thus , conditions on the second term are that it also satisfy Laplace's equation and the boundary conditions as given 5-12 Solutions to Chapter 5 (c) Because of the homogeneous boundary conditions in the y = 0 and y = d planes, the solution is selected as being sinusoidal in the y direction. Because the region extends to infinity in the -z direction, exponential solutions are used in that direction, with the sign of the exponent arranged to assure decay in the -z direction. 00 iWo. ~ A • (n1l"Y) rnrz/d (2) ....b = LJ n sIn d e n=l The coefficients are determined by the requirement on this part of the poten­ tial at z = o. Vy ~ . (n1l"Y)-d = LJAnsIn d (3) n=l Multiplication by sin(m1l"y/d), integration from y =a to y = d, solution for Am and replacement of Am by An gives 2V 2VAn = -cosn1l" = _(_I)n (4)n1l" n1l" The sum of the potentials of (1) and (2) with the coefficient given by (4) is (e). (d) The equipotential lines must be those of a plane parallel capacitor, (1), far to the left where the associated field lines are y directed and uniform. Because the boundaries are either at the potential V or at zero potential to the right, these equipotential lines can only terminate in the gap at (z, y) = (0, d), where the potential makes an abrupt excursion from the zero potential of the right electrode to the potential V of the top electrode. In this local, the potential lines converge and become radially symmetric. The boundaries are themselves equipotentials. The electric field, which is perpendicular to the equipotentials and directed from the upper electrode toward the bottom and right electrodes, can then be pictured as shown by Fig. 6.6.9c turned upside down. 5.5.5 (a) The potential far to the left is that of a plane parallel plate capacitor. It takes the form Az + B, with the coefficients adjusted to meet the boundary conditions at z =0 and z = a. Cb(y -. -00) -. Cba = Va (1-2z) (1)2 a (b) With the total potential written as (2) the potential Cbb can be used to make the total potential satisfy the boundary condition at y = O. Because the first part of (2) satisfies Laplace's equation and the boundary conditions far to the left, the second part must go to zero there. Thus, it is taken as a superposition of solutions to Laplace's equation Solutions to Chapter 5 5-13 that are zero in the planes y = 0 and y = a (so that the potential there as given by the first term is not disturbed) and that decay exponentially in the -y direction. 00 ... ~ A . (n1fz) rury/a. (3) ....b = L..J "SIn -- e ,,=1 a Aty= 0, ~(z, 0) = ~d(Z), Thus, ~b(Z, 0) = ~d(Z) -~a.(z) and evaluation of (3) at y = 0, multiplication by sin(m1fz/a) and integration from Z = 0 to Z= a gives la. [ () Vo ( 2z)]. m1fZ a~d Z - -1--sID--dz=A m - (4) o 2 a a 2 from which it follows that 21a. n1fZ { ~. evA" =- ~d(Z) sin (--)dz -~tr' n en (5)a 0 a 0, nodd Thus, the potential between the plates is ~ = Vo (1- 2z) + t A" sin (~)e"try/a. (6) 2 a ,,=1 a where A" is given by (5). 5.5.6 The potential is taken as the sum of two, the first being zero on all but the boundary at z = a where it is Voy/a and the second being zero on all but the boundary at y = a, where it is Voz/a. The second solution is obtained from the first by interchanging the roles of z and y. For the first solution, we take 00 • h(~) ~I = L A" sin (~) SID. a. (1) ,,=1 a sIDhn1f The coefficients follow by evaluating this expression at z = a, multiplying by sin(m1fy/a) and integrating from y = 0 to Y = a. la. Voz . (n1fz)- SID - dz= A,,(a/2) (2)o a a Thus, A" =- 2Vo (_1)" (3) n1f The first part of the solution is given by substituting (3) into (1). It follows that the total solution is ... ~ 2Vo (-1)" [ . (n1fz) . h (n1fY) . (n1fY) . h (n1fz)].... = L..J-- SID -- SID -- +SID - SID -- (4) ,,=1 n1f sinh(n1f) a a a a Solutions to Chapter 5 5-14 5.5.'1 5.6 5.6.1 (a) The total potential is sero at y = 0 and so also is the first term. Thus, ~1 must be zero as well at y = O. The first term satisfies the boundary condition at y = b, so ~1 must be zero there as well. However, in the planes :I: =0 and :I: = a, the first term has a potential Vy/b that must be cancelled by the second term so that the sum of the two terms is zero. Thus, ~1 must satisfy the conditions summarized in the problem statement. (b) To satisfy the conditions at :I: = 0 and :I: = a, the y dependence is taken as sin(ml"y/b). The product form :I: dependence is a linear combination of exponentials having arguments (R'JI"y/b). Because the boundary conditions in the :I: = 0 and :I: = a planes are even about the plane :I: = a/2, this linear combination is taken as being the cosh function displaced so that its origin is at:l: = a/2. DO " • (R'JI"Y) [R'JI"( a)] () = L...J An SIn -b- cosh T :I: -2' (1) n=l Thus, if the boundary condition is satisfied at :I: = a, it is at :I: = 0 as well. Evaluation of (1) at :I: = a, multiplication by sin(m'Jl"y/b) and integration from y = 0 to Y = b then gives an expression that can be solved for Am and hence An A _ 2V(-1)n () n-R'JI"cosh(R'JI"a/2b) 2 In terms of these coefficients, the desired solution is then DO Vy L • (R'JI"Y) [R'JI" a)]~ =-+ AnsIn -- cosh -(:1:-- (3) b n=l b b 2 SOLUTIONS TO POISSON'S EQUATION WITH BOUNDARY CONDITIONS The potential is the sum of two homogenous solutions that satisfy Laplace's equation and a third inhomogeneous solution that makes the potential satisfy Pois­ son's equation for each point in the volume. This latter solution, which follows from assuming ~p = ~p(y) and integration of Poisson's equation, is arranged to give zero potential on each of the boundaries, so it is up to the first two to satisfy the bound­ ary conditions. The first solution is zero at y = 0, has the same :I: dependence as the wall at y = d and has a coefficient that has been adjusted so that the magnitude of the potential matches that at y = d. The second solution is zero at y = d (the displaced sinh function is a linear combination of the sinh and cosh functions in column 2 of Table 5.4.1) and so does not disturb the potential already satisfied by the first term at that boundary. At y = 0, where the first term has been arranged to make no contribution, it has the same y dependence as the potential in the y = 0 plane and has its coefficient adjusted so that it has the correct magnitude on that boundary as well. Solutions to Chapter 5 5-15 5.6.2 The particular solution is found by assuming that the particular potential is only a function of 11 and integration of Poisson's equation twice. With the two integration coefficients adjusted to make the potential of this particular solution zero on each of the boundaries, it is the same as the last term in (a) of Prob. 5.6.1. Thus, the homogeneous solution must be zero at 11 = 0, suggesting that it has a sinh function 11 dependence. The z dependence of the potential at y = d then suggests the z dependence of the potential be made sin(kz). With the coefficient of this homogeneous solution adjusted so that the condition at y = d is satisfied, the desired potential is . sinh k1l Po ( ) .=.0smhkz . hkd --2 11 y-d (1)sm f o 5.6.3 (a) In the volume, Poisson's equation is satisfied by a potential that is independent of y and z, 2 2 8• Po ( )= --p = --cosk z-6 (1) V .p 8z2f o Two integrations give the particular solution (2) Ep= PO sin k(z -6)ix (3) fo k (b) The boundary conditions at y = ±d/2 are (4) Because the configuration is symmetric with respect to the z -z plane, use cosh(ky) as the 11 dependence. Thus, in view of the two z dependencies, the homogeneous potential is assumed to take the form .h= [A sin kz + B cos k(z ­6)1 cosh ky (5) The condition of (4) then requires that ElIJh = -[Acoskz ­B sin k(z ­6)lkcoshky (6) and it follows from the fact that at 11 = d/2 that (3) + (6) = (4) A = -Eo/kcosh(kd/2)j B = -Po/f ok2 cosh(kd/2) (7) so that the total potential is as given by (d) of the problem statement. 5-16 Solutions to Chapter 5 (c) First note that because of the symmetry with respect to the z plane, there is no net force in the y direction. In integrating pEs over the volume, note that Es is Po • ( ) cosh leh [ Po •( )]Es =-lesmlez-8 + (kd) Eo coslez--lesmle z -8 (8) f o cosh "2 f o In view of the z dependence of the charge density, only the second term in this expression makes a contribution to the integral. Also, P = Po cos le(z -8) = Po[cos le8 cos lez -sin le8 sin kz] and only the first of these two terms makes a contribution also. 12../10 jd/2 cosh leyfs = Pocosle8coskz (kd) Eocoskzdydz o -d/2 cosh "2 (9) = [211"poEocosleHanh(lcd/2)Jlle2 5.6.4 (a) For a particular solution, guess that () = Acoslc(z -8) (1) Substitution into Poisson's equation then shows that A = Po/fole 2 so that the particular solution is ()p = Ple°2 cos le(z -8) (2) f o (b) Aty = 0 (3) while at y = d, ()h. = Vocoslez -P° cosle(z -8) (4) f o le2 (c) The homogeneous solution is itself the sum of a part that satisfies the condi­ tions (5) and is therefore sinh ley ()1 = Vocos lcz sinh led (6) and a part satisfying the conditions (7) which is therefore ..... _ Po le( ~) cosh le(y -~) '\11'2 ---- cos z-(} lc2 f o cosh (led/2) (8) Solutions to Chapter 5 5-17 Thus, the total potential is the sum of (2), (6) and (8). Po [COSh k(y -~)] sinh ky ~ = -k2 cosk(x -6) 1- (led) +Vocoskx . hkd (9) fo cosh 2" sm (d) In view of the given charge density and (9), the force density in the x direction is Po . [ cosh k(y -~)] Fz = -k smk(x -6) cos k(x -6) 1- (led) f o cosh 2" (10) . sinhky+ PokVo sm kx cos k(x -6) sinh kd The first term in this expression integrates to zero while the second gives a total force of P kV: / (11) /z = s~h k~ i0 2fr 1e i0 d sin kx cos k(x -6) sinh kydydx With the use of cos k(x -6) = cos kx cos k6 + sin kx sin k6, this integration gives - v: (cosh kd -1) sin k6 (12) fz-Po1f 0 ksinhkd 5.6.5 By inspection, we know that if we look for a particular solution having only a y dependence, it will have the same y dependence as the charge distribution (the second derivative of the sin function is once again a sin function). Thus, we substitute Asin(1fy/b) into Poisson's equation and evaluate A. (1) The homogeneous solution must therefore be zero on the boundaries at y = band y = 0 and must be -Pob2 sin(1fy/b)/f o1f2 at x = ±a. This latter condition is even in x and can be matched by the solution to Laplace's equation (2) if the coefficient, A, is made (3) Thus, the solution is the sum of (1) and (2) with A given by (3). 5-18 Solutions to Chapter 5 5.6.6 (a) The charge distribution follows from Poisson's equation. _~ =V2." => P = foV sin~:l:Sin ~'I (~2 + ;) (1) (b) To make the total solution satisfy the lero potential conditions. the homo­ geneous solution must also be lero at 11 = 0 and 11 = b. Atz =0it must also be lero butatz= a the homogeneous solution must be ." = -V sin('lI'1Ijb) sin~a. Thus. we select the homogeneous solution .... _ A' 'lI'1I sinh('lI'zjb) (2) "It'll. - sm b sinh('lI'ajb) make A = -Vsin ~a and obtain the potential distribution if, V. ('lI'1I) [. Q • Q sinh(,rzjb)] "It' = sm T sml'Z -sml'a sinh('lI'ajb) (3) 5.6.'1 A particular solution is found by assuming that it only depends on z and integrating Poisson's equation twice to obtain Pol2 z z3 ." =-6Eo (, -"is) (1) The two integration constants have been assigned so that the potential is lero at z =0and z= I. The homogeneous solution must therefore satisfy the boundary conditions .",(z =0) =.",(z = I) =0 pI2 Z z3 .",(y= ±d) =-~o (, -"is) (2) The first two of these are satisfied by the following solutions to Laplace's equation. ~ . n'll'z cosh (7) ~", = LJ An sm (-,-) h (!!!rJ!) (3) 71.=1 COS, This potential has an even y dependence. reflecting the fact that the boundary conditions are even in y. To determine the coefficients in (3). note that the second pair of boundary conditions require that .f: A sin n'll'z = _por C~ _Z3) (4) n=l 71. I 6Eo I 13 Multiplication of both sides of this expression by sin(m'll'zjl). and integration gives I poll'· (m'll'z) Po l' 3 . m'll'zAm -=-- zsm -- dz+ -- z sln--dz (5)2 6Eo 0 I 6Eol 0 I Solutions to Chapter 5 5-19 or Thus, the required potential is w= Po l2 (=-_ X3) + ~ ~(_l )3 PO (_1)n sin n'll'x cosh (T) (6) 6e l ZS L-l n'll' e l cosh (mrd)o n=l 0 I 5.6.8 (a) The charge density can be found using Poisson's equation to confirm that the charge density is that given. Thus, the particular solution is indeed as given. (b) Continuity conditions at the interface where y = 0 are (1) 8wa 8wb 8y = 8y (2) To satisfy these conditions, add to the particular solution a solution to Laplace's equation in the respective regions having the same x dependence and decaying to zero far from the interface. (3) wb = (fj2Po -0 2) cos fjxe OlIJ + B cos fjxe f11J (4)eo Substitution of these relations into (1) and (2) shows that A = e {fj2 Po _ 02)2(1-Ii0) (5) o -Po ( 0) B = e (fj2 _ 02)2 1 + Ii (6) o and substitution of these coefficients into (3) and (4) results in the given potential distribution. 5.6.9 (a) The potential in each region is the sum of a part due to the wall potentials without the surface charge in the plane y = 0 and a part due to the surface charge and having zero potential on the walls. Each of these is continuous in the y = 0 plane and even in y. The x dependence of each is determined by the respective x dependencies of the wall potential and surface charge density distribution. The latter is the same as that part of its associated potential so that Gauss' continuity condition can be satisfied. Thus, with A a yet to be determined coefficient, the potential takes the form w= {V~~:~~: cosfjx -Asinhfj(y -a) sinfj(x -xo ); 0 < y < a (1) V~~:~~ cosfjx -A sinh fj(y + a) sinfj(x -xo ); -a < y < 0 5-20 Solutions to Chapter 5 The coefficient is determined from Gauss' condition to be 8iI>a 8iI>b] -u (2) -Eo [ -8 y --8Y y=O = uo sin P(z -zo) => A = 2EoP cos°hPa (b) The force is (3) From (1), -)-VQ sinf3z _ uosinhf3a Q( _)Ez (Y - 0 -'"cosh f3a 2Eocosh f3a cos", z Zo (4) The integration of the second term in this expression in (3) will give no con­ tribution. Substitution of the first term gives duoVf31z+2fr/{1 • . d1r cosf3zo fz = hf3 smp(z -zo) smpzdz = uoV f3( Q) h f3 (5)cos ao ",cos a (d) Because the charge and wall potential are synchronous, that is U = w/f3, the new potential distribution is just that found with z replaced by z -Ut. Thus, the force is that already found. The force acts on the external mechanical system (acts to accelerate the charged particles). Thus, Ufz is the mechanical power output and -Ufz is the mechanical power input. Because the system is loss free and the system is in the steady state so that there is no energy storage, -Ufz is therefore the electrical power output. . 1rcosf3zo ()Electrical Power Out = -Ufz = -UduoVf3- hf3 6f3 cos a (e) For (6) to be positive so that the system is a generator, ~ < pzo < 3;. 5.7 SOLUTIONS TO LAPLACE'S EQUATION IN POLAR COORDINATES 5.1.1 The given potentials have the correct values at r = a. With m = 5, they are solutions to Laplace's equation. Of the two possible solutions in each region having m = 5 and the given distribution, the one that is singular at the origin is eliminated from the inner region while the one that goes to infinity far from the origin is eliminated from the outer solution. Hence, the given solution. Solutions to Chapter 5 5-21 5.7'.2 (a) Of the two potentials have the same 4J dependence as the potential at r = R, the one that is not singular at the origin is (1) Note that this potential is also zero on the y = 0 plane, 80 it satisfies the potential conditions on the enclosing surface. (b) The sunace charge density on the equipotential at y = 0 is (2) and hence is uniform. 5.7'.3 The solution is written as the sum of two solutions, ~a and ~b. The first of these is the linear combination of solutions matching the potential on the outside and being zero on the inside. Thus, when added to the second solution, which is zero on the outside but assumes the given potential on the inside, it does not disturb the potential o~ the inside boundary. Nor does the second potential disturb the potential of the first solution on the outside boundary. Note also that the correct combination of solutions, (rlb)3 and (blr)3 in the first solution and (ria) and (air) in the second solution can be determined by inspection by introducing r normalized to the radius at which the potential must be zero. By using the appropriate powers of r, this approach can be used for any 4J dependence of the given potential. 5.7'.4 From Table 5.7.1, column two, the potentials that are zero at 4J =0 and 4J =a are r±m sin m4J (1) with m= mr/a, n = 1,2, ... In taking a linear combination of these that is zero at r = a, it is convenient to normalize the r dependence to a and write the linear combination as (2) where A and B are to be determined. It can be seen from (2) that to make ~ = 0 at r= a, A = -Band the solution becomes (3) Finally, the last coefficient and n are adjusted so that the potential meets the condition at r = b. Thus, (4) 5-22 Solutions to Chapter 5 5.1.5 To make the potential zero at 4J = 0, use the second and fourth solutions in the third column of Table 5.7.1. cos[pln(r)] sinh p4J, sin[pln(r)] sinh p4J (1) The linear combination of these solutions that is zero at r a is obtained by simply normalizing r to a in the second solution. This can be seen by using the double-angle formula to write that solution as Asin[pln(r/a)]sinhp4J = Asin[pln(r) -pln(a)]sinhp4J = A{sin[pln(r)] cos[pln(a)] (2) -cos[pln(r)] sin[pln(a)]} sinh p4J This solution is made to be zero at r = b by making p = n1r/ln(b/a), where n is any integer. Finally, the last boundary condition at 4J = 0 is met by adjusting the coefficient A and selecting n = 3. A = V / sinh[311"a/ln(b/a)] (3) 5.1.6 The potential is a linear combination of the first two in column one of Table 5.7.1. V 311" 24J ~ = A4J + B = --- (4J --) = V (1--) (1)(311"/2) 2 311" This potential and the associated electric field are sketched in Fig. 85.7.6. Figure S5.7'.6 Solutions toChapter5 5.8EXAMPLES INPOLAR COORDINATES5-23 5.8.1 Eitherfrom(5.8.4)orfromFig.5.8.2,itisclearthatoutsideofthecylinder, thez=0planeisonehavingthesamezeropotential asthesurfaceofthecylinder. Therefore, thepotential andfieldasrespectively givenby(5.8.4)and(5.8.5)also describe thegivensituation. Intuitively, wewouldexpectthemaximum electricfieldtobeatthetopof thecylinder, atr=R,q,=1r/2.From(5.8.5),thefieldatthispointis Emax=2Eo (1) andthismaximum fieldisindeedindependent ofthecylinder radius.Tobemore rigorous, from(5.8.5),themagnitude ofEis (2) where e==V[1+(R/r)2]2 cos2fJ+[1-(R/r)2]2 sin2fJ ITthisfunctionispictured astheverticalcoordinate inathreedimensional plot wherethefloorcoordinates arerandq"itsextremes arelocatedat(r,q,)where thederivatives intherandq,directions arezero.Thesearethelocations where thesurfacerepresented by(2)islevelandwherethesurfaceiseitheramaximum, aminimum orasaddlepoint.Thus,tolocatethecoordinates whicharecandidates forgivingthemaximum, notethat and ~;=~o2~2{[I+(R/r)2]2 cos2fJ+[1-(R/r)2]sin2fJ}=0 (4) Locations where(3)issatisfied areeitherat orat withrnotequaltoRorat~=o q,=1r/2 r=R(5) (6) (7) withq,notgivenby(5)or(6).Putting(5)into(4)showsthatthereisnosolution forrwhileputting(6)into(4)showsthattheassociated valueofrisr=R.Finally, putting(7)into(4)givesthesamelocation,r=Randq,=1r/2.Inspection of(5) showsthatthisisthelocationofamaximum, notaminimum. 5-24 Solutions to Chapter 5 5.8.2 Because there is no 4> dependence of the potential on the boundaries, we use the second m = 0 potential from Table 5.7.1. ~= Alnr+B (1) Here, a constant potential has been added to the In function. The two coefficients, A and B, are determined by requiring that Vb = Alnb+B (2) Va = Alna+B (3) Thus, A = (Va -Vb)/ln(a/b) B = {Vblna- Valnb}/ln(a/b) (4) and the required potential is ~=v. In(r/b) _Vlln(r/b) VI a In(a/b) b In(a/b) + b (5) = lValn(r/b) -Vbln(r/a)Jlln(a/b) The electric field follows as being (6) and evaluation of this expression at r = b shows that the field is positive on the inner cylinder, and everywhere else for that matter, if Va < Vb­ 5.8.3 (a) The given surface charge distribution can be represented by a Fourier series that, like the given function, is odd about 4> = 4>0 U. = L00 Un sin mr(4) -90 ) (1) n=l where the coefficients Un are determined by multiplying both sides of (1) by sin mll"(4) -4>0) and integrating over a half-wavelength. Thus, 4uo Un = -j nodd (3)nll" 5-25 Solutions to Chapter 5 and u'" = 0, n even. The potential response to this surface charge density is written in terms of solutions to Laplace's equation that i) have the same rP dependence as (I), ii) go to zero far from the rotating cylinder (region a) and at the inner cylinder where r = R and are continuous at r = a. ~ {[(a/R)'" -(R/a)"'](R/r)"'}' ( ) a<r (4)cP = ~ CP", (R/a)"'[(r/ R)'" -(R/r)"'] sm n rP -00 R < r < a odd The coefficients CP", are determined by the "last" boundary condition, requir­ ing that acpa aCPb]u.(r=a)=-f o ---- (5) [ar ar r=a Substitution of (I), (3) and (4) into (5) gives (6) (b) The surface charge density on the inner cylinder follows from using (4) to evaluate u.(r = R) = -foaa~b Ir=R = -f~2 f: CP",n(R/a)"'sinn(rP -90) (7) ,,=1 odd Thus, the total charge on the electrode segment in the wall of the inner cylin­ der is q = w lQ u.(R)RdrP =-Lco Q",[cosn9 0 -cosn(a -Do)] (8) o ..._1 odd where (c) The output voltage is then evaluated by substituting 90 -Ot into (8) and taking the temporal derivative. Vo = -Ro ~: = -ORo f nQ",[sin nOt +sin n(a -Ot)] (9) ,,=1 odd 5-26 Solutions to Chapter 5 5.8.4 The Fourier representation of the square-wave of surface charge density is carried out as in Prob. 5.8.3, (1) through (3), resulting in 00 u, = L u" sin mr(1fl -(0 ) (1) ...01 odd where 4uou" = -j nodd n7l'" The potential between the moving sheet at r= R and the outer cylindrical wall at r = a, and inside the moving sheet, are respectively ~ { (a/R)"[(r/R)" -(R/r)"j}. ( ) a <r< R ~ = =:~" = (r/R)"[(a/R)" -(R/a)" smn Ifl -00 r < a (2) odd where the coefficient has been adjusted so that the potential is zero at r = R and continuous at the surface of the moving sheet, where r = a. The coefficients are determined by using Gauss' continuity condition with the surface charge density written as (1) and the potential given by (2)j ( a~a a~b) n n -Eo -a --a = u, => -Eo~,,(a/R)"[ -(a/R)" + -(R/a)"] rr r=a a a (3) + ~(a/R)"[(a/R)" _ (R/a)"] = 4uo a n7l'" which implies that ~ =_ 2uoa (4)"n 2 71'"Eo The surface charge on the detection segment is u, = Eo aa~a I =-f: 4uo(a/R)"+l sin n(1fl -(0 ) (5) r r=R ..=1 7I'"n odd and so the total charge on that segment is (6) where Q" = 4UowR(a/R)"+l~ 71'" n 2 Finally, with 00 = Ot, the detected voltage is therefore tlo = -R o ~: = -ORof nQ,,[sin nOt+sinn(a-Ot)] (7) .._1 odd Solutions to Chapter 5 5-27 5.8.5 Of the potentials in the second column of Table 5.7.1, the requirement that the potential be zero where <p = 0 selects the two that vary as sin(m<p) while the fact that the space of interest extends to the origin precludes those with negative exponents, for m > 0, the last two. The potential will be zero at <p =a ifm = n1l"Id, n = 1,2, ... Thus, candidate potentials are (1) Evaluated at r = R, this potential takes the form of a Fourier series, used here to represent the uniform potential. v = f: An sin (n:<p) (2) m=l Multiplication by sin(q1l"<Pla) and integration from <p =0 to <p = a gives an expres­ sion which can be solved for the coefficients in (2). _ Va cos (q1l"<P)]Q = A ~ =* An = 4V {lin; n odd q1l" a 0 q 2 11" OJ n even (3) Thus, (1) and (3) are the given answer. 5.8.6 Far from r = R, the field becomes that of a pair of electrodes extending from the origin to infinity in the planes <p = 0 (with zero potential) and <p = a (with potential V). The associated electric field is <p directed and simply the voltage V divided by the distance ar between the electrodes, following lines of constant r. ~(r -+ 00) = V! =* E(r -+ 00) = ~i4> (1)a ar Although this potential satisfies the boundary conditions on the "wedge" electrodes, it does not satisfy the boundary conditions over the surface at r = R. On that surface, the potential should be the constant V. To satisfy this boundary condition, we add to (1) a potential that is zero on the surfaces <p = 0 and <p = a where (1) already satisfies the boundary conditions and that goes to zero at r -+ 00, where (1) is also the correct potential. (2) The coefficients An are determined from evaluating (2) on the electrode at r = R, where V<p ~ A . (n1l"<P)V = -+ LJ n SlD -- (3) a n=l a Solutions to Chapter 5 5-28 The first term on the right in (3) is tra.nsferred to the left, both sides of the expres­ sion multiplied by sin(m1/"~/a) a.nd both sides integrated from ~ =0 to ~ =a to obtain a (m1/"~) a [. m1/"~ m1/"~ (m1/"~)]}Q AmaV { --cos -----sln-----cos --=-- (4)m1/" a (m1/")2 a a a 02 This expression can be solved for the coefficient, which (with m -n) is _ 2V A1'- (5) n1/" Evaluated using this coefficient, (2) is the desired potential. 5.8.'1 (a) From the four equations in the second column of Table 5.7.1, the sin functions satisfy the boundary conditions that Cb = 0 at ~ = 0 and ~ = 21/" if m= n/2, n = 1,2, ... With the understanding that n is positive, the solutions with exponents -m are excluded so that the potential is finite as r -O. Thus, the remaining potential is the superposition of the modes DO Cb= LAn(r/R)n/2sin(~~) (1) 1'=1 (b) The boundary condition at r = R requires that 00 Vo = L A,. sin (~~) (2) 1'=1 Multiplication of both sides of this expression by sin(p~/2) and integration gives (3) or 2 --Volcos(m1/") -11 = 1/"Am (4)m so that it follows that An = 0, n even and for n odd _ 4VoAI' -n1/" (5) Substitution of this coefficient into (1) then gives the desired potential. (6) 5-29 Solutions to Chapter 5 (c) The associated electric field follows from this expression as 14Vo ~ 1[ nr!-l. n .nrt-n] E = ---;:-~;; Il'i Rn/2 Sln (if) +l<I»i Rn/2 cos (if) (7) odd r (b) (a) (e) Figure S5.8.' A sketch of the lead term in (6) and (7) is shown in Fig. 85.8.7a. The potential is finite at the tip of the fin but the electric field intensity varies as 1/..;r at the tip. On the surface 81 shown in Fig. 85.8.7b, the surface charge density follows from (7) as oo4foVo L1 r!j-1 f E-t-(r ..I. = 0) = --- --- (8)o ." ,Y' 1/" 2 Rn/2 ..=1 odd On the circular cylindrical surface 82 at radius a, also shown in Fig. 85.8.7b, 4foVo ~ 1 a!-l. n foEr(r = Q, f) = --1/"-L.J "2 Rn/2 Sin (if) (9) ,.,=1 odd while on surface 83 , 4 v: 00 1 n-1 -f E-t-= -~ ~_.!:.:... - o ." 1/" L.J 2 Rn/2 (10) ,,=1 odd Solutions to Chapter 5 5-30 The total charge represented by the first mode in the series is therefore 2EoVo[_ fR r-1/2dr _ rrra-l/2sin(~/2)ad~ _ fR r-1/2dr] = 8EoVo (11) 'frVR Ja Jo Ja 'II" (d) The potential and field distribution is sketched in Fig. S5.8.7b. 5.8.8 The potential takes the form of (5.8.15) with azimuthal coordinate displaced so that ~ -+ ~o -~. 4> =; An sin [n'll" ;:~:~:~] sinh [,n(aib) (~o -~)] (1) Evaluated at ~ = 0, this expre88ion is then the same as (5.8.15) evaluated at ~ = ~o' Thus, the coefficients are the same as given by (5.8.17). For n even, An = 0 and for n odd (2) 5.8.9 The radial distribution Rn(r) is governed by (5.7.5). d (dR n ) 2 r dr r"d;" +PnRn =0 (1) Multiplication of this expression by another of the eigenfucntions and the weighting factor l/r and integration results in the expression r [R-r.!!(rdRn ) +p2 !RnRm]dr'= 0 (2)Ja rdr dr nr With the identification udtl = d(Uti) -vdu where ( dRn) du = d r"d;"' tI = R- (3) Eq. (2) can be integrated by parts a la dRn ]a ( dRndRm) 21 1 r-R m -r--- dr+Pn -RnRmdr=O (4)dr b b dr dr br This same procedure can be repeated with the roles of n and m reversed. Substrac­ tion of the resulting expression from (4) gives dRn dRm]a (2 2) fa 1 r[---;I;:Rm-Rnb b + Pn-P m J b ;RnRmdr=O (5) lH boundary conditions require that the first term is zero, or in particular that Rn(a) = 0 and Rm(b) = 0, then the orthogonality condition follows. a1 (p~ -p~) -RnRmdr =0 (6) br 5-31 Solutions to Chapter 5 5.9 THREE SOLUTIONS TO LAPLACE'S EQUATION IN SPHERICAL COORDINATES 5.9.1 (a) The given surface potential has the same fJ dependence as for the uniform field potential of (5.9.4) and the dipole field potential of (5.9.3). With the coefficients of these potentials adjusted to match the given potential at r = a, ~ _ {v(r/a) cos fJj r<a -V(a/r)2 cos fJj a<r (1) (b) A sketch of ~ and E is shown in Fig. 6.3.1. 5.9.2 (a) The surface charge density has the same fJ dependence at r = a as the discon­ tinuity in the normal derivative of the potential. This suggests representing the potentials inside and outside the sphere with the same fJ dependence as the given surface charge distribution. In addition, these potentials must be finite at the origin and at infinity. The natural choices are the uniform field potential given by (5.9.4) inside the sphere and the dipole potential of (5.9.3) outside the sphere. ~ _ {A(a/r)2 cos fJj a < r (1) -A(r/a)cosfJj r<a The coefficients have already been adjusted so that the potential is continuous at r = a. Gauss' continuity condition then requires that :"'fO(a~a -a~b) r=a =0"0 cos fJ * -fo [~+ ~] A = (2) 0"0 so that A = 0"0a/3f o and the potential is as given with the problem. (b) In Example 6.3.1, the potentials inside and outside the sphere take the same form as in (1) /(6.3.9) and (6.3.8)] and satisfy boundary conditions which take the same form as used here /(6.3.6) and (6.3.7)]. Indeed, we will see in Sec. 6.3 that with the polarization density given the polarization charge density is specified- and the determination of the associated potential and field is much the same as in this chapter when the charge is specified. Hence, Fig. 6.3.1 portrays the potential and field. 5.9.3 Because the given charge density does not depend on 1/1, the potential is also independent of 1/1. In that case, Poisson's equation in spherical coordinates reduces to -!. ~ (~ a~) + _1 _ ~ ( sin fJ a~) =_ Po cos fJ (1) r2ar ar r2sin fJ afJ afJ f o First, given the dependence of the charge density on fJ, look for a particular solution having the form ~p = ArP cos fJ. Substitution into (1) then shows that p = 2 and A = -Po/4fo so that a particular solution is ~p = -4Po r 2 cosfJ (2) f o 5-32 Solutions to Chapter 5 The sum of this potential and a solution to Laplace's equation must satisfy the condition that the potential be zero at r = a. Again, for the fJ dependence of the particular solution, it is natural to take a uniform field as the homogeneous solution. Thus, with B an adjustable coefficient, •= _.f!!!...,-2 cos fJ + Br cosO (3)4Eo and by requiring that the total potential be zero at r = a, it follows that B= poa/4Eo so that the potential is as given with the problem statement. 5.9.4 Because the given charge density does not depend on q" the potential is also independent of q,. In that case, Poisson's equation in spherical coordinates reduces to 1a(2 a.) 1 a('fJa.) Po(/)m ()r2 ar r ar + r2 sin fJ afJ sm tii =-Eora cos fJ 1 First, given the dependence of the charge density on fJ, look for a particular solution having the form (r/a) cos fJ. Substitution into (1) then shows that p = m+ 2 and A = -poa2 /Eo(m + 1)(m + 4) so that a particular solution is 2 A;. po a ( / )m+2 fJ 'Jt'p = Eo(m+ 1)(m+4) r a cos (2) The sum of this potential and a solution to Laplace's equation must satisfy the condition that the potential be zero at r = a. Again, for the fJ dependence of the particular solution, it is natural to take a uniform field as the homogeneous solution. Thus, with B an adjustable coefficient, • =.p + B(r/a) cos fJ (3) and by requiring that the total potential be zero at r = a, it follows that the required potential is _P a2 • = Eo(m+ l)(m+ 4) (rfa) [(rfa)m+l -l]cosfJ (4) 5.10 THREE-DIMENSIONAL SOLUTIONS TO LAPLACE'S EQUATION 5.10.1 Given the zero potential surfaces at y = 0 and y = b and at z = 0 and z = w, it is natural to construct the solution from product solutions having the form A;. X()' m1rY • n1rZ 'Jt'= z sln-b-sm~ (1) Solutions toChapter 5 where,tosatisfyLaplace's equation X(x)={sinhkm"x coshkm"x5-33 and km"=.J(m'lr/b)2+(R'Ir/w)2 Theboundary conditions onthesurfacesatx=0andx=aarethesame.Thus,if X(x)ischosentobeevenaboutanoriginatx=a/2,thepotential thatsatisfiesthe condition ofbeingtIatx=0willalsobetIatx=a.Thus,X(x)ismadealinear combination ofthesolutions givenwith(1)whichisthecoshfunction displaced so thatitsargument iszerowherex=a/2. X(x)=Am"coshkm,,(x -i) (2) Thesolution therefore takestheformof(a)givenwiththeproblem. Atx=0,the condition atx=0requiresthat ~~Ah(km"a).(m'lr1l)•(R'lrll) (3)tI=L..JL..Jm"COB-2-sm-b-sm--;;- m=l,,=l Notethatthisexpression isthesameas(11)ifthesinh(kmnb) isreplaced by cosh(km..a/2)andx/a-1I/b.Theevaluation ofthecoefficient usingtheorthogo­ nalityoftheproductsolutions istherefore essentially thesameasgivenby(5.10.11)­ (5.10.15), resulting in(b)asgivenwiththeproblem. 5.10.2 Giventhe:I:andIIdependence ofthesurfacechargedensity, whichisthe sameasthatofthecomponents ofEintheIIdirection oneithersideofthesurface y=a/2,lookforsolutions oftheform ~=Y(y)sin(~)sin(II) (1)aw where andY(y)={Sinhklly coshk1l1l (2)kll=../(1IJa)2+(7I"/b)2 Tosatisfythecontinuity conditions aty=b/2,thepotential function isgivena piece-wise representation. Thefunction intheupperregionmustbezeroaty=b, soY(y)ischosenasasinhwithitsargument displaced toy=b.Inthelowerregion, thesinhfunction withitsoriginaty=adoesthejob.Thus, ....._{Asinhkll(y-b)}. (71":1:). (~)"*"-B.hk sm sinsmllY a w At11=b/2,thepotential mustbecontinuous andGauss'continuity condition must besatisfied. -Asinh(k llb/2)=Bsinh(kllb/2) -fokll(A -B)cosh(kllb/2) =Uo Itfollowsthatthecoefficients in(2)are A=-B=-Uo/2fokllcosh(kllb/2)(3) (4) (5) 5-34 Solutions to Chapter 5 5.10.3 In each case, the solution can be regarded as the superposition of a particular solution to Poisson's equation and a homogeneous solution to satisfy the boundary conditions. The determination of representation begins with the selection of the former. As a first solution, select a particular solution that is only z dependent. Then, Poisson's equation reduces to d2~ Po (1) dz2 =-€o and the particular solution that (for convenience) is also zero at z = 0 and z = a is (2) With this potential satisfying the boundary conditions on two of the surfaces, the homogeneous solution must assure satisfying the conditions on the remaining four surfaces. This is done by adding to (2) solutions designed to satisfy the conditions at Y = 0 and Y = b while being zero at all the other surfaces and therefore neither disturbing the already satisfied conditions at z = 0 and z = a nor those to be satisfied by the next homogeneous solution. To satisfy both the conditions at Y = 0 and y = b, the y dependence is taken as even about y = b/2. A second homogeneous solution is then added to this one to assure satisfaction of the conditions at Z = 0 and Z = w/2 while not disturbing the potential at the other four surfaces. Thus, the potential takes the form 00 00 ~ = -2Po z(z -a) + LL Bmn coshkmn(y --) b sin (m'1l" z) sin C~!z) Eo m=ln=l 2 a w (3)00 00 +L LOmncoshkmn(z- ;) sin (:'1I"z) sin (n'1l"y)bm=ln=l The coefficients Bmn and Omn are determined by requiring that the potential indeed be zero on the surfaces y = 0 and Z = 0 (and hence also at y = b and Z = w). Po 00 00 kmnb . m'1l" • n'1l" ~z(z -a) = LL Bmn cosh (-2-) sm (-;-z) sm (-;-z) (4) o m=ln=l Po 00 00 kmnw . m'1l" • n'1l" -2z(z-a) = L LOmncosh(--)sm(-z)sm(-b y) (5) Eo 2 am=ln=l The coefficients therefore follow from the same procedure as illustrated by (5.10.11) through (5.10.15). For m or n even the coefficients are zero. For m and n odd, (6) 5-35 Solutions to Chapter 5 emn = P(Ic) (4/m!") fa z(z -a) sin (~z)dz 2Eocosh ~ 10 a (7) = -Po (4/m!") ~ 2Eocosh (Icm;W) (mll")3 Two more solutions are obtained by replacing the role of z with that of y and of z. As a fourth solution, expand the charge distribution in a three dimensional Fourier series 00 0000 "" """"R . (mll"z) . (nll"Y) . (qll"z)Po = LJ LJLJ mnq sm--sm -b-sm -- (8)a wm=l n=lq=l The coefficients Rmnq follow by multiplying by . (rll"z) . (slI"Y) . (ulI"z)sm--sin--sm-­ a b w integrating over the volume and solving for Rr,u. Then, with rsu -+ mnq, (9) for m and nand q odd and zero for m or n or q even. Given this (z, y, z) dependence and given that the second derivative of each of the sinusoids results in the same sinusoidal function, we are motivated to look for a particular solution having the same form. 00 0000 ...... "" "" ""...... . (mll"z) . (nll"Y) . (qll"z) (10) -.r = LJ LJLJ -.rmnq sm--sm -b-sin -­ m=l n=l q=l a w Substitution of this expression into Poisson's equation shows that term by term it is not only a solution to Poisson's equation (and therefore a particular solution) if (11) but satisfies the boundary conditions as well. SOLUTIONS TO CHAPTER 6 6.1 POLARIZATION DENSITY 6.1.1 (a) From (6.1.6), the polarization charge density is P1' = -V·p = Popsinfh; (1) (b) The polarization surface charge density at the respective surfaces follows from (6.1. 7) evaluated at the resp ective interfaces. u 81' = -n. (pa _ pb) (2) _{-(O-Pocospz) =PoCOSPZj y=d --(Pocospz -0) = -PoCOSPZj y=O 6.2 LAWS AND CONTINUITY CONDITIONS WITH POLARIZATION 6.2.1 (a) Given the polarization density, the polarization current density follows from (6.2.9). ap dPo .) Q (.J l' = lit = "dt cos fJZ Ix + I~ (1) The polarization charge density is as found in Prob. 6.1.1. (b) Substitution of these quantities into (6.2.10) gives ap1' VJ apo • dPo • at + . l'= --atpsmpz -"dtpsmPz = 0 (2) 6.3 PERMANENT POLARIZATION 6.3.1 (a) The polarization charge density between the electrodes is P1' = -V . P = Popsinpz (1) Thus, at each point between the electrodes, V2W= _P1' =_PoP sinpz (2) Eo Eo 1 6-2 Solutions toChapter6 andaparticular solution isgottenfrom 82~PofJ. Q ....Po. Q--2=--sm,.,z=>'VI'p=-sm,.,z8z Eo fJEo Tosatisfytheboundary conditions, usethehomogeneous solutions Az ~=Az+~osinfJz ,.,Eo whichsatisfies ~(z=0)=o.Tomake ~(z=a)=-V, AVPo. Q=-----sm,.,aafJEoa sothat(6.3.4)becomes ~=Po(sinfJz-=sinfJa)-V= fJEo a a(3) (4) (5) (6) /t/J=-V a f"o--.~ y (b)Inthiscase, andP=Pocos{3xix~=0 Flpre98.3.1 Pp=-V·P=fJPosinfJy (7) 82~p fJPo• Po• () --=--smfJy=>~=-smfJy 88y2 Eo pEofJ Boundary conditions atz=0andz=a,aresatisfied by~=-Vz/a.Thus, welet ~=PosinfJy-V=+~1 (9) EofJ a Then ~1mustbe-(Po/EofJ) sinfJyatz=0andz=a.Suchasolution to Laplace's equation issymmetric aboutz=a/2; ~=P~sinfJY- VZ+AcoshfJ(z-~)sinfJY (10) Eo,., a 2 Tosatisfyboundary conditions A=_Po1 () EofJcosh(~) 11 Thus,from(6.3.4)and(6.3.5), ....Po. Q[coshfJ(z-~)]VZ 'VI'=-sm,.,y1- -- EofJ cosh(~) a 6-3 Solutions to Chapter 6 6.3.2 The polarization charge density inside the rectangular region is (1) The potential is therefore a solution to V 2"", P. 11" • 11""*" =-o-Sln-X (2) aEo a that satisfies the zero potential boundary conditions. Two of these conditions are satisfied by the particular solution to (6.3.2) that follows from assuming that it, like the charge distribution,. only depends on X. tP~p 11" • 11"--=-Po-sm-x (3)dx2 aEo a Two integrations, with the integration constants adjusted to make the potential at x = a and x = 0 zero then give the particular solution "'" p. a . 11""*" = o-sm-x (4)p 1I"Eo a The homogeneous solution must also be zero on these boundaries and cancel this particular solution when evaluated at y = ±b. (5) Because these conditions are even in y, and because of the former boundary con­ ditions, the potential is therefore taken as having a cosh dependence on x and the potential distribution suggested by the conditions of (6.3.5) at y = ±b. a 11" cosh ~y ~h =-Po-sin -X--:-'=:-:- (6) 1I"Eo a cosh ~b The required potential is then the sum of the particular and homogeneous solutions, (6.3.4) and (6.3.6). a 11" [ cosh :zr..y ] (7) ~ =Po-sin -x 1_ G 1I"Eo a cosh;b 6.3.36-4 (a)First, ap"Pp=-v·p=--=0aySolutions toChapter6 (1) and (2) r p=Pocos[(211"/A)xJill-x (3)Figure88.8.8 (b)TheCJ>aboveandbelowmustsatisfyLaplace's equation andtheboundary conditions thataty=0 EoE;-Eo~=-Eoa:aI+EoaaCJ>bI=Pocos[(2;)z] Y",=0 Y",=0 Totheseends,andtomake CJ>-+0aty-+±oo,make(4) (5) 6.3.4 Intheregion-a<y<0,thedivergence ofthepolarization densityiszero andsothepolarization chargedensityiszeroaswell.Thus,inbothregions(a) and(b),thepotential mustsatisfyLaplace's equation. Boundary conditions onthe potential arethatitbethegivenvaluesaty=±a,thatitbecontinuous aty=0 andthatitsatisfyGauss'continuity condition aty=O.Thiscondition requires that -n.[aCJ>a_aCJ>b] =O',payay,,=0 wherethepolarization sudacechargedensityfollowsfrom(1) Thus,thefieldsarethesameasiftherewereanunpaired sudacechargedensity o',u=Posin,8(z -zo)intheplaney=o.Withtheidentification of0'0-+Po, thephysical situation isthesameasconsidered inProb.5.6.12andthesolution as outlined there. Solutions to Chapter 6 6-5 6.3.5 The given polarization density is uniform, so there is no volume polarization charge density. The polarization surface charge density at the cavity interface is (1) Thus, the boundary conditions at r = R are ~a = ~b (2) (3) On the right in this last expression is the sum of the polarization and unpaired surface charge densities. Superposition can be used to find the potentials due to the respective terms on the right and then their sum can be taken. Symmetry and Gauss' integral law give the electric field due to the uniform unpaired surface charge density. 1 411"E or2 E~ = 411"R20'o => E~ = -(R/r)20'o (4) Eo There is no electric field intensity inside, so the potential there is what it is on the surface. Thus, { ~. r<R ~_ Eo' ­ - R~"'Q. r> R (5) r' ­ To find the potential from the second term in (6.3.3), assume that .... _ {AiCOS() (6) ~- R~Arrcos() where the coefficient has been adjusted to satisfy (6.3.2). Substitution of these expressions into (6.3.3) then gives A =_PoR (7)3Eo The sum of (6.3.5) and (6.3.6) with the latter evaluated using (6.3.7) is the given potential. 6.3.6 In polar coordinates, the uniform y directed polarization density is p= PJ)' = Polcos <Pi.. -sin <Pi",) (1) Because the divergence of P is zero in the volume, the only polarization charge is a surface charge density at r =R. This is O'.p = -n . cos(pa -pb) = -Po cos <p (2) 6-6 Solutions to Chapter 6 The equipotential boundary condition in the plane 11 = 0 is met by assuming solutions ibQ =-A cos ;i ibb = Brcos; (3) r Boundary conditions at r = R are (1'-[ -BibQ --Bibb] = ~ '*(-A + B) cos; = -Pocos; (4)Br Br r=R Eo W A-=BR (5)R Simultaneous solution for the coefficients gives A __ PoR2. Po -2' B=-­2 (6) and hence, Q PoRR b PoR rib = ----cos;· ib = ----cos; (7)2 r' 2 R 6.3.1 The fields in regions (a) and (b), respectively above and below the interface, are taken as uniform. Because the line integral of E between the electrodes is zero, aE:+b~ =0 (1) At the interface, there is a polarization surface charge density (2) Thus, Gauss) continuity condition requires that (3) Solution of (6.3.1) and (6.3.3) then gives E'!. =_Po 1 (4) :I: Eo (~ + 1) 6-7 Solutions to Chapter 6 6.3.8 (a) The polarization charge density is Pp = -V· P = -V· V,p = _V2,p (1) where ,p = Por cos(¢> -a) is a solution to Laplace's equation. Thus, PP = O. (b) The surface polarization charge density at r = b is It is assumed that there is no unpaired surface charge density on this interface, so the boundary conditions are cpI(r = a) = 0 (3) cpI(r = b) = cpII(r = b) (4) foE; -foB;I = Pocos(¢> -a) (5) Solutions to Laplace's equation that have the same dependence as the right hand side of (6.3.5) take the form cp _ {A[(rla) -(air)] cos(¢> -a) (6)- B(rlb) cos(¢> -a) Here, the solution that is infinite at the origin has been omitted and the two contributions to the outer potential adjusted to satisfy (6.3.3). Substitution of (6.3.6) into (6.3.4) and (6.3.5) then gives b a Pob2b a B = A(- --) =--(- --) (7)a b 2foa a b Thus, (6.3.6) and (6.3.7) are the given potentials. 6.3.9 (a) Note that the scalar function inside the gradient operator is a solution to Laplace's equation. Thus, (1) and there is no polarization charge density in the volume of the rotor. However, at the interface there is a surface polarization charge density given by (2) (b) Boundary and continuity conditions are (2) 6-8 Solutions to Chapter 6 awa aWb ] -Eo [--a;:-(r = b) -ar (r = b) = Pomcos mtP (3) Given the tP dependence of a.p , solutions to Laplace's equation are assumed to take the form wa = A[(r/a)m -(a/r)mJcos mtP (4a) wb = A(r/b)m[(b/a)m -(a/b)m] cosml/l (4b) where a linear combination ofsolutions has been selected in the annular region, (a), that satisfies the zero potential condition at r = a and the coefficients have been arranged so that the potential is continuous at r = b. The last of the boundary conditions then determines A. Thus, the potential is b<r<a r<b (6) (c) With the substitution tP -+ tP -Ot, at a given instant in time there is only a shift in the origin of tP. Because the field laws do not involve a time rate of change, they are satisfied by the new solution. To stay at a point of constant tP -Ot and hence constant P requires being at the angular position tP = Ot+ constant. Thus, the new solution is one that represents the fields associated with a rotor having the angular velocity O. (d) From (6.3.6), the surface charge density on the wall at r = a is awa bPo 2m a.u = Eo- (r= a) = --(b/a)m-cos m(tP -Ot) (7)ar 2 a The net charge on the segment is then q = l r a.uadtP = -lbPo(b/a)m[sin(-mOt) -sinm( -~ -Ot)] ~~m m (8)=lbPo(b/a)m[sin(mOt) -sin(1l" + mOt)] = 2lbPo(b/a)m sin(mOt) and hence the output voltage is (9) 6-9 Solutions to Chapter 6 6.3.10 (a) The potential in regions (a) and (b), respectively 0 < z and z < 0, take the form of (6.6.26) and (6.6.27) with V = 0 in the latter because both of the electrodes are grounded over their full length in the z direction and a -. d. 00L Vne-A[ssin n;y ~a = (1) n=l ~b = 00L Vne.llfs sin n;y (2) n=l The coefficient in these expressions, Vn, has been adjusted so that the po­ tential is continuous at the interface. Given V(y), these coefficients follow by evaluating either of these expressions at y= 0 00 LVnsin n;y=V(y) (3) n=l multiplying by sinemll'y/ d) and integrating from y = 0 to Y = d. (4) Given V(y), this integral can be evaluated and the coefficients needed to complete (1) and and (2) determined. (b) In addition to the continuity of potential which is already satisfied by (1) and (2), the continuity condition at z =0 is (5) (c) With Po now the given quantity, substitution of (1) and (2) into (5) gives (6) The coefficients are evaluated in this case by the same procedure as leading to (4). (7) Evaluated using this coefficient, (1) and (2) become the given potential. Solutions to Chapter 6 6-10 6.8.11 In region (b), the potential must satisfy Poisson's equation with the charge density found in Prob. 6.1.1. (1) while in region (a) it satisfies Laplace's equation. At the interface, the potential must be continuous and satisfy Poisson's continuity condition for the polarization surface charge density found in Prob. 6.1.1. (2) Finally, the potential must go to zero as 'II -+ 00 and be zero in the plane 'II = O. The particular solution to (1) is taken as depending only on z. Thus, two integrations give ~:= P~sinfJz (3) Eo,", The z dependence of the potential due to the surface charge density is cos(fJz) while that due to the volume charge denisty is sin(fJz). The potential is taken as the sum of potentials due to these two sources, ~ =~. + ~". The potential due to the surface charge satisfies Laplace's equation in each region and takes the form A.e-I'(u-d) cos fJz ~. = Binb (4){ A. Binb I'd cos fJz Here, the coefficients have been adjusted to make the potential continuous at 'II =d, while the sinh function satisfies the zero potential boundary condition at 'II = O. The coefficient is determined by requiring that Gauss' continuity condition be satisfied with the surface charge density given by (2). -EoA.[-fJ -fJcoth(fJd)]cos(fJz) = Pocos(fJz) => A. Po (5)= -:-:--....:;....".....,~:'":' EofJ[l + coth(fJd)] The part of the potential due to the bulk charge takes the form A"e-I'(u-d) sin(fJz) ~,,= { £i;[1 + B" sinh(fJ'Y) + C" cosh(fJ'Y)] sin(fJz) (6) where the solution in the lower region has been taken as the sum of the particular solution, (3), and two solutions to Laplace's equation. This·part of the potential must also be zero at 'II = 0, so C = -1. In addition both the potential and its normal derivative must be continuous at 'II = d. A" = P~[l + B" sinh(fJd) -cosh(fJd)] (7) Eo,", Solutions to Chapter 6 6-11 -fJA u = Po [Bu cosh(fJd) -sinh(fJd) (8) Eo Simultaneous solution of these expressions gives (sinh2 Z -cosh2 Z = -1). Au = Po [cosh(fJd) -1] EofJ [cosh(fJd) + sinh(fJd)J (9) B = cosh(fJd) +sinh(fJd) -1 (10) u cosh(fJd) +sinh(fJd) Finally, the total solution is the sum 'of (4) and (6) with the coefficients given by (5), (9), and (10). 6.4 POLARIZATION CONSTITUTIVE LAWS 6.4.1 In terms of the number density N, the polarization density is given by P= Nqd = (E -Eo)E (1) It follows that the, separation d of single electronic charges needed to account for the given polarization is d = (E-Eo)E = (1.5)(8.85 x 10-12)(10'1) = 11 10-1. (2)Ng (6 x 1026/8)103(1.6 X 10-19) . X This is less than 1/1000 of a dimension typical of an atom. 6.5 FIELDS IN THE PRESENCE OF ELECTRICALLY LINEAR DIELECTRICS 6.5.1 (a) Tlte divergence of EE is zero a [V·EE= -E(Z)-] =0a1/ t1 d (1) and the curl of E, a unifoJ:ID field, is as well. Given that the field is normal to the perfectly conducting boundaries, which extend to infinity, it follows that the solution, which does indeed satisfy the relevant field laws, is uniquely specified. (b) On the upper surface of the lower electrode in the regions to right and left, (2) 6-12 Solutions to Chapter 6 It follows that the net charge on the lower electrode is (3) and hence the capacitance is as given. (c) In this case, the surface charge density on the lower electrode is (4) and so the net charge on the lower electrode is (5) so that C is as given. 6.5.2 A uniform electric field, E = (v/d)i)' is irrotational and satisifies Gauss' law with the permittivity varying with z, a direction perpendicular to the proposed electric field. a [ v (1) V· fE = oy fall + acos,8z)d] = 0 Thus, E is indeed uniform and (2) This is also the density of unpaired surface charge on the lower electrode, so the total charge on that electrode is f' v q = C1 fa(l + acos,8z)ddz 0 (3a) Cfa [ a. ]'=d z+ psm,8z Ov = Cv Cfa [ a. ]C == d 1+ psm,81 (3b) 6.5.S (a) Because the field is independent of z and z, aDyV·D=-=O (1)ay and from this it follows that Dy = DII(t). Solutions to Chapter 6 6-13 (b) In terms of the given distribution of permittivity, (2) This expression can be solved for E" and hence for the y dependence of E". To determine the unknown D", that expression is integrated from the lower to the upper electrode and the result equated to the voltage. The total charge on the lower electrode, and hence G, follows from this result = AD = [€oXa A /l (1 + 2Xa)] q " l n (1 + Xa) tI (4) 6.5.4 Because the field is independent of x and z, (1) and from this it follows that D" = D,,(t). This means that (2) is independent of y and can be solved for E". The voltage is then (3) and this expr.ession can be solved for D", which is the surface charge density on the lower electrode. q = AD" = Gtlj G= -d(1 -A€p e-l/ d) (4) 6.5.5 (a) For each, the electric field intensity in each region takes the form E = irA/r [the potential takes the form cP = Aln(r)]. In the first case, the integral of this field between the electrodes must be the same whether it is taken in the dielectric or in the free space region. Thus, in the first case, laA tI = -dr = Aln(a/b) => E = irtl/rln(a/b) (1) br Note that this solution satisfies the conditions that the tangential field be continuous at the dielectric-free space interfaces and that the normal D be Solutions to Chapter 6 6-14 continuous (there is no normal D). The field is normal to the circular cylinder electrodes and so these are equipotentials, as required. In the second case, the coefficient A has a different value in each of the regions. The two coefficients are found by requiring that (2) and that at the interface (3) Thus, in the second case, E_ irv {1; b < r< R (4)-[In(R/b) + E: In(a/R)]r E/Eo; R<r< a (b) The capacitance follows from integrating the surface charge density over the inner electrode. In the first case, q = l[abEEr(r =6)+ (27r-a)6EoEr(r = 6)] =Cv; (5a) C == l[aE + (27r -a)Eol/ln(a/6) (56) while in the second case e q = l27rbDr (r = b) == Cv; C == 27rlE/[ln(R/b) + -In(a/R)) (6) Eo 6.5.6 Based on experience in the special case where the wedge is of uniform permit­ tivity (so that the spatial variations in permittivity are bumps at the interfaces) postulate that the electric field is no different than if the dielectric wedge were not present. E= [v/rln(a/6»)i r (1) Because the electric field is perpendicular to the gradient in permittivity, there is no induced polarization charge, (6.5.9), and hence no distortion of this field by the dielectric. The field of (1) has no divergence (and of course no curl) and hence does satisfy the bulk conditions throughout the volume. It also has no tangential value on the boundaries, as required. The given capacitance follows from integrating the unpaired surface charge over the surface of the inner electrode. 6-15 Solutions to Chapter 6 6.6 PIECE- WISE UNIFORM ELECTRICALLY LINEAR DIELECTRICS 6.6.1 Given that the imposed potential takes the form .(r -(0) = -Eor cos 0 (1) assume postentials of the form A•=-EorcosO + r2 cosO; r> R (2) r A) r.=BRcos6=(EoR+ WRcosO; r<R (3) Here I the coefficients have already been adjusted to make the potential continuous at r = R. The remaining condition is that (4) from which it follows that A = EoR3(Ea -Eo} (&)(2Eo + Ea) The given potentials follow from substitution of (5) into (2) and (3). 6.6.2 (a) Assume a potential within the cavity that is consistent with the dipole being at the origin with the addition term satisfying Laplace's equation while having the same 0 dependence as the dipole and being finite at the origin. Outside the cavity, the potential again has the 6 dependence of the dipole and goes to zero at infinity. ...L-co;9 + Brcod' r < a • _ 4t1'Eor , (1)-{ Acos 9• a <r r2 I Potential .continuity and continuity of normal D at r = a requires that the coefficients A and B satisfy [i: ~a] [~] = [~] (2) (is" 0 4'1l"~i Thus, A- 3p • (3)-41r(Eo +2E)' so that ~he required potential is { ~ 2~ r<a • = pcos 0 ;:t" -aa Tl+Br; (4)41rEo 3 1. 1+2-'- ;:;, a<r '0 Solutions to Chapter 6 6-16 The electric field follows as 2 2 ~] • [1 2-f--l 3 3 3 E= --.!!..- [r + a ~ cos 01.. + -a3 l+~ r<a r 4'11"Eo { 6 c088i + 3 8in8i. a<r1+*rr.. 1+*rr 8, (5) (b) In the limit E --+ 00, the tangential electric field at r = a becomes 1 2(.£-1) 1 1 lim-- =---=0 (6) E-CO a3 a3(1 Eo + ~:) 3 a3a and the potential inside the cavity, (1a), becomes lim ~b = pcosO (.!.. _~) (7) E-CO 4'll'Eo r2 a3 (c) IT the cavity is regarded as an equipotential at the outset, it follows from (ta) that P1 P---+Ba=O=>B=--­ (8) 4'11"Eo a2 4'11"E oa3 in agreement with what was obtained by taking the limit, (7). 6.6.3 Feom (5.8.4), the potential around a perfectly conducting rod of radius R in a uniform electric field is ~ = -EaR(!:... -R) costIJ (1)R r The potential for a two-dimensional electric dipole is given by (a) of Prob. 4.4.1. ~ = Aid costIJ (2) 2'11"Eo r Comparison of these expressions shows that the induced two-dimensional dipole moment is Aid = (2'11"E oR 2 )Ea (3) The density of the rods is (1/82) perunit area andtherefore thepolarization density IS (4) 6.6.4 The dielectric spheres have induced dipole moments that follow from (a) of Prob.6.6.1. n3 (E,-Eo) p= 4'11"E oEo n.-(E, + 2Eo) (1) Using the arguments of (6.6.6)-(6.6.9), it follows that the equivalent permittivity is E= 1 + 4'll'(R/8)3 (E, -Eo) (2) (E, + 2Eo) Solutions toChapter6 6-17 6.6.5 Writing thepotential intheupperregionasthatofthepointchargeqat y=handitsyettobedetermined imageaty=-h,bothontheyaxis,wehave (Sec.4.4) ~=_1_{~-~j0<y (1) 4'll"Eor+jY<0 where r+=yz2+(y-h)2+z2jr_=yz2+(y+h)2+z2 Intheirrespective regions, thesehavebeenchosentosatisfyPoisson's equation (in theupperregion)andLaplace's equation. Attheinterface, wherey=0in(1),the potential mustbecontinuous forallzandz (2) (3)andthenormalelectricfluxdensitymustbecontinuous (thereisnounpaired surface chargedensity) Simultaneous solution oftheseexpressions givestherelations forqaandqbsum­ marized by(b)intheproblem. Todetermine theforceonthechargecausedby thesurfacepolarization chargeitinducesattheinterface ofthedielectric, compute theelectricfieldaty=h,z=0,z=0using(1)andignoring theselffield(itcan produce nonetforceonitself)andmultiply byq. f=i)'qEI/(z=O,y=h,z=0) Thus,theforceisoneofattraction, asgiven.(4) 6.6.6 Intheupperhalfspace,theparticular solution isthatofalinecharge.Because ithasthesamezdependence ofitspotential inthey=0plane,ahomogeneous solution isaddedtothiswhichisthepotential ofanimagelinechargeat(z,y)= (0,-h). (1) Inthelowerhalfspace,thepotential istakenasthatduetoalinechargelocated at(z,y)=(0,h). (2) Thecoefficients, ~aand~barenowadjusted tosatisfythecontinuity conditions on thepotential andthenormaldielectric fluxdensityinthey=0plane. (3) (4) 6-18 Solutions to Chapter 6 Because each of the terms in one or the other of these expressions has (by design) the same :z; dependence, the boundary conditions can be satisfied by adjusting the coefficients. Simultaneous solution gives ,xa = ,x(€a -€b)/(€a + €b) (5) ,xb = 2€b,x/(€a + €b) (5) In the limit where €b -+ 00 the field in the upper region, (5), becomes that of a line charge over a ground plane, where the image line charge is equal in magnitude and opposite in sign to that of the line charge and the field lines are perpendicular to the surface. In the opposite extreme where the upper region has a very large €, the field lines in the upper region tend to have no normal component. One way to see this is to observe that in the limit €a -+ 00 the image line charge becomes equal to the line charge. 6.6.1 (a) The uniform electric field that would exist if the permittivities were equal is written in polar coordinates as iP = -Eorcos 4J => E = Eo(cos 4Jir-sin 4Ji",) (1) (b) The surface polarization charge density induced by this imposed field is a.v = -P: + P: = -(fa -€b)E r = €b(1- €a)Er = lt€bEocos4J (2) €b (c) The potential induced by this surface charge density is of the form iP = {A cr1"J j r > R A(r/R) cos4Jj r< R (3) where the outer solution leaves the field as that imposed at infinity and the coefficients have been adjusted to insure continuity of the potential at the surface. The continuity condition from Gauss' law then gives BiPa BiPb) -( €a a;: -€b Br = lt€bEo cos 4J (4) hence A = ItEoR (5) 2 Substitution of this coefficient into (3) confirms the given potential. (d) The exact solution given by (6.6.21) and (6.6.22) is first written in terms of It. iPa = -REocos4J[':" _ !!:,_It_] (6)R r2-1t iPb = -REocos4J[':" 2(1- It)] (7)R 2-1t To linear terms in It, note that _It_ -+~. 2(1 -It) -+ 1_ ~ (8) 2-1t 2' 2-1t 2 Using these expressions in (6) and (7) gives the same approximate expressions for the potential as given with the problem. Solutions toChapter6 6.6.8 (a)Ifthedielectric isuniform, thensoistheelectricfield.6-19 (1) (b)From(6.6.25), if(1)approximates theelectricfieldthentheapproximate polarization surfacechargedensityis (2) (c)ForEb<Eel,trap<O.Thus,thedistribution ofsurfacechargedensityand henceelectricfieldisasshowninFig.S6.6.8. (a)jf\\¥W++ .....±+=++ (b) Flprese.e.8 (d)Forthesecondcase,thepolarization surfacechargedensityisassketched in Fig.S6.6.8b. 6.6.9 (a)Thepotential isrepresented asapiece-wise continuous function.Inregions (a)and(b)wherethepermittivity isuniform, itisexpanded insolutions to Laplace's equation thathavezeropotential ontheboundaries. Tosatisfythe potential boundary condition totheleft,Yisaddedtothepotential inregion (b). !~sinh[n;(z-a)]. rur _~-An sinh(nll'a/d) sm(d"Y)i ~-~Bsinh[n;(z+a)] .(nll')y. LJn.h(/d)smdY+,n=l smnll'a Continuity ofDzattheinterface requiresthatO<z<a -a<z<0(1) (nll')(nll'a) (nll')(nll'a)EaAndcothT=-BnEbdcothT whichgives(2) (3) 6-20 Solutions to Chapter 6 To make the potential continuous at z = 0, the constant V is expanded in the same Fourier series as representing the y dependence in the other terms in (1). V = L00 On sin (n~y) (4) n=l Multiplication by sin(m'll'"y/d) and integration on y from 0 to d then gives an expression that can be solved for the coefficients. 4V. On = rnr'n odd (5){ o·, n even The potential continuity condition is then satisfied by each term in the series. 4V An = Bn +-j n odd (6) n'll'" It follows from (3) and (6) that the coefficients in (1) are An = Bn = 0 for n even and for n odd. A _ 4V 1 . (7)n -n'll'" 1 + fa/ fb ' (b) In sketching ~ and E, as shown in Fig. S6.6.9a for the case where the permit­ tivities are equal, note that the potential varies from V to 0 across the gaps. Every other point on the boundaries is either at potential V or potential o. Thus, equipotentials all terminate and originate in the gaps. The equipoten­ tial ~ = V/2 is inthe z = 0 plane. Thus, the potential and field lines in each region are as shown in Fig. 5.5.3. (c) The surface charge density is given by using (6.6.25) with E approximated by what it would be if the permittivities were equal. (8) Inthe case where fa/fb > 1, Uap < 0, as illustrated by Fig. S6.6.9b. Some of the field lines originating to the left terminate in the negative U ap on the interface. Thus, the dielectric to the right tends to shield out the field. With fa/ fb < 1, the surface charge density is positive, and the field tends to be shielded out of the material to the left. (d) With fa :> fb, the surface becomes an equipotential and the field is concen­ trated in the region to the left, as shown in Fig. S6.6.9c. Solutions toChapter6 c'[)=v-- ('l) -,, (b)6-21 (e) FIKure98.8.9 (e)With Eb:>Ea,thefieldisshielded outoftheregiontotheleft.Thefieldlooks muchasinFig.S6.6.9cexceptthatthefieldsareontherightratherthanthe left.Theequipotential ~=V/2isinthez=0plane.Thus,thepotential and fieldlinesineachregionareasshowninFig.5.5.3. 6.7SMOOTHLY INHOMOGENEOUS ELECTRICALLY LINEAR DIELECTRICS 6.7.1 Farfromthelowerend,thesystembecomes apairofparallelplateshaving thepotential difference Vseparated byadielectric havingitspermittivity gradient intheydirection. Thus,asy-00,thepotential becomes simply z ~(y-00)-V- (1)a Theproduct solutions throughout theregionbetween theplatesareasdeveloped inExample 6.7.1.Fromthose,weaddto(1)thosethatarezeroatz=0and 6-22 Solutions to Chapter 6 x = a (so as not to disturb the fact that (1) already satisfies the conditions on the potential there) and that go to zero as y -+ 00 [again, so that the potential there becomes (1)]. (2) To determine the coefficients, Vn , (2) is evaluated at y= 0 and set equal to the potential there. V = L00 Vn sin n1r x +V:: (3) a a n=l Multiplication by sin(m1rx/a) and integration from x = 0 to x = a then gives the coefficients and hence the potential. a21 x. n1rVn =- V(I- -) sm -xdx = (2/n1r)V (4) a 0 a a 6.1.2 The solutions to (6.7.2) for the given distributions of permittivity are as found in Example 6.7.1 with the roles of x and y interchanged. In the region to the left, {3 -+ -{3. Because the system extends to infinity in the ±x directions, exponential solutions are selected in each of the regions that decay to zero at infinity. (1) Continuity of D~ gives one condition on the coefficients. [€pe{t~ E~l~=o = [€pe-{t~ E;I~=o => Bn = -An (2) To match the potential in the x = 0 plane, the first term in the solution to the left, in (lb), is expanded in the same series as the other terms. V(a_y)= f:Cn sin(n1r y) (3)a a n=l The coefficient is found by multiplying this expression by sin(m1rY / a) and integrat­ ing from y= 0to Y= a. (4) It follows from (2) and (4) that (5) Solutions toChapter6 6.7'.3 (a)Thepolarization chargedensityisapproximately6-23 (1) I(b)ForEoXp>0,thefieldinducespositiveandnegative regionsofchargedensity intheupperandlowerregionsrespectively. Thesearecentered ontheyaxis wherethefunction yexp(-y2/ a2)peaks,aty=a/-/iThus,someofthe fieldlinesentering frombelowinFig.86.7.3terminate onthenegative charge whilesomeleavingatthetoporiginate onthepositive charge.Thefieldis thatofadiffusedipole. y \) +++++ (\ Figure88.'.S SOLUTIONS TO CHAPTER 7 7.1 CONDUCTION CONSTITUTIVE LAWS 'T .1.1 H there are as many conduction electrons as there are atoms, then their num­ ber density is N _ ~ _ (6.023 X 1026(8.9 X 103) _ 4 1028 electrons (1)--MP - 63.5 -8. X rn3 o The mobility is then 0' 5.8 x 107 -3 IJ-= N_q_ = (8.4 X 1038)(1.6 X 10-19) = 4.3 X 10 (2) The electric field required to produce a current density of lA/cm2 is E= -J = 104 = 1.7 X 1O-4v/rn (3) 0' 5.9 X 107 Thus, in copper, the velocity of the electrons giving rise to this current density is only (4) 7.2 STEADY OHMIC CONDUCTION 'T .2.1 Boundary conditions on the conducting region are that Cl> = 0, Cl> = v on the perfectly conducting surfaces at r = a and r= b respectively and that there is no normal current density on the insulating surfaces where z = 0, z = d. The latter are satisfied by a potential that is independent of the axial coordinate, so an appropriate solution to Laplace's equation, arranged to be zero on the outer electrode, is Cl> = Aln(r/ a) (1) The coefficient is adjusted to make the potential v on the inner electrode so that A = v/ln(b/a) and (1) becomes Cl> = vln(r/a)/ln(b/a) (2) The current density is vO' 1 (3)In(b/a) ;: and so the total current is 211'bdO' 1 211'0'd vi = 21rbdJr = -= v=- (4)In(b/a) b In(a/b) R Thus, R is as given. 1 7-2 Solutions to Chapter 7 1.2.2 The net current passing through the wire connected to the inner spherical electrode, " must be equal to the net current at any radius r. i = {J. da = 41rr2uEr => E r = ' 2 (1) -4Js 1rur Thus, , dr ill t} = la Erdr =-la -2=-[---] (2) b 41ru br 41ru b a By definition 1I =iR 80 R = <t-~)/41ru. 1.2.3 (a) Associated with the uniform field is the potential 1I (p = --(y -d) (1)d IT the surrounding region is insulating relative to that between the elec­ trodes, the normal component of the current density on the conductor surfaces bounded by the insulating surroundings is zero. The potential is constrained on the remainder of the surface enclosing the conductors, so the solution is uniquely specified. Provided the laws are satisfied everywhere inside the con­ ducting region, the solution is exact. The given solution does indeed satisfy the boundary conditions on the surfaces of the conducting region. In the case of (a), the potential and normal component of current density must be contin­ uous across the interior interface. Further, in the uniformly conducting regions of (a), Laplace's equation must be satisfied, as it is by a uniform field. In the case of (b), (7.2.4) is satisfied by the given potential. (b) The total current is related to 1I by integrating the current density over the surface of the lower conductor. (c) A similar calculation gives the resistance in the second case. ,. cl'-zdz = -UallC l'(1 +-z)dz = 2ua lc = GlI (3) = Ull --1I odd 0 1 2d 1.2.4 The potential in each of the uniformly conducting regions takes the form (1) where the four coefficients are adjusted to make the potentials zero and 1I on the respective electrodes, and make both the potential and the normal current density continuous at the interface between the conductors. On the surfaces at r = a and 7-3 Solutions to Chapter 7 r = b, the current density must be zero, as it is for the potential of (1) because the electric field E __!. 8~ _ {(Alr)i~ (2)-r 8q, -(B/r)l~ has no radial component. Rather than proceeding to determine the four coefficients in (1), we work directly with the electric field. The integration of E from one electrode to the other must be equal to the applied voltage. "" A "" B-r-+ -r-= fJ (3)2 r 2 r Further, the current density must be continuous at the interface. (4) It follows from these relations that (5) The current through any crosB-section of the material [say region (a)1 must be equal to that through the wire. Thus, .la la [2dO'a dr] s = d O'aE~dr = (/) -fJ == G'lJ (6) b "" 1 + O'aO'b b r and the resistance is 2dO'a (/) G =( a) Ina b (7) "" 1+:<.A.a. '1.2.5 (a) From (7.2.23) (1) (b) We need the electric field, which follows from (7.2.19) by using the result of (1) to evaluate Jo = i/A = GfJ/A' (2) Thus, the unpaired charge density is evaluated using (7.2.8). (3) 1-4 Solutions to Chapter 1 1.2.6 (a) The inhomogeneity in permittivity has no effect on the resistance. It is there­ fore given by (7.2.25). (b) With the steady conduction laws stipulating that the electric field is uniform, the unpaired charge density follows from Gauss' law. (v) v BE EaV 1pu=V·EE=V· E-i =--=--------::c (1)d Y d By da (1 + ~)2 1.2.7' At a radius r, the area of the conductor (and with r = a and r = b, of the outer and inner electrodes, respectively) is (1) Consistent with the insulating surfaces of the conductor is the requirement that the current density and associated electric field be radial. Current conservation (fundamentally, the requirement that the current density be solenoidal) then gives as a solution to the field laws CTEr [2,",2(1- cos i)] =i (2) and it follows that (3) The voltage follows as v = r Erdr = i(aS -63 )/611"CT o(1- cos ~)6Sa (4)lb 2 and this relation takes the form i = vG, where G is as given. 7'.2.8 There can be no current density normal to the interfaces of the conducting material having normals in the azimuthal direction. These boundary conditions are satisfied by an axially symmetric solution in which the current density is purely radial. In that case, both E and J are independent of q,. Then, the total current is related to the current density and (through Ohm's law) electric field intensity at any radius r by i = 211"QdrJ r = 211"QdCT oaEr (1) Thus, , Er = 211"QdCT oa (2) and because Erdr = i(a -b) = vla (3) b 211"QdCT oa G = 211"QdaCT o/(a -b) (4) Solutions to Chapter 7 7-5 7.3 DISTRIBUTED CURRENT SOURCES AND ASSOCIATED FIELDS '1.3.1 In the conductor, the potential distribution is a particular part comprised of the potential due to the point current soruce, (6) with ip -+ I and In order to satisfy the condition that there be no normal component of E at the interface, a homogeneous solution is added that amounts to a second source of the same sign in the lower half space. Of course, such a current source could not really exist in the lower region so if the field in the upper region is to be given some equivalent physical situation, it should be pictured as equivalent to a pair of like­ signed point current sources in a uniform conductor. In any case, this second source is located at r = vz2 + (y+ h)2 + z2 and hence the potential in the conductor is as given. In the lower region, the potential must satisfy Laplace's equation everywhere (there are no charges in the lower region). The field in this region is uniquely specified by requiring that the potential be consistent with (a) evaluated at the interface (1) and that it go to zero at infinity in the lower half-space. The potential that matches these conditions is that of a point charge of magnitude q = 2Ie/u located on the y axis at y= h, the given potential. '[.3.2 (a) First, what is the potential associated with a uniform line current in a uniform conductor? In the steady state (1) and for a surface S that has radius r from the line current, K,K, = 27rrJ r = 27rruEr => Er =-­ (2)27rur Within a constant, the associated potential is therefore K, ~ = --In(r) (3)27rU To satisfy the requirement that there be no normal current density in the plane y = 0, the potential is that of the line current located at y = h and an image line current of the same polarity located at y = -h. 7-6 Solutions to Chapter 7 Note that the normal derivative of this expression in the plane y =0 is indeed zero. (b) In the lower region, the potential must satisfy Laplace's equation everywhere and match the potential of the conductor in the plane y =o. (5) This has the potential distribution of an image line current located at 11 = h. With the magnitude of this line current adjusted so that the potential of (5) is matched at z = 0, (6) the potential is matched at every other value of z as well. 1.3.3 First, the potential due to a single line current is found from the integral form of (2). (1) Thus, for a single line current, (2) For the pair of line currents, spaced by the distance d, X, X, [ dCOS 4J] K,dcos4J~ = --lln(r -dcos4J) -InrI = --In 1--- --. --':::------'- (3) 2~q 2~q r 2~qr 7.4 SUPERPOSITION AND UNIQUENESS OF STEADY CONDUCTION SOLUTIONS 1.4.1 (a) At r = b, there is no normal current density 80 that (1) while at r = a, (2) 7-7 Solutions to Chapter 7 Because the dependence of the potential must be the same as the radial deriva­ tive in (2), assume the solution takes the form cos(} Cb = Arcos(} + B-2­ (3)r Substitution into (1) and (2) then gives the pair of equations 1 -2b-~] [A] =[ 0 ] (4)[ 0' -20'a S B Jo from which it follows that (5) Substitution into (3) results in the given potential in the conducting region. (b) The potential inside the hollow sphere is now specified, because we know that the potential on its wall is (6) Here, the origin is included, so the only potential having the required depen­ dence is Cb =Crcos(} (7) Determination of C by evaluating (7) at r = b and setting it equal to (6) gives C and hence the given interior potential. What we have carried out is an "inside-outside" calculation of the field distribution where the "inside" region is outside and the "outside" region is inside. '1.4.2 (a) This is an example of an inside-outside problem, where the potential is first determined in the conducting material. Because the current d~nsity normal to the outer surface is zero, this potential can be determined without regard for the geometry of what may be located outside. Then, given the potential on the surface, the outside potential is determined. Given the tP dependence of the normal current density at r = b, the potential in the conducting region is taken as having the form (1) Boundary conditions are that 8Cbb Jr = -0'-- =0 (2)8r at r = a, which requires that B = asA/2 and that (3) 7-8 Solutions to Chapter 7 at r= b. This condition together with the result of (2) gives A = Jo/CT[(a/b)3­ 11. Thus, the potential in the conductor is (4) (b) The potential in the outside region must match that given by (4) at r = a. To match the 0 dependence, a dipole potential is assumed and the coefficient adjusted to match (4) evaluated at r = a. a 3Joa ( / )2 (5)~ = 2CT[(a/b)3 _11ar cos 0 1.4.3 (a) This is an inside-outside problem, where the region occupied by the conductor is determined without regard for what is above the interface except that at the interface the material above is insulating. The potential in the conductor must match the given potential in the plane y = -a and must have no derivative with respect to y at y = o. The latter condition is satisfied by using the cosh function for the y dependence and, in view of the x dependence of the potential at y = -a, taking the x dependence as also being cos(,Bx). The coefficient is adjusted so that the potential is then the given value at y = o. ~b _ V cosh ,By ,B - cosh f3a cos x (1) (b) in the upper region, the potential must be that given by (1) in the plane y = 0 and must decay to zero as y --> 00. Thus, ~a = V cos f3x e-{3Y (2)cosh f3a 1.4.4 The potential is zero at 4J = 0 and 4J = 11"/2, so it is expanded in solutions to Laplace's equation that have multiple zeros in the 4J direction. Because of the first of these conditions, these are solutions of the form ~ ex r±n sin nO (1) To make the potential zero at 4J = 11"/2, 11"n2" = 'If, 2'1f, ... => n= 2,4, ... 2m; m = 1,2,3, ... (2) Thus, the potential is assumed to take the form 2m + Bm r-2m~ = L00 (Amr ) sin 2m4J (3) m=l 7-9 Solutions to Chapter 7 At the outer boundary there is no normal current density, so a.-(r= a) =0 (4)ar and it follows from (3) that (5) At r = b, the potential takes the form • =Eco Vm sin2mtP = V (6) m=l The coefficients are evaluated as in (5.5.8) through (5.5.9). Jr 2 VVn -11"l/ v sin 2nOdO =-i nodd 4 0 n Thus, Am = 4v/m1rb2m[l-(a/b)4mJ (8) Substitution of (8) and (5) into (8) results in the given potential. '1.4.5 (a) To make the tP derivative of the potential zero at tP = 0 and tP = Q, the tP dependence is made cos(n1l"tP/ Q ). Thus, solutions to Laplace's equation in the conductor take the form where n = 0, 1,2, ... To make the radial derivative zero at r = b, (2) so that each term in the series (8) satisfies the boundary conditions on the first three of the four boundaries. Solutions to Chapter 7 7-10 (b) The coefficients are now determined by requiring that the potential be that given on the boundary r = a. Evaluation of (3) at r = a, multiplication by cos(m'lrt/J/oe) and integration gives a2 a tI 1 / m'lrt/J tI1 m'lr -- cos (-)dt/J + - cos (-t/J)dt/J2 0 oe 2 a/2 oe = fa f An [(a/b) (ntr/a) + (b/a)(ntr/a)]10 n=O (4) cos (n'lrt/J) cos (m'lrt/J)dt/J oe oe 2oe. m'lr=--sm(-)m'lr 2 and it follows that (3) is the required potential with (5) '1.4.6 To make the potential zero at t/J = 0 and t/J = 'Ir/2, the t/J dependence is made sin(2nt/J). Then, the r dependence is divided into two parts, one arranged to be zero at r = a and the other to be zero at r = b. 00 ~ = 2: {An[(r/a)2n -(a/r)2nJ + Bn{(r/b)2n -(b/r)2nJ} sin(2nt/J) (1) n=l Thus, when this expression is evaluated on the outer and inner surfaces, the bound­ ary conditions respectively involve only Bn and An. ~(r = a) = tla = L00 Bn[(a/b)2n -(b/a)2nJ sin 2nt/J (2) n=l ~(r = b) = tlb = L00 An[(b/a)2n -(a/b)2nJ sin2n.p (3) n=l To determine the Bn's, (2) is multiplied by sin(2mt/J) and integrated (4) and it follows that for n even Bn = 0 while for n odd (5) Solutions to Chapter 7 7-11 A similar usage of (3) gives (6) By definition, the mutual conductance is the total current to the outer electrode when its voltage is zero divided by the applied voltage . 'a.I "..=0 = d fr / _':11'_1 _ ad4J =_ d fr 2 co (7) G = _-.!!.. l2 a""'" ua l /E An 4n sin 2n4Jd4J Vb Vb 0 ar r-a Vb 0 n=l a and it follows that the mutual eonductance is (8) 7.5 STEADY CURRENTS IN PIECE-WISE UNIFORM CONDUCTORS 7.5.1 To make the current density the given uniform value at infinity, Jo • -+ --rcosOj r -+ 00 (1) Ua At the surface of the sphere, where r = R (2) and (3) In view of the 0 dependence of (I), select solutions of the form Jo cosO .a=.b .a = --rcosO + A--' .b = BrcosO (4) Ua r 2' Substitution into (2) and (3) then gives A __ JoRs (ua -Ub) • - Ua (2ua + Ub) , (5) and hence the given solution. ---7-12 Solutions to Chapter 7 1.5.2 These are examples of inside-outside approximations where the field in region (a) is determined first and is therefore the "inside" region. (a) HUb:::> 0'.., then ~a(r =R) $lid constant =0 (1) (b) The field must be -(Jolu ..)I. far from the sphere and satisfy (1) at r= R. Thus, the field is the sum of the potential for the uniform field and a dipole field with the coefficient set to satisfy (1). ~.. $lid RJo [ -r -(IR r)2] cosO (2) 0'.. R (c) Atr = R, the normal current density is continuous and approximated by using (2). Thus, the radial current density at r = R inside the sphere is A solution to Laplace's equation having this dependence on 0 is the potential of a uniform field, ~ = Brcos(O). The coefficient B follows from (3) so that ~b $lid _ 3JoR(rIR) cosO (4) Ub In the limit where Ub :::> 0'.., (2) and (4) agree with (a) of Prob. 7.5.1. (d) In the opposite extreme, where 0'.. :::> Ub, (5) Again, the potential is the sum of that due to the uniform field that prevails at infinity and a dipole solution. However, this time the coefficient is adjusted so that the radial derivative is zero at r = R. .. $IId---RJo [ -+-r 1( R1)2] (6) ~ r cosO 0'.. R 2 To determine the field inside the sphere, potential continuity is used. From (6), the potential at r = R is ~b = -(3RJ o/2u..)cos 0 and it follows that inside the sphere b 3 RJo 1 )~ $lid ---(r R cosO (7)2 0'.. In the limit where 0'.. :::> Ub, (a) of Prob. 7.5.1 agrees with (6) and (7). Solutions to Chapter 7 7-13 '1.5.3 (a) The given potential implies a uniform field, which is certainly irrotational and solenoidal. Further, it satisfies the potential conditions at z = 0 and z = -I and implies that the current density normal to the top and bottom interfaces is zero. The given "inside" potential is therefore the correct solution. (b) In the "outside" region above, boundary conditions are that t»(z = O,z) = -vz/'j t»(z, 0) = OJ zt»(a, z) =OJ ~(-l, z) =v(1- -) (1)a The potential must have the given linear dependence on the bottom horizontal interface and on the left vertical boundary. These conditions can be met by a solution to Laplace's equation of the form :sz. By translating the origin of thez axis tobeatz=a, the solution satisfying the boundary conditions on the top and right boundaries is of the form v t» = A(a -z)z =-la (a -z)z (2) where in view of (1a) and (1c), setting the coefficient A = -v/l makes the potential satisfy conditions at the remaining two boundaries. (c) In the air and in the uniformly conducting slab, the bulk charge density, Pu, must be zero. At its horizontal upper interface, CTu = faE: - fbE~ = -fovz/'a (3) Note that z < 0soifv >0, CTu > 0 as expected intuitively. The surface charge density on the lower surface of the conductor cannot be specified until the nature of the region below the plane z = -b is specified. (d) The boundary conditions on the lower "inside" region are homogeneous and do not depend on the "outside" region. Therefore the solution is the same as in (a). The potential in the upper "outside" region is one associated with a uniform electric field that is perpendicular to the upper electrode. To satisfy the condition that the tangential electric field be the same just above the interface as below, and hence the same at any location on the interface, this field must be uniform. H it is to be uniform throughout the air-space, it must be the same above the interface as in the region where the bounding conductors are parallel plates. Thus, v. v. ()E =;Ix + yl. 4 The associated potential that is zero at z = 0 and indeed on the surface of the electrode where z = -za/l is v v ~ = -z+ -z (5)a I Finally, instead of (3), the surface charge density is now 7-14 Solutions to Chapter 7 '1.5.4 (a) Because they are surrounded by either surfaces on which the potential is con­ strained or by insulating regions, the fields within the conductors are deter­ mined without regard for either the fields within the square or outside, where not enough information has been given to determine the fields. The condi­ tion that there be no normal current density, and hence no normal electric field intensity on the surfaces of the conductors that interface the insulating regions, is automatically met by having uniform fields in the conductors. Be­ cause these fields are normal to the electrodes that terminate these regions, the boundary conditions on these surfaces are met as well. Thus, regardless of what d is relative to a, in the upper conductor, E = -ix!j ~ = !Zj J = -O'!i x (l) a a a while in the conductor to the right • tJ .... tJ J tJ. ()E = -I)'-j 'W' = -1Ij = -0'-1 2 a a a )' (b) In the planes 11 = a and Z= a the potential inside must be the same as given by (l) and (2) in these planes, linear functions of Z and of 11, respectively. It must also be zero in the phmes Z =0 and 11 =O. A simple solution meeting these conditions is tJ ~= AZ1I= -Z1I (3)a2 Figure Sf.& •.( (c) The distribution of potential and electric field intensity is as shown in Fig. S7.5.4. '1.5.5 (a) Because the potential difference between the plates, either to the left or to the right, is zero, the electric field there must be zero and the potential that of the respective electrodes. (l) (b) Solutions that satisfy the boundary conditions on all but the interface at z = 0 are (2a) Solutions toChapter7 DO ~b=V+LBnenrrs/4 sinn1l'y an=1 (c)Attheinterface, boundary conditions are 8~4 8~b -0'4--=-0'1>--8z 8z ~4=~1> (d)Thefirstoftheserequires of(2)that Writtenusingthis,thesecondrequiresthat 00 DO ""'A.n1l' ""'O'aA.n1l' LJnsm-y= v-LJ-nsm-y n=1 a n=10'1> a7-15 (26) (3) (4) (5) (6) Theconstant termcanalsobewrittenasaFourierseriesusinganevaluation ofthecoefficients thatisessentially thesameasin(5.5.3)-(5.5.9). DO4v.n1l'v=L-sm-y1I'nan=1 Thus, (O'a)4vAn1+-=- 0'1>n1l' anditfollowsthattherequired potential is(7) (8) (9) =~=I ==Ql[)~'~== Figure57'.&.& 7-16 Solutions to Chapter 7 (e) In the case where Ub ::> U a, the "inside" region is to the left where bound­ ary conditions are on the potential at the upper and lower surfaces and on its normal derivative at the interface. In this limit, the potential is uniform throughout the region and the interface is an equipotential having ~ = tI. Thus, the potential in the region to the right is as shown in Fig. 5.5.3 with the surface at 11 = b playing the role of the interface and the surface at 11 =0 at infinity. In the case where the region between electrodes is filled by a uni­ form conductor, the potential and field distribution are as sketched in Fig. 87.5.5. In the vicinity of the regions where the electrodes abut, the potential becomes that illustrated in Fig. 5.7.2. By symmetry, the plane z = 0 is one having the potential ~ =tI/2. (f) The surface at 11 = a/2 is a plane of symmetry in the previous configuration and hence one where E" = O. Thus, the previous solution applies directly to finding the solution in the conducting layer. 7.6 CONDUCTION ANALOGS '1.6.1 The analogous laws are E=-V~ E=-V~ (1) V'UE=8 V'EE=pu (2) The systems are normalized to different length scales. The conductivity and per­ mittivity are respectively normalized to U c and E£ respectively and similarly, the potentials are normalized to the respective voltages Vc and V£. (z, 11, z) = (~'ll' &:)l£ (3) ~=Vc~ ~=V£~ (4) E = (Vc/lc)~ E = (V£/l£)~ (5) 8=(ucVc/~)I. (6) Pu = (E£Vdl~)p (7)'-U By definition, the normalized quantities are the same in the two systems Q:(r) = .€(r) (8) I.(r) =!!.u(r) (9) so that both systems are represented by the same normalized laws. E=-~ (10) Solutions to Chapter 7 7-17 (11) Thus, the capacitance and conductance are respectively C = fEIE£fE . d1!/LE. d! (12) G = uele £d .dll/ L::&· dll (13) where, again by definition, the normalized integral ratios in (12) and (13) are the same number. Thus, ~G=~~=!~ (~ U e Ie U Ie Note that the deductions summarized by (7.6.3) could be made following the same normalization approach. 7.7 CHARGE RELAXATION IN UNIFORM CONDUCTORS 1.1.1 (a) The charge is given when t = 0 .1r.1r ()P = Pi sm ~ xsm,? 1 Given the charge density, none of the bulk or surface conditions needed to determine the field involve time rates of change. Thus, the initial potential distribution is determined from the initial conditions alone. (b) The properties of the region are uniform, so (3) and hence (4) apply directly. Given the charge is (c) of Prob. 4.1.4 when t = 0, the subsequent distribution of charge is • 1r • 1r tff' f (2)P = Po ()t sm ~zsm "bY; Po = Pie- j T == ; (c) As in (a), at each instant the charge density is known and all other conditions are independent of time rates of change. Thus, the potential and field distri­ butions simply go along with the changing charge density. They follow from (a) and (b) of Prob. 4.1.4 with Po(t) given by (2). (d) Again, with Po(t) given by (2), the current is given by (6) of Prob. 4.1.4. 1.1.2 (a) The line charge is pictured as existing in the same uniformly conducting ma­ terial as occupies the surrounding region. Thus, (7.7.3) provides the solution. AI = AI(t = O)e-tff'; T = f./u (1) (b) There is no initial charge density in the surrounding region. Thus, the charge density there is zero. (c) The potential is given by (1) of Probe 4.5.4 with AI given by (1). Solutions to Chapter 7 7-18 1.1.3 (a) With q < -qc, the entire surface of the particle can collect the ions. Equation (7.7.10) becomes simply i = -p.p61fR2 Ea r (cos6 +!!...) sin6d6 (1)Jo qc Integration and the definition of qc results in the given current. (b) The current found in (a) is equal to the rate at which the charge on the particle is increasing. dq p.p -=--q (2)dt E This expression can either be formally integrated or recognized to have an exponential solution. In either case, with q(t =0) = qo, (3) 1.1.4 The potential is given by (5.9.13) with q replaced by qc as defined with (7.7.11) ~ = -EaRcos6[..!:.. _ (R/r)2] + 121rEoR2Ea (1)R 41rE o r The reference potential as r -00 with 6 = 1f/2 is zero. Evaluation of (1) at r = R therefore gives the particle potential relative to infinity in the plane 6 = 1r/2. ~= 3REa (2) The particle charges until it reaches 3 times a potential equal to the radius of the particle multiplied by the ambient field. 7.8 ELECTROQUASISTATIC CONDUCTION LAWS FOR INHOMOGENEOUS MATERIAL 1.8.1 For t < 0, steady conduction prevails, so a( )/at =0 and the field distribu­ tion is defined by (7.4.1) v ·(uV~) =-8 (1) where ~ = ~I: on S'j -uV~ =3I: on S" (2) To see that the solution to (1) subject to the boundary conditions of (2) is unique, propose different solutions ~a and ~b and define the difference between these solu­ tions as ~d = ~a -~b (3) Solutions to Chapter 7 7-19 Then it follows from (1) and (2) that (4) where ~d =0 on S'; (5) Multiplication of (4) by ~d and integration over the volume V of interest gives Gauss' theorem converts this expression to (7) The surface integral can be broken into one on S', where ~d = 0 and one on SIt, where UV~d = O. Thus, wha.t is on the left in (7) is zero. If the integrand of what is on the right w.ere finite anywhere, the integral could not be zero, so we conclude that to within a constant, ~d = 0 and the steady solution is unique. For 0 < t, the steps beginning with (7.8.11) and leading to (7.8.15) apply. Again, the surface integration of (7.8.11) can be broken into two parts, one on S' where ~d = 0 and one on SIt where -UV~d = O. Thus, (7.8.16) and its implications for the uniqueness of the solution apply here as well. 1.9 CHARGE RELAXATION IN UNIFORM AND PIECE­ WISE UNIFORM SYSTEMS 'T.9.1 (a) In the first configuration, the electric field is postulated to be uniform through­ out the gap and therefore the same as though the lossy segment were not present. E= irv/rln(a/b) (1) This field is iITotational and solenoidal and integrates to v between r = b and r= a. Note that the boundary conditions at the interfaces between the lossy-dielectric and the free space region are automatically met. The tangential electric field (and hence the potential) is indeed continuous and, because there is no normal component of the electric field at these interfaces, (7.9.12) is satisfied as well. (b) In the second configuration, the field is assumed to take the piece wise form R<r<a (2)b <r< R 7-20 Solutions to Chapter 7 where A and B are determined by the requirements that the applied voltage be consistent with the integration of E between the electrodes and that (7.9.12) be satisfied at the interface. Aln(a/R) + bln(R/b) = 4) (3) ;w [EoA _ Ebb] _ ub =0 (4)RR R It follows that A = (;WEb + u)v/Det (5) !J =;wEov/Det (6) where Det is as given and the relations that result from substitution of these coefficients into (2) are those given. (c) In the first case, the net current to the inner electrode is 'l • 1[( )bb I v lo:buv ,=1w 271" - 0: Eo + 0: E bln(a/b) + bln(a/b) (7) This expression takes the form of the impedance of a resistor in parallel with a capacitor where i = vG + jwCv (8) Thus, the C and G are as given in the problem. In the second case, the equivalent circuit is given by Fig. 7.9.5 which implies that 'l v(;wC a )(1 + ;wRCb),= (9)1 + jwR(C a + Cb) In this case, the current to the inner electrode follows from (6) as 27l"l~·WE (1 + i!!!!)'l InaR tT ,= ---_...>....:.!.~----'---~ (10) 1 + jwln(Rlb) [~ + E ] tT InlalR) 'n(rlb) Comparison of these last two expressions results in the given parameters. 1.9.2 (a) In the first case, where the interface between materials is conical, the electric field intensity is what it would be in the absence of the material. (1) This field is perpendicular to the perfectly conducting electrodes, has a contin­ uous tangential component at the interface and trivially satisfies the condition of charge conservation at the interface. 7-21 Solutions to Chapter 7 In the second case, where the interface between materials is spherical, the field takes the form E_.R { 1/,-2j R < r < a -I.. e A (2)D/r2j b<r<R The coefficients are adjusted to satisfy the condition that the integral of E from r = b to r = a be equal to the voltage, .. 11 1'.11 ..A(---)+D(---)=v (3)R a bR and conservation of charge at the interface, (7.9.12). 0' A •(1 ~) 0 --D+3W f --f- == (4)R2 Ow W Simultaneous solution of these expressions gives 1= (0' +jWf)fJ/Det (5) iJ = jWfo/Det where a-R . [a-R R-b] Det:= O'(~) +3W f(~) +fo(---,;Il) which together with (2) give the required field. (b) In the first case, the inner electrode area subtended by the conical region oc­ cupied by the material is 271"b2 [1- cos(a/2)J. With the voltage represented as v = Re fJexp(jwt), the current from the inner spherical electrode, which has the potential v, is (6) Equation (6) takes the same form as for the terminal variables of the circuit shown in Fig. S7.9.2a. Thus, (7) abO' G = 271"[1 -cos(a/2)J­a-b . (271"-a)]abf oCa =271"[1-cos-2- a _ b (8) 7-22 Solutions toChapter 7 abfCb=211"[1-cos(a/2)] a_b t-+ v (a)t-•+ v (b) FlpreSf.9.J (9) (11) (12)(10)=411'Rba_,'w(411'RbE+411'aREg)R-b R-b a-R Thistakesthesameformastherelationship between theterminal voltage andcurrentforthecircuitshowninFig.S7.9.2b. "I;wCa(G+;wCb)~t= vG+;w(Ca+Cb) Thus,theelements intheequivalent circuitare G=4Rbu.C=41rRbf o•Cb=41rRbf 1rR_b,aR-b IR-bInthesecondcase,thecurrentfromtheinnerelectrode is "I2(ufJ. fJ) t=41rbb2+,wfb2 ,'w(411'aREg)(4l1'Rba+,'w411'Rbf)a-R R-b R-b 7.9.3 Intermsofthepotential, v,oftheelectrode, thepotential distribution and hencefielddistribution are (3)(2)(1)va ~=v(a/r)=>E=il'2r Thetotalcurrentintotheelectrode isthenequaltothesumoftherateofincrease ofthesurfacechargedensityontheinterface between theelectrode andthemedia andtheconduction currentfromtheelectrode intothemedia. i=l[:tfEr+uEr]da Inviewof(1),thisexpression becomes.(fadvua) dvt=21ra2--+-v=21rfa-+(21rua)va2dta2 dt Theequivalent parameters arededuced bycomparing thisexpression toonede­ scribingthecurrentthrough aparallelcapacitance andresistance. 7-23 Solutions to Chapter 7 J '1.9.4 (a) With A and B functions of time, the potential is assumed to have the same ; dependence as the applied field. cos; ~a =-Ercos;+A-­ (1)r ~b = Brcos; (2) The coefficients are determined by continuity of potential at r = a (3) and the combination of charge conservation and Gauss' continuity condition, alsoatr=a (UaE: -ubE~) + :t(f:aE: -f:bE~) = 0 (4) Substitution of (1) and (2) into (3) and (4) gives A A --Ba = Ea => B =--E (5)a a2 A d A CTa(E -a2) + Ub B + dt [f:a(E + a2) + f:bB ] = 0 (6) and from these relations, With Eo the magnitude of a step in E(t), integration of (7) from t = 0-when A = 0to t = 0+ shows that (8) A particular solution to (7) for t > 0 is A = (Ub -ua)a2Eo (9)Ub+Ua while a homogeneous solution is exp(-t/T), where (10) Thus, the required solution takes the form 7-24 Solutions to Chapter 7 where the coefficient Al is determined by the initial condition, (8). Thus, [(fb-fa) (Ub -Ua )] 2E -t/" + (Ub -Ua) 2EA --(fb +fa) - (Ub +O'a) a oe (Ub +ua) a 0 (12) The coefficient B follows from (5). AB= --Eo (13) a2 In view of this last relation, and then (12), the unpaired surface charge density is A A U au = fa (Eo + R2) +fb( R2 -Eo) (14) = 2(faUb -fbUa) E (1 _ e-t /,,) Ua +Ub o (b) In the sinusoidal steady state, the drive in (7) takes the form Re t'exp(jwt) and the resonse is of the form Re ..4exp(jwt). Thus, (7) shows that ..4 -[(Ub -ua) +jW(fb -fall 2 t (15) -(O'b +ua) +jW(fb +fa) a p and in turn, from (5), lJ =_ 2(ua + jWfa ) (16) (Ub +ua) +jW(fb +fa) This expressions can then be used to show that the complex amplitude of the unpaired surface charge density is A 2(Ubfa -Uafb) () U au = (Ub +Ua) +jW(fb +fa) 17 (c) From (1) a.nd (2) it is clear that the plane <p = 7f/2 is one of zero potential, regardless of the values of the drive E(t) or of A or B. Thus, the z = 0 plane can be replaced by a perfect conductor. In the limit where Ua --+ 0 and W(fa +fb)/Ub <: 1, (15) and (16) become (18) iJ --+ -2jwfa E (19) Ub p Substitution of these coefficients into the sinusoidal steady state versions of (1) and (2) gives ~a = -Re EA a[r---a] cos<peJw. t (20)Par A ...b R 2jwfa EJ·wt .... =-e--pe (21) Ub These are the potentials that would be obtained under sinusoidal steady state conditions using (a.) and (b) of Prob. 7.9.5. Solutions to Chapter 7 7-25 1.9.5 (a) This is an example of an "inside-outside" situation. The "inside" region is the one where the excitation is applied, namely region (a). In so far as the field in the exterior region is concerned, the surface is essentially an equipotential. Thus, the solution given by (a) must be constant at r = a (it is zero), must become the uniform applied field at infinity (which it does) and must be comprised of solutions to Laplace's equation (which certainly the uniform and dipole fields are). (b) To approximate the interior field, note that in general charge conservation and Gauss' law (7.9.12) require that (1) So long as the interior field is much less than that applied, this expression can be approximated by (2) which, in view of (a), is a prescription for the normal conduction current den­ sity inside the cylinder. This is then the boundary condition on the potential in region (b), the interior of the cylinder, and it follows that the potential within is b 2fo dE Cl> = Arcosq, = --;;rcosq,"dt (3) Note that the approximation made in going from (1) to (2) is valid if f dE fo~:> f~ ~ f02cosOE:> -f02cosq,-d (4) 0' t Thus, if E(t) = Eocoswt, the approximation is valid provided 1 >­Wf (5) U V 1.9.6 (a) Just after the step, there has been no time for the relaxation of unpaired charge, so the system is still behaving as if the conductivity were zero. In any case, piece-wise solutions to Laplace's equation, having the same 0dependence as the dipole potential and having the dipole potential in the neighborhood of the origin are (1) b P cosO Cl> =---- + Br cos 0 (2) 411"f o r2 At r = a, Cl>a = Cl>b (3) 7-26 Solutions to Chapter 7 a~a a~b -f--=-f - (4)ar 0 ar Substitution of (1) and (2) into these relations gives [ -!s--a] [A] P [a] (5)!t fo B = 411"foa3 1 Thus, the desired potentials are (1) and (2) evaluated using A and B found from (5) to be A= ~ (6) 411"(f o + 2f) B = 2(fo-f)p ( ) 411"foa3(fo + 2f) 7 (b) After a long time, charge relaxes to the interface to render it an equipotential. Thus, the field outside is zero and that inside is determined by making Bin (2) satisfy the condition that ~b(r =a) =O. A=O (8) B=--P­ (9) 411"foa3 (c) In the general case, (4) is replaced by a~a a a~a a~b ua;:-+ at (fa;:--fo ar )= 0 (10) and substitution of (1) and (2) gives 2uA + dd [2fA + fo( -2p + B(3)] =0 (11)t 411"fo With B replaced using (5a), dA A 3 dp 2f+f o -+-- . T=-­ (12)dt T -411"(2f + fo) dt' 2u With p a step function, integration of this expression from t = 0-to t = 0+ gives (13) It follows that (14) and in tum that TPo / 1) (3e-t B = 411"a 3 2f+f -f (15)o o As t -0, these expressions become (6) and (7) while as t -00, they are consistent with (8) and (9). 7-27 Solutions to Chapter 7 1.9.1 (a) This is an fIIinside-outside- situation where the layer of conductor is the fIIin_ side- region. The potential is constrained at the lower surface by the electrodes and the y derivative of the potential must be zero at the upper surface. This potential follows as ebb -V cosh fjy Q (1) - cosh fjd cos 1JZ The potential must be continuous at the upper interface, where it follows from (1) with y =0 that it is ebG(y =0) =V cos fjz (2) coshfjd The potential that matches this condition in the plane y = 0 and goes to zero as y goes to infinity is ebG= V cosfjz e-{J" (3) coshfjd Thus, before t = 0, the surface charge density is ebG ebb a. u =-[€o aa _€aa] = €ofjV cosfjz (4) y y 1/=0 cosh fjd (b) Once the potential imposed by the lower electrodes is zero, the potentials in the respective regions take the form ebG= Ae-{JI/ cos fjz (Sa) .....b _ Asinhfj(y +d) Q (Sb) 'I/' - sinhfjd cos1Jz Here, the coefficients have been adjusted so that the potential is continuous at y = o. The remaining condition to be satisfied at this interface is (7.9.12). b) bata (€oE; -€E" -aE" = 0 (6) Substitution from (S) shows that ata[(€ofj + €cothfjd) cos <pA]+ afj cothfjdcos <pA = 0 (7) The term inside the time derivative is the surface charge density. Thus, (7) can be converted to a differential equation for the surface charge density dt1.a a.a dt+~=O (8) where T= (€otanhfjd+€)/a Thus, given the initial condition from part (a), the surface charge density is _ €ofjV cos fjz -tIT (9) a.a - cosh fjd e Solutions to Chapter 7 7-28 '1.9.8 (a) Just after Qhas been turned on, there is still no surface charge on the interface. Thus, when t =0+, ~<I(y =0) = ~b(y = 0) (1) B~<I B~b Eo-(Y =0)=E-(Y =0) (2)By By It follows from the postulated solutions that (3) -EoQ -Eoqb = -Eq<l (4) and finally that q<l(O+) and qb(O+) have the given values. (b) As t -+ 00, the interface becomes an equipotential. It follows from the postu­ lated solution evaluated at the interface, where the potential must be what it is at infinity, namely zero, that (5) (c) Throughout the transient, (1) must hold. However, the condition of (2) is generalized to represent the buildup ofthe surface charge density, (7.9.12). At y=O (6) When t > 0, Q is a constant. Thus, evaluation of (6) with the postulated solutions gives dqb dq<lE0-;U -Edt -uq<l = 0 (7) Using (1) to eliminate qb, this expression becomes (8) where T = (Eo + E)/U. The solution to this expression is Aexp(-t/T), where A is the initial value found in part (a). The other image charge, qb, is then given by using (1). '1.9.9 (a) As t -+ 00, the surface at z = 0 requires that there be no normal current density and hence electric field intensity on the (b) side. Thus, all boundary conditions in region (b) and Laplace's equation are satisfied in region (b) by a uniform electric field and a linear potential. (1) 1-29 Solutions to Chapter 1 The field in region (a) can then be found. It has a potential that is zero on three of the four boundaries. On the fourth, where z = 0, the potential must be the same as given by (1) (2) To match these boundary conditions, we take the solution to Laplace's equa­ tion to be an infinite sum of modes that satisfy the first three boundary conditions. ~" =_ ~ An sinh ~(z -b) sin (nll"Y) (3) LJ sinh (mrb) a n=l " The coefficients are determined by requiring that this sum satisfy the last boundary condition at z = o. co v =L'Ansm-) (nll"Y (4) -(a -y)a a n=l Multiplication by sin(mll"y/a) and integration from z = 0 to z= a gives 21" v . mll"Y) 2vAm =- -(a- y)sm(- dy=- (5)a 0 a a mll" Thus, it follows that the potential in region (a) is (6) (b) During the transient, the two regions are coupled by the temporal and spatial evoluation of unpaired charge at the interface, where z = o. So, in region (b) we add to the asymptotic solution, which satisfies the conditions on the potential at y = 0, y = a and as z -+ -00, one that term-by-term is zero on these boundaries and as z -+ -00 and that term-by-term satisfies Laplace's equation. co b v '"' /. nll"y~ = -(a -y) + LJ Bn en1l'''' "sm(-) (7) a n=l a The result of (4)-(5) shows that the first term on the right can just as well be represented by the same Fourier series for its y dependence as the last term. A;,b ~ 2v . (nll"Y) ~ B n1l''''/" • (nll"Y)"It' = LJ -sm --+ LJ ne sm-- (8) n=l nll" a n=l a The potential in region (a) can generally take the form of (3). There remains finding An(t) and Bn(t) such that the continuity conditions at z = 0 on the potential and representing Gauss plus charge conservation are met. Evaluation 7-30 Solutions to Chapter 7 of (3) and (8) at z = 0 shows that the potential continuity condition can be satisfied term-by-term if 2vA,. = -+B,. (9) n'1l" The second condition brings in the dynamics, (7.9.12) at z = 0, (10) Substitution from (3) and (8) gives an expression that can also be satisfied term-by-term if n'1l" (n'1l"b) n'1l" n'1l" (n'1l"b dA,.-ua -coth - A,. -ub-B,. -Ea -coth -)-­a a a a a dt (11)n'1l"dB,.-Eb---=Ob dt Substitution for B,. from (9) then gives one expression that describes the temporal evolution of A(t). (12) where E coth ("1l'b) + E T = a a b -Ua coth (n:b) +Ub To find the response to a step, the volue of A,. when t o is found by integrating (12) from t = 0-when A,. = 0 to t = 0+. (13) The solution to (12), which takes the form of a homogeneous solution exp(-tIT) and a constant particular solution, must then satisfy this initial condition. A,. = A,.1 e -tiT + Ub (-2) ( Vob) (14) n'1l" Ua coth ,.: + Ub The coefficient of the homogeneous term is adjusted to satisfy (13), and (14) becomes (15) 7-31 Solutions to Chapter 7 There is some insight gained by writing this expression in the alternative form Given this expression for An, Bn follows from (9). In this specific situation these expressions are satisfied with U a = 0 and Ea= Eo. (c) In this limit, it follows from (15) that as t -00, An -Vo{2/n'll") and this is consistent with what was found for this limit in part (a), (5). With the permittivities equal, the potential and field distributions just after the potential has been turned on and therefore as there has been no time for unpaired charge to accumulate at the interface, is as shown in Fig. 87.9.9a. To make this sketch, note that far to the left, the equipotentials are equally spaced straight lines (surfaces) running parallel to the boundaries, which are themselves equipotentials. All of these must terminate in the gap at the origin. In the neighborhood of that gap, the potential has the form familiar from Fig. 5.7.2 (except that the equipotential. =V is at tP = 'II" and not at tP =211-)' (a) (b) Figure Sf.9.9 In the limit where t-00, the uniform equipotentials in region (b) extend up to the interface. Just as we could solve for the field in region (b) and then for that in region (a), we can also draw the fields iIi th~t order. In region (a), the potential is linear in y in the plane z = 0 and zero on the other two boundaries. Thus, the equipotentials that originate on the boundary at 7-32 Solutions to Chapter 7 z = 0 at equal distances, must terminate in the gap, where they converge like equally spaced spokes on the hub of a wheel. The transient that we have described takes the field distribution from that of Fig. 87.9.9a, where there is a conduction current normal to the interface from the (b) region side supplying surface charge to the interface, to that of Fig. S7.9.9b, where the current density normal to the surface has subsided because charges on the interface have created just that field necessary to null the normal field in region (b). SOLUTIONS TO CHAPTER 8 8.1 THE VECTOR POTENTIAL AND THE VECTOR POISSON EQUATION 8.1.1 (a) Ampere's differential law inside the solenoid gives VxB=O (1) The continuity law of magnetic flux gives V"lo'oB=O (2) Therefore, H is the gradient of a Laplacian potential. A uniform field is, of course, one special case of such a field. At the boundary, representing the coil as a surface current . NiK =14>­d we have n X(Ba-B b) = K (4) where n = -i.., the outside region is (b). Further we have n"1o'0(Ba -Bb) = 0 (5) (b) An axial z-directed uniform field inside, zero field outside, automatically sat­ isfies (1), (2) and (5). On the surface we get from (4) • H a• • Ni -I.. x zl. = 14>d and since I.. xI. = -I.,. H a= Ni z d (c) Ais tP directed by symmetry. From the integral form of V x A = 1o'0B we obtain Taking a radius r we find 2'11"rA (r) = {'1I"r21o'0H; for r < a 4> '1I"a211 Ha for r > a "'0 z Therefore r Ni £A _ '210'07 lor r < a 4> -{ a' !li £ 2r 10'0 d lOr r > a 1 8-2 Solutions toChapter8 8.1.2 Usingthecoordinates definedinFig.P4.4.3,superposition oflinecurrent vectorpotentials (8.1.16)gives (1) where Tolineartermsin(d/2)2,thenumerator ofthisexpression is (2) where r="';z2+112 Similarly, thedenominator is (3) Thus,tolineartermsin(d/2r)2, (1)becomes Observethat(4) =.=cosq,j~=sinq,. r r 'Z2_'; 2..1..22=cos."-smq,=cos2q,r(5) anditfollowsthat(4)isthegivenvectorpotential. Solutions toChapter8 8-3 8.1.3 Wecantakeadvantage oftheanalogofasolution ofPoisson's equation for atwodimensional chargeproblem, andforatwodimensional currentproblem (because thestructure islong,l::>wandl::>dwetreatitastwodimensional). The analogchargeproblem isonewithtwochargesheetsofopposite signs,producing auniform field,andapotential Cbexy.Thus(seeFig.S8.1.3) -y FlsureS8.1.3 inside,A.=constoutside,andweadjustAsothatwegettheproperdiscontinuity of8AII/8ytoaccountforthediscontinuity ofH", Therefore and Ni.A.=J.'o-y insidew _±Ndi.{top -1'02wbottom 8.2THEBIOT-SAVART SUPERPOSITION INTEGRAL 8'.2.1 TheBiot-Savart integral, (7),isevaluated recognizing that (•.) rII/)XIr'r11=Vz2+r2(1) 8-4 Solutions to Chapter 8 Thus, Jo li!:t.12 11' r (2) fa dzrdtPdr H. = 4,," 0 0 J" vr + r2 (z2 + r2) The integration on z amounts to a multiplication by Ii. while that on tP is simply a multiplication by 2,,". Thus, (2) becomes H -Ii.Jo r i2dr () .-2 J" (z2+ r2)3/2 3 and integration gives Ii.Jo[ -r vI]aH. =-- + In(r + r2 + z2) (4) 2 vr2 + z2 " which is the given result. 8.2.2 We use the Biot-Bavart law, H-i-fds X lr'r (1) -4,," Ir-r'12 The field due to the turns within the width RdO, and length sin ORd,p which produce a differential current ids = Kosin2 OR2 dOd,p, is (Note: Ir'r = -I...) 2 2 dH, = Kosin OR dO d,p (2) 4,,"R2 I. I, Fleure 58.2.2 The field along the. axis adds as one integrates around one tum, the components normal to the axis cancel dH. =-sin 0 rll'dH, = KodO sin30 (3)Jo 2 The total field is obtained by adding over all the currents H. = Ko (II' sin3 OdO = 2Ko (4) 2 J,=o 3 8-5 8.2.3 8.2.4 8.2.5 Solutions to Chapter 8 We replace K o sin9 by K o in Prob. 8.2.2. We can start with the integral in (4), where we drop one factor of sin 9. We get H., = K o r sin29d9 = 'lfKo 2 18=0 4 We can use the result of 8.2.3 for a single shell. The total current distribution can be thought of as produced by a concentric set of shells. Each shell produces the field ~JodR. Thus the net field at the center is la 'If 'If H., = -Jo dR = -JoR 40 4 No matter where the vertices of the loop, (8.2.22) can be used to determine the field. However, the algebra is simplified by recognizing that the triangle not only has sides of equal length, d, but that the z axis is at the center of the triangle. Thus, each leg makes the same contribution to the z component of the field along the z axis, and along that axis the z and 'Y components cancel. To see that the sides are of length equal to that of the one paralleling the z axis, note that the distance from the center of the leg to the vertex on the 'Y axis is V3f4d and that based on the base d/2 and this distance, either of the other leg lengths must be of length J(d/2)2 + (V3f4d)2 = d. Further, if the z axis is at the center of the triangle, then the distance from the origin to either of the legs not parallel to the z axis must be the distance to the parallel leg, V3f4d/3. Thus, we should have 2V3f4d/3 = J(d/2}2 + (v'3f4d/3}2, as indeed we do. For the leg parallel to the z axis, a = dix b= -~i _! ~di -zi 2x 3V"4 ~ • (1) e = -1d. --l/fd'1-ZI• -2 x 34 ~ • Thus, e X a= -zdi~ + ~/fd2i. ~ Iexal = d(..!-d2+ z2) 1/2 (2)12 a.e=d2/2 lei = (~/3 + z2) 1/2 a.b= -d2/2 Ibl = lei 8-6 Solutions to Chapter 8 and the given result follows from (8.2.22), multiplied by 3 to reflect the contributions from the other two legs. This same result is obtained using either of the other legs. For example, using the back leg, (3) 8.2.6 From (8.2.22) B i e x a (a.ea·b)= 41r Iexal2 ~-1bI we can find the H-field produced by a current stick! We look at one stick in the bottom layer of wires, extending from the position vector b (' ). d i I. = z -ZIx-2"Y "-2"1. to the position vector with a:= e-b= U. Thus exa = l[(z' -z)i" + ~ixJ Ie xal2 = 12[(z' -z)2 + (d/2)2J Ibl = lei = \I'(z' -z)2 + (d/2)2 + (1/2)2 r r a .e =- a .b =-­2 2 Therefore, B due to one stick, carrying the differential current Ni dz' is w B = Nidz'l [(z' -z)i" + gixl 12 4'1rw 12 [(z' -z)2 + (d/2)2J \I'(z' -z) + (d/2)2 + (1/2)2 Nidz' [(z' -z)i" + gixl ~ 2'1rw (z' -z)2 + (d/2)2 Solutions toChapter 8 8-7 inthelimitwhenIisverylongcompared withdandw.Thisverysameresult couldhavebeenobtained fromAmpere's lawandsymmetry considerations foran infinitely longwire(seeFig.88.2.6) H=_Nidx'_1_i=Nidx(x'-x)l7'+~ix W211'rtP211'W(x'-x)2+(d/2)2 17 aa9e999999e9999e899e8e xIxx' r I...... I I I .,.,••.,••••• II•••., .,.,.,,=-d/2 Figure S8.~.8 Thetotalfieldisobtained byaddingthecontribution fromasymmetrically located setofwiresatthetop,whichcancelsthey-component anddoublesthex-component, andbyintegrating overthelengthofthecoil fVJ/2Nidx' dH--- -:---~--:-...,.....,- :J:--VJ/2211"w(x'-x)2+(d/2)2 Ni1[2(w)] 1[2(W)] =-tan----x+tan---+X 1I"W d2 d2 sincefdx 2-l(/)x2+(d/2)2=dtan2xd Wemaytestthisresultbyhaving W-+00.Then H_Ni :J:- W QEDasiscorrectforsheetsofaninfiniteset. 8.2.1 From(8.1.8)integrated overthecross-section ofthestick, 1-'0IJ(r')dtl'I-'o.,e·ade (1) A=411"Ir-r'!=411"'1hy;i"jIr-r'l wherea/lalisaunitvectorinthedirection ofthestickandhence[a/lalJde isa differential lengthalongthestick.Usingtheexpression forIr-r'lfollowing (8.2.17), (1)isconverted toanexpression readyforintegration. A""0.a,e·de=411"~lh";e+r~ (2) 8-8 Solutions to Chapter 8 Integration gives (3) Finally, substitution from (8.2.21) makes this expression the given result. 8.3 THE SCALAR MAGNETIC POTENTIAL 8.3.1 From the Biot-Savart law B = i-. f ds' X ir'p 4", Ir-r'12 we find the axial field H!II 2 i 1... Rd4l .6 i 211" . 36H =- sm =--sm 11 411" 0 R2/sin26 411" R i Jt3 =- 32RJKJ +32 For large %, . R2 H -2~ 11 ­ 411"z3 which is consistent with the axial field of a dipole (see Fig. S8.3.1). Flpre 81.S.1 Solutions toChapter 8 8.3.2 Thepotential ofonewirecarrying thecurrentiinthe+zdirection is ~1/J=--tP 211" Thesuperposition gives(Fig.88.3.2) ThelinesW=constaredescribed by8-9 (1) (2) tantPl-tantP2tan(tPl-tP2)=const= A- A-1+tan'f'ltan'f'2 Therefore-lI.-_-lI.­x-a x+a 1.....1i!.-+x2-a2 2ya=const[x2-a2+y2] Thisistheequation ofcirclesthatgothrough thepointsx=±a,y=o. ,pI aTZIy I I.--- ixRsinfJ z Figure 88.3.2 Figure 88.3.3 8.3.3 Assumethatthecoilextends fromz=-l/2toz=+l/2.Thepotential ofa loopis ~W(r)=-0 411" r'\1()211"RsinBRdB ( B)( (z-z')) u= =211"1-cos=211"1----;:======~=oR2 yI(z-z')2+R2 Theindividual differential loopsoflengthdz'carrycurrentslfidz'.Therefore the totalpotential is W(x)-_Nir'=1/2dz'(1_------r.~(=z =-:==:z':f::)===-=) -21}%1=-1/2 yI(z-z')2+R2 =~li[l+V(~-Z)2+R2-V(~+z)2+R2] 8-10 Wecanchecktheresultforalongcoil,I-+00.ThenSolutions toChapter8 VI Iv(2Z)2(2R)2 I(2z)(-=fz)2+R2=-1=f-+-~-1=f-2 2 I I2I andwefindHi\Ii(z)=-[1-2z] 21 givingafield _a\Ii_H_Hi az- %-I whichiscorrect. 8.4MAGNETOQUASISTATIC FIELDS INTHEPRESENCE OFPERFECT CONDUCTORS ,.j' 8.4.1 From(8.3.13), 'T'()riR2cos(J'!I!'r-+O-+---- 411'r2 andatr=6 a\IiI-0arr=b Tomeettheseconditions, takethesolutions toLaplace's equation riR2cos(J \Ii=----2-+Arcos(J 411'r(1) (2) (3) wherethefirstautomatically satisfies (1)andthecoefficient Aofthesecondis determined byrequiring (2).Thus, i1l'R2(12r)\Ii=---+- cos(J 411'r263 Thenegative gradient ofthismagnetic potential isthegivenfieldintensity. 8.4.2 Themagnetic fieldofthedipoleisgivenby(8.1.21) Hid(.-I.. -I..)=--2-sm'PII'+cos'PI",21l'r Thiscorresponds toascalarpotential of(4) Solutions to Chapter 8 8-11 The conductor acts like a perfect conductor cancelling the normal component of H, Hr. Thus we must have the total scalar potential id . (a r)W=--sml/> -+­2,ra r a with the field 8.4.3 (a) Far from the half-cylinder, the magnetic potential must become that of a uniform magnetic field in the -z direction. (1) Thus, to satisfy the condition that there be no normal component of the field intensity at the surface of the half-cylinder, a second solution is added to this one having the same azimuthal dependence. cos I/>W= Horcos I/> + A-­ (2) r Adjusting A so that 8w-(r = R) = 0 (3)8r results in the given potential. (b) As suggested, the field intensity shown in Fig. 8.4.2 satisfies the requirement of being tangential to the perfectly conducting surfaces. Note that the surface current density has the polarity required to exclude the magnetic field from the perfectly conducting regions, in accordance with (3). 8.4.4 The potential Wof the uniform field is The sphere causes H to be tangential. The normal component Hr must be cancelled: We obtain for the H field 8-12 Solutions to Chapter 8 8.4.5 (a) An image current is used to satisfy the condition that there be no normal component of the field intensity in the plane y = O. Thus, the solution in region y < 0 is composed of a particular part due to the line current at z= 0, y = -h and a homogeneous part equivalent to the field of a line current at z = 0, y = h flowing in the opposite direction. To write these fields, first note that for a line current on the z axis, (1) Translation of this field to represent first the actual and then in addition the image line current then results in the given field intensity. (b) The surface current density that must exist at y =0 if the region above sustains no field intensity is K = nx H => K. = Hz(y = 0) (2) This is the given function. 8.4.6 (a) The scalar potential produced by one segment of length dz' is d,T. Kod:l:' -1 ( Y ) Kodz' -1 (:1:'- z)... =---tan -- =--cot -- (1)211' z -Z, 211' Y The integral over the strip is lz'=a K { a-z'If = d'lf = ---.£ (a -z) cot-1 (--) ~=b 211' Y 1(b -z) y [ a-z 2]-(b -z) cot---+ -log 1 +(--) (2) y 2 y _ ~ log [1 + (b ~ z)2 ] } where the integral is taken from: B. O. Pearce, R. M. Foster, A Short Table of Integrals, 4th Ed., Ginn and Co. (1956). To this potential must be added an image potential that causes a'lf/ az=0atz= O.This isachieved by adding to (2) a potential with the replacements Ko--K o a--a, b--b Solutions to Chapter 8 8-13 (b) The field H = -V'If and thus from (2) Ko [ 1 (a -x) 1 (b -x)Hx =-- -cot--- + cot--­ 2~ y y + (a -x)jy _ (b -x)jy 1 + (a;x)2 1 + (b;X)2 _ (a -x)jy + (b -x)jy ] 1 + (a;x)2 1 + (b;X)2 Ko [ l(a-x) l(b-X)] = -- -cot- -- +cot- -­ 2~ y y = Ko [tan-1 (-y-) _ tan-1 (-y-)] 2~ a -x b -x To this field we add Hx = Ko [tan-1 (-y-) - tan-1 (-y-)] 2~ x + a b + x 8.5 PIECE-WISE MAGNETIC FIELDS (a) The surface current density is N .. cP'K = -tsm 1 (1)2R z so that the continuity conditions at the cylinder surface where r = Rare a H b Ni. H</>- </>=2RsmcP (2) (3) Looking forward to satisfying (2), the cP dependence of the scalar potential is taken to be cos cP. Thus, the appropriate solutions to Laplace's equation are (4) 'lfb = Cr cos cP (5) so that the field intensities are H a -_ A( cos 2 cP lr• + sin 2 cP. I</> ) (6) r r 8-14 Solutions to Chapter 8 Hb = -C(cos ¢ir -sin ¢i<f» (7) Substitution of these fields into (2) and (3) then gives ~ _ C = Ni (8) R2 2R A R2 + C = 0 (9) from which it follows that A= RNi. C=_Ni (10)4' 4R Substitution of these coefficients into (4)-(7) results in the given expressions for the magnetic scalar potential and field intensity. (b) Because the flux density is uniform over the interior of the cylinder, the flux linked by a turn in the plane x = x' = R cos ¢' is n.. HR· A-.' Ni R . A-.' (11) 'I!',). = /-Lo .,2 sm,+, = /-Lo 4R 2 sm '+' Thus, the total flux is \ 0111" /-LoN • A-.'( N) • A-.'RdA-.';'\ =~ --sm,+, - sm,+, '+' o 2 2R N21r = ~0/-Lo--N2 111" . 2 '+'A-.'d-l.' [/-Lo--- (12) SIn '+' = ] ~ 0 4 0 8 and thus the inductance is identified as that given. 8.5.2 (a) At r = b, there is a jump in tangential H: (1) with region (a) outside, (b) inside the cylinder carrying the windings. Thus n= i r and at r= b (2) Further the normal component of W must be continous at r = b. aW(a) aW(b)---+--=0 (3)ar ar At r= a, the normal component of H has to vanish: aWl = 0 (4)ar r=a 8-15 Solutions to Chapter 8 (b) We have a "square-wave" for the current distribution. Therefore, we need an infinite sum of terms for q;: q;(b) = Eco A,.(r/b)" cos n(<p -<Po); 0< r< b n=l q;(a) = Eco B,.(a/r)" cos n(<p -<Po); b< r < a (5) n=l +Lco 0,.(r/a)n cosn(<p -<Po) ,.=0 We picked the normalization of the coefficients so that the boundary condi­ tions are most simply stated. From (4) we have and thus (6) From (3) we have (7) and using (6) (8) From (2) we obtaiR: The expansion of the square wave K.(<p) is K.(<p)=K o L ~sinn(<p-<po) (10) " n1l" ft.-odd Thus, using (6), (8) and (10) in (9) we obtain, for n odd: and 0,. = 0 for n even. Thus (11) 8-16 Solutions to Chapter 8 and A" = ~b Ko[(b/ar~" -1] (12)n1r for n odd, zero for n even. We should check a few limits right away. When a -+ 00, we get (for n odd) 2b A" = --2-Kon1r and 2b Web) =-L -2-Ko(r/b)" cos n(1/> -1/>0) ,,-odd n 1r 2b W(a) = L -2-Ko(b/r)" cos n(1/> -1/>0) ,,-odd n 1r which gives the field due to the cylinder alone. For a -+ b, we get A" = 0 "" 4b ~ L,.., -2-Kocosn(1/> -1/>0) ,,-odd n 1r There is a I/> directed field in the region between the coil and the shield of magnitude 1 aW "" 4Ko • ( )H~ ~ --- ~ L,.., -- sm n I/> -1/>0 a al/> ,,-odd n1r which is approximately square-wave-like. These checks confirm the correctness of the solution. (c) The inductance of the rotor coil is computed from the flux linkage of an individual wire-loop, 1 -~'+1I' I co A 1-~'+1I' ~.A = l 1I0Hrbdl/> = L -lilo nb" cos n(1/> -I/>o)bdl/> ~=-~' r=b ,,_1 ~=-~' odd (b)2"].- "')= ~ L.." l110--4Kob [1 - smn'f'("" -'f'o,,_1 n1r a odd where l is the length of the system. The flux linkage is obtained by taking the number of wires per unit circumference N/1rb, multiplying them by ~.A and integrating from 1/>' = 1/>0 to 1/>' = 1/>0 +1r >. = f N bdl/>'~.A = IN f: 110 4Kob[1 -(!)2"] fdl/>' sin n(¢" -1/>0) 1rb 1r "=1 n1r a odd 8N'2 . (~ 1[ ( b)2"]) = l--lIo'& L.." -2 1-­ 2:11" ,,=1 n a odd 8-17 Solutions to Chapter 8 where we use the fact that K _ Ni 0- 2b The inductance is L = ~ = 8~2 Jl-ol f :2 [1 _ (~) 2n] n.=l "'-odd The inductance is, of course, <Po independent because the field is "tied" to the rotor and moves with <Po. 8.6 VECTOR POTENTIAL AND THE BOUNDARY VALUE POINT OF VIEW 8.6.1 (a) For the two-dimensional situation under consideration, the magnetic field in­ tensity is found from the vector potential using (8.1.17) H = ~(~ 8Az i-8Azi,p) (1) Jl-o r 8<p r 8r Thus, if the vector potential were discontinuous at r R, the azimuthal magnetic field intensity would be infinite there. (b) Integration of (1) using the fields given by (1.4.7) gives ! {R3/9R+ h; r R < < R r Az = -Jl-o H,pdr + f(<p) = -Jl-oJ o ~2 In(r/ R) + 12; (2) A - {gl (r); r < R z - g2(r); R < r (3) Because the integrations are performed holding rand <p-constant, respectively, the integration "constants" are actually functions of the "other" independent variable, as indicated. From (3) it is clear, however, that there is no depen­ dence of hand 12 on <p. Given that the vector potential is zero at r = 0 and that Az is continuous at r = R, h = 0 and h = R2 /9. Thus, the vector potential is as given. (c) In terms of the vector potential, the flux is given by (8.4.12). Because there are no contributions on the radial legs and because Az (r = 0) has been defined as zero, .A = fA. ds = I[Az(O) -Az(a)] = -IAz(a) 0' (4) .= Jl-0 1R 2 Jo [In(ar/ R) + ~] 3 3 This illustrates how the use of A to represent the field makes it possible to evaluate the flux linkage without carrying out an integration. 8-18 Solutions to Chapter 8 8.6.2 A must be z-directed and must obey Poisson's equation (1) Now "\72 = ~ i.(ri.) r dr dr in the special symmetry of the problem. Thus (2) and r< b (3) Outside this region b < r < a, Az obeys Laplace's equation A", ex Cln(r/b) + const At r= b we must have continuous A", and dAz/dr (continuous Hq,). Thus, b2 const = -l-'oJz4 and Thus ~- " X direction ./' offield / / / \ \ ~ ............. --­ positive direction of loop Figure 88.6.2 Solutions to Chapter 8 8-19 The flux is, according to (8.6.5) [see Fig. 88.6.2] >. = l(A~ -A~) and thus >. = -lA~ because A~ =0 For c < b For c > b b2 >. = lJLo Jz "4[1 + 2ln(cjb)] Note that A z ¥= 0 for r > O. This should be remedied by adding a constant to Az• It does not affect the flux linkage. 8.6.3 (a) In cylindrical coordinates where there is no <p dependence, the vector potential has only a () component A= A/;/(r, z)i/;/ (1) and the flux density is found from JLoH = VxA * JLoH = ir( -aaA/;/) +iz[~aa (rA/;/)] (2)z rr For reasons that are apparent in part (b), it is convenient to write A as A = Ac(r, t) (3) r in which case, (2) becomes H_ ~ [_ aAc• aAc .] JLo -a lr + a lz (4) r z r (b) For any surface S enclosed by the contour C, the net flux can be found from the vector potential by >. = £A. ds (5) In particular, consider a surface enclosed by a contour C having as the first of four segments a contour spanning 0 < <p < 211" at the radius, a, from the z axis. The second segment connects that circular contour with a second at the radius b by a segment connecting the two in a plane of constant <p. The contour is closed by a second contour in an adjacent <p = constant plane joining these circular segments. Integration of (5) gives contributions only from the circular contours. The segments joining the circular contours are perpendicular to the direction of A, and in any case make compensating contributions because they are in essentially the same <p = constant planes. Thus, the flux through the surface having outer and inner radii, a and b respectively, is as given. 8-20 Solutions to Chapter 8 8.6.4 (a) The vector potential, A., BatisfieB Laplace'B equation. The first three condi­ tionB of (8.6.18) are met by the solution · hR'I" • R'I"A•= ARBm -yBm-z (1)a a The last condition is met by BuperimpoBing these solutions 00 ~A' hR'I" . R'I"A•= LJ I' BIn -yBm-z (2) 1'=1 a a and evaluating the coefficientB by requiring that this function satisfy the fourth boundary condition of (8.6.18). 00 ~ A • h R'I"b . R'I"A=LJ RBm -sm-z (3) 1'=1 aa Multiplication by Bin(m'l"zja) and integration giveB Aa m'l"] a Ama. m'l" --COB-Z = --Bmh-b (4)m'l" a 0 2 a which therefore giveB the coefficients as (5) so that (2) becomes the given solution. (b) The total current in the lower plate is i = 1a K.dz =-1a Hs(Y = O)dz =-1a -1 8A I dz (6) _. o 0 0 Jjo 8y 1/=0 Evaluation using the given vector potential gives . ~ 8A ~ IBinwt ,= -~ IL R'I"sinh (RtI'b) = -LJ 2RBinh (RtI'b) (7) .._1 ,..0 a 1'=1 a odd (c) In the limit where bja::> I, . h (Rd) 1 RtI'blasm ---e (8)a 2 8-21 Solutions to Chapter 8 and (7) becomes 00 i --. -E !-e-n/rrbIG sinwt --. _Ie- wbIG sinwt (9) ,,_1 n odd Taking In of the magnitude of this expression gives In( I~I) == -7I"(b/a) (10) which is the straight line portion of the plotted function. (d) In the limit b/a <: 1, (7) becomes i --. _-!.-SA ~ E -!.----!.-A~ (11)Po r b n2- Po b This is the same as what is obtained if it is assumed that the field is uniform and simply Hz --. A/bpo so that K. --. -Hz => i --. K.a --. -aA/bpo (12) 8.6.5 The perfectly conducting electrodes force H to be tangential to the electrodes. Thus 8A",/8z == -P o HlI vanishes at y == 0, y == d except for the gap at z == 0 and 8A",/8y == PoH z vanishes at z == ±a. The magnetic vector potential jumps by A as one goes from z ==0_toz ==0+,at y ==0 and y == d. Thus A. is constant around the c shaped contour as well as the :J shaped one. Denoting by the superscripts (a) and (b) these two regions respectively, we have for Laplacian solutions of A", A~b) == E00 Bn sinhn;(z-a)sinn;y + Bo(z -a) n=l At z == 0, the constants Ao and Bo account for the jump of A"" Bo == -A/2 == Ao• The vector potential and its curl must be continuous for 0 < y < d atz ==O.We thus have An == -Bn for all n except n == O. The sinusoidal series has to cancel that jump for 0 < y < d. We must have "A . h n7l" • n7l" " 4Ao • n7l"L.J n sm -a sm -y == -L.J -- sm -y n d d n-odd n7l" d and similarly for the series in region b. We obtain A (G) _ " 2Asinh 7(z + a) . ml' A( )•-L.J . hRtr sln- y-- z+a n-odd n7l" sm da d 2 Solutions to Chapter 8 8-22 (b) _" 2A sinh !!f(z -a) . mr _!.( _)A. -L..J • h mr sm d y 2z a n-odd n1l' sm (f"a (b) See Fig. S8.6.5. Ftsure 58.6.& 8.6.6 (a) We must satisfy Poisson's equation for the vector potential everywhere inside the perfectly conducting boundaries (1) and make the normal flux density and hence A. zero on the boundaries. A. =0 at z= ±a,y = 0, y = b (2) A particular solution to (1) follows by looking for one that depends only on z. (3) Then the homogeneous solution must satisfy Laplace's equation and the con­ ditions A.h =0at z= ±a; (4a) 2 . a• 1I'Z A.h = I-'o'ln0"2sm - at Y = O,b (4b) 11' a The first ofthese conditions, can be met by making the z dependence sin(1I'Z/ a). Then, the y dependence must be comprised of a linear combination of exp(+ky) and exp(-ky). IT the y coordinate were at y = b/2, the second of the condi­ tions of (4) would be even in y. So, make the linear combination cosh k(y-k)] and for convenience adjust the coefficient so that the second of conditions f4) are met, divide this function by its value at y =b/2. This makes it clear that the coefficient is the value given on the boundary from (4). Thus, the desired solution, the sum of the particular and homogeneous parts, is A = A + A -I-'oinoa 2 [COSh Hy -~) -1] . (~) (5) • !liP .h - 2 h (ft'b)" sm11' cos 20 a Solutions toChapter8 (b)Thefluxlinkedbyoneturnis .~=-l[A.(z,y) -A.(-z,y)! =_2poiRoa2l[COShi(y-l)_1]sin!! ,,"2 cosh(;:) a andthetotalfluxofallofthewindings inseriesis8-23 (6) (7) + 8.6.7@@ Figure98.8.8 (C)Asketchofthelinesofconstant vectorpotential andthusBfortheparticular, homogeneous andtotalsolution (thesumofthese)isshowninFig.88.6.6. Itisperhapseasiesttoenvision thesumbypicturing theaddition ofcontour mapsofthetwoparts,theaxesoutofthepaperbeingtheheightA.ofthe respective surfaces. (a)Thisisaproblem involving aparticular andahomogeneous solution ofthe vectorPoissonequation. Theparticular solution isduetouniform current densityJo=Roi Z2-a2 Ap=-PoRoi 2i. Alternatively, wemayfindthehomogeneous solution bycomparison with Prob.8.6.6.InthatproblemthewiredensityWassinusoidal. Nowitisuniform. A.Wasantisymmetric, nowitissymmetric. Wecanexpandthesymmetric wiredistribution asasquarewave. ().'"4noin7rJ.3:,y=Ro'=L..J--cos-Z ..RlI"2a ..-odd 8-24 Solutions toChapter 8 Theparticular solution ofthevectorpotential isthus .""42a)2(n1r)Ap=-i.JLono~ L...J-(-cos-x..n1rn1r 2a ",-odd Thecomplete solution is •.""42a2n1r[COSh~:(y-~)A=1.JLono~L...J-(-) cos(-x) hmfb..n1rn1r 2a cos4a odd (b)Thefluxlinkageofawireatx,yis andthus-1] 8.6.8 (a)Herewehaveasolution verymuchlikethatofProb.8.6.6,exceptthatthe particular solution hastobereplaced byaninfinitesumwhosel'econdderivative reproduces the squarewaveofmagnitude ino.Thus A•.""4(a)2.(n1rx)b=-l.J.'otn oL...J--sm--n1rn1r an-odd x=o -a FigureS8.6.8 Thecomplete solution is(compare Prob.8.6.6)a A_..""~(~)2.(n1rx)[cosh(n1r/a)(y -~)_]-1.JLo~no L...J sm (b) 1n1rn1r a cosh n211"n-odd a 8-25 Solutions to Chapter 8 (b) The inductance is computed from where 21Az is the Hux linkage of one turn nod:z;' dy' is the wire density. Thus integrating one typical term: r d:z;' sin (mr:z;') r[COSh 7 (y -£) _1] dy' = 2( ~)[2~ tanh mrb -b]10 a 10 cosh mrb n7l" n7l" 2ao 0 2a and the inductance is -21 ~ 16( a )4[n7l"b h(n7l"b)] L -p.on LJ ----tan ­ o d n7l" n7l" 2a 2a,.-od SOLUTIONS TO CHAPTER 9 9.1 MAGNETIZATION DENSITY 9.2 LAWS AND CONTINmTY CONDITIONS WITH MAGNETIZATION 9.2.1 M= Mo cos p:z:(i x + i)') The volume charge density and thus there is positive surface charge density on top y=d and a charge density of opposite sign at the bottom, y = -d. 9.2.2 (a) The magnetization is uniform, with the orientation shown in Fig. P9.2.1. Thus, it is solenoidal and the right hand side of (9.2.2) is zero and therefore equal to the left hand side, which is zero because B= o. Certainly a zero H field is irrotational, so Ampere's law is also satisfied. Associated with M inside is a magnetic surface charge density. However, this is cancelled by a surface charge density of opposite sign induced in the infinitely permeable wall so as to prevent there being an B outside the cylinder. (b) In view of the direction defined as positive for the wire, the Hux linked by the coil is (1) Thus, with the terminus of the right wire defined as the + terminal and 1 = Ot, the voltage is (2) 1 Solutions to Chapter 9 9-2 9.2.3 (a) From Ampere's law £l J H .ds = .da we find f H·ds=O because there is no J present. This means that H = -V"\If and "\If is a scalar potential that satisfies Laplace's equations, since H is divergence-free. The only possible solution to this problem, subject to "\If = const at y = 0 and y = a, is "\If = constj and hence H = O. (b) Since B = JLo(H + M) (1) we have B= iyJLoM o cos (3(x -Ut) (2) The flux linked by the turn is >. = JLol i:~: Mocos(3(x -Ut)dx = ldM {sin((3d -(3Ut) sin((3d +(3Ut) } JLo (3d + (3d 0 = ldM {sin(3dCOS(3Ut -cos (3dsin (3Ut JLo (3d 0 sin (3d cos (3Ut + cos (3d sin (3Ut } + (3d sin (3d= 2JLoldM0--rid cos (3Ut The voltage is d>' sin (3d . v = dt = -2(3UJLoldM0--rid sm(3Ut 9.3 PERMANENT MAGNETIZATION 9.3.1 The given answer is the result of using (4.5.24) twice. First, the result IS written with the identification of variables ao JLoM o --+ --j Xl = a, x2 = -a, Y -+ Y -b (1) Eo JLo Solutions to Chapter 9 9-3 representing the upper magnetic surface charge. Second, representing the potential of the lower magnetic surface charge, -Uo ---+ -Mo; Xl = a, x2 = -a, Y ---+ Y + b (2) J1.o The sum of these two results is the given answer. 9.3.2 In the upper half-space, where there is the given magnetization density, the magnetic charge density is Pm = -V· J1.oM = J1.oMoa. cos fJxe-ay (1) while at the interface there is the surface magnetic charge density U m = -J1.oM z (Y =0) = -J1.oMocosfJx (2) In the upper region, a particular solution is needed to balance the source term, (1) introduced into the magnetic potential Poisson's equation (3) given the constant coefficient nature of the Laplacian on the left, it is natural to look for a product solution having the same x and y dependence as what is on the right. Thus, if (4) then (3) requires that F[_fJ2 + a.2] = -Moa. ~ F = Moa.f(fJ2 -a.2) (5) Thus, to satisfy the boundary conditions at y = 0 aWG aWb -J1.o ay + J1.o ay = -J1.oMo cos fJx (6) we take the solution in the upper region to be a superposition of (5) and a suitable solution to Laplace's equation that goes to zero at y ---+ 00 and has the same x dependence. Gw= [Ae-,8y + Moa. e-ay] cos fJx (fJ2 -a.2) (7) Similarly, in the lower region where there is no source, Wb = Ce,8y cos fJx (8) Substitution of these solutions into the two boundary conditions of (6) gives A= Mo (9)2(a.- fJ) C=- 2(a. M+ o fJ) (10) and hence the given solution. 9-4 Solutions to Chapter 9 9.3.3 We have This is Poisson's equation for W with the particular solution: f3M o wp = 2 2cos f3x exp aya -(3 The homogeneous solution has to take care of the fact that at y = 0 the magnetic charge density stops. We have the following solutions of Laplace's equation Wh ={ A cos f3xe-13v y > 0 B cos f3xe 13v 11 < 0 There is no magnetic surface charge density. At the boundary, wand awlay must be continuous and exf3Mo + f3B = -f3A ex2 -(32 Solving, we find Mo ( ex)B =-2(ex _ (3) 1 + 73 and 9.3.4 The magnetic volume charge density is 1 a 1 a Pm = -'\1. 1-£0M = -1-£0;: ar (rMr) -1-£0;: at/! M.p = -1-£0 Mop(rlR)p-l cos p(t/! -')') + 1-£0 Mop(rlR)P-l cos p(t/! -')') r r =0 There is no magnetic volume charge density. All the charge density is on the surface am = 1-£0Mrlr=R = 1-£0Mocosp(t/! -')') This magnetic surface charge density produces 1-£0H just like a. produces foE (EQS). We set r> R r< R Solutions to Chapter 9 9-5 Because there is no current present, 9 is continuous at r= R and thus A=B On the surface a9 a'iJ! -~Oa;lr=R+ + ~Oa;lr=R_ = am = ~oMocosp(~ -1) We find A RA=-M o 2PR =M o 2p (b) The radial field at r = d + R is ~oHr(r = d+ R) = ~o ~o cosp(~ -1) (R~ d) pH The flux linkage is 2 ~oN2Mo ( R )P+l (11' )A= ~oN Hral = 2 al R + d cos P '2 -Ot The voltage is dA _ pO~oN2Moal(-.!!:-)p+l 0 dt - 2 R + d cos P t (c) If p is high, then unless d is made very small 9.4 MAGNETIZATION CONSTITUTIVE LAWS 9.4.1 (a) With the understanding that Band H are collinear, the magnitude of B is related to that of H by the constitutive law B = ~olH + Motanh(aH)] (1) For small argument, the tanh function is approximately its argument. Thus, like the saturation law of Fig. 9.4.4, in the neighborhood of the origin, for aH <: 1, the curve is a straight line with slope ~o(1 + aMo). In the range of aH Il:$ 1 the curve makes a transition to a lesser slope ~O. (b) It follows from (9.4.1) and (1) that B = ~o [~l~ + Mo tanh (~:~)] (2) and in turn from (9.4.2) that A= 1I'w2N2~O [Nli M. h(QNli)] 2 4 211'R + otan 211'R (3) Thus, the voltage is v = dA2/dt, the given expression. 9-6 Solutions to Chapter 9 9.4.2 The flux linkage is according to (9.4.2) (1) The field intensity is according to (9.4.1) Therefore dA2 _ 1rW2 N dB dt --4- 2di" where we need the dispersion diagram to relate H. (i.e. i) to B (see Fig. 89.4.2). tB(t)B dBdi ex v(t) Figure 99.4.2 9.5 FIELDS IN THE PRESENCE OF MAGNETICALLY LINEAR INSULATING MATERIALS 9.5.1 The postulated uniform H field satisfies (9.5.1) and (9.5.2) everywhere inside the regions of uniform permeability. It also satisfies the continuity conditions, (9.5.3) and (9.5.4). Finally, with no H outside the conductors, (9.5.3) is satisfied. The only way in which the permeable materials can alter the uniform field that exists in Solutions to Chapter 9 9-7 their absence is by having a component collinear with the permeability gradient. As shown by (9.5.21), only then is there induced the magnetic charge necessary to altering the distribution of H. Here, such a component would be perpendicular to the interface between permeable materials, where it would produce a surface magnetic charge in accordance with (9.5.22). Because H is simply i/w throughout, the total flux linking the one turn circuit is simply and hence, because A= Li, the inductance is as given. 9.5.2 From Ampere's law applied to a circular contour around the inner cylinder, anywhere within the region b <r< a, one finds t H<f>=­21rr where i<f> points in the clock-wise direction, and z along the axis of the cylinder. The flux densities are B _ IJ-at and <f> -21rr in the two media. The flux linkage is A= l{ {R IJ-bi dr + r IJ-ai dr}Jb 21rr JR 21rr = 2l1r[IJ-bln(R/b) + IJ-aln(a/R)]i The inductance is 9.5.3 For the reasons given in the solution to Prob. 9.5.1, the H field is simply (i/w)i •. Thus, the magnetic flux density is (1) and the total flux linked by the one turn is A= ( Bzdydx = djD (-IJ-m X ) 3:..dx = IJ-:;ld i (2)Js -I l w _w By definition, A= Li, so it follows that L is as given. 9-8 Solutions to Chapter 9 9.5.4 The magnetic field does not change from that of Prob. 9.5.2. The flux linkage is i (a-b)ia >. = l b I-'m(r/b) 21lT dr = I-'m l -b- i The inductance is a-b L = I-'ml-­b 9.5.5 (a) The postulated fields have the r dependence of the H produced by a line current i on the z axis, as can be seen using Ampere's integral law (Fig. 1.4.4). Direct substitution into (9.5.1) and (9.5.2) written in polar coordinates also shows that fields in this form satisfy Ampere's law and the continuity condition everywhere in the regions of uniform permeability. (b) Using the postulated fields, (9.5.4) requires that l-'a A = I-'bC ~ C = I-'a A (1) r r I-'b (c) For a contour that encloses the interior conductor, which carries the total current i, Ampere's integral law requires that (fJ == 2'11" -a) 1 H4>rdr = i = ar~ + fJr C = aA + fJC (2)J'a r r Thus, from (1), (3) (d) The inductance follows by integrating the flux density over the gap. Note that the same answer must be obtained from integrating over the gap region occupied by either of the permeable materials. Integration over a surface in region a gives >. = 'ia l-'a A dr = ll-'aAln(a/b) = ll-'aln(a/b)i (4) b r a+(2'11"-a)(l-'a/l-'b) Because>' = Li, it follows that the inductance of the shorted coaxial section is as given. (e) Since the field inside the volume ofthe inner conductor is zero, it follows from Ampere's continuity condition, (9.5.3), that A/b = i/b[a + fJ~]j region (a) K. =H ~K. = b (5) 4> { C/b = i(l-'a/l-'b)/b(a+ fJ~)i region (b) Solutions to Chapter 9 9-9 Note that these surface current densities are not equal, but are consistent with having the total current in the inner conductor equal to i. (6) 9.5.6 The H-field changes as one proceeds from medium J.'G to the medium J.'fI. For the contour shown, Ampere's law gives (see Fig. 89.5.6): z=-w Figure S9.5.8 The flux continuity gives Therefore and the flux linkage is and the inductance is A dl L=i=...!!...+!!!..=.!! "'.. "'~ 9.6 FIELDS IN PIECE· WISE UNIFORM MAGNETICALLY LINEAR MATERIALS 9-10 Solutions to Chapter 9 9.6.1 (a) At the interface, Ampere's law and flux continuity require the boundary con­ ditions (1) (2) The z dependence of the surface current density in (1) suggests that the magnetic potential be taken as the solutions to Laplace's equation w_ {Ae-Ifll sinpz - Celfll sin pz (3) Substitution of these relations into (1) and (2) gives [-13 13] [A] _[Ko] (4)P.oP p.p C -0 and hence A __l!-Ko • -P.o 13[1 + ::]' (5) Thus, the magnetic potential is as given. (b) In the limit where the lower region is infinitely permeable, the boundary condjt:;on at y= 0 for the upper region becomes awG H:(y =0) =-az (y =0)= Ko cos pz (6) This suggests a solution in the form of (3a). Substitution gives (7) which is the same as the limit p./P.o -+ 00 of (5a). (c) Given the solution in the upper region, flux continuity determines the field in the lower region. In the lower region, the condition at y =0 is aWb ( ) P.o awG ( ) P.o . --y=O =---y=O = -Kosmpz (8)ay p. ay p. and it follows that PC sin pz = P.o Kosin pz ~ C = P.o Ko/p (9) p. p. which agrees with (5) in the limit where p./P.o > 1. Solutions to Chapter 9 9-11 9.6.2 (a) The H-field is the gradient of a Laplacian potential to the left and right of the current sheet. Because D x D =0 at y = ±d,qI = const. (b) At the sheet D x (HG-Db) =K (1) and thus aqlG aqlb ~y --+ -= Kosin (-) (2)ay ay 2d From flux density continuity we obtain aqlG aqlb ~o as =~o as (3) From (2) we see that qlG and qlb oc cos(~y/2d) and thus qlG = A cos (~1I)e-JI'Z/2d (4a)2d qlb = Bcos (""1I)eJl'Z/2d (4b)2d This satisfies qI = const at 11 = ±d. We have from (3) ~ ~ --A=-B 2d 2d and from (2) ~ ~ 2dA -2dB =Ko giving KoA=-B=-­ (~/d) Therefore qI: =± Ko cos (~Y)e'FJI'Z/2d (~/d) 2d 9.6.3 (a) Boundary conditions at r = R are G b 1 aqlG 1 aqlb Ni. H. -H. =-R a4J + R a4J = 2R sm 4J (1) aqlG aqlbBG_Bb =-~-+~o- =0 (2)rr ar ar To satisfy these, it is appropriate to choose as solutions to Laplace's equation outside and inside the winding qI ={(Air) cos 4Jj R <r Crcos4Jj r < R (3) Solutions to Chapter 9 9-12 Substitution of these relations in (1) and (2) shows that the coefficients are NiR 1£ AA= ; 0=--­ (4) 211 +(1£/1£0)1 , 1£0 R2 and substitution of these into (3) results in the given expressions for the magnetic potential. (b) The magnetic field intensity inside is uniform and ~ directed. Thus, the in­ tegration over the area of the loop amounts to a multiplication by the area. The component normal to the loop is Hz cos a, Hz = -0. Therefore, ~ =nl£oH z cos a(2al) = -nl£oO cos a(2al) (5) With no current in the rotating loop, the flux linkage-current relation reduces to ~ = Lmi, so the desired mutual inductance multiplies i in (5). 9.6.4. (a) It is best to find the H-field first, then determine the vector potential. The vector potential can then be used to find the flux according to 8.6.5. Look at stator field first (r = a). The scalar potential of the stator that vanishes at r=bis (1) On surface of stator nxH· =K (2) where n = -il.. K= i.i1N. sin ~ (3) where the stator wire density N. is N _ N1 •- 2a with N1 the total number of turns. Since H• 1 a'iJI I. 1 A. .I. (a b).n X =--_ I. =--sin 'I' ---I. r a~ r-a a b a We find A = -'-N 1i1 ab (5)2 a2 -b2 The H field due to stator windings is: (6) The rotor potential is 'iJlr = Bcos(~ -6) (:. -~) (7)a r 9-13 Solutions to Chapter 9 We find similarly, (8) The H -field is N . b 2 2 B r = ~Z2 a2 _ b2 [(1 + :2) cos(~ -O)ir-(1-:2) sin(~ -O)i<f>J (9) Fluxes linking the windings can be obtained by evaluating Is B .da or by use of the vector potential Az• Here we use Az• The vector potential is z-directed and is related to the B field by "V XA=B= 1-'0B = ! aAz i_ aAz i<f> (10) r a~ r ar From the r-components of B we find by inspection N1i1 ab (r b).Az = 1-'0-- 2 b2 -b +-sm cP2a - r (11) N2i2 ba ( r a).+ 1-'0--a2 b2 -+-sm(cP -0)2 -a r Of course, the cP component gives the same result. (b) The inductances follow from evaluation of the flux linkages. The flux of one stator turn, extending from cP = -cPo to cP = 1r -cPo is (12) The inductance is obtained by computing the flux linkage (13) The inductance is (14) In a similar way we find (15) The mutual inductance is evaluated from ~>.., the flux due to the field pro­ duced by the stator, passing a turn of the rotor extending from -~o + 0 to 1r -cPo + 0 ~>.. = l[A:(1r -cPo + 0) -A:(-~o + O)]r=b (16)= l-'olN1i1 22abb2 sin(cPo -0) a­ 9-14 Solutions toChapter9 Themutualfluxlinkageis A21=1'" Nb2~';.·bd<P. =~o1NIN2il 22abb2coso (17) 4>0=02 a- Asimilaranalysis givesL12whichisfoundequaltoL21.Fromenergyargu­ mentspresented inChap.11,itcanbeproventhatL12=L21isanecessity. Notethat 9.6.5 (a)Thevectorpotential ofthewirecarrying acurrent1is where(1) andaisareference radius. ITwemountanimageofmagnitude ibatthe position z=0,1/=-h,wehave (2) where r2=V(y+h)2+z2 Thefieldinthe~material isrepresented bythevectorpotential y>Owhereiaistobedetermined. WefindfortheB=~Hfield HVA•8Aa•8Aa ~o=X=Ix8y-1~8z __~o{.(1Y-h+.Y+h) -271"IxV(y_h)2+z23'bV(y+h)2+z23 .(1 z . z )-l~ S+'b SiV(y-h)2+z2V(y+h)2+z2(3) (4a) H~oia 1{.(h)•}~=--2- 3IxY--I~Zi 71"V(1/-h)2+z21/<0 (4b) 9-15 Solutions to Chapter 9 At Y= 0 we match Hz and p.HII obtaining (5) (6) By adding the two equations we obtain: (7) and thus (8) (b) When p. ~ p'o, then H tan ~ 0 on the interface. We need an image that cancels the tangential magnetic field, i.e. (c) We have a normal flux as found in (4a) for ib = I This normal flux must be continuous. It can be produced by a fictitious source at y = h of magnitude ia = 21. The field is (compare (4b)) (d) When p. ~ p'o, we find from (2) and (8) in concordance with the above! 9-16 Solutions to Chapter 9 9.6.6 The field in the upper region can be taken as the sum of the field due to the wire, a particular solution, and the field of an image current at the position y = -h, z = 0, a homogeneous solution. The polarity of this latter current is determined by which of the two physical situations is of interest. (a) IT the material is perfectly conducting, there is no flux density normal to its surface in the upper region. In this case, the image current must be in the -z direction so that its y directed field is in the opposite direction to that of the actual current in the plane y = O. The field at y = h, z = 0 due to this image current is J.&oH = (2~(~h) i x (1) and therefore the force per unit length is as given. The wire is repelled by a perfectly conducting wall. (b) In this case, there is no tangential magnetic field intensity at the interface, so the image current is in the same direction as the actual current. As a result, the field intensity of the image current, evaluated at the position of the actual current, is the negative of that given by (1). The resulting force is also the negative of that for the perfect conductor, as given. The wire is attracted by a permeable wall. 9.6.'1 (a) In this version of an "inside-outside" problem, the "inside" region is the highly permeable one. The field intensity must be H~. in that region and have no tangential component in the plane z = O. The latter condition is satisfied by taking the configuration as being that of a spherical cavity centered at the origin with the surrounding highly permeable material extending to infinity in the ±z directions. At the surface where r = a, the normal flux density in the highly permeable material tends to be zero. Thus, the approximate field takes the form cos(Jwa = -Horcos(J + A- ­ (1)2r where the coefficient A is adjusted to make 8wa n.Blr=a =0 =* a;:-(r =a) =0 (2) Substitution of (1) into (2) gives A = _a3 Ho/2 and hence the given magnetic potential. (b) Because there is no surface current density at r = A, the magnetic potential (the tangential field intensity) is continuous there. Thus, for the field inside Wb(r = a) = Wa(r = a) = -3H oa/2 (3) To satisfy this condition, the interior magnetic scalar potential is taken to have the form Wb=Crcos(J=Cz (4) Substitution of this expression into (3) to evaluate C = -3H o/2 results in the given expression. Solutions to Chapter 9 9-17 9.6.8 The perfectly permeable walls force the boundary condition ff = 0 on the surfaces. The bottom magnetic surface charge density is neutralized by the im­ age charges in the wall (see Fig. 89.6.8). The top magnetic surface charge density produces a magnetic potential ff that is ff = A sinh ,8(y -a) cos,8z y > d/2 (1a) and ff = Bsinh,8(y+~) cos,8z y < d/2 (16) At the interface at y = d/2, ff is continuous Asinh,8(~ -a) = Bsinh,8d (2) and thus B __ sinh,8(a -~) -A sinh,8d (3) The magnetic surface charge density at y = d/2 is O'm = p.oM o cos,8z (4) It forces a jump of off/oy at y= d/2: --off I +-off I= Mocos,8x (5) oy y=d/2+ oy y=d/2_ and we find -Acosh,8(~ -a) + Bcosh,8d = Mo (6)2 ,8 Using (3) we obtain A =_ Mo sinh,8d ,8 cosh,8(~ -a) sinh,8d -cosh,8dsinh,8(~ -a) ~ ~~ m = -Tsinh,8(~ + a) The vertical component of B, By, above the tape, for y > d/2, is off sinh,8d ,By = -P.o-;- = p.oM o . (d ,) cosh,8(y -a) cos,8x (8)uy smh,8 2" + a Note that in the limit a --+ d/2, the flux is simply p.oM o as expected. IT the tape moves, cos,8z has to be expressed as cos,8(z' -Ut). The flux is sinh,8d d j'll'},A=wNp.oM o . (d ) cosh,8(h+- -a) X' cos,8(x'-Ut)dz' (9) smh,8 2" + a 2 -1/2 The integral evalues to ~ [sin ,8(~ -Ut) + sin,8(~ + Ut)] = ~ sin,8~ cos ,BUt (10) and from here on one proceeds as in the Example 9.3.2. dA "0 = dt 9-18 Solutions to Chapter 9 9.6.9 In terms of the magnetic scalar potential, boundary conditions are w(x, b) = OJ w(x, 0) =0 (1) a'll 1rY a'll 1rYHy = --a (0, y) = -K o cos -j -a(b, y) = K o cos - (2) yay a To satisfy the first pair of these while matching the y dependence of the second pair, the potential is taken as having the y dependence sin(1ry/a). In terms of'll, the conditions at the surfaces x = 0 and x = b are even with respect to x = b/2. Thus, the combination of exp(±1rx/a) chosen to complete the solution to Laplace's equation is even with respect to x = b/2. '11 = A cosh [~(x -~)] sin (1rY) (3)a 2 a Thus, both of the relations (2) are satisfied by making the coefficient A equal to A= aKo (4)1rcosh(1rb/2a) 9.6.10 The solution can be divided into a particular part due to the current density in the wire and a homogeneous part associated with the field that is uniformly applied at infinity. Because of the axial symmetry in the absence of the applied field, the particular part can be found using Ampere's integral law. Thus, from an integration at a constant radius r, it follows that H",p21rr = 1rr2Jo; r < R H",p21rr=1rR2Jo; R<r (1) so that the particular field intensity is r< R (2)R <r in polar coordinates H =.! (.!aAz i aAz i",) _ p. r atP r ar (3) and it follows from (2), integrated in accordance with (3), that r < R (4)R< r In view of the applied field, the homogeneous solution is assumed to take the form A _{Dr sintPj . r< R zh - -P.aHo r sin tP + CS1~ 1> j R<r (5) 9-19 Solutions to Chapter 9 The coefficients C and D are adjusted to satisfy the boundary conditions at r = R, (6) 1 8Aa 1 8Ab --_%+-_% = 0 (7) J1.a 8r J1.b 8r The first of these guarantees that the flux density normal to the surface is continuous at r = R while the second requires continuity of the tangential magnetic field intensity. Substitution of (5) into these relations gives a pair of equations that can be solved for the coefficients C and D. (8) The coefficients which follow are substituted into (5) and those expressions respec­ tively added to (4) provide the given expressions. J 9.6.11 (a) Given the magnetization, the associated H is found by first finding the distri­ bution of magnetic charge. There is none in the volume, where M is uniform. The surface magnetization charge density at the surface, say at r= R, is (1) Thus, boundary conditions to be satisfied at r = R by the scalar magnetic potential are (2) (3) From the () dependence in (3), it is reasonable to assume that the fields outside and inside the sphere take the form -H r cos () + A co. 9 ~ ={ a -Hrcos() r2 (4) Substitution of these expressions into (2) and (3) gives 1 H = Ha -3M ~ M = 3(Ha -H) (5) Thus, it follows that B == J1.a(H + M) = J1.a(-2H + 3Ha) (6) (b) This relation between Band H is linear and therefore a straight line in the B -H plane. Where B = 0 in (6), H = 3Ha/2 and where H = 0, B = 3J1.aHa' Thus, the load line is as shown in Fig. S9.6.11. 9-20 05 2468" H(unitsofIdamps/m)-Solutions toChapter9 Figure99.8.11 (c)ThevaluesofBandHwithinthespherearegivenbytheintersection ofthe loadlinewiththesaturation curverepresenting theconstitutive lawforthe magnetization ofthesphere. (d)Forthespecificvaluesgiven,theloadlineisasshowninFig.89.6.11. The valuesofBandHdeduced fromtheintersection arealsoindicated inthe figure. 9.6.12 Weassumethatthefieldisuniforminsidethecylinder andthenconfirmthe correctness oftheassumption. Thescalarpotentials insideandoutsidethecylinder are 'Ii-{-HoRcos4J(r/R) +Acos4J(R/r)r>R -Ccos4J(r/R) r<R Because'Iiiscontinuous atr=R ITthereisaninternaluniformmagnetization M=Mix,then n·M=Mcos4J Theboundary condition forthenormalcomponent of#LoBatr=Rgives Therefore, from(2)and(4) C M-=-H+-R 02(1) (2) (3) (4) (5) Solutions toChapter 9 9-21 andtheinternal (r<R)Hfieldis(weusenosubscripts todenotethefieldinternal tocylinder): (6) Themagnetization causesa"demagnetization" fieldofmagnitude M/2.Wecan construct "loadline"tofindinternalBgraphically. 8ince B=11-0(H+M) wefindfrom(6)forthemagnitude oftheinternalHfield H=(H_M+H+H)=H_~+H o22 0211-02 orBH=2Ho-- 11-0 Thetwointersection pointsare(seeFig.89.6.12) H=2HoforB=O and B=211-0Ho forH=0 Wereadoffthegraph:B=0.67tesla,H=2.5X105amps/m.(7) (8) (9) I B (Ieslol 0.5 2468 ~H(unitsof10omps/ml- Figure59.6.12I B (tesla)h!!iR Ni/2R / 2468 H(unitsof,domps/m)- Figure59.6.13 9.6.13 Therelation between thecurrentinthewinding andHandMinthesphere aregivenby(9.6.15). NiM=3(--H)3R Fromthis,theloadlinefollowsas NiB==11-0(H+M)=11-0(Ii"-2H)(1) (2) Theintercepts thatcanbeusedtoplotthisstraight lineal<;showninFig.89.6.13. Thelineshownisforthegivenspecificnumbers. Thus,withinthesphere,B~0.54 andH~1.8. Solutions to Chapter 9 9-22 9.7 MAGNETIC CIRCUITS 9.1.1 (a) Because of the high core permeability, the fields are approximated by taking an "inside-outside" approach. First, the field inside the core is approximately subject to the condition that n 'B =0 at r =a and r = b (1) which is satisfied because the given field distribution has no radial component. Further, Ampere's integral law requires that 2ft' 12ft' Ni H",rd~ =Ni = -rd~ = Ni (2)1o 0 21/"r In terms of the magnetic scalar potential, with the integration constant ad­ justed to define the potential as zero at ~ = 1/", 18'if! Ni Ni --- = - => 'if! = --~+const r 8~ 21/"r 21/" (3) Ni ~ = 2(1-;J This pot.ential satisfies Laplace's equation, has no radial derivative on the inside and outside walls, suffers a discontinuity at ~ = 0 that is Ni and has a continuous derivative normal to the plane of the wires at ~ = 0 (as required. by flux continuity). Thus, the proposed solution meets the required conditions and is uniquely specified. (b) In the interior region, the potential given by (3), evaluated at r = b, provides a boundary condition on the field. This potential (and actually any other potential condition at r = b) can be represented by a Fourier series, so we represent the solution for r < b by solutions to Laplace's equation taking the form 00 'if! = L ,pm sin m~ (~) m (4) m=l Because the region includes the origin, solutions r-m are omitted. Thus, at the boundary, we require that N' ~ 00 -'(1--) = '" ,pm sin m~ (5)2 1/" L-m=l Multiplication by sin n~ and integration gives 2ft' N' ~ 12ft' 001-;(1-;) sin(n~)d~ = L ,pmsinm~sinn~d~ o 0 m=l (6) = ,pn1/" Thus, N'12 ft' ~ N',pm =-' (1--) sin m~d~ =-' (7) 21/" 0 1/" m1/" Substitution of this coefficient into (4) results in the given solution. Solutions to Chapter 9 9-23 9.7.2 The approximate magnetic potential on the outer surface is 00 W= L -'N" sinmfji (1) m1l" m=1 according to (b) of Prob. 9.7.1. The outside potential is a solution to Laplace's equation that must match (1) and decays to zero as r ~ 00. This is clearly 00 W= L -'N" (a/r)m sin mfji' (2) m=1 m1l" 9.7.3 Using contours C1 and C2 respectively, as defined in Fig. S9.7.3, Ampere's integral law gives Haa = Ni => Ha = Ni/a (1) (2) ~~-------~ w r/ Figure S9.1.3 From the integral form of flux continuity, for a closed surface S that intersects the middle leg and passes through the gaps to right and left, we know that the flux through the middle leg is equal to the sum of those through the gaps. This flux is linked N times, so (3) Substitution of (1) and (2) into this expression gives (4) where the coefficient of i is the given inductance. 9-24 Solutions toChapter9 9.'1.4 ThefieldinthegapduetothecoilofNturnsisapproximately uniform becausethehemisphere issmall.FromAmpere's law Hh=Ni (1) whereHdirected downward isdefinedpositive. Thisfieldisdistorted bythesphere. Thescalarmagnetic potential aroundthesphereis Niq;=Rhcos6[(r/R)-(R/r)2] where6istheanglemeasured fromtheverticalaxis.Thefieldis H=-~i{II'cos6[1+2(R/r)2]- i9sin6[1-(R/r)2]}(2) (3) (4) (5)Figure89.7'.4 Thefluxlinkedbyonetumatangleais(seeFig.89.7.4) ~A=1a lJoHr21rR2sin6d6 N"fa =-3IJoT 21rR210sin6cos6d6 3IJoNi2( =---1rR 1-cos2a)2h But1-cos2a=2sin2awhichwillbeusedbelow.Thefluxlinkageis'>'21where1 standsforthecoilonthe1r/2legofthe"circuit", 2forthehemispherical coilr/2n '>'21=1 0~ARsinaRda 3Nnr/2 =-"4lJoTi1rR2 1 0sin3ada NnR2'=-1J02h:1r, Themutualinductance is (6) Solutions to Chapter 9 9-25 9.1.5 In terms of the air-gap magnetic field intensities defined in Fig. S9.7.5, Ampere's integral law for a contour passing around the magnetic circuit through the two windings and across the two air-gaps, requires that (1) Figure S9.1.5 In terms of these same field intensities, flux continuity for a surface S that encloses the movable member requires that (2) From these relations, it follows that (3) The flux linking the first winding is that through either of the gaps, say the upper one, multiplied by N1 (4) The second equation has been written using (3). Similarly, the flux linking the second coil is that crossing the upper gap multiplied by N2 • (5) Identification of the coefficients of the respective currents in these two relations results in the given self and mutual inductances. 9-26 Solutions toChapter9 (1)9.1.6 Denoting theHfieldinthegapofwidthzbyHsandthatinthegapgby Hg,Ampere's integrallawgives fH.ds=zHs+gHg=Ni wherefluxcontinuity requires (2) Thus (3) Thefluxis Theinductance is L=N~A=-=,IJ_o_N_2-;;-­ s+-'-, tra32trad 9.1.1 Wepicktwocontours (Fig.89.7.7)tofindtheHfieldwhichisindicated inthe threegapsasHa,HbandHc.Thefieldsaredefinedpositiveiftheypointradially outward. Fromcontour01: (1) /" I/I I I I C2 I""•••ITI'" I IH.. H, H. -~-I-f--d- I-e- d Figure89.7'.7' FromcontourO2 (-Ha+Hc)g=N1i1+N2i2 Thefluxmustbecontinuous sothat(2) (3) 9-27 Solutions to Chapter 9 We find from these three equations (4) d -eN1i1 eN2i.Hb=-------- (5)2d g 2d g He = d-eN1i1 + 2d-eN2 i2 (6)2d g 2d g The flux linkage of coil (1) is: The flux linkage of coil (2) is: The inductance matrix is, by inspection 9.1.8 (a) 1J! must be constant over the surfaces of the central leg at x = Tl/2 where we have perfectly permeable surfaces. In solving for the field internal to the central leg we assume that a1J!/an = 0 on the interfaces with fJ-o. (b) If we assume an essentially uniform field HI-' in the central leg, Ampere's integral law applied to a contour following the central leg and closing around the upper part of the magnetic circuit gives (1) Therefore (2) Solutions to Chapter 9 9-28 '1'(x = 1/2) = N1i1 + 2 N2i 2 (3) (c) In region a, at y = 0, '1' must decrease linearly from the value (2) to the value (1) (4) At y= a, '1'=0 (5) At x = ±1/2,0 < Y< a, '1' must change linearly from (2) and (3) respectively, to zero '1'(x = _~, y) = N1i1; (6) N2i 2 (a: y) '1'( x= 2'y') =-N1i1 + 2 N2i2 (a -a y) (7) (d) '1' must obey Laplace's equation and match boundary conditions that vary linearly with x and y. An obvious solution is '1' = Axy + Bx + Cy We have, at y= 0 and thus B =_ N1i1 + N2i2 l In a similar way we find at y= a Aax + Bx + Ca =0 and thus C=O, Aa=-B which gives 9.7.9 From Ampere's integral law we find for the H fields (1) where K is the ("surface-") current in the thin sheet. This surface current is driven by the electric field induced by Faraday's law 2~ (3a + w) = fE .ds =_!!. fJ.'oD . daua dt (2)dH 1=-J.'aw-­dt Solutions toChapter9 9-29 Finally,thefluxiscontinuous sothat J1-H13aw=J1-H2aw (3) and H2=3H1 Whenweintroduce complex notation anduse(4)in(1)wefind Ht{l1+3l2)=Nio+Kh(4) (5) andK=-JWJ1-awa6.H12(3a+w) Introducing (6)into(5)yields ~Nio1H1=-;-:---=--:--:- -----,--­(i1+3l2)1+jWTm(6) (7) awh Tm=J1-a6.( )( )h+3126a+2wwhere 9.7.10 Thecross-sectional areasofthelegstoeithersidearehalfofthatthrough thecenterleg.Thus,thefluxdensity,B,tendstobethesameoverthecross­ sections ofallpartsofthemagnetic circuit.Forthisreason,wecanexpectthat eachpointwithinthecorewilltendtobeatthesameoperating pointonthegiven magnetization characteristic. Thus,withHgdefinedastheair-gapfieldintensity andHdefinedasthefieldintensity ateachpointinthecore,Ampere's integrallaw requires that 2Ni=(l1+l2)H+dHg (1) Inthegap,thefluxdensityisJ1-oHgandthatmustbeequaltothefluxdensityjust insidetheadjacent polefaces. J1-oHg=B (2) Thegivenload-line isobtained bycombining theserelations. Evaluation ofthe intercepts ofthislinegivesthelineshowninFig.89.7.10.Thus,inthecore,B~0.75 TeslaandH~0.3X104A/m. f B (tesla)---- 0.5xI04 H(omps/m)-E ........ IIIa. E o \Da 20.5 61 1.5 Hb(unitsof10omps/m) FigureS9.7.10 FigureS9.7.11 Solutions to Chapter 9 9-30 9.1.11 (a) From Ampere's integral law we obtain for the field Hb in the J.L material and Ha in the air gap: bHb +aHa = Ni (1) Further, from flux continuity (2) and thus (3) Now Bb = J.Lo(Hb + M) and thus (4) or Ni bHb = -----M (5)a+b a+b This is the load line. (b) The intercepts are at M = 0 Ni Ni 6Hb =--=-= 0.25 X 10 a+b 2a and at Hb = 0 M = bNi = 0.5 X 106 We find M = 0.22 X 106 Aim Hb = 0.13 X 106 Aim The B field is J.Lo(Hb + M) = 411" X 10-7 (0.13 + 0.22) x 106 = 0.44tesla SOLUTIONS TOCHAPTER 10 10.0INTRODUCTION 10.0.1 (a)Thelineintegraloftheelectricfieldalong01isfromFaraday's law: becausenofluxislinked(seeFig.S10.0.la). Therefore -t/+iR=a becausethevoltagedropacrosstheresistorisiR.Hence t/=iR R v(1) (2) + v ThelineintegralalongO2is whichleadstoF1sureBIO.O.la,h 4iR=d.A dtc (3) (4) 1 Solutions to Chapter 10 10-2 Therefore, we find for the voltage across the voltmeter 1 dw.\ v=--- (5)4 dt (b) With the voltmeter connected to 2, (1) becomes v =2iR Using (2), and similarly for the other modes . [1 dW.\]v(3) = 3[IR] = 3-­4 dt v(4) = 4iR = 4[!dW.\] = dw.\ 4 dt dt For a transformer with a one turn secondary (see Fig. S10.0.lb), v = 1 E· dl = !.... !B .da= !!.w.\fa at dt 10.0.2 Given the following one-turn inductor (Figs. S10.0.2a and S10.0.2b), we want to find (a) tI2 and (b) VI. The current per unit length (surface current) flowing along the sheet is K = i/d. The tangential component of the magnetic field has to have the discontinuity K. A magnetic field (the gradient of a Laplacian potential) HIlS = di inside (1) = 0 outside has the proper discontinuity. This is the field in a single turn "coil" of infinite width d and finite K = i/ d. It serves here as an approximation. (a) tI2 can be found by applying Faraday's law to the contour O2, Using (I), and the constitutive relation B = PoD, l(B) l(A) d 1 i(t)E·ds+ E·ds=-- Po-dxdy (2) (A)a2 (B)a2 dt 82 d Solutions toChapter10 10-3 Sincetheinductor waDsareperfectly conducting, E=0forthesecondintegral ontheleftin(2).Therefore, or, slJjodiet) ~U2=---­ddt -- Isurfacecurrent,K, flowsthrough. inductor walls8 P,o~I'"----- ......'--+:z:---of"one-turn ~ inductor d~;""------'--.----"?I /;/--- flows through thissurfaceK=i(t)/d __ y__ ---~:::.=;.=--------_ ..~.. Flsure810.0.3 (b)Now,tl1canbefoundbyasimilarmethod. WritingFaraday's lawon01, (3) Since01doesnotlinkanyflux,(3)canbewritten d-til=--(0)=0dt Solutions to Chapter 10 10-4 10.1 MAGNETOQUASISTATIC ELECTRIC FIELDS IN SYSTEMS OF PERFECT CONDUCTORS 10.1.1 The magnetic field intensity from Problem 8.4.1 is B i1fR2 [ 1I(1 1). . 1I(1 2).]= 4;-2cos 11 ,.s -b3 Ir+sm 11 ,.s + b3 16 The E-field induced by Faraday's law has lines that link the dipole field and uniform field. By symmetry they are tP-directed. Using the integral law of Faraday's law using a spherical cap bounded by the contour r = constant, 9 = constant, we have . 6fE· ds= 21frsin9E~ =-:t 1 J.'oB r21frsin9rd9 di 1fR216 11=-J.'o-- 21f~2sin9cos9d9(- --)dt 41f 0 ,.s b3 di 1fR2 2 ( 1 1). 2 = -J.' ---1fr- ---2sm 9 o dt 4,.. ,.s b3 Thus: 10.1.2 (a) The H-field is similar to that of Prob. 10.0.2 with K specified. It is z-directed and uniform H. = {K inside (1)o outside Indeed, it is the gradient of a Laplacian potential and has the proper discon­ tinuity at the sheet. (b) The particular solution does not need to satisfy all the boundary conditions. Suppose we look for one that satisfies the boundary conditions at 11 =0, Z =0, and 11 = a. IT we set (2) with Ezp(O, t) = 0 we have satisfied all three boundary conditions. Now, from Faraday's law, (3) Integration gives (4) 10-5 Solutions to Chapter 10 x=o x=a x=o x=a (a) (b) Figure SlO.l.~a,b The total field has to satisfy the boundary condition at y = -l. There, the field has to vanish for almost all 0 ~ x ~ a, except for the short gap at the center of the interval. Thus the E",-field must consist of a large field : E",p, over the gap 9, and zero field elsewhere. The homogeneous solution must have an E",-field that looks as shown in Fig. SlO.1.2a, or a potential that looks as shown in Fig. SlO.1.2b. The homogeneous solution is derivable from a Laplacian potential cI>h (5) which obeys all the boundary conditions, except at y = -l. Denote the potential cI>h at y = -l by cI>h(y = -l) = aE",pf(x) (6) so that the jump of /(x) at x = a/2 is normalized to unity. Using the orthogonality properties of the sine function, we have -sinh ( m1l" l) ~ Am = aE",p fa / (x) sin (m1l" x) dx (7) a 2 }",=o a It is clear that all odd orders integrate to zero, only even order terms remain. For an even order, except m = 0, a 2lm1l" la/ x m1l"/(x) sin (-x) = 2 -sin (-x)dx ",=0 a ",=0 aalmfr 2 2a /= -()2 usinudu mll' u=o (8) 2 = (~;)2 [ -ucosul;;'fr/2 +lmfr / COSUdU] = ~(_l)-'f+l mll' Therefore m-even (9) m-odd Solutions to Chapter 10 10-6 The total field is dK {[ ~ / sinh!M y m1l"]E = lJo-ix Y-I L.J 2(-1)m 2 . h'::''/I' I cos (-z)dt m sma a (10) • ~ m/2cosh ~'/I'y • (m1l" )]}-1)'1 L.J 2(-1) . h !Ml sm -z m sIn 2 a ~ ... ~n 10.1.3 (a) The magnetic field is uniform and z-directed B= i.K(t) (b) The electric field is best analyzed in terms of a particular solution that satisfies the boundary conditions at tP = 0 and tP = a, and a homogeneous solution that obeys the last boundary condition at r = a,. The particular solution is tP­ directed and is identical with the field encircling an axially symmetric uniform H-field (1) and thus r dK E~ = -"2IJ0dt (2) The homogeneous solution is composed of the gradients of solutions to Laplace's equation (3) At r= a, these solutions must cancel the field along the boundary, except at and around tP = a/2. Because 8 < a, we approximate the field E</>h at r = a as composed of a unit impulse function at tP = a/2 of content a dK aE</>p = -"2alJ0dt (4) and a constant field a dK E</>h = "21J0dt over the rest of the interval as shown in Fig. S10.1.3. Feom (3) 1 aCbh 1 L n1l"tPE</>h I_ =--- =-- (n1l"/a) An cos(-) (5)r_G a atP a a n l_ T-E~p Figure SI0.1.8 10-7 Solutions to Chapter 10 Here we take an alternative approach to that of 10.1.2. We do not have to worry about the part of the field over 0 < ~ < a, excluding the unit impulse function, because the line integral of E~ from ~ =0 to ~ =a is assured to be zero (conser­ vative field). Thus we need solely to expand the unit impulse at ~ = a/2 in a series of cos (~tr ~). By integrating 1 a--(m7f/a)A m-= cos(m7f/2)aE<flp (6) a 2 where the right hand side is the integral through the unit impulse function. Thus, (7) Therefore (8) and E=-~o d: i{~ + f: 2(_1)m/2(r/a)~-1 m_3 m-eYeD. (9) 10.1.4. (a) The coil current produces an equivalent surface current K = Ni/d and hence, because the coil is long (1) (b) The (semi-) conductor is cylindrical and uniform. Thus E must be axisym­ metric and, by symmet~, ~-directed. From Faraday's law applied to a circular contour of radius r inside the coil dB. 227frE~ =---7frdt and r Ndi E~ = -2~od dt (c) The induced H-field is due to the circulating current density: where we have set i(t) = I coswt Solutions to Chapter 10 The H field will be axial, z-and ~independent, by symmetry. (The z-"inde­ pendence" follows from the fact that d::> b.) From Ampere's law 10-8 VxH=J we have dHz--=J.,dr and thus r2 N Hz induced = -wC1'4IL0"dlsinwt For Hz induced <: Hz imposed for r ~ b 10.1.5 (a) From Faraday's law aVxEp=--Bat (1) and thus aElIP N di --=-IL -­az ° ddt (2) Therefore, (3) (b) We must maintain E·n =0 inside the material. Thus, adding the homogeneous solution, a gradient of a scalar potential., we must leave E z = 0 at z = 0 and z= b. Further, we must eliminate ElI at y =0 and y = a. We need an infinite series .h =L An cos (~'Ir z) sinh (nb'lr y) (4) n with the electric field At y = ±a/2 (6) Solutions toChapter10 f(x)=x-~ -b/2 (a)10-9 E"yEta E (7)Set-P.~~=p<lIIitive number (b) FlpreS10.1.1 Wemustexpandthefunction showninFig.S10.1.5a intoacosineseries.Thus, multiplying (6)bycos":,tI'zandintegrating fromz=0toz=b,weobtain m1l"b (m1l")Ndilb (b)m,.. ---A cosh-·-a=I/o-- z--cos-zdzb2m 2b 0ddt02b {NIJi(b)2=-1/007dt2m;rm-oddOm-even SolvingforAm m-even m-odd(8) TheE-fieldis E-_Ndi{(z_~)i_~4bj(m'll")2 -1/00ddt 2'11LJcosh(m'll"aj2b)n-odd [sin(~,..z)sinh(~'Il"y)lx _cos(n;z)cosh(~'Il"Y)ly]} (c)SeeFig.SlO.1.5b.(9) Solutions to Chapter 10 10-10 10.2 NATURE OF FIELDS INDUCED IN FINITE CONDUCTORS 10.2.1 The approximate resistance of the disk is R= !211"a~ (J 2 at. where we have taken half of the circumference as the length. The fiux through the disk is [compare (10.2.15)1 A=J.'oi2a 2 This is caused by the current i2 so the inductance of the disk L22 is (using N = 1): The time constant is This is roughly the same as (10.2.17). 10.2.2 Live bone is fairly "wet" and hence conducting like the surrounding fiesh. Current lines have to close on themselves. Thus, if one mounts a coil with its axis perpendicular to the arm and centered with the arm as shown in Fig. 810.2.2, circu­ lating currents are set up. IT perfect symmetry prevailed and the bone were precisely at center, then no current would fiow along its axis. However, such symmetry does not exist and thus longitudinal currents are set up with the bone off center. Flsure 810.2.2 10-11 Solutions to Chapter 10 10.2.3 The field of coil (1) is, according to (10.2.8) (1) The net field is with Hind = K~ where K~ is the ¢J directed current in the shell. The E-field is from Faraday's law, using symmetry (2) But (3) and thus, for r = a 2Hind d d--+-Hind = --H o (4)/Aou!i.a dt dt In the sinusoidal steady state, using complex notation (5) and (6) where /Aou!i.a 1"m=-­2 At small values of W1"m (7) 10.3 DIFFUSION OF AXIAL MAGNETIC FIELDS THROUGH THIN CONDUCTORS 10-12 Solutions to Chapter 10 'J 10.3.1 The circulating current K(t) produces an approximately uniform axial field H. = K(t) (1) As the field varies with time, there is an induced E-field obeying Faraday's law 1 E.ds=-~ r#LoB .da (2)10 dt 18 The E-field drives the surface current K= AuE (3) that must be constant along the circumference. Hence E must be constant. From (1), (2), and (3) K d 24aE = 4a-= -_IL Ka (4)Au dt""o and thus d 4-K+--K=O (5)dt lJoUAa Thus (6) with p-ouAa 1"m= -4- (7) 10.3.2 (a) This problem is completely analogous to 10.3.1. One has and, because K be constant Therefore or with H. = K(t) (1) = AuE must be constant along the surface, so that E must d d2 (2d + V2d)E = --d#LoK(t)- (2)t 2 ~ K dd(2 + v2)-= --(lJoK)- (3)Au dt 2 dK K-+-=0 (4)dt 1"m #LouAd 1"m = 2(2 + V2) (5) Solutions toChapter10 ThesolutionforJ=K/li.is10-13 (6) (b)Since 1E.ds=O101 andthelineintegralalongthesurfaceisV2dE,wehave (c)AgainfromFaraday's law(7) (8) (9) (10) 10.3.3 (a)Wesetuptheboundary conditions forthethreeuniform axialfields,inthe regionsr<b,b<r<a,r>a(seeFig.S10.3.3). Ho(t)-H1(t)=-Kout(t)=-Joutli.=-uEoutli. (1) H1(t)-H2(t)=-KID(t)=-JIDli.=-uEiDli. (2) 1 positive direction~ ofK FleureSI0.S.S 10-14 Solutions to Chapter 10 From the integral form of Faraday's law: 21l"aE out = -11-0 dtd [H1(t))1l"(a2 -b2) + H2(t)d2] (3) 21l"bEin = -11-0:t [H2(t)d2] (4) We can solve for Eout and Ein and substitute into (1) and (2) (1/:1 [a2 -b2 dHdt) b2 dH2(t)]Ho()t -H1(t) -_ 11-0""2 a dt + -;----;;u- (5) _ (1/:1b dH2(t)H 1()t -H2(t)-1I-0-2-----;;u- (6) We obtain from (6) (7) where lI-o(1/:1b Tm ==-­2 From (5), after some rearrangement, we obtain: => ~~ dH2 + ~ (~-~) dHdt) + H1(t) =H (t) (8) m a dt mba dt 0 (b) We introduce complex notation Ho =Hm coswt = Re {Hme;wt} (9) Similarly H1 and H2 are replaced by H1,2 = Re IH1,2e;wtj. We obtain two equations for the two unknowns III and II2: -Ill + (1 + iWTm)II2 = 0 1+1.WTm(abb)] A b . A[ -~ H1+ ~1WTmH2 = Hm They can be solved in the usual way 1+iwT m I fI=IH0 m ~iWTm = _ (1+ iWTm)H m1 LJet lJet m II2= 11 + WT,:t~ -~) JI=_Hm LJet LJet where LJet is the determinant. LJet == -{[1 + iWTm(i -~)](l + iWTm) + iWTm~} Solutions toChapter 10 10.3.4 (a)Totheleftofthesheet(seeFig.810.3.4), B=Koi-. Totherightofthesheet B=Ki. AlongthecontourGl,useFaraday's law 1E.ds=_!!.rB·da101 dtJs10-15 (1) (2) (3) IIK-K oIt, I/ !J,.1;1 I(To (T=----..:----,= 1+acos!'f FigureSI0.a.4 Alongthethreeperfectly conducting sidesoftheconductor E=O.Inthesheetthe currentK-Koisconstant sothat V·J=O~V·(uE)=O ilb(K-Ko) dKE·ds= Ii.dy=-I-'oab-d 01 1/=0 00 t K-Kor( 1fY) dKli.uoJI/=o1+acosbdy=-I-'oab""dt Theintegralyieldsbandthus(4) (5) (6) (7) (8)From(7)wecanfindKasafunction oftimeforagivenKo(t). (b)They-component oftheelectricfieldat:t:=-ahasauniformpartanda y-dependent partaccording to(5).They-dependent partintegrates tosero andhenceispartofaconservative field.Theuniformpartis K-Ko dKEwb=-Ii.b=I-'oab-d 000 t 10-16 Solutions toChapter10 Thisistheparticular solutionofFaraday's law withtheintegral dKEyp=-I-'ozdi" andindeed,atz=-a,weobtain(8).Thereremains K-Ko('lrY) Ellh=-!:iu oacosb(9) (10) (11) Itisclearthatthisfieldcanbefoundfromthegradient oftheLaplacian potential ~=Asin<'7)sinh(~z) (12) thatsatisfiestheboundary conditions ontheperfectconductors. Atz=-a andthus8~I 'Ir'lrY.<'IraK-Ko'lry-- =-Acos-smh -)=-acos-8y:1:=-4bb b !:iuo b(13) (14) 10.4DIFFUSION OFTRANSVERSE MAGNETIC FIELDS THROUGH TmNCONDUCTORS 10.4.1 (a)Letusconsider anexpanded viewoftheconductor (Fig.810.4.1). Aty=!:i, theboundary condition onthenormalcomponent ofBgives (1) 11 (a) (e)~(IT,lL) (b) F1sure910.4.1 10-17 Solutions to Chapter 10 Therefore (2) At y= 0 (3) Since the thickness, 11, of the sheet is very small, we can assume that B is uniform across the sheet so that, (4) Using (3) and (4) in (2), BG-Bb=O (5)11 11 From the continuity condition associated with Ampere's law Since K=K.I., n =I,., _HG+Hb = K (6)III III • The current density J in the sheet is J _ K. (7)•-11 And so, from Ohm's law E _ K. (8)•-l1a Finally from Faraday's law BDVxE=-­ (9)Bt Since only BII matters (only time rate of change of flux normal to the sheet will induce circulating E-fields) and E only has a z-component, BE. BBII -Bz =-lit From (8) therefore, and finally, from (6), (10) (b) At t =0 we are given K =I.Kosinpz. Everywhere except within the current sheet, we have J =0 => B = -V\If 10-18 Solutions to Chapter 10 So from V . ,",oH = 0, we have Boundary conditions are given by (5) and (10) and by the requirement that the potential mut decay as y -+ ±oo. Since Hz will match the sinfJz dependence of the current, pick solutions with cos fJz dependence w(a) = A(t) cos fJze-fJ1I (l1a) web) = O(t) cos fJzefJ1I (l1b) H(a) = fJA(t) sin fJze-fJ1I i x +fJA(t) cos fJze-fJ1I i y (12a) H(b) = fJO(t) sin fJzefJ1I i x-fJO(t) cos fJzefJ1Iiy (12b) From (5), Therefore, A(t) = -O(t) (13) From (10), :z [fJA(t) sin fJze-1J1I11I=0 -fJO(t) sin fJze{J1I1 1I=0] dA(t)= -dO',",ofJ cos fJze-fJ1I 1=0 dt"11 Using (13) dA(t)2fJ2 A(t) cos fJz = -dO',",ofJ cos fJzdt" The cosines cancel and dA(t) + ~A(t) = 0 (14)dt dO',",o The solution is A(t) = A(O)e-t/ r (15) So the surface current, proportional to Hz according to (6), decays simila.rly as Solutions toChapter10 10-19 10.4.2 (a)ITthesheetactslikeaperfectconductor (seeFig.S10.4.2), thecomponent of Bperpendicular tothesheetmustbesero. y y=d )--~-c~~-{i()------z IL--+00 K(t)=i.K(t)cos{jz Figure SlO.4.~ Aty=0themagnetic fieldexperiences ajumpofthetangential component withnIIi)'andB2=0, Hz=-K(t)cospz Thefieldinthespace0<y<disthegradient ofaLaplacian potential 'ilf=AsinpzcoshP(y-d) ThecoshischosensothatHIJisseroaty=d:(1) (2) (3) B=-AP[cospzcoshP(y -d)ix+sinpzsinhP(y-d)i)'] (4) Satisfying theboundary. condition aty=0 -ApcospzcoshPd =-K(t)cospz Therefore A=K(t) pcoshPd 'ilf=K(t)sinpzcoshP(y -d) pcoshPd(5) (6) (7) (b)ForK(t)slowlyvarying, themagnetic fielddiffusesstraight through 80the sheetactsasifitwerenotthere.Thefield"sees-IJ-00material and, therefore, hasnotangential H 'ilf=Asinpzsinhf3(y-d) (8) 10-20 Solutions to Chapter 10 which satisfies the condition Hz = 0at y = d. Indeed, B =-AP[cosp:r:sinhp(y -d)ix +sinpzcoshP(y -d)l~1 Matching the boundary condition at y = 0, we obtain A =_ K(t) (9)P sinh Pd q; =_ K(t) sin pzsinh Ply -d) (10)P sinh Pd (c) Now solving for the general time dependence, we can use the previous results as a clue. Initially, the sheet acts like a perfect conductor and the solution (7) must apply. As t - 00, the sheet does not conduct, and the solution (10) must apply. In between, we must have a transition between these two solutions. Thus, postulate that the current 1.K, (t) cos pz is flowing in the top sheet. We have . K,(t)cospz =ut::..E. (11) Postulate the potential .q; = O(t) sin pzcosh P(y -d) _ D(t) sinp:r:sinh P(y -d) (12) pcoshPd psinhpd The boundary condition at y =0 is 8q; -8z 11/=0 =Hz 11/=0 = -K(t)cospz (13) = -O(t) cospz -D(t) cos pz Therefore O+D=K (14) At y= d 8q; 1 I cospz -8z lI=d = Hz lI=d = K, (t) cos pz = -O(t) cosh Pd (15) The current in the sheet is driven by the E-field induced by Faraday's law and is z-directed by symmetry 8E. __ !... H _ cospzcoshP(y -d) dO 8y - 8t IJo z -lJo cosh Pd dt (16)cos p:r:sinh Ply -d) dD -lJo sinh Pd dt Therefore, E _ lJo cos p:r:sinhp(y -d) dO cos pzcosh P(y -d) dD • - pcoshPd dt -lJo psinhPd dt (17) 10-21 Solutions to Chapter 10 At y= d 1 dD K. cos [3x Ez = -1-'0 [3 sinh [3d cos [3xdI = u!J. (18) Hence, combining (14), (15), and (18) I-'ou!:::& dD cosh [3dK. = -C(t) = -K + D = --[3- coth[3ddI (19) resulting in the differential equation I-'ou!:::& h RddDD K --cot l' -+ = (20) [3 dt With K a step function (21) where I-'ou!:::& 1"m = -- coth [3d (22)[3 and C =Koe-tlrm At t = 0, D = 0 and at t = 00, C = O. This checks with the previously obtained solutions. 10.4.3 (a) If the shell (Fig. 810.4.3) is thin enough it acts as a surface of discontinuity at which the usual boundary conditions are obeyed. From the continuity of the normal component of B, Br a -Br b = 0 (1) 1Ifo tH o (T ~ (a) (b) Figure 810.4.3 10-22 Solutions to Chapter 10 the continuity condition associated with Ampere's law (2) use of Ohm's law J KE=-=- (3) U !1u results in H: -Hg = Kif> = !1uEIf> (4) The electric field obeys Faraday's law aBVxE=-­ (5)at Only flux normal to the shell induces E in the sheet. By symmetry, E is <p-directed 1 a ( .) aBr(v x E)r= -'-0 ao Elf> Sin 0 = --a (6) rSIn t And thus, at the boundary 1 a [. O[H G H b] A aH r RsinO ao Sin 9- 9 = -J.&ouu---;jt (7) (b) Set Ho(t) = Re {Hoeiwt}[cosOi .. -sinOi 9 ] (8) The H-field outside and inside the shell must be the gradient of a scalar potential .9. Acos 0 Wa = -HorcosO +-2­ (9) r Wb = GrcosO (10) iio= -HosinO + ~ sinO (11) r iig = GsinO (12) 2Aii: = HocosO + 3'" cosO (13) r ii~ = -GcosO (14) From (1) a b 2A '" Br = Br ~ Ho + R3 = -0 (15) Introducing (11), (12), and (13) into (7) we find 1 a {. 2( 1 "')} . { 21cos 0 }RsinO ao Sin 0 -Ho+ R3 -0 = -JWJ.&o!1u HocosO+ R3 (16) 10-23 Solutions to Chapter 10 from which we find A, using (15) to eliminate O. .A =_ iWIJot::..uR4 H o (17)2(iwIJot::..uR + 3) .A provides the dipole term m=.A = -iwIJot::..uR4Ho 411" 2(iwIJot::..uR + 3) (18) and thus (19) with IJout::..R1'= :.....:...._­ 3 (c) In the limit WT -+ 00, we find as in Example 8.4.4. 10.4.4 (a) The field is that of a dipole of dipole moment m =ia iaW= --cosO (1)411"r2 (b) The normal component has to vanish on the shell. We add a uniform field saW= Ar cos 0 + --2 cosO (2) 411"r The normal component of Hat r = R is aWl (ia ) -- =0= -A-2-- cosO ar r=R 411"R3 and thus (3) and (see Fig. SI0.4.4). 10-24 Solutions to Chapter 10 Flpre 810.4.4 (c) There is now also an outside field. For r < R ia qr = 411T2 cos 8 + A(t)r cos 8 (5) For r > R, qr = O(t) cos 8 (6) r 2 The 8-components of B are H(J = 4~:S sin(J + A sin (Jj r < R (7a) and H(J = 0 sin (Jj r > R (7b),.s The normal component at r = R is 2ia )Hr = (-Rs -A cos(J (8a)41f and 20 Hr = RS cosO (8b) With the boundary condition (7) of Prob. 10.4.3, we have 1 a [. 2 (0 ia )] 2p.ol1u dO RsinO ao Sin 0 RS -41fRs -A = ---w-cos0d; (9) From the continuity of the normal component of B, we find (10) 10-25 Solutions to Chapter 10 The equation for 0 becomes 1 a [. 2 e(o _ ia 20 _ 2ia)] =_ 2poli.u edO R4 sine ae sm 411" + 411" R3 cos dt (11) or dO ia T: -+0=- (12) m dt 411" with 'Tm = PouIi.R/3. IT we consider the steady state, then 0= Re [Cei"'tj (13) C= 1 ia (14)(1 +iw'T m ) 411" A= 2ia _ 20 = 2ia iW'T m (15)411"R3 R3 411"R3 1+;W'T m Jointly with (5) and (6), this determines \li. (d) When W'Tm -+ 00, we have C-+ 0, no outside field and A= 2ia/411"R3 which checks with (3). When W'Tm -+ 0, we have no shield and A-+ O. The shell behaves as if it were infinitely conducting in the limit W'Tm -+ 00. 10.4.5 (a) IT the current density varies so rapidly that the sheet is a perfect conductor, then it imposes the boundary condition (see Fig. 810.4.5), D'PoB=O at r=b ., : ." .... . .: .": -.. .... .",. . ..... ... :,'·..· " 0.··.. ·' .... K= K(t) sin 2,pi• .--:-. -'"7',-.~ ' .." •0" : 0••• ':" .. ..... " . .: • .' .' .~ " .' ~.' : I '0' • ' .. : ",: "'- ': " ' :." ";. p. -+ 00 -.•f.' Figure 810.4.5 10-26 Solutions to Chapter 10 Inside the high Il. material H = 0 to keep B finite. So at r = a, nxH=K Therefore -i.H~ = K(t) sin 24>i. Thus, the potential has to obey the boundary conditions 8'iJ1 -=0 at r=b (1)8r _!8'iJ1 =-K(t)sin24> at r=a (2) r 84> In order to satisfy (2), we must pick a cos 24> dependence for 'iJI. To satisfy (1), one picks a [(r/b)2 + (b/r)2] cos 24> type solution. Guess Indeed, 28'iJ1 [2r 2b]a;: = A b2 --;:3 cos24>=0 at r= b ~: = -A[(r/b)2 + (b/r)2J2sin24> From (2), ~[(a/b)2 + (b/a)2J2sin 24> = -K(t) sin 24> a Therefore, 'iJI _ K(t)a [(r/b)2 + (b/r)2J 24> (3)---2- [(a/b)2 + (b/a)2] cos (b) Now the current induced in the sheet is negligible, so all the field diffuses straight through. The sheet behaves as if it were not there at all. But at r = b we have J.' -co material, so H = 0 inside. Also, since now there is no K at r= b, we must have H~ = 0at r = b It is dear that the following potential obeys the boundary condition at r = b 'iJI = A[(r/b)2 -(b/r)2J cos 24> H~ = _! 8'iJ1 = ~[(r/b)2 -(b/r)2]2sin24> = 0at r = b r 84> r Again, applying (2) A [(a/b)2 _ (b/a)2J2 sin 24> = -K(t) sin 24> a 10-27 Solutions to Chapter 10 Thus, \11 K(t)a l(r/b)2 -(b/r)2] 2~ (4)= --2-I(a/b)2 -(b/a)2] cos (c) At the sheet, the normal B is continuous assuming that l::.. is small Also, from Faraday's law I dB VxE=-­ (5)dt Since only a time varying field normal to the sheet will induce currents, we are only interested in (V X E)r By symmetry there is only a z-component of E 1 aE _ aBr -;a~ • --at: (6) One should note, however, that there are some subtleties involve in the deter­ mination of the E-field. We do not attempt to match the boundary conditions on the coil surface. Such matching would require the addition of the gradient of a solution of Laplace's equation to Ep = i.E•. Such a field would induce surface charges in the conducting sheet, but otherwise not affect its current distribution. Remember that in MQS Eo BE is ignored which means that the charging currents responsible for the bUfCI-up of charge are negligible com­ pared to the MQS currents flowing in the systems. Feom Ohm's law, J = uE. But, J = K/l::... 1 a K. aBr -; a~ l::..u =-at (7) Applying the boundary conditions from Ampere's law, nX IHgaplr=b -H,.._oo] =K.i. Soat r = b (8) Now guess a solution for \11 in the gap. Since we have two current sources (the windings at r = a and the sheet at r = b) and we do not necessarily know that they are in phase, we need to use superposition. This involves setting up the field due to each of the two sources individually 10-28 Solutions to Chapter 10 Here, A represents the field due to the current at r = b, and G is produced by the current at r = a. Apply the boundary condition (2), at r = a. We find from the tangential H-field 2G(t) [(a/b)2 _ (b/a)2] = -K(t) a Thus, -aK(t) G(t) = 2[(a/b)2 -(b/a)2] (10) The normal and tangential components of H at r= b are 2b 2a2 4 Hr = -{A(t)[a2 + 63] + G(t),)COS2¢ (11) H", = {A~t) [(b/a)2 -(a/b)2]}2sin2¢ (12) From (8) lSo~Ub :¢ [A~t) [(b/a)2-(a/b)2]2sin 2¢] = {(:: + 2;2) a~~t) +~ ~~} cos2¢ Using (10), dA(t) A 2 [«(1/6)2 -(6/a)21 ---;j,t + (t) lSobAu [(a/b)2 + (b/a)2] a dK(t)= [(a/b)2 + (b/a)2][(a/b)2 -(b/a)2] dt Simplifying, aA(t) + A(t) = DdK(t) (13)at r at lJobAu [(a/b)2 + (b/a)2] r = -2 -[(a/b)2 _ (b/a)2] (14) a D = [(a/b)2 + (b/a)211(a/b)2 _ (b/a)2] (15) dK/dt is a unit impulse function in time. The homogeneous solution for A is A(t) ex e-t/r (16) and the solution that has the proper discontinuity at t= 0 is A=DK o (17) 10-29 Solutions to Chapter 10 'If _ -aKa [(r/b)2 + (b/r)2] 2. -2 (a/b)2 + (b/a)2 cos ~ It is the same as if the surface currents spontaneously arose to buck out the field. At t -+ 00, e-t /.,. -+ 0 -aKa [(r/b)2 -(b/r)2] 'If = -2-(a/b)2 _ (b/a)2 cos 2~ This is when the field has enough time to diffuse through the shell 80 it is as if no surface currents were present. 10.4.6 (a) When w is very high, the sheet behaves as a perfect conductor, and (see Fig. 810.4.6) ,T. _ bK[(r/a) + (a/r)J A. (1) '.I!' -[b a] cos."ii+;; Then, indeed, a'If/ ar = 0 at r = a, and -t~ accounts for the surface current K. K{t) = KD{t) sin 41 i ". ", '.p-+oo '.• I... ... , '. ., :' . ..." .. '. : .' . 0.. • :'.:' .~ .: : •:.:: 0_ ',.. Figure 810.4.6 10-30 Solutions to Chapter 10 (b) When w is very low, then a'JI/at/J = 0 at r = a and ,T. _ bK!(rla) -(aIr)] A. (2)'J!'- [~_~} cos¥' (c) As before in Prob. 10.4.5, we superimpose the field caused by the two current distributions 'JI = {A(t)[ ~ -~} +O(t)[!: -~J} cos ~ (3) arb r The r-and ~-components of the field are: Hr =-{A(t)[~ + ~] +O(t)[~ + :2]} cos~ (4) H4> = {A(t) [!: _~] + O(t) [!: _ ~]}sin~ (5)rar rbr At r = b, (6) and thus A(t) = ~o(t~ (7) ii-b At r = a, -H<t>lr=a = K. where K. is the current in the sheet. From (7) of the preceding problem solution, we have at r = a 1 a H", aHr ----- = -}Jo- (8) a a~ Au at Thus, using (4) and (5) in (8): O(t) [~ _ !} = -}JoAua{ ~ dA(t) + dOlt) [! + ~)} (9) aba adt dt ba2 Replacing A through (7) we obtain dO + [~-~}O(t) = 2b dKo(t) (10)dt l-'oAua[~ +~] (a/b)2 -(6/a)2 dt Thus (11) with (12) Solutions to Chapter 10 10-31 D= 2b (alb)2 -(bla)2 The solution for a step of Ko(t) is C = DKoe-t/f' (13) DK -t/f' _ 2bKo -t/f'C(t)- - oe -(alb)2 _ (bla)2 e Combining all the expressions gives the final answer: 'Ii = ;C0b{[!: -~] -2 [!-~] e-t/f'}cos tP --a- a r [-+-]a b b a For very short times tiT <: 1, one has which is the same as (1). For very long times exp -tiT = 0 and one obtains (2). 10.5 MAGNETIC DIFFUSION LAWS 10.5.1 (a) We first list the five equations (10.5.1)-(10.5.5) VxB=J (10.5.1) J =O'E (10.5.2) a vx E= -atp.B (10.5.3) V·p.B=O (10.5.4) V·J=O (10.5.5) Take the curl of (10.5.3) and use the identity (1) also note that v.J =V. O'E=o'V.E =0 (2) because 0' is uniform. Therefore, 2 a -V E=--Vxp.B (3)at Solutions to Chapter 10 10-32 or -V2(Jju) = -IJ~J (4)at (b) Since J= i.J., equation (b) follows immediately from (4). We now use (10.5.S) a vx (Jju) =-atlJH But vx (Jju) = ! v x (i.J.(z, y)) = ! (ix a8 J. -i,. a8 J.)u u y z and thus aH a (J.). 8 (J.).at =-8y UIJ Ix+ 8z UIJ I,. 10.6 MAGNETIC DIFFUSION TRANSIENT RESPONSE 10.6.1 The expressions for H. and JfJ obey the diffusion equation, no matter what signs are assigned to the coefficients. The summations cancel the field -K"zjb and current density K"jb respectively, at t= aand eventually decay. IT one turns off a drive from a steady state, the current density is initially uniform, equal to K"jb and the field is equal to -K"zjb and then decays. But, the symmations with reversed signs have precisely that behavior. 10.6.2 (a) The magnetic field is H=i.H. = K" (1) and there is no E-field, nor J within the block. (b) When the current-source is suddenly turned off, the H-field cannot disappear instantaneously; the current returns through the conducting block, but still circulates in the perfect conductor around the block. For this boundary value problem we must change the eigenfunctions. At z = 0, the field remains finite, because there is a circulation current terminating it. Thus we have, instead of (10.6.15), 00 H.= I: OnCOS(~:z)e-t/T" (2) n-odd with the decay times 4IJub2 Tn = (ml")2 (S) Initially, H. is uniform, and thus, using orthogonality 10 m1l" 2b • m1l" bH. cos -zdz = K,,- sm-= -Om (4)-b 2b m1l" 2 2 Solutions to Chapter 10 10-33 and thus m-I( 4: ) C". = (-1)"--m", Kp; mood (5) ..-I 4: (R"') tlH. = L: (-1)-'---K cos -:r; e-r ..R'" 26 pn-odd The current density is aHa 2 ~ . (R"') tlJ1I =--- = -1::(-1) , Kpsm -:r; e-r .. a:r; b 2b H we pick a new origin at :r;' =:r; +b, then . (R"') . (R"', R"') R'" , . (R"')sm -:r; = sm -:r;--= -cos -:r; sm ­2b 2b 2 2b 2 ~ (R"') = -(-1) :I cos -:r;' for R odd 2b Interestingly, we find At t = 0 this is the expansion of a unit impulse function at :r:' = 0 of content -2K p• All the current now :Hows through a thin sheet at the end of the block. The factor of 2 comes in because the problem has been solved as a SYMmetric problem at :r:' = 0, and thus half of the current ":Hows· in the "imagined· other half. 10.1 SKIN EFFECT 10.1.1 (a) In order to find the impedance, we need to know the voltage tI, the complex current being k •. The voltage is (see Fig. 10.7.2) . (1) and, from Faraday's law (2) From (2) and (10.7.10) (3) Solutions to Chapter 10 10-34 and thus the impedance is at :t = -b (4) But the factor in front is iawp.o _ a(l +i) (5)d(1 + j) - duo (b) When b <: 0, we can expand the exponentials and obtain Z = a(1 +i) 1 + (1 + iH + 1-(1 +iH duo 1+ (1+ iH -1+ (1+iH (6) a(1 + i) 1 a = duo (1 +iH = dub (c) When b ,. 0, then we need retain only the exponential exp[(1 + i)b/ol with the result: z = a(l + j) (7)duo so that Re(Z) =­a duo This looks like (6) with b replaced by o. 10.7'.2 (a) When the block is shorted, we have to add the two solutions exp±(1 + i)f so that they add at the termination. Indeed, if we set (1) then the E-:6.eld is, from (2) and thus through integration (3) and is indeed zero at z = o. In order to obtain Hz = k. at z= -b we adjust Aso that (4) Solutions to Chapter 10 10·35 (b) The high frequency distribution is governed by the exp -(1 +i)i(:I: < 0) and thus II "'" k e-(1+" f= k -(1+i) £j! (5) II -• e(l+i) k .e 6 This is the same expression as the one obtained from (10.7.10) by neglecting exp -(1 +i)i and exp(1 +i)b/o. (c) The impedance is obtained from (3) and (4) aE a(1 +i) e(1+i)b/6 -e-(1+i)b/6 11 I--'-:-~. ~~-;:------;-::~=dK. z=-b -duo e(1+i)b/6 +e-(1+i)b/6 SOLUTIONS TO CHAPTER 11 11.0 INTRODUCTION 11.0.1 The Kirchhoff voltage law gives di v=v + L-+ R'~ (1)c dt where i= C dvc (2)dt Multiplying (1) by i we get the power flowing into circuit vi = vi + ~(~Li2) + Ri2 (3)c dt 2 But (4) and thus we have shown . d '2Rtn = -w+~ (5)dt where (6) Since w is under a total time derivative it integrates to zero, when the excitation i starts from zero and ends at zero. This indicates storage, since the energy supplied by the excitation is extracted after deexcitation. The term i2 R is positive definite and indicates power consumption. 11.1 INTEGRAL AND DIFFERENTIAL CONSERVATION STATEMENTS 11.1.1 (a) IT S = S",ix , then there is no power flow through surfaces with normals per­ pendicular to x. The surface integral tS·da 1 11-2 Solutions toChapter11 becauseSsisindependent ofyandz. (b)BecauseWandPd,arealsoindependent ofyandz,theintegrations transverse tothex-axisaresimplymultiplications byA~Hencefrom(11.1.1) assaw--=-+Pd,axat Wehavetousepartialtimederivatives, becauseWisalsoafunction ofx. (c)Thetimerateofchangeofenergyandthepowerdissipated mustbeequalto thenetpowerflow,whichisequaltothedifference ofthepowerflowingin andthepowerflowingout. 11.2POYNTING'S THEOREM 11.2.1 (a)Thepowerflowis y==-b TheEQSfieldisExB=-EsH.l~ FigureSll.Z.! E_Vd,s-a aHaaEs--=e--ayat(1) (2) (3) 11-3 Solutions to Chapter 11 and thus 8EsH. = 7Jf0/it (4) since H. = 0 at 7J = O. From (I), (2), and (4) • Vet d (Vet) 7Jfo dYetEx H = -l)"Yf o-; dt -; = -i)"~Vet dt (5) (b) The power input is: -/ ExH·da over the cross-section at 11 = -b where da = -i)" and therefore, bfo dYet d (1 2) -Ex H .da = -awV et -=--CVet (6) / a2 dt dt2 with C= fobw a (c) The time rate of change of the electric energy is (7) (d) The magnetic energy is (8) Now d Vet-V et .... ­dt 1" where 1" is the time of interest. Therefore, if 11-4 11.2.2 (a) FromFaraday's lawSolutions toChapter11 (1) (2) andtherefore (3) ,.------------oo-~. iy Fleur.811.3.3 (b)Theinputpoweris-fS·da,integrated overthecrosB-section at'J=-bwith da"-1,..Theresultis -fS·da=Pobaw!!!~=!!!L~ufJdt2dt2 with L=poab w (c)Themagnetic energyis withthesameLasdefinedabove.Thusthemagnetic energybyitseHbalances theconservation equation. (d)Theelectricenergystorageis Solutions to Chapter 11 11-S where dld/dt ~ Id/r, with r equal to the characteristic time over which Id changes appreciably. Thus, as long as 11.3 OHMIC CONDUCTORS WITH LINEAR POLARIZATION AND MAGNETIZATION 11.3.1 (a) The electric field of a dipole current source is E= ipd_ [2 cosfJir+sin Oi9] (1)411'0'rq The H-field is given by Ampere's law vx H=J =O'E (2) Now, by symmetry it appears that H must be t/J directed (3) and thus 1 a . 1 a v xH =i r-.-0 aO(H~smO) -i9--(rH~) (4)rsm r ar By inspection of the O-component of (4), with the aid of (1) and (2), one finds ipd. 0H.; = --sm (5)411'r2 The same result is obtained by comparing r components. Therefore, (6) The density of dissipated power is Pd =E· J = O'E2 =(ipd)2_1_[4cos2 0+sin20] 411' ur6 (7)ip d)2 1 2= (- --[1 + 3cos 0]411' O'r6 Solutions to Chapter 11 11-6 (c) Poynting's theorem requires (8) Now V . S in spherical coordinate is 1 a(2) 1 a( .)V· S = 2"-a r Sr + -.-0 ao SfJ smO rr rsm Now V· (E x H)= (ip d)2 ~1-3 sin2 0 -4cos2 0 +2sin2 OJ 4'1t' ar (9) = (i d)2 1 2 OJ _..£- -11 + 3cos 4'1t' ar6 Thus, (8) is indeed satisfied according to (7) and (9). (d) V· (~J) = (ip d)2V . ~ 12cos2 Oi.. +sin 0cos OifJJ4'1t' a.,­ (i d)2 1 2 2. 2 = _..£- -616cos 0 -2 cos 0 + sm OJ41f ar =_(ip d)2_1_ 11 + 3cos2 OJ = v· (Ex H) 4'1t' ar6 (e) We need not form the cross-product to obtain flow density. The power flow density is the current density weighted by local potential ~. 11.3.2 (a) The potential is a solution of Laplace's equation t) ~ = --, l! In(r/a) (1)nb E= t) i.. (2)In(a/b) r at) I.. VxH= J =aE = In(a/b) r (3) from Ampere's law. By symmetry (4) and aH. at) 1 -az = In(a/b) r (5) 11-7 Solutions to Chapter 11 and thus H __ av ~ (6)'" -In(a/b) r + v(t) z =-l Figure 511.3.2. (b) The Poynting vector is av2 z S=E XH= -illln2(a/b) r2 (7) (c) The Poynting flux is fS.da = -(r=a Sz21rrdr\ Jr=b z=-I (8) 21rav2 l 21ral 2 = -ln2 (a/b)ln(a/b) = -In(a/b)V (d) The dissipated power is ! ! lr=a av2 21rr (9) 2dvPd = dvaE=/0 1 2( /b) 2 drdzz=-I r=b n a r 21ral 2 v In(a/b) (e) The alternate form for the power flow density is (10) f S·da = -[Sr(r = b) -Sr(r = a)]21rbl (11)21rC1l 2 =- v In(a/b) This is indeed equal to the negative of (9). 11-8 ExH Cl>JSolutions toChapter11 ------ (f)SeeFig.SU.3.2b. (g)Atz=-I, ThusFlpre811.3.tb f21fO'lv. B.ds=In(a/b)=I (12) 11.S.S (a)Theelectricfieldis FromAmpere's law:.21fO'I 2 VI=In(a/b)vQ.E.D. (13) (1) z z=dI I V IVrI- Z=O+ + Figure811.3.3 (2) 11-9 Solutions to Chapter 11 2'11"rH _ {'II",-2 [~ + Etc (tl/d)] for r < b( •-'Il"b2[0'~ + Etc (tl/d)] + 'II"(r2 -62)Eott (tI/d) for b <r< a 3) and thus . forr<6 (4)for6< r< a The Poynting flux density Ex H = i. X i.EIIH. -iI'HO'~ + Etc(tI/d»~ forr<6 (5) = { -il'ir { ~ [Eb2 + Eo(~ -62)] ~ (tI) + ~~3 tI} ~ for6< r< a (b) (6)r<6 6<r<a Forr< 6, For6< r< a: Q.E.D. (76) (e) (8) Solutions to Chapter 11 11-10 The potential ~ is given by tI ~= --(z-d)d and (9) Therefore, s = {-i.(a; + ~~~)(z -d)~ forr<b (10)-i !A.U(z -d)!!. forb<r<a •ddt d (d) The integral is r-f S .da =l211"rdr[S,.(z =0) -S,.(z = d)] (11) For r < b: O'U E dU) U _2 O'U E dU)=lr 211"rdrd(-+-- -=11""--(-+-- tI (12a)o d ddtd d ddt For a < r < b: Equations (12) agree with (6). (e) The power input at r = a is from (12b) 2(O'tI Edtl) (2 2)EO dtl .1I"b -+--tI + 11" a -b --tI = til (13)d ddt ddt where . [O'tI d] 2 d,= d 2d +Edt (tI/d) + 1I"(a2 -b )Eo dt (tI/d) which is the sum of the displacement current and convection current between the two plates. 11.3.4 (a) From the potentials (7.5.4) and (7.5.5) we find the E-field E =-V~ =i rEoCOSf(1 + (R)20'b -O'a) r eTb + eTa -i4>Eo sin f (1-(R) 2eTb -O'a) r < R (la) r eTb + eTa Solutions to Chapter 11 11-11 and 2ua E (. -1."-1.)o II' cos Y' -I", sIn Y' Ub + Ua r< R (1b) Figure Sl1.3.4 The H-field is z-directed by symmetry and can be found from Ampere's law using a contour in a z -x plane, symmetrically located around the x-axis and of unit width in z-direction. If the contour is picked as shown in Fig. 811.3.4, then £H. ds = 1J .da= 2Hz = 21'" Jrrd4J 2ru E sin-l.(1+ (!1.)2 CTb -CTfl ) for r > R (2) == a 0 'f' r O"b+O'a. { 2rUbEo 2+CT4 sin 4J for r < R O'b (T a. The Poynting vector is Ex H = E",Hzi r _ ErHzi", = -irruaE; sin2 1jJ [1 _ (R)4 (Ub -Ua) 2] r Ub + Ua _ iq,ruaE; sin IjJ cos IjJ [1 + (R)2 (Ub -Ua )] 2 r> R r Ub + Ua • E2 ' 2 -I. ( 2ua )2= -lrrUb 0 SIn Y' Ua + Ub _ i",rub E; sin IjJ cos IjJ ( 2ua )2 r< R Ua + Ub (b) The alternate power flow vector S = <I>J follows from (7.5.4)-(7.5.5) and (1) <I>J = -iruaE;rcos2 1jJ [1-(!!)4 (Ub -ua)2] 4 Ub + Ua + i",uaE;r sin IjJ cos IjJ [1 _ (R)2 Ub -Ua]2 r> R r Ub + Ua (4) • 22( 2ua )2= -lrUbEo r cos IjJ Ub + Ua +i4>UbE;rsin4JcosljJ( 2ua )2 r < R Ub + Ua 11-12 Solutions to Chapter 11 (e) The power dissipation density Pet is Pet =O'E2= O'oE~ eos2 ~ [1 +(R)20'b -0'0]2 r O'b +0'0 (Sa) +O'o~ sin2 ~ [1 _ (R)2 O'b -0'0]2 r> R r O'b +0'0 r.12( 20'0 )2 = O'bb= r<R (5b) o 0'0 +O'b (d) We must now evaluate V· (E XB) and V· ~J and show that they yield -Pd. (6a) for r > R, V. S = -20'bE~ sin2 ~( 20'0 )2 0'0 +O'b -(eos2~-Sin2~)O'bE~( 20'0 )2 (6b) 0'0 +O'b = -O'b~ ( 20'0 ) 2 0'0 +O'b for r < R. Comparison of (5) and (6) shows that the Poynting theorem is obeyed. Now take the other form of power flow. The analysis is simplified if we note that V .J = O. Thus V· ~J =J. V~ = Jr!.-~ +J.!~~ = -O'E2 ar ra~ = -O'oE~ eos2 ~ [1 + (R)2 O'b -0'0)] 2 (7a) r O'b +0'0 _ O'aE~ sin2 ~ [1-(R) (O'b -O'a)]2 r> R r O'b +O'a and r<R (7b) Q.E.D. 11-13 Solutions to Chapter 11 11.4 ENERGY STORAGE v' 11.4.1 From (8.5.14)-(8.5.15) we find the H-fields. Integrating the energy density we find where we have used 171" sinOdO(4cos2 0+sin2 0) = -171" d(cos 0)(3 cos2 0 + 1) = /1 dx(3x2 +1) = (x3 + x)I~1 = 4 -1 Because we find that Q.E.D. 11.4.2 The scalar potential of P9.6.3 is r> R r< R The field is H_ .!!-i cos ~ {(ir cos ~ + i", sin ~)(R/r)2; r> R -2R 1 + J!. J!. (ir cos ~ -i", sin ~); r< R 1-'0 1-'0 Solutions to Chapter 11 11-14 The energy is 11.4.3 The vector potential is from (8.6.32) r<a ,",oB=VxA = -i. x VA.. = Ni i. x {[2(rj a) -1] sin cPi.. + (.!: ­3a a = -,",oNi [(~ _ 1) cos cPlp ­(2~ -1) sin cPi",]3a a a The energy is 1) cos cPi",} (1) Therefore, 11.4.4 The energy differential is The coenergy is dw'm = d(i1.\d + d(i2.\2) -dWm = .\ldi1 + .\2di2 = (Llli1 + L12i2)di1 + (~lil + L22i2)di2 (1) (2) Solutions toChapter 11 with -11-15 (3) FigureSI1.4.4 Ifweintegrate thisexpression alongaconveniently chosenpathinthei1-i2plane asshowninFig811.4.4, weget !:~oLlli1di 1+1':2=0 (L21i1+L22i2)di2 '2=0 ,}=contt 1L'2L..1L'2=2ll~l+21~1~2 +222~2 (4) 1(L'2L..L" L'2) =2ll~l+12~1~2 +21~2~1 +22~2 1L(N2'22NNo..N2'2) =2 0l~l+ 12~1~2+ 2~2 whenthelastexpression iswrittensymmetrically, using(3). 11.4.5 Ifthegapissmall(a-b)<:a,thefieldisradialandcanbeevaluated using Ampere's lawwiththecontour showninFig.811.4.5.Itissimplest toevaluate the fieldofstatorandrotorseparately andthentoadd.Thefieldvanishes at¢J=1f/2 andthus£H·dB=-(a-b)Hr(¢J) r/J__ lengthIalong zcontour o® FigureSI1.4.5(1) Solutions to Chapter 11 11-16 For the stator field, the integral of the current density is 1 lrt/2 Nlil . N1i1J .da=---sm tPadtP =---cos tP (2) s ~ 2a 2 where N1 is the total number of terms of the stator winding. Therefore, the stator field is given by (3) The rotor coil gives the field (4) where N2 is the total number of turns of the rotor winding. In a linear system, coenergy is equal to energy, only the independent variables have to be chosen prop­ erly, i.e. the energy expressed in terms of the currents, is coenergy. When expressed in terms of fluxes, it is energy. The coenergy density is (5) The coenergy is (6) We find (7) and 11.4.6 al ) D = (v1+a2E2 +Eo E The coenergy density in the nonlinear medium is [note E· dE = d(iE21 w; =lED . dE= I i(V1 :~2E2 + EO) dFfJ = al V1 + a2E2 + -21Eo~ a2 Solutions toChapter 11 Inthelinearmaterial I12W=-€Ee20 Integrating thedensities overtherespective volumes onefinds(E2=tJ2/a2) [aV tJ21tJ2] 1tJ2w'=-!.1+a2-+-€o-eca+-€o-(b -e)ca ea2 a22a2 2a2 Q.E.D. 11.4.1 (a)H=i.i/winbothregions. Therefore, B=i.P,oi/w11-17 inregion(a) inregion(a) inregion(b) 11.5ELECTROMAGNETIC DISSIPATION 11.5.1 From(7.9.16) wefindanequation forthecomplex amplitudeEa: E_ ,"WEb+O'b A a-(jW€a+O'a)b+(jW€b+O'b)atJ(1) andsince wefind(2) E- J'W€a+O'a A () b-(jW€a+O'a)b+(jW€b+O'b)atJ 3 (Another wayoffindingEbfrom(1)istonotethatEaandEbarerelatedtoeach otherbyaninterchange ofaandbandofthesubspcripts.) Thetimeaveragepower dissipation is 1E21E2(Pd)="2O'alalaA+"2O'bIblbA =~aO'a(w2€~+O'~)+bO'b(w2€~+O'~)1°12 2(bO'a+aO'b)2+w2(b€a+a€b)2 Solutions to Chapter 11 11-18 11.5.2 (a) The electric field follows from (7.9.36) .. ..(... ) O'a. + jWfa.Eb=-V4>=3EpcosOlr-smOI 6 . (2 )i r < R (lb)2O'a. + O'b + JW fa. + fb Therefore () 21O'b I" Eb12 29,Ep12 O'~ +W2 f~ (2b)Pd = = O'b (2O'a. + O'b)2 +W2(2fa. + fb)2 j r < R The electric field in region (a) is IT we denote by A= O'a. -O'b + jW(fa. -fb) -(2O'a. + O'b) + jw(2fa. + fb) we obtain 2(Pd) =iO'a.IP;a.12 = IEp I{ cos2 0[1- 4(R/r)3Re A+ 4(R/r)6IAI2J + sin2 6[1 + 2(R/r)3Re A+ (R/r)6IAI2j} (b) The power dissipated is 4'1l"R3 (Pd) = -3-(Pd) (3) where (Pd) is taken from (2b). 11.5.3 (a) The magnetic field is z-directed and equal to the surface current in the sheet. In region (b) (1) in region (a) it is H=i.K (2) The field at the sheet is, from Faraday's integral law (3) The field at the source is (4) 11-19 Solutions to Chapter 11 The power dissipated in the sheet is, using (3) dHb 2 Pd =!(1E;dV = (1LlWdb2J-l~( ----;It) (5) The stored energy is rW dv = !J-lO(Ha )2adw + !J-lo(Hb )2bdwlv 2 2 (6) = !J-lodw[b(H b )2 + aK2]2 (b) The integral of the Poynting vector gives dK dHbf Ex H· da = -EyHzwd = -(aJ-l0 dt + bJ-l0----;It )Kwd (7) Now dHbHb = K -E y(1Ll = K -bJ-lo----;It(1Ll (8) When we introduce this into (7) we get f 1 dK2 1 dHb2ExH· da =-{-all wd-+ -bll wd--}2 r-O dt 2 r-o dt (9) dHb 2 -(1b2 wdJ-l~ (----;It) (1Ll But the last term is Pd; and the term in wavy brackets is the time rate of change of the magnetic energy. 11.5.4 Solving (10.4.13) for ..4, under sinusoidal, steady state conditions, gives 1 [ 1-&] A • I-' 2A =(. )-JWTm + ---T m a H o JWTm + 1 J-loLl(1a (1) 1 [. J-l-J-lo] 2 =. -JWTm + --- a H o (JWTm + 1) J-l + J-lo From (10.4.11), we obtain 6 A ~ 6=_J-lO(Ho+~)= 1-'+1-'0 H (2) J-l a2 1+ iWTm 0 The discontinuity of the tangential magnetic field gives the current flowing in the cylinder. From (10.4.10) A ( A ..4)LlH", =-Ho -a2 sin¢> -Csin¢> ·. J-l-J-lo 2J-lo] Hosin¢>=-[1 + JWTm + JWTm ---- ---- . (3) J-l + J-lo J-l + j:.,; 1+ JWTm JWTm . A =-2 . sm¢>Ho = Kz1 + JWTm Solutions to Chapter 11 11-20 Note the dependence of the current upon w: when WTm ::> 1, then the current is just large enough (-2Hosin<p) to cancel the field internal to the cylinder. When WTm -+ 0, of course, the current goes to zero. The jump of H", is equal to K. The power dissipated is, per unit axial length: 2 Pd. = -1/ulEI2dv =1-ul:1a1 ". It.. 12d<p (4)2 2 0 But (5) and thus (6) 11.5.5 (a) The applied field is in the direction normal to the paper, and is equal to Hocoswt = Niocoswt/d (1) The internal field is H o + K where K is the current Howing in the cylinder. From Faraday's law in complex formf E· ds = -iwp.(Ho + K)b2 (2) Because K must be a constant, Etangential to the surface of the cylindrical shell must be constant. The path length is 4b. We have K = ul:1t = _iwp.ul:1b (Ho + K) (3) 4 and solving for K K =-jWTm Ho (4)1 +jWTm where p.ul:1b Tm =-- (5)4 The surface current cancels Ho in the high frequency limit WTm -+ 00. In the low frequency limit, it approaches zero as WTm approaches zero. Thus Pd. = ~ / ulEI2dv = ~ 4bl:1du IKI2 = ~N2i2 w2T~ (6)2 2 u21:12 uAd 01 + w2T~ (b) The time average Poynting Hux is -Re fE x :A: .da = -Re i4bdtb* = -Re {2bdH;(-jWT m)(Ho + K)} * ,. (7) = Re 2bdjWTmHoK = 2bd w2T~ IH l2 = ~ w2T~ N 2i2 ouAl+w2T~ uAdl+w2T~ 0 which is the same as above. 11-21 Solutions to Chapter 11 11.5.6 (a) When the volume current density is zero, then Ampere's law in the MQS limit becomes VxB=O (1) and Faraday's law is (2) IT we introduce complex notation to describe the sinusoidal steady state E = Re t(r)ejWT etc., then we get from the above VxB=O (3) v XE= -jw~o(B + M) (4) IT tf is linearly related to Ii we may write (5) where Xm is, 'in general, a function of w, we may define (6) and write for (4) vx t= -jwfJ (7) with B == P.B (8) Because V· ~o(B + M) = 0, we have (9) (b) The magnetic dipole moment is, according to (20) of the solution to PI0.4.3. A nSlI jwrm= -21('4- 0 • (10)1+ 3wr with r = ~oO't::..R/3. As wrm -+ 00, this reduces to the result (9.5.16). The susceptibility is found from (5): A 2 (R/)3 jwrXm =-1(' 8 1+'3wr where 1/s3 is the density of the dipoles. (c) The magnetic field at z = -l is (14) Solutions to Chapter 11 The electric field follows from Faraday's law: applied to a contour along the perfect conductor and current generator 11-22 (15) and thus (16) The power dissipated is f A APd =-21Re E XH*.da 1I A A= -Re EyH; lad (17)2 x=­ = ~Re jwJi.lkl2adl Introducing (12) and (13) we find (18) 11.5.7 From (10.7.15) we find A A (x+b)Hz = K. exp -(1 + j) -5- (1) so that Hz = K. at the surface at x = -b. The current density is A A .... H • aHz • (1+ j) K ( ') (x + b)J!::::!. v X = -1)' ax = 1)' --5- •exp-1 + J -5- (2) The power dissipation density is (3) and thus the power dissipated per unit area is x1=0 ,k.,21°O 2(x+b) Ik.12 Pddx!::::!. -- exp- dx =-- wattsjm2 x=-b a x=-b 5 2a5 11-23 Solutions to Chapter 11 11.5.8 (a) From (10.7.10) we find Hz everywhere. The current density is The density of dissipated power is: __1 I 12 cosh T2", + cos T2", (2) A - K.a02 cosh ~ -cos 2b6 6 The total dissipated power is 0 1 A 20 sinh 2c'" + sin 26'" 1° Pd = ad1Pddx = ad-----c2IK.1 - 2b 2b ",=-b au 2 cosh T -cos T -b IK 12 sinh ~ + sin ~ (3) = ad-'- 6 6 2ao cosh ~ -cos ~ 6 6 (b) Take the limit 0 ~ b. Then sinh ¥ '::::! cosh ~b le2b/ 6 and the sines and2 cosines are negligible. ad 2 A 1 Pd = 2ao 1K• (4) which is consistent with P11.5.7. When 2b/o ~ 1, then 2b 2b 1 2b 2( 1 2b 2) 2b 2cosh (-) - cos (-) ~ 1 +-(-) - 1--(-) = (-) (5)o 0 20 20 0 .h (2b) . (2b) 4bsm -+sm - '::::!- (6)o 0 0 and thus = ad-1-lk 12~ = adlk.12 (7)Pd 2ao' b 2ab The total current is (8) The resistance is a R= abd (9) and (10) Q.E.D. 11-24 Solutions to Chapter 11 11.5.9 The constitutive law aM -='YH (1)at gives for complex vector amplitudes (2) and thus A 'YXm=-. (3) 3W and (4) The flux is A ( 'Y)A B=AH=I-'o 1+-:-H (5) 3w The induced voltage is d>' • ~ A 1J =-=> 1J = 3WA (6)dt and (7) But (8) and thus N 2 2 ~ A 1 W ~ A=I-'--' (9)8R and thus A • ~. N1 2 W 2 ~ 'YN2 1 W 2~ , (. L R)~ 1J = 3WA = 3WI-'0""8il"' +1-'0 8R = 3W + m' (10) Thus R -l-'o'YN1 2 W 2 m-8R (11) Solutions toChapter11 11-25 11.5.10 (a)ThepeakHfieldis (1) Thus(seeFig.SU.5.10a). H---tBIIB I --i II II I I II I-2Hc FJsureS11.I.I0a (b)Theterminal voltageis d1rW2dB,,=-N1--B ex-dt4dt(2) TheBfieldjumpssuddenly, whenH=He.ThisisshowninFig.SU.5.10b. Thevoltageisimpulse likewithcontentequaltothefluxdiscontinuity: N~B21of.•• (c)Thetimeaveragepowerinputisfvidtintegrated overoneperiod.Contribu­ tionscomeonlyatimpulses ofvoltageandareequalto (3) But (4) andthus (5) 11-26 IMPULSESolutions toChapter11 --H IMPULSEt=tof----""""'-f----H(t) t~ FlpreSIl.S.IOb (d)Theenergyfedintothemagnetizable material perunitvolumewithintime dtisa adtH·-p.(H+M)=dtH·-B=H.dBat0 at Asonegoesthrough afullcycle,fH·dB=areaofhysteresis loop Thisis4HeB•.Thusthetotalenergyfedintothematerial inonecycleis f1rW2volume H·dB=(21rRT)4B.He 11.6ELECTRICAL FORCES ONMACROSCOPIC MEDIA Thecapacitance ofthesystemis 0=Eo(b-e)d IJ Theforceis(6) (7) (8) Solutions to Chapter 11 11-27 / 11.6.2 The capacitance per unit length is from (4.6.27) C = 1rfo (1)In(-k + ..j(l/R)2 -1) where the distance between the two cylinders is 2l. Thus replacing l by e/2, we can find the force per unit length on one cylinder by the other from 1 2 dC 1 2 d [ 1rfo ] Ie = 2"v de = 2"v de In[~ + ..j(e/2R)2 -1] --L + ~ 1 (2) 1 2 1rfo 2R (2R)2 V(E/2Rj2-1= --v 2 ln2[(e/2R) + ..j(e/2R)2 -1] ~ + ..j(e/2R)2 -1 This expression can be written in a form, in which it is more recognizable. Using the fact that >./ = Cv we may write f -_~ 1 + (e/2R)/..j(e/2R)2 -1 (3) o-41rf o R ~ + ..j(e/2R)2 -1 When e/2R ~ 1, and the cylinder radii are much smaller than their separation, the above becomes f--~ (4)e -21rf2eo This is the force on a line charge >./ in the field >.,j(21rf o 2e). V 11.6.3 The capacitance is made up of two capacitors connected in parallel. C = 21rf o (l-e) + 21rfe In(a/b) In(a/b) (a) The force is I -~ 2 dC _ 2 1r(f-fo) e -2 v de -v In(a/b) (b) The electric circuit is shown in Fig. S11.6.3. Since R is very small, the output voltage is Vo = iR l vo : RL3v + + -+ V Figure 811.6.3 11-28 Solutions toChapter11 FromKirchoff's voltagelaw iR+V=tI Now q=Otl .VdO -,~-dtand.dqd dO dtl,=-=-(Otl)=-tl+O-dtdt dt dt IfRissmall,thentIisstillalmostequaltoVanddtl/dtismuchsmallerthan (tldO/dt)/O. Then and tlo=Ri=-211'RV(E -Eo)~;/In(a/b) 11.6.4 Thecapacitance isdetermined bytheregioncontaining theelectricfield 0=211'Eo(l-e) In(a/b) (a)Theforceis DA-6--------Q--- c -1I'Eov2e=I In(a/b) 0Bft -­vc ADB q Figure911.8.4 (b)SeeFig.S11.6.4.Whene=0,thenthevalueofcapacitance ismaximum. GoingfromAtoBinthef-eplanechangestheforcefrom0toafinite negative valuebyapplication ofavoltage.TravelfromBto0maintains the forcewhileeisincreasing. Thuseincreases atconstant voltage.Themotion from0toDisdoneatconstantebydecreasing tovoltagefromafinite valuetozero.FinallyasonereturnsfromDtoAtheinnercylinderispushed Solutions to Chapter 11 11-29 back in. In the q -tJ plane, the point A is one of zero voltage and maximum capacitance. As the voltage is increased to Vo , the charge increases to 21l"f o l q = avo = In(a/b) Vo The trajectory from B to a keeps the voltage fixed while increasing e, de­ creasing the capacitance. Thus the charge decreases. As one moves from a to D at constant edecreasing the voltage to zero, one moves back to the origin. Changing eto zero at zero voltage does not change the charge so that D and A coincide in the q -tJ plane. (c) The energy input is evaluated as the areas in the q -tJ plane and the e-f plane. The area in the e-f plane is 1l"fo lV 2 In(a/b) 0 and the area in the tJ -q plane is ~ 21l"f olV 2 2ln(a/b) 0 which is the same. 11.6.5 Using the coenergy value obtained in P11.4.6, we find the force is 2aw' [alV a tJ2 1 tJ2] 1 f tJ2Ie = _e I =-( 1+-- -1) + -f -ca-__o_c ae v a2 a2 2o a2 2 a 11.7 MACROSCOPIC MAGNETIC FORCES 11.7.1 The magnetic coenergy is I1(L '2 2L .. L '2)Wm = 2 utI + I2t It2 + 22t 2 The force is Since we have 11-30 Solutions to Chapter 11 11.7'.2 The inductance ofthe coil is, according to the solution to (9.7.6) 1m = !i2dL = _!i2 l-'oN2 1 2 dx 2 [II: + -lL...-]2 7I"a2 ;as 2J1'ad J 11.7'.3 We first compute the inductance of the circuit. The two gaps are in series so that Ampere's law for the electric field gives 1/(H1 + H2) = ni (1) where HI is the field on the left, H2 is the field on the right. Flux conservation gives (2) Thus n, x H1 =-­ 1/ a The flux is ... I-'oni (a -x) d"*">'=----x 1/ a The inductance is 2L = n~>. = l-'on xd(a -x) 1/ a The force is f. =!'2(aL I aLI) =! '21-'0n2d{ (a -2x) I_ x(a -x) I} m 2' a x + a:l 2' 1/ x 2:1X 1/ a 1/ 11.7'.4 Ampere's law applied to the fields Ho and H at the inner radius in the media 1-'0 and 1-', respectively, gives b b Ho la -dr = Hla -dr = Ni (1) b r b r and thus Ni Ho = H = bin!! (2) b The flux is composed of the two individual fluxes Ni ~>. = 271" In!! 11-'0(1-e) + I-'el (3) b The inductance is L = N~>./i = ln~~b} N2{l-'e + 1-'0(1-en (4) The force is 1(' ~) =! '2 dL = 71"(1-' -1-'0) N2'2 (5)',,. 2' de In(a/b) , 11-31 Solutions to Chapter 11 11.7.5 The H-field in the two gaps follows from Ampere's integral law 2H6. = 2Ni (1) The flux is ~A = l-'oHd(2a -O)R = l-'oNid(2a -O)R/6. (2) and the inductance (3) The torque is T = -,1·2 -dL = -11 dRN2 ,~ '2/ A (4)2 dO ,..0 \/ 11.7.6 The coenergy is w:n =f[Aadi a + Abdib + Ardir] = 21L'.'a2 + 21L'.'b2 + 21Lr'r'2 (1) +M cos Oiair + M sin Oirib where we have taken advantage of the fact that the integral is independent of path. We went from ia =ib = ir = 0 first to ia, then raised ib to its final value and then ir to its final value. (b) The torque is 8w:" .( M . 9' M lI')T = ao = 'r - Sln 'a + cos uSb (c) The two coil currents ia and ib produce effective z-directed surface currents with the spatial distributions sin<p and sin(<p -~) = -cos<p respectively. IT they are phased as indicated, the effective surface current is proportional to cos(wt) sin <p - sinwt cos <p = sin(<p -wt) Thus the rate of change of the maximum of the current density is d<p/dt = w. (d) The torque is T = /r[-M sin(Ot -'1)/coswt + M cos(Ot -'Y)/sinwt] = /r/(-M sin(Ot -'1 -wt) But if 0=w, then Solutions to Chapter 11 11-32 11.8 FORCES ON MACROSCOPIC ELECTRIC AND MAGNETIC DIPOLES 11.8.1 (a) The potential obeys Laplace's equation and must vanish for y --+ 00. Thus the solution is of the form e-~" cos pz. The voltage distribution of y = 0 picks the amplitude as Vo. The E field is E = PVo(sin pzi x + cos pzi)')e-~" (b) The force on a dipole is f =p 0 VE = 411'E oR3 (E 0 V)E It behooves us to compute (E . V)E. We first construct the operator Eo V = pVoe-~"(sinPz :z + cospz:y) Thus Eo VE = pVoe-~"{ sinpz :z [pVo(sinpzi x + cospzi)')e-~"] + cos pz:y [pVo(sinpzi x + cos pzi)')e-~"] = p2 Vo2p[(sin pz cos pzi x -sin2pzi)')e-~" -(cos,8zsin,8zi x +cos2,8z1)')e-~"] = _p2Vo2pi)'e-~" and thus 11.8.2 Again we compute, as in PH.S.1, (Eo V)E in spherical coordinates (1) and the gradient operator is )2) Thus, (3) 11-33 Solutions to Chapter 11 and (4) and the force is 3 2Q2 2Q2R3 f =p .VE= -41rf oR (41rf )2r5 -41rf r5 (5) o o Note that the computation was simple, because (a / ar)ir= O. In general, derivatives of the unit vectors in spherical coordinates are not zero. 11.8.3 The magnetic potential 'If is of the form 'If = {ACOS{3xe-/lY y> 0 A cos {3xe/lY y< 0 At Y = 0, the potential has to be continuous and the normal component of ILoB has to be discontinuous to account for the magnetic surface charge density Pm =V. ILdM. :=)0 ILoM ocos {3x Thus 'If =-Mo cos {3xe-/lY 2{3 This is of the same form as ~ of PH.B.l with the correspondence Vo +-+ Mo /2{3 The infinitely permeable particle must have H = 0 inside. Thus, in a uniform field Hoi., the potential around the particle is (We use, temporarily, the conventional orientation of the spherical coordinate, () = 0 axis as along z. Later we shall identify it with the orientation of the dipole moment.) 'If = -HoR cos ()[ ~ -{R/r)2] The particle produces a dipole field 3 H oR (2 (). . ()') m ( ()' . ().)--3 - cos Ir + sm 10 = --3 2 cos Ir + sm 10 r 41rr Thus the magnetic dipole is ILom = 41rILoHo~ This is analogous to the electric dipole with the correspondence ILo +-+ f o Since the force is f= ILom· VB we find perfect correspondence. 11-34 Solutions to Chapter 11 11.8.4 The field of a magnetic dipole I-'om II i. is H = I-'om,.s (2 cos (Jil' +sin (Ji8) 41r1-'0 The image dipole is at distance -Z below the plane and has the same orientation. According to P11.8.S, we must compute f =I-'om . VB =I-'om . V I-'om,.s (2 cos (Jil' +sin (Ji8 ) 41r1-'0 where we identify r=2Z after the differentiation. Now il' and i8 are independent of r and thus since (J = O. But and thus 11.9 MACROSCOPIC FORCE DENSITIES 11.9.1 Starting with (11.9.14) we note that J = 0 and thus f= IFdv = -I ~H2Vl-'dv (1) The gradient of I-' of the plunger is directed to the right, is singular (unit impulse­ like) and of content I-' -1-'0' The only contribution is from the flat end of the plunger (of radius a). We take advantage of the fact that I-'H is constant as it passes from the outside into the inside of the plunger. Denote the position just outside by z_, that just inside by z+. 11 2 .21Z+--H V I-'dv = -lx1ra 2 z_ 2 ~ -ix-1ra [ I-'H2 2dl-'H -dz dz 21z + Id 2 ] (2) -I-'-H dz z_ dz 11-35 Solutions to Chapter 11 where we have integrated by parts. The integrand in the second term can be written d 2 dHp.-H = 2p.H- (3)dx dx and the integral is 1"'-"'+ dH p.H-= p.HHI"'+ = -p.o~1 (4)dx "'- "'­ where we have taken into account that p.H is x-independent and that H(x+) = O. Combining (2), (3), and (4), we find . x '/fa2 2 H2 (5) f = -I -Jl ,..0 Using the H-field of Prob. 9.7.6, we find (6) This is the same as found in Prob. 11.7.2. 11.9.2 (a) From (11.9.14) we have F=JxB (1) Now B varies from p.oHo to P.oHi in a linear way, whereas J is constant (2) where la+.o. a drJ =K (3) Now, both J and Hi are functions of time. We have from (10.3.11)-(10.3.12) Solutions to Chapter 11 11-36 11.9.3 (a) Here the first step is analogous to the first three equations of P11.9.2. Because J is constant and H varies linearly • T. K (Ho + Hi) (. .)Ir r= 1-'0 2 I. X 14> (1) (b) If we introduce the time dependence of A from (10.4.16), with I-' = 1-'0' A = -Hma2e-t/Tm (2) and of Kz from (19) -HO rri -2 A . A.. -2H . A.. -tiT", K Z - 4>-fl4> -2 sm 'I"--m sm 'l"e (3) a Further note that H~ =0 at t = O. Therefore from (3) and (2) H~ = -2Hm sin¢> at t = 0 (4) At t = 00 H~=-Hmsin¢> (5) because the field has fully penetrated. Thus H~=-Hmsin¢>[1+e-t/Tml (6) From (6) and (3) we find H~=-Hmsin¢>[1-e-t/T",] (7) Thus we find from (1), (3), (6), and (7) irTr = -ir~0[(H4:)2 -(H~)2] = -ir~o H~ sin2 ¢>[(1 + e-t / T",)2 -(1-e-t / T",)2] = -ir21-'0H;' sin2 ¢>e-t/Tm -- Figure 811.9.2 The force is inward, peaks at t = 0 and then decays. This shows that the cylinder will get crushed when a magnetic field is applied suddenly (Fig. 811.9.2). SOLUTIONS TO CHAPTER 12 12.1 ELECTRODYNAMIC FIELDS AND POTENTIALS 12.1.1 The particular part of the E-field obeys Hwe set then or Because of (2), a v x Ep =-atB (1) v .EoEp = 0 (2) B=VxA (3) v x (Ep + aa~) = 0 (4) Ep = a -atA ­V.p (5) a 2at v .A + V .p = 0 (6) But, because we use the Coulomb gauge, V·A=O (7) and thus V2 • p= 0 (8) There is no source for the scalar potential of the particular solution. Further (9) Conversely, (10) and v X E,. =0 (11) Therefore, E,. =-V.,. (12) and from (10) (13) Thus (9) and (13) look like the inhomogeneous wave equation with a2 jat2 terms omitted. 1 12-2 Solutions toChapter 12 12.1.2 %t22Aisoforderl/r2A,V2AisoforderA/£2.Thus,J1.f.%t2 2Aisoforder'!f£2 compared withV2A.ItisnegligibleifJ1.f.£2/r2=£2/c2r2~1.Thesameapproach showsthatJ1.f.(a2/at2)cpcanbeneglected compared withV2CPif£2/c2r2~1. 12.2ELECTRODYNAMIC FIELDS OFSOURCE SINGULARITIES 12.2.1 Thetimedependence ofq(t)isthesameasthatofFig.12.2.5,exceptthatit nowextends overonefullperiod. t=1'/2 tTrq(---) --20t.Tr--, q(---)-,--/ 20~,­", /' ........."-, \ \ \, ""t=l' ," ' '---~ ql(T-~) o /"l-(---"I'l" _,"II1\---r E-Iine. ~ / / / '"E-Iine. FigureS12.2.1a PlotofElectricDipoleField.Anysetoffieldlinesthatcloseuponthemselves -----12-3 Solutions to Chapter 12 may be considered to be lines of equal height of a potential. The potential does not necessarily reproduce the field intensity at every point. i.e. ........-... E =-(i<l> X V~). f(r,9) (1) The "underbrace" gives the pattern. The "overbrace" is the multiplier. It does not change the direction of the field. Take II [ r/ ]'I..+ + 2r/ sm ll'ul8 E= -d{2 3"q +2 [_qq +""2q"]. } (2) cos u 411"E r cr r cr c r where q = q(t -.!:.)2 IT one defines (3) Then V~ = (~) [2sin9( -!r-3/ 2 -q' !r-1/ 2 + !r/!r1/2 _ r/' r-1/ 2)i.. 411"E 2 c 2 C c2 (4) +ie2 cos 9(qr-3/ 2 + ~ r-1/ 2) ] One constructs a vector perpendicular to V~, i<l> X V~, by interchanging the 9 and r components and reversing the sign of one of them Thus if we choose f(r, 9) = r-3/ 2, we reproduce the E-field ofthe dipole by expres­ sion (1). We can sketch the function ~ for 9 = 11"/2. 12-4 Solutions toChapter12 t=2T (J,.-- ..... '" ...-/ "."<"" ,'....'.- "'... /P-t, /(II1, ....--,'III III III III III I FigureSU.J.lbTrq'(---)2c -r 12.2.2 Interchange E-H,H--Eand1'0-Eo.From(23) di=iwqd-iwqmd=iWlJom whereqmisthemagnetic charge.Weobtain Ok0" -;1cr E..11wIJom•lie 4>=- smu-- 411" r(l) (2) Solutions to Chapter 12 12-5 and from (24) QED (3) 12.2.3 Because Io'om(t) = qmd -qd in the electric dipole case, the time dependence of q(t)d and Io'om(t) correspond to each other. With E-H and H--E we must obtain mutually corresponding field patterns. 12.2.4 We can use the field sketch of Problem 12.2.1 with proper interchange of variables. 12.3 SUPERPOSITION INTEGRAL FOR ELECTRODYNAMIC FIELDS 12.4: ANTENNAE RADIATION FIELDS IN THE SINUSOIDAL STEADY STATE 12.4.1 From (4) tPo(O) = sin 0 (' e-jlc.',:ilc.'cOIBdz' l 10 = sin 0 1 {e-jlc(1-co8B)' _ I} (I) I jk(cos0 -1) = sinO 2 . [kl(l_ n)] -jlc(1-co8B)'/2 l k(I-cosun) sm2 cos u e The radiation pattern is (2) With kl = 271" .T.(n) _ sin20 (. 22 . 2 0)... u = Sln 7I"sm- (3) 471"2 sin4(Oj2) 2 The radiation pattern peaks near 0 = 60°. Solutions to Chapter 12 12-6 1jJ(O) Figure 812.4.1 12.4.2 By analogy with (3) one replaces H<f> -E<f>, IJ +-+fand i(z')dz' = jw(qz)dz' ­ J'w(qmd) dz' = J'wIJIJ(z')dz' where we interpret qd and qmd as assigned to unit length. Thus, from (2) of Prob. 12.2.2, with IJo -IJ, fo -f, 2 jkr I¥! ., E<f> = -sinO--k e- - M(z')eJkr'lrdz' 4'11' r f 2 jkr = kI . ~e- M ejOl.°f (0) 4'11' V~ 4 0 0 where 12.4.3 tPo(O) = _sinO (' sin~(z' -I) ejkz'cos8dz' I Jo sm {3I =_ si~O {' ~{(ej~(z'-I) _ e-j~(z'-I))ejkzlcOs8d({3z') {3lsm{3I Jo 2J sinO 1 {ej(IJ+kCOS8)1 -1 _. I e-j(IJ-kcos8)1 -1 e3IJ. ,}=- - e J~ ­ (3lsin{3I2j j(l+~cosO) -j(l-~cosO) sin 0 2 {{3I' . {3I k jk cos 81}= {31' {3I k2 cos + J sm -(3 cos 0 -e sm 1- "ji'i cos2 0 12.4.4 (a) From (12), and with an = n~ix, tPa = L3 ejka".lrei(OI.,,-OI.o) (1)n=O = 1+ ej(f cos <f> sin 8+01.1-01. 0 ) + ej (7I' cos <f>sin 8+01.2-01. 0 ) 12-7 Solutions to Chapter 12 (b) Since tPo = sin 0, and Qi = 0 ItPolltPal = 11 + 2 cos (i cos e; sin 0) IsinO (2) (c) tPa =1+ ejf(C08~8iD9+1) + ejll'(Co8~8in9+1) = ejf(Co8~8in9+1){e-jf(c08~8iD9+1) + 1+ ejf(C08~8iD9+1)} (3) = ejf(Co8~8in 9+1) [2 cos i(cos e; sin 0+ 1) + 1] 12.4.5 (a) tPa(O) = L1 ejlc....lrej(a ..-ao) = 1+ ej [lI'co8/1+al- a o] (1) n=O (b) (2) (c) G = 411"cos2 (~cosO) sin2 0 I; dO 1:11' de; sin 0cos2 (~cos 0) sin2 0 (3) Define cosO = u (4) r dO sin3 0cos2 (~cos 0) = j1 du(I- u2) cos2 (~u) (5)Jo 2 -1 2 Now consider integral Id 22 1( 1. 2) 2 z3 2z 1. zz cos "z = '2 z + '2sm zz -3" + "8cos2z -8" sm 2% (6) The integral is Solutions to Chapter 12 12-8 The gain is 411'" cos2 (~ cos 0) sin2 0G----7~-;:;''-;:-- (8) - 211'"U + ;2} (d) We find for '11(0) of array '11(0) = {I1/10(0) I11/11(0)111/12(0)1}2 (9) with 1/12(0) = 1- eikasinOcos'" (10) In order to get maximum superposition in the direction 4> = 0, one needs ka = 11'" or a = >../2. Thus 11/12(0) I= 12 sin (~sin 0 cos 4» I 12.5 COMPLEX POYNTING'S THEOREM AND RADIATION RESISTANCE 12.5.1 The radiation field Poynting vector of the antenna is from 12.4.2, 3.4.5 ~(EoH;) = ~ ((:~): filloI2(1/Io(0))2 (1) where 1/10(0) is from 12.4.28 _ 1 cos( 3;)-cos(3;cos 0) 1/10(0) -e1\') . e1\') . 0""2 sm ""2 sm (2) ~ cos (~cosO) 311'" sin 0 The radiated power is 2 ~ 1 1\' 121\' 1 ~-11 0 1Rrad = dO sin 0 d4>-E oH; 2 10 0 2 _! (311'")2. ~/ II 12(~)2 11\' cos2(~cosO) . (3) -2(4 )2V J.Lo/ fa a 3 211'" .2 sm OdO 11'" 11'" 0 sm 11 = !II 12 VJ.Lo/f o( )l1\'dO' cos2 e; cosO)a 2 211'" smO 2 2 411'" 0 sin 0 Therefore VJ.LO/f O11\' . cos2(3; cos 0)Rrad= 2 dO sm 0 .2 11'" 0 sm 0 _1 11 cos2(321\' x) (4)--VJ.Lo/f o dx 2 211'" -1 1-x = 1040 Solutions to Chapter 12 12-9 12.5.2 The scalar potential of the spherical coil is (see Eq. 8.5.17) (1) This identifies (2) We have for the 0 component of the H-field (3) and thus the radiation field is k2 A A mHo ~ ---sinO 411"r (4) The power radiated is (5) Therefore, Rrad = ~; VlLo/foN 2 (kR)4 (6) The inductance of the coil is from (8.5.20) (7) and therefore (8) 12.6 PERIODIC SHEET-SOURCE FIELDS: UNIFORM AND NONUNIFORM PLANE WAVES Solutions to Chapter 12 12-10 12.6.1 (a) From continuity: ak:c . A 0 az + 3wO'. = Taking into account the z-dependence: (2) and therefore (3) and (b) The boundary condition on the tangential B is: nil I)' Since B II i. (4) and thus b: -b: = k:c (5) H. is antisymmetric, of opposite sign on the two sides of current sheet. (6) and thus (7) From (12.6.6) and (12.6.7) E = Re[ix ( -f30'0) + i)'( ± 0'0 )]e'Fillllei(wt-k.,:c) (8) 2Eok:c 2Eo (c) As in Problem 12.2.1, a plot of a divergence-free field can be done by defining a potential. and obtaining the field (9) Now, it is clear that the potential necessary to produce (8) is 12-11 Solutions to Chapter 12 Then • ~;o,. • 8q, • 8q, -I" X v 'li' = Ix 8y -I)' 8x and is found to be equal to (8) with f(x, y) equal to unity. By visualizing the potential, one may plot E lines. ky imaginary: H-lines E-lines lines of equal height of ~ Figure S12.6.1a At wt = 0, the potential is ky real: E-line ../ H-line L Figure S12.6.1b x Solutions to Chapter 12 12-12 At wt = 0, the potential is 12.6.2 (a) The E-field will be z-directed, the H-field is inthez-yplane t. = Asin(kzz)e'fi1c1l1l (1) From (12.6.29) I'r 1 8E. 1( ·k)A· kn z = --.---= --.-T' SIB zZ (2) 'WIJ 8y ,wIJ II The discontinuity of tangential H gives: D X (DB-Db) =K (3) in z -z plane. And thus, combining (2) and (3) (4) and therefore A=_wIJK o (5)2lell From (2) and (5) (6) and from (12.6.30) II II = ,.lek z K2 o cos(kz z)e'fi1c1l1l (7) II (b) Again we can use a potential ~ to which the H lines are lines of equal height. IT we postulate Then ~ = (~) Ko sink ze'fi1c1l1likll 2 z • VA;. • 8~ Ko kz k -I. X '* = Ix 8y T ik cos zZ ll • 8~ • Ko • k • Ko kz -I)" 8z = TlxT SIB zZ -I)" T ik cos ll The potential hill at wt =0 is Re[~J = T~o sin kzz sin klly (8) (9) k zZ (10) Solutions toChapter12 wt=0 o00 ooH-Iine E-Iine12-13 o o (c)Wemaywrite(1)o00 FigureSn.6.2 andfor(6)and(7) :H:=iKo{±ix(eik.."'=fikYI/_e-ik.."'=fikyl/) 4 +:'"ill(eik.."'=fikYI/+e-ik.."'=fikyl/)} 1/(11) (12) 12.6.3 (a)AtfirstitisbesttofindthefieldEzduetoasinglecurrentsheetaty=O. Wehave From(12.6.29)(1) (2) 12-14 Solutions to Chapter 12 From the boundary condition (3) we get 2LAe-:iksf/ll = _Ke-:iksf/ll WI-' and thus A =_ wI-'K (4) 2fJ Now we can add the fields due to each source (5) (b) When (6) Then K b = -Kae-j(ltl (7) there is cancellation at 11 < -d/2 (c) (8) (d) In order to produce maximum radiation we want the endfire array situation of fJd = 'If/2. (Indeed, sin fJd = 1 in this case.) Because (9) we have 1 [ ]1/2 w=-Viii ~_(~)2 (10) f/Il 2d The direction is Solutions to Chapter 12 12-15 12.6.4- (a) If we want cancellations, we again want (compare P12.6.3) Ub = -ua.e-;klld (1) (b) A single sheet at y = d/2 gives H. = ±Ae-;k ..ze'F;k ll(lI-t) (2) Now, akz .,. --+JW(T=O (3)az gives k z= kW,.(Ta. (4) z and 2h;I II=0+ = ~ (Ta (5) Therefore A = 2kW,. (To. (6) z and the field of both sheets is H. =j~ua.e-;k"Ze-;kll(lI+t) sinkzd (7)kz (c) klld = 11'/2. Therefore, as in P12.6.3, W = _1_[k 2 _ (.!.)2] 1/2 ..;iiE z 2d (8) 12.7 ELECTRODYNAMIC FIELDS IN THE PRESENCE OF PERFECT CONDUCTORS 12.7.1 The field of the antenna is that of a current distribution Icoskzl. We may treat it in terms of an array factor of three antennae spaced >../2 apart along the z-axis. From 12.4.12 3 l,pa(O)1 = 1L:e'k t COB91 = 11+eiJrcoB9 +e2;JfCOB91 ,=0 (1) = le-;JfCOB9 + 1 + e;JfCoB91 =1+ 2cos(11' cos 0) The function ,polO) follows from 12.4.8 with kl =11' ,polO) = ! cos (~cos 0)/ sin 0 (2) 11' 2 Combining (1) and (2) we complete the proof. Solutions to Chapter 12 12-16 12.7.2 The current distribution, with image, is proportional to Isin kz I. The point at which the current is fed into the antenna calls for sero current. Since the radiated power is finite, Rrad is infinite. In practice, because of the finite losses, it is not infinite but much larger than VIl-o/Eo. 12.7.3 (a) We have a surface current kz akz ... az + 1WO'. = 0 (1) Therefore .. jw. (1l"Z)Kz = -TO'o SID - (2) 'II'" a a The H-field is z-directed and antisymmetric with respect to y. .. ('II'"z) '/I:HM = ±Asin -. e~' IIY (3)a From the boundary condition n x (ila -ilb) = it. (4) with n II ir- A . (z) jw . (z)2 SID - = --O'oSID - (5)a 'II'"/a a jwA= ---0' (6)2'11'"/a 0 The E-field is from (12.6.6) t 1 all. 1( iw ) . .(z) '/I:z = -.--- = ±-,- ---0'0 (=f1k ) SID -e~' IIY 1WEo ay 1WEo 2'11'"/a Y a jkyO'o . ('II'"z) ~j/l: y (7) = Sln-e II Eo(2'11'"/a) a and from 12.6.7 E 1 all. ( 1)( iWO'o) 'II'" ('ll") '/I:y = --.--- = --,- =f-- -cos -z e~' II" 1WEo az 1WEo 2('II'"/a) a a (8) 0'0 () '/I: =±-cos -z e~' II" 2100 a (b) On the plate at z = -a/2 ... t I jk"O'o ~'/I: 0'. =Eo z z=-a/2 =-2'11'"/a e' II" (9) Solutions toChapter12 Atz=a/2itisofopposite sign.Thesurfacecurrentis ~IiI-iwuoTileyAy=-II11.=-0./2 -±-/-e 1/2'11"a andisthenegative ofthatatz=a/2. (c) k2k2 2 II.+y=W/-&oEo andthus12-17 (10) (11) (12) ky=VW2/-&oEo_{~)2 Againwemayidentifyapotential whoselinesofequalheightgiveE.Indeed, (13) gives (14) (d)Forkgimaginary andwt=0 wt=0 wt=rr/2 displacement currentdensity Flsure812.7'.3. 12-18 Solutions to Chapter 12 For kv real, wt =0 Re[.J ==f sin cr:l:) cos kv!lto/) Eo 211'a a wt =0 wt = 71"/7. disphu:ement convection current displacement ftux \Ins Figure S12.f.lb 12.1.4 (a) We now have a TE field with (1) From (12.6.29) 18E16 1 (. (1I':I:)'F'/o Iia: =--.---=--.-=f1k )Acos -e IIV:J 1w~ 8y 1w~ 11 a (2) =± k1l A cos (11':1:) e'Fj/ollv w~ a Solutions toChapter12 andtheboundary condition weobtainrelation forA: k" '11":& '11":&-2-Acos -=Kocos-wp a a12-19 (3) (4) or andthus From(12.6.60)A=_wpKo 2k"(5) (6) (7) (b)SincetheE-fieldisz-directed, itvanishes atthewallsandthereisnosurface chargedensity.Onwallat:&=-a/2 andthusFlsureSlJ.T.(a k-.'II"/aK'fi"~" II-32koe "Ontheotherwall,thecurrentisopposite.(8) (9) (c) k"=JW2poEo-('II"/a)2 (10) sincekill:='II"/a.Againwehaveapotential~, thelinesofequalheightofwhich giveB. 1Ko('11":&)",""~=--cos -e"~"jk"2a(11) 12-20 (d)Forkyimaginary: wt=0Solutions toChapter12 wt=1r/2 ,_--..--H-lield o~---r- E-field 00G x 000 o FigureSU.f.4b forkyreal:o Eo0~000 o 00(;) (;)00 o FigureSU.f.4.:Hwt=0 SOLUTIONS TO CHAPTER 13 13.1 INTRODUCTION TO TEM WAVES 13.1.1 (a) From (13.1.3): aEz [ () (O)J aHaay = f3Re A cos f3y exp 3wt = Wat: (1) = f3IAI cosf3ycos(wt +~) where ~ is the phase angle of A. Integrating the above yields H. = LIAlcosf3ysin(wt +~) = -Re jLAcosf3ye;wt (2) w~· w~ Introducing (2) and the expression for Ez into (13.1.2) gives -f32 IAI sin f3y sin(wt + ~) = -wflAI sin f3y sin(wt + ~) (3) w~ from which the dispersion relation follows f32 = W2~f. (b) From (13.1.13) This gives, using (2), (4) and thus A= _j w~/(o = -j/(o. ~_1_ (5)f3 cos f3b V-; cos f3b Using (2) we find - -R K'" cos f3y ;wtH•- e 0 cos",Qb e (6) and putting the value of A from (5) into the expression for Ez gives = -R oK f;~ sin f3y ;wtEz e3 0 f cos", Qb e (7) 1 Solutions to Chapter 13 13.1.2 (a) The standing wave H. = Re A sin {iyeiwt satisfies the boundary conditions of sero H. at y =O. From (13.1.2) aHa . t aE:I'--={iRe A cos {iye'W =E-- (1) ay at Integrating to find Ell' gives E:I' = -!!...Re jA cos {iyeiwt (2) WE From (13.1.3) we find aE:I' = {i2 Re jAsin{iyeiwt =J.& aHa =wJ.&Re jA sin {iyeiwt (3) ay WE at and thus (i2 = W2 J.&E (4) (b) Turning to the boundary conditions, E:I'(-b, t) = Re ~deiwt /a (5) and thus from (2) -!!...Re jAcos{ibeiwt = Re ~deiwt /a (6) WE and hence A-.WEVd _1__ . ~Vd_1_ (7)-3 (i a cos{ib -3yp. a cos{ib We find 13.1.3 Using the identity (1) one finds from (13.1.17) l'liIE:I' = "'!¥ 1 • -Rej.n. o -----;(e'''I- e-'''")e'Wiii 't Ecos{ib 23 (2) = -Re !Ko ~[ei(wt-{JII) -e-i(wHfJlI)J/cos{ib2 y; The exponentials in the brackets represent waves that retain constant amplitude when dy = ±idtexhibiting the (phase) velocities ±w/{i = ±1/..,fiii. 13-3 Solutions to Chapter 13 -L/ 13.1.4 (a) The EQS potential in a coax is a solution of Laplace's equation. The field with rotational symmetry is ~= Aln­r (I)a satisfying ~ = 0 on outer conductor of radius a. The field is z-independent with a constant potential difference. The potential difference is Aln(b/a} =V (2) The field is E= -V~ = -i.. :rA1n(r/a} = -i.. ~ = i.. rln~/b} (3) (b) The field has cylindrical symmetry with field-lines parallel to ill>. The potential "\Ii' is (4) The H field is (5) Ampere's integral law gives !H.dS= fJ·da=I (6) Since H is z independent, I = constant and at z = -I A --211"r = -211"A = I (7)r Therefore (8) (c) The preceding analysis suggests that E=i .. V(z,t} (9a)In(a/b}r and (9b) can be solutions of Maxwell's equations. To show this it is advantageous to separate the V operator into (10) Solutions to Chapter 13 13-4 where T"7 • a 1. a vT = I r ­+ -1",­ar r a</> is the transverse part of the operator. Then (ll) v X E = VT X E +i. X :zE (12) Now VT differentiates only rand </>. The EQS field, which is z independent, has VT X E = O. Hence we conclude that the same holds for the "Ansatz" (9). But i. X ir = i", and i. xi", = -ir . We obtain from Faraday's law 1 !~V=_JL_l_a1 In{a/b) r aa 21Tr at (13) The common r-dependence can be eliminated, and we find (14) where L = JLln{b/a) 21T A similar reasoning applied to V X H and Ampere's law yields (15) • 1 a1. € av -Ir =Ir -21Tr az In{b/a)r at -- (16) or with a1 = _cay az at c = 21T€ In{b/a) (17) (18) V 13.1.5 (a) With the time dependence exp iwt, we get for the transmission line equations of (14) and (17) of Prob. 13.1.4 dV dz = -iwLJ (I) where dJ A -= -iwCVdz v = Re Veiwt (2) 13-5 Solutions to Chapter 13 and 1 = Re ieiwt Eliminating V from (1) and (2) one obtains tPV . di 2-=-3wL- =-w LCVA (3)dz2 dz with the solutions (4) with fi =w";LC (5) We pick the solution v= Asinfiz (6) because the short forces V to be zero at z = O. From (1) we find i dV ifi1A =-- = -Acosfiz (7)wLdz wL and since 1 = Re 10 eiwt at z = -I, wL Acosfil = -i 1 (8)fi 0 or A=-iVL/C~ (9)cos PI where we used (5). We find for the current and voltage as functions of z and t: . t 101(z, t) = Re-- cos pze'w (10)cosfil V(z, t) = -Re iVL/C10 sin~z,eiwt (11) cOSfJ (b) At low frequencies cosfiz!::::! 1 for all -I < z < 0 and sinfiz !::::! fiz = w";LCz. Using (9) of the preceding problem, (12) For the E-field we find from the preceding problem and (11) above . R' L 1 iwt _. R. JJ 1 iwt z oeE - --II' e3w In(a/b)r --II' e 3w 211" z oe (13) 13-6 Thisgivesthevoltageatz=-ISolutions toChapter13 (14) Theinductance isLlbecauseL,asdefinedhere,istheinductance perunit length.Thuswehaveshownthat,inthelimitoflowfrequencies, thestructure behaves asasingle-turn inductor. (c)TheH-fieldinthespacebetween theconductors isthegradient ofapotential 'IiextPthatisasolution ofLaplace's equation. Thus, 10 •tH=Re-ie'w21fr.p WeobtainEfromFaraday's law VEpaHR'10•iwtX=---=-pe:Jw-1.peat 21rr(15) (16) c I f--------.-_.-t-- z«0) z=o FllPlreSII.1.5 Withthelineintegral alongthecontourCshownFig.S13.1.5, wemayfindfrom theintegralformofFaraday's law (17) Integrals overtheradialcoordinate appearonbothsides.Thus,comparing the integrands wefind whichisthesameas(13).ERjwp10 iwtr=- e~ze (18) 13-7 Solutions to Chapter 13 ~ 13.1.6 (a) From the solutions (4) in Prob. 13.1.5 we pick the cospz dependence, because the magnetic field, proportional to 1, is zero at z = 0 according to (7) of the same problem. Indeed, ifV = A cos pz, then j eW jP.1A =-- = --Asmpz (1)wLdz wL Since Re[Acospzexpjwt] ..=_1 = ReIVoexpjwt] (2) we find A=~ (3) cos Pi and 1=-jyC/L V°/.llsinPz (4) cos~ Therefore, V(z, t) = Re [c:opl cospzexPjwt] (5) 1(z, t) = -Re jyC/L V°/.l sin (3zeiwt (6) cos~l In(r/a) • V(z, t) 1 E= V(z, t)VT In(a/b) = I"ln(a/b) r (7) where VT is the transverse gradient operator, • a . 1 aVT = 1..-+1",-­ar r a4J and we use the result of Prob. 13.1.4. In a similar vein (8) (b) At low frequencies, cos (3z !:::! I, sin (3z !:::! (3z and V(z, t) !:::! Re Voexp jwt. Then, assuming Vo to be real, i.. 1 () E= In(a/b) r Vocoswt H= ;:ryC/L(3zVosinwt =i"'rln'(:/b)zVosinwt (10) (c) At low frequencies, using EQS directly i.. 1 E = In(a/b) r Vocoswt (11) 9 13-8 Solutions to Chapter 13 namely the gradient of a Laplacian potential ex In(r/a). The H-field follows from aEVXH=f­ (12)at with A H = 14>-z (13) r introduced into (12) • a H I A II" 1V. .VxH = -II" az 4> =-1"-;:- = -WE In(a/b) r oSlnwt and therefore A = In(:/b) Vosinwt (14) which gives the same result as (10). 13.2 TWO-DIMENSIONAL MODES BETWEEN PARALLEL­ PLATES 13.2.1 We can write mr 1( .n", .n'll" )cos-:r; = -exp3-:r;+exP-3-:r;a 2 a a and • n'll" 1 ( .n", .n'll" )Sln-:C= --; exp3-:r;-exp-3-:r;a 23 a a Introducing these expressions into (13.2.19)-(13.2.20) we find four terms of the form 'Q .n'll" '(Q n'll" )exp =f3fJnyexp =f3 -:r; = exp =f3 fJnY ±-:r; =exp-jk .r a a where k n'll". Q • = ±-I x ±fJnl~ a and r = Ix:r;+I~y This proves the assertion that the solution consists of four waves of the stated nature. These waves are phased so as to yield :r;-dependences of the form cos n: :r; and sin nat!' :c to satisfy the boundary conditions. 13-9 Solutions to Chapter 13 13.2.2 We can start with the solutions (13.2.19) and (13.2.20) shifting z so that , a :1;=:1;-­2 Considering TM modes first we note that n", .(n",z' n",)Hz ex cos-z = cos --+­a a 2 n"':I;' n", . n"':I;' . n",=cos--cos- -sm--sm­a 2 a 2 ={ (-1)~~ cos (m::) n even (-1)-~- sin (,,~S) n odd We see that the modes with even n are even with respect to the symmetry plane of the guide, the modes with n-odd are odd. Next studying the TE-modes, . n", . (n",z') n", (n",z') . n",Ezexsm-z=sm -- cos-+cos -- sm­a a 2 a 2 = {(-1):'~1 sin~, n even (-1)-~- cos "~s n odd We find that Ez is even for n odd, odd for n even. (a) When z, = ±a/2 and the modes are odd, Hz = (_1)(,,-1)/2 sin "2ft , Ez = (-1)"/2 sin "2ft ; in the first case n is odd and Hz is an extremum at z' = ±a/2, and in the second case n is even and Ez is zero at both boundaries. (b) When z, = ±a/2 and the modes are even then Hz = (-I)"/2cos(;ft) and Ez = (_1)("-1)/2 cos ;'11' we see that both boundary conditions are in both cases, because n is odd in the first case and Hz is an extrenum, n is even in the second case, and Ez is zero. 13.3 TE AND TM STANDING WAVES BETWEEN PARALLEL PLATES (1) 13.3.1 Solutions to Chapter 13 where we have integrated by parts. Because dh.n/dz = 0at z=0 and z= a, the integral of the integrand containing the total derivative vanishes. Next take the complex conjugate of (13.3.1) applied to h.m multiply by 'h.n and integrate. The result is 13-10 (2) Subtraction of (1) and (2) gives Thus (G A. '" 10 h.mh.ndz = 0 when p~ -=I p~ and orthogonality is proven. The steps involving 2.n are identical. The only difference is that la d ~("* de.n)zd e.m do z z vanishes because 2:m vanishes at z = 0 and z = a. 13.3.2 (a) The charge in the bottom plate is q = lUI 1(a+A)/2 EElIdzdz (1) o (a-A)/2 Using (13.3.15) ""' 4mrE f) 1 ;wt] 2wa ( )!!=.!.. n",~ q= Re [ LJ -------e ---1 sm-- (2) 3 ..=1 a fJn a sinfJnb n", 2a 044 lwhere we have used the fact that UI j(a+A)/2 mr wa [ mr a + f1 mr a -b.] sin (-z)dzdz = -- cos(---) -cos(---) o (a-A)/2 a n", a 2 a 2 wa . n", . n",f1 = 2-sm-sm-­n", 2 2a _ 2wa ( 1).!!;.l . mrb. --- - ~ sm-­ nll" 2a (3) Va = -Re iwqejwtR ( 1) !!=.!. • n!tA. ] (4)= -Re iw8EWRv L - 3 s~ 2a ejwt[ fJn a sm fJn b n Solutions toChapter13 WhenP",b=1rwehavearesonance. Now P",=JW2~E_(n;)2 Theresonance frequency ofthen-thmodeoccursat p",!a=J4w2~Ea2 -(n1r)2=1r a or w..jiii.a=J(n2+1)~213-11 (5) (6) (7) (b)(b)Forn=1thisisat1r.Thenextmoderesonates atV5f41r. Thus,inthis range,tworesonances occurforwhichtheresponse goestoinfinity.Ofcourse, inthislimit,losseshavetobetakenintoaccount whichwillmaintain the response finite.Thelowfrequency limitiswhen· 1rw..jiii.<:n­ a Then R•Rbn1r.hn1rb f''''Slnf'''' -+--sm-aa and _R[.R"~(-1)!!jl sin~jwt] Va-e3w81rEW VL..J MsinhMbe '" G G (c)From(13.3.13), whenonlyonemodepredominates, HR[4iWEfJcosp",yn1r]jwt .!:::!e-R-- .RbCos- ef'",asmf'",a wheren=1atwy'iifa=1randn=2atwy'iifa=V5f41r. Togeta finiteanswer,weneedv/sinp",btoremainfiniteastheresonance frequency isapproached. 13.3.3 (a)H.atx=0andx=agivesthesurfacecurrents inthebottomandtop electrodes. Because thevoltagesourcespushcurrents intothestructure in opposite directions, thesurfacecurrents, andH.,havetovanishatthesym­ metryplane. Thex-component oftheEfieldcanbefounddirectlyfrom(13.3.14), replac­ ingthesinp",y/sinp",b bycosp",y/cosP",b totakeintoaccount thechanged symmetry ofthefield ER[~4vcosp",y n1r],·wi II:=eL..J-- cos-xe,,_1acosP",b a odd Because8H./8y=iWEEswefind H-R[~4jWEfJsinP",yn1rJjwt •-eL..JR Rbcos"'Je,,=1f'",acosf''''a "odd Solutions to Chapter 13 13-12 13.3.4 (a) The flux linkage>. is (1) and the voltage is dH. Va = IJA--;It (2) (b) From (13.3.13) we find that liI.1 is a maximum for :& = 0 and :& = a. (c) From the detailed expression (13.3.13), using (2) _ -R [~4W2IJEUA_1_ n1rX] iwt Va - e L.J Q • Q bcos e "=1 IJn a sIn IJn a "odd (d) The loop should lie in the 11 -z plane. Then it links Hz that is tangential to the bottom plate. 13.3.5 The Ez field is derivable from a potential that is a square wave as shown in Fig. S13.3.5. We have (1) v -,,---:Or----'--­ x=o J \ x=a /I-II tl4- x=T X - 2 FIKure SIS.S.1 and using orthogonality, multiplication of both sides by sin n; :& and integration gies U4 a • n1r av [ (n1r a +d n1r a -d)] 1 2-An = V sIn-zdz =-- cos ---) -cos(--­ 2 ~ a n1r a 2 a 2 2 av . (n1r) . (n1rd) = 2-SIn - SIn ­n1r 2 2a We find 4v. (n1r) . (n1rd) An = -SIn - SIn ­ n1l' 2 2a We may adapt (13.3.13)-(13.3.15) for this case by replacing 4v/n1r by . (n1r) . n1rd4v/ n1r SIn "2 sm 2a Solutions to Chapter 13 13-13 ~ 4jwdj • (n1r) . (n1r d) C08 fJnY n1r] iwtH• = Re [ L..J -- sm - sm - cos-z e .._1 fJn a 2 2a sin fJnb a odd ~ 4v. (n1r) . (n1rd)sinfJnY n1r] iwtE:I: = Re L..J -- sm - sm-- cos-z e[ .._1 a 2 2a sin fJnb a odd ~ 4n1r . (n1r) . (n1rd) cos fJnY . n1r ] iwtE = Re L..J --sm - sm-- sm-z e y [ ..=1 a 2 2a sinfJnb a odd 13.3.6 In (13.3.5) we recognized that E. at Y = b must be the derivative of a po­ tential that is a square wave. This, of course, is equivalent to the statement that E. possesses two impulse functions. In a similar manner, Hy can be considered the derivative of a flux function f: pHydz. Note the analogy between (13.3.19) and (13.3.14). We may, therefore, adapt the expansion of P13.3.5 to this problem, be­ cause the flux function of Example 13.3.2 is the same as the potential of example 13.3.1. From (13.3.17)-(13.3.19): ~ 4jAw. (n1r) . (n1rd) sinfJmY . m1r ] iwtE •= Re [ L..J ---sm - sm -- sm-z e m=l m1r 2 2a sin fJmb a odd ~ 4fJm A . (n1r) . (n1rd)cos fJmY . m1r ] ,·wtH•= Re L..J -- sm - sm - sm-z e[ m=l pm1r 2 2a sin fJmb a odd ~ 4A. (n1r) . (n1r d) sinfJmY m7l'] ,·wt1l = Re [ L..J --sm - sm - cos-z e y m=l pa 2 2a sinfJmb a odd 13.4 RECTANGULAR WAVEGUIDE MODES 13.4..1 The loop in the Y -z plane produces H-field lines along the z-direction. IT placed in the center of the waveguide, at z = a/2, these field lines have the same symmetry as those of the TEIO mode and thus excite this mode. The detection loop links these fields lines as well Of course, the position of tl.~ exciting loop must be displaced along Y by one quarter wavelength compared to the capacitive probe for maximum excitation. 13-14 Solutions to Chapter 13 13.4.2 The cutoff frequencies are given by The dominant mode has n =0 and thus it has the (lowest) cutoff frequency :cIm=l,n=o = (;) The higher order modes have cutoff frequencies The cuttoff frequencies are in the ratio to that of the dominant mode: TEol 1.33 TEll and TMll 1.66 T~o 2.0 T~l and TM~u 2.4 13.4.3 (a) TM-modes have all three E-field components. They approach the quasistatic fields of Ex. 5.10.1 which imposes the same boundary conditions as this exam­ ple. Hence the modes are TM. From (9) we find that ez oc -jle,/;: = ;;~~ .. ·Ie ~ ,,3al! S· ~ d ~ . h to;' tand e. oc -1 11 ". = "11 .' mce Gz an e. must van18 a y= ,e" mus behave as a cosine function of y, so that 2z and 2. are sine functions of y. Therefore, E = Re '""'(A+ e-:ilJm...1I + A- eilJm...") sin ~:z:sin ~ze·;wt" L.J L.J mn mn a w m n (1) = Re EE2A~ncosPmnysin m1r :z:sin~:z:sin~zeiwt a a w m n where (2) From (13.4.9): Ez = Re "'''' -jPmn(7) (A+ e-ilJm ..." -A- eilJm ...")L.J L.J W2uE _ .02 mn mn m n ,.. f'mn cos ~:z:sin ~zeiwt (3) a w _ R '"'" fJmn 7- 2A+ . R m1r. n1r iwt-e L.JL.J 2 R2 mnsmf'mnycos-:z:sm-ze m n W I"E -f'mn aw Solutions toChapter13 andsimilarly, R""""13m",~A+.II •m7f n1l"jwtE.=eL..JL..J 2 1122m",smpm",ysm-zcos-ze m'"W/Sf-Pm", a w13-15 (4) (b)Aty=b,Esasafunction ofzmustpossesstwoequalandopposite unit impulsefunctions ofcontentv(t)/atogivethepropervoltagedropatthe edges.Theintegral ofEs,-f;Esdzmustbeasquarewavefunction of amplitude v.ThesameholdsWithregardtotheintegralofE.withrespect toz.Insummary, EsandE.aty=bmustbederivable fromapotential thatisatwo-dimensional squarewavewiththeFourierexpansion (5.10.15) (coIQ.pare 5.10.11): 0000 4>(z,y)=ReLL __1,,==1 modd ,,"odd x=o16v.m7f•n1l",·wt--sm-zsm-zemn1l"2 aw impulse function(5) impulse function Thus,aty=b­X F1cure913.4.3x=a Comparison with(3)gives m1l"/(22). 16vm1l"2Am",13m",- W/Sf-13m",sm13m",b =--2-a mn1l"a(7) formandnodd.Thisgivesthequotedresult.Ananalogous relation maybe obtained forE.whichyieldsthesameresult. (c)Theamplitudes gotoinfinitywhensin13m",b=0or or 13.4.413-16 (d)Solutions toChapter13 wVjii.a=1rJm2+(:n)2+(ip)2 Wehavealreadyusedthefactthatthedistribution ofEsandE.intheY=b planeisthesameasinthequasistatic case.Theonlydifference liesinthe y-dependence which,forlowfrequencies givesthepropagation constant andispureimaginary. TheEQSsolutionaccording to(5.10.11) and(5.10.15) IS .".=Re~~16&sinhkmny.m1r•n1riwt "Ii!L.JL.J--2 .hkbSin-ZSIn-ze m=l..=1mn1rsinmn aw oddodd andgivesforEs: E•-_-Re~~16&(m1r)sinhkmnY m1r. n1riwt- L.JL.J--- cos-zsm-ze 11mn1r2asinhkmnb awm. . ",odd odd Thisisthesameexpre88ion astheEQSresult. z=w a-tl2a+tl2w+~ 2 w-~ 2 ___ ...l--_..L-_.L...- __-'-_ x=o %=0x=4/2x=a FlsureSIS.4.4. Theexcitation produces aH,,"ItlookslikeTE-modes aregoingtosatisfyallthe boundary conditions. H"mustbezeroatY=0andthusfrom(25)oftext 0000 H"=ReLL(O~ne-ilfm .."+O';neilfm ..")cosC:1rz)cosC::z)eiwt m=On=O =-ReLL2iO~nsinPmnYcos(~z)cos(~z)eiwt m=On=O a w(1) AtY=bwemustrepresent thetwodimensional square-wave inthez-direction and inFig.S13.4.4b inthez-direction asshowninFigS13.4.4c. Solutions toChapter13 13-17 a-d-2- z=o w-d-2-z=w x=o a/2 x=a FigureSIS.4.4b Wehave,settingFigureSIS.4.4e (2) 1w14",'" m1l" (n1r(P1l") (q1l" 1 L-L-Amncos(-X)cos -z)cos-xcos-z)dxdz= --Apq(aw)00 a w a w 4 w+oIil. A12i2p1l" q1l"=-Hodzdxcos(-x)cos-z J!!:A ..-I! a w 2 2 w±oIil. ~1212p1l" q1l"+Hodz dxcos(-x)cos-z .!!!=A A a w2 2 Ho[.(P1l"a).(P1l"a-d)]=-(7)(~)sm72-sm7-2- [.(q1l"W+A).(q1l"W-A)]sm--- -sm---w2 w2 Ho[.p1l"a+d.p1l"a] +(7)(~) sm7-2--sm72 [.(q1l"w+A) .(q1l"w-A)sIn--- -sm--- w2 w2 Ho[.(P1l"a)] =-(7)(~) sm72 .(P1I"a-d).(P1l"a+d).(P1l"a]-sIn--- -sm--- +sm--)a2 a2 a2 [.q1l"w+A .q1l"W-A]sm----sm---w2 w2 Ho[.(P1l").(P1l")(P1l")] q1l".q1l"A=-()("1r)2sm--2sm-cos-A2cos-sm-- P!! .0.::.. 2 22a 22w 4w 4Ho•(P1I")(q1l").q1l"A[P1I"]=-(p~)(~) sm"'2COS"'2sm2w1-cos2aA (3) WefindthatPmustbeoddandqmustbeevenforafiniteamplitude toresult. A=Ho(-l)P-l(-l)f- l[l-cosp1I"A]pqpq1l"2 2a(4) 13-18 Solutions to Chapter 13 The case q = 0 must be handled separately. 1 () Ho [• (P1f' a) . (P1f' a -d)] -A 0 aw =--- sm-- -sm--- w 2 p (p~) a 2 a 2 H0 [ • P1f'( a+ d) . (P1f' a)] +-- sm--- -sm --w (5)(7) a 2 a 2 2wHo [ P1f'] . P1f' =-(p~) 1-cos 2a sm"2 and thus Apo = Ho [1- cos P1f' .6](-1)P-l P1f' 2a From (13.4.7) and (13.4.8), one finds '"'" 2;C;:;'nPmn (m:) . m1f' n1f' ;wtHz = Re LJLJ 2 p2 cospmnysm-xcos-ze (7) mn W JJ.€ -mn a w (8) with C;:;'n expressed in terms of the Amn's by (2) 13.5 DIELECTRIC WAVEGUIDES: OPTICAL FIBERS 13.5.1 (a) To get an odd function of x for e", one uses the Ansatz Ae-a",(z-d) d<x<oo e - Asink",z -d < x < d (1)'" -{ sink",d _Aea",(z+d) -00 < x <-d which has been adjusted so that e", is continuous at x = ±d. Since (2) and thus -azAe-a",(z-d)k -_1_ kA cosk ..z (3) l/-. z sink",d 'WJJ. { -azAea",(z+d) kl/ and e", are continuous at x = d. The continuity of e", has already been established. From the continuity of hl/: (4) Solutions toChapter13 13-19 Thecutoffsareatk",d=(2n-1H(seeFig.S13.5.1a). I I I I I---1-._ k",d (2n-1)11" 211"kd--+---- '" 2 FigureSI3.6.la (c)Whenaccording to13.5.3 k",d=JW2IJfi-k~=(2n-1)~ andwgoestoinfinity,thenkllmustapproachw.,fiifiasymptotically. (d)SeeFig.S13.5.1h (Fig.6.4fromWavesandFieldsinOptoelectronics, H.A.Haus,Prentice-Hall, 1984). Asymptote ~..w\I~ 1.0 0.5 2.-T·wv~•••---Asymptote tJ=(oJ",;;t; 0.05010.0 1.... FigureSI3.5.lb 13-20 Solutions to Chapter 13 13.5.2 The antisymmetric mode comes in when Q", 'II" -=0 and k",d= ­k", 2 and from (13.5.8) or or 1 '11"/2 [E;c . ~ '11"/2 w = VJifid V(l--:J = V7"dV -;; ";1- E/Ei 8 =3 X 10 _1_ '11"/2 = 3.85 X 1010 10-2 yI2.5. /1 ---LV 2.5 f = .!!!.... = 6.1 X 109 Hz 211" 13.5.3 (a) For TE modes Ae-a.,(",-d) x>d e-A cos k.,'" or A sin k a'" -d < x < d'" -{ cos k.,d sin k.,d Aea.,(",+d) or -Aea.,(",+d) x< -d where we have allowed for symmetric and antisymmetric modes. Continuity of ellS has been assured on both boundaries. The magnetic field follows from h = _I_de", (2) y jwJ1. dx and thus _.!!.£ Ae-a.,(",-d) x>d ~ 1 k"'· kh =- -='A~ or -d < x < d (3)Y jw "'i cos k.,d{ .!!.£Aea.,("'+d) or x< -d '" Continuity of hy gives k", -Q", = -tank",d (4a) J1. J1.i Solutions toChapter13 forevenmodes,and foroddmodes.Here andthus,eliminating k",as=J1c'tJ-W21-'E kIlO=JW2lJiEi -k~13·21 (4b) (5) (6) and (7) (b)Cutoffoccurswhenas/kIlO=0andksdisfixed.WefindthatwhenIJiis increased above1-',Wmustbelowered. (c)Theconstitutive law(a)forsymmetric modeshasthegraphicsolutionofFig. 13.5.2.Theonlychangeistheexpression foras/kIlObutitsksddependence isqualitatively thesameias/kIlOincreases whenIJi/I-'increases atconstant w.Thismeansthattheintersection pointmovestogreaterksdvalues. k~ increases directlywithincreasing1Ji/I-'according to(6)anddecreases with increasing kIlO'Theintersection pointofksddoesnotchangeasfast,inpartic­ ular,athighfrequencies itdoesnotmoveatall.Hence,thedirectdependence onIJipredominates, k"goesupandAdecreases. 13.5.4 (a)Thefieldsarenow :.r:>d -d<:.r:<d :.r:<-d(1) wherewehaveallowedforbothsymmetric andantisymmetric solutions. Cointi­ nuityofh.hasbeenasuredonbothboundaries. Further, Since A1dhae=--- "iWEd:.r:(2) (3) (4) 13-22 Solutions to Chapter 13 _!!a.Ae-a,.(.-d) 1 E2 = --_!..A a1n 1l:.. or !..coa1l:•• (5)" iw E; coa1l:,.d E; aln1l:,.d{ ~Aea,.(.+d) or _ !!a.Aea,.(.+d) E Continuity of 2" at z = ±d gives (6a) for even modes, and (6b) for odd modes. Further, (7) Thus (8) and (9) (b) The cutoff frequencies are determined by k.d = m~ and a. =o. From (9) or SOLUTIONS TO CHAPTER 14 14.1 DISTRIBUTED PARAMETER EQUIVALENTS AND MODELS 14.1.1 The fields are approximated as uniform in each of the dielectric regions. The integral of E between the electrodes must equal the applied voltage and D is con­ tinuous at the interface. Thus, (1) and it follows that VEa =-.,.------,-------:---:-:­ [a + b(€a/€b)] (2) so that the charge per unit length on the upper electrodes is (3) where C is the desired capacitance per unit length. Because the permeability of the region is uniform, H = I/w between the electrodes. Thus, >. = (a + b)J.LoH = LI; L == J.Lo(a + b)/w (4) where L is the inductance per unit length. Note that LC t J.L€ (which permittivity) unless €a = €b· 14.1.2 The currents at the node must sum to zero, with that through the inductor related to the voltage by V = Ldiconductor/ dt L a D.z at [I(z) -I(z + D.z)] = V (1) and C times the rate of change of the voltage drop across the capacitor must be equal to the current through the capacitor. C a D.z at [V(z) -V(z + D.z)] = I (2) In the limit where D.z -0, these become the given backward-wave transmission line equations. 14.2 TRANSVERSE ELECTROMAGNETIC WAVES 1 14-2 Solutions to Chapter 14 14.2.1 (a) From Ampere's integral law, (1.4.10), H", =1/21ff' (1) and the vector potential follows by integration H", = _-!:.. aaA• => A.(r) -A.(a) = -IJ201'n(r/a) (2) IJo r 1t' and evaluating the integration coefficient by using the boundary condition on A. on the outer conductor, where r = a. The electric field follows from Gauss' integral law, (1.3.13), Er = ).,f21t'Er (3) and the potential follows by integrating. Er =-alb => lb(r) :-lb(a) = ~'n(~) (4)ar 21t'E r Using the boundary condition at r = a then gives the potential. (b) The inductance per unit length follows from evaluation of (2) at the inner boundary. L ==! = A.(b) -A.(a) = IJo In(a/b) (5) 1 1 21t' Similarly, the capacitance per unit length follows from evaluating (5) at the inner boundary. 0== A, = 21t'E (6)V In(a/b) 14.2.2 The capacitance per unit length is as given in the solution to Prob. 4.7.5. The inductance per unit length follows by using (8.6.14), L = 1/0c2• 14.3 TRANSIENTS ON INFINITE TRANSMISSION LINES 14.3.1 (a) From the values of Land 0 given in Prob. 14.2.1, (14.3.12) gives Zo= ~ln(a/b)/21t' (b) From IJ = IJo = 41t' X 10-7 and E = 2.5Eo = (2.5)(8.8.5 X 10-12), Zo = (37.9)ln(a/b). Because the only effect of geometry is through the ratio alb and that is logarithmic, the range of characteristic impedances encoutered in practice for coaxial cables is relatively small, typically between 50 and 100 ohms. For example, for the four ratios of alb, Zo =26,87,175 and 262 Ohms, respectively. To make Zo = 1000 Ohms would require that alb = 2.9 x lOll! Solutions toChapter14 14-3 14.8.2 Thecharacteristic impedance isgivenby(14.3.13). Presuming thatwewill findthat1/R::>1,theexpression isapproximated by Zo=..;;[iln(21/ R)/fr andsolvedfor1/R. l/R=iexp[frZo/..;;[iJ =exp[fr(300)/377J Evaluation thengives1/R=6.1. 14.8.8 Thesolutionisanalogous tothatofExample 14.3.2andshowninthefigure. v I 14.3.4Figure814.8.8 From(14.3.18) and(14.3.19), itfollowsthat V::I:=Voexp(-z2/2a2)/2 Then,from(9)and(10) 1V=iVo{exp[-(z -ct)2/2a2J+exp[-(z+ct)2/2a2]} 14-4 Solutions to Chapter 14 14.3.5 In general, the voltage and current can be represented by (14.3.9) and (14.3.10). From these it follows that 14.3.6 By taking the ao/at and ao/az of the second equation in Prob. 14.1.2 and substituting it into the first, we obtain the partial differential equation that plays the role played by the wave equation for the conventional transmission line (1) Taking the required derivatives on the left amounts to combining (14.3.6). Thus, substitution of (14.3.3) into (I), gives By contrast with the wave-equation, this expression is not identically satisfied. Waves do not propagate on this line without dispersion. 14.4 TRANSIENTS ON BOUNDED TRANSMISSION LINES 14.4.1 When t = 0, the initial conditions on the line are I =0 for 0 < z < I From (14.4.4) and (14.4.5), it follows that for those characteristics originating on the t = 0 axis of the figure For those lines originating at z = I, it follows from (14.4.8) with RL = oo(rL = 1) that V_ =V+ Similarly, for those lines originating at z = 0, it follows from (14.4.10) with rg = 0 and Vg = 0 that V+ =0 Combining these invarlents in accordance with (14.1.1) and (14.1.2) at each location gives the (z, t) dependence of V and I shown in the figure. Solutions toChapter14 14-5 :II Vj~V. 1,=0 / V+=0" V I 14.4.2J=-V"/2Z,, Figure814.4.1 Whent=0,theinitialconditions onthelineare 1==V./2Z.!II V.==V./2..V=Oj .I==v./z.1=Vo/Zofor0<z<l Figure814.4.2 From(14.4.4)and(14.4.5),itfollowsthatforthosecharacteristics originating on 14-6 Solutions toChapter14 thet=0axisofthefigure V+=Vo/2;V_=-Vo/2 Forthoselinesoriginating atz="itfollowsfrom(14.4.8)withRL=0(rL=-1) that V_=-V+ Similarly, forthoselinesoriginating atz=0,itfollowsfrom(14.4.10) that V+=0 Combining theseinvarients inaccordance with(14.4.1)and(14.4.2)ateachlocation givesthe(z,t)dependence ofVandIshowninthefigure. 14.4.S Hthevoltageandcurrentonthelineareinitiallyzero,thenitfollowsfrom (14.4.5)thatV_=0onthosecharacteristic linesz+ct=constant thatoriginate onthet=0axis.BecauseRL=Zo,itfollowsfrom(14.4.8)thatV_=0forallof theotherlinesz+ct=constant, whichoriginate atz=,.Thus,atz=0,(14.4.1) and(14.4.2)become V=V+; I=V+/Z o andtheratiooftheseistheterminal relationV/1=Zo,therelationforaresistance equalinvaluetothecharacteristic impedance. Implicittothisequivalence isthe condition thattheinitialvoltageandcurrentonthelinebezero. 14.4.4 •t 21lc lie Figure814.4.4~-,....--~---....,.--'''--------- ..tz. Solutions toChapter14 14-7 Thesolutionisconstructed inthez-tplaneasshownbythefigure.Because theuppertransmission lineisbothterminated initscharacteristic impedance and freeofinitialconditions, itisequivalent toaresistance Raconnected totheter­ minalsofthelowerline(seeProb.14.4.3).ThevaluesofV+andV_followfrom (14.4.4)and(14.4.5)forthecharacteristic linesoriginating whent=0andfrom (14.4.8)and(14.4.10) forthoserespectively originating atz=Iandz=o. 14.4.5 Whent<0,asteadycurrentflowsaroundtheloopandtheinitialvoltage andcurrentdistribution areuniformoverthelengthofthetwoline-segments. v.,-RaVo•L'_Vo '-Ra+R b''-Ra+Rb Intheuppersegment, showninthefigure,itfollowsfrom(14.4.5)thatV_=O. Thus,fortheseparticular initialconditions, theuppersegment isequivalent toa termination onthelowersegment equaltoZa=Ra.Inthelowersegment, V+and V_originating onthezaxisfollowfromtheinitialconditions and(14.4.4)and (14.4.5)asbeingthevaluesgivenonthez-tdiagram. Theconditions relatingthe incidenttothe,reflected waves,givenrespectively by(14.4.8)and(14.4.10), arealso summarised inthediagram. Useof(14.4.1)tofindV(O,t)thengivesthefunction oftimeshownatthebottomofthefigure. ..tv..=01'_=0 / +­",\'=!:!(R.,-Ro).12(P....Ro) z. \'=!:!1R.,-Ro) -2(R.,.,.Ro) 1'.11",(t) __.!...---I- --:' ':"""'" .~t Id1'(0,1) : 1~~-.L~____ I ____ 1t=Jz. I'.fR.,-R.) TIR.,.,.R,}l/e 21fe \'.IR.-R.}' T(R.,-R,l' Figure81-&.4.5 14.4.6 From(14.4.4)and(14.4.5),itfollowsfromtheinitialconditions thatV+and V_arezeroonlinesoriginating onthet=0axis.ThevalueofV+onlinescoming 14-8 Solutions toChapter14 orfromthez=0axisisdetermined byrequiring thatthecurrents attheinput terminal sumtozero. ig (l/Zo-1/Rg) V+=(1/Rg+l/Zo)+V_(l/Zo+1/Rg) Itfollowsthatfor0<t<T,V+=loRu/2whileforT<t,V+=o.Atz=I, (14.4.8)showsthatV+=-V_.Thus,thesolutionisassummarized inthefigure. v;=-v+=o z~f::::::::t:=lt:=====::::;::=========:::P • i gV(o.t)t IoRg/2-q-J...,----2l-/e-ci:Jr--2l-/e-+-T--...t I~/2 l/e Fleur.814.4.8 14..4..7 (a)Byreplacing V+/Zo-+1+,V_/Zo-+-L,thegeneralsolutions givenby (14.4.1)and(14.4.2)arewrittenintermsofcurrentsratherthanvoltages. 1V=Y o(I+-L) (1) whereYo==l/Zo.Whent=0,theinitialconditions arezero,soonchar­ acteristic linesoriginating onthet=0axis,1+andLarezero.Atz=" itfollowsfrom(lb)that1+=L.Atz=0,summation ofcurrents atthe terminal gives ig=(GgjYo)(l+-L)+(1++L) which,solvedforthereflected waveintermsoftheincidentwavegives(2) (3) Solutions toChapter14 where14-9 (4) Feomtheserelations, thewavecomponents 1+andLareconstructed as summarized inthefigure.Thevoltageattheterminals ofthelineis (5) +.4.N=iDL tIQ(l+fQI:t,(ItIqtIql 2 3 Figure814.4.7' Itfollowsthatduringthissameinterval, theterminal currentis (rN -1 )1(0,t)=101-(1+;"o/Og)(6) (b)Intermsoftheterminal currentI,thecircuitequation forthelineinthelimit whereitbehavesasaninductor is ig=lLOg~~+I Solution ofthisexpression withig=10and1(0)=0is 1(0,t)=10(1-e-t/T);l'==lLOg(7) (8) 14-10 Solutions toChapter14 (c)InthelimitwhereGg/Yoisverylarge (9) Thus, I2Nl2(N-l)-<t<-c c(10) Following thesamearguments asgivenby(14.4.28)-(14.4.31), gives I I2(N-1)-<t<2N-c c(11) whichinthelimithere,(14.4.31) holdsthesameas(8)where(YofGg)(cfl) = .JC/L/h/LCG g=l/lLG g•Thus,thecurrentreponse(whichhasthesame stair-step dependence ontimeasfortheanalogous example represented by Fig.14.4.8)becomes theexponential response ofthecircuitinthelimitwhere theinductor takesalongtimeto"charge" compared tothetransit-time ofan electromagnetic wave. 14.4.8 :: lie-2J__f_v_o---1-7 _ / 1V(0,')(10)=V.=V.12 ~.v. oVo/21 -------hl:- 2lleJ Vo=(1-~e-{I-l/c}/r)z~ +Vg- Figure914.4.8 Theinitialconditions onthevoltageandcurrentarezeroanditfollowsfrom (14.4.4)and(14.4.5)thatV+andV_oncharacteristics originating onthet=0 Solutions to Chapter 14 14-11 axis are zero. It follows from (10) that on the lines originating on the z ~ 0 axis, V+ = Vo/2. Then, for 0 < t < llc, the incident V+ at z = I is zero and hence from the differential equation representing the load resistor and capacitor, it follows that V_ = 0 during this time as well. For llc < t, V+ = Vo/2 at z = I. In view of the steady state established while t < 0, the initial capacitor voltage is zero. Thus, the initial value of V_(I,O) is zero and the reflected wave is predicted by OL(RL + Zo) ~; +V_ = ~o U-l(t -llc) The appropriate solution is V_ (I, t) = iVo(l -e-lt-I/c)/")j 'f' == OL(RL +Zo) This establishes the wave incident at z = O. The solution is summarized in the figure. 14.5 TRANSMISSION LINES IN THE SINUSOIDAL STEADY STATE 14.5.1 From (14.5.20), for the load capacitor where ZL = IfjWOL, Y(,81 = -11"/2) Yo Yo Yo = YL = jwOL Thus, the impedance is inductive. For the load inductor where ZL = jwLL, (14.5.20) gives Z(,81 = -11"/2) Zo Zo = jwLL and the impedance is capacitive. 14.5.2 For the open circuit, ZL = 00 and from (14.5.13), r L = 1. The admittance at any other location is given by (14.5.10). Y(-l) 1-rLe-2#1I 1-e-2i~1 ---y;:-- = 1+rLe-2:i~1 = 1+ e-2:i~1 where characteristic admittance Yo = 11Zoo This expression reduces to Y(-l) -- =jtan,81Yo which is the same as the impedance for the shorted line, (14.5.17). Thus, wi~h the vertical axis the admittance normalized to the characteristic admittance, the frequency or length dependence is as shown by Fig. 14.5.2. 14-12 Solutions toChapter14 14.5.3 Thematched linerequiresthat9'_=0.Thus,from(14.5.5)and(14.5.6), v=v+exp(-j,8Z)j z Yo=~SiIlIP(~+1)1 I,=fff.silll.B(~+1)1 Figure814.1.1 Atz=-I,thecircuitisdescribed by Vg=i(-l)lig+V(-l) where,incomplex notation, Vg=Re9'gexp(.iwt),9'g==-jVo'Thus,forRg=Zo, andthegivensinusoidal steadystatesolutions follow. 14.5.4 Initially, boththecurrentandvoltagearesero.Withthesolutionwrittenas thesumofthesinusoidal steadystatesolutionfoundinProb.14.5.3andatransient solution,v=V.(z,t)+Vt(z,t)jI=I.(z,t)+It(z,t) theinitialconditions onthetransient partaretherefore, l't(z,O)=-V.(z,O)=~osinl,8(z+I)l Solutions to Chapter 14 14-13 It{z,O) = -I. (z, 0) = VZo sin[,B(z + l)] 2 ° The boundary conditions for 0 <t and with the given driving source are satisfied by V•. Thus, Yt must satisfy the boundary conditions that result if Vg = O. In terms of a transient solution written as 14.3.9 and 14.3.10, these are that V_ =0 at z = 0 and [from (14.5.10) with Vg = 0 and Rg = Zol that V+ = 0 at z = -l. Thus, the initial and boundary conditions for the transient part of the solution are as summarized in the figure. With the regions in the x -t plane denoted as shown in the figure, the voltage and current are therefore, V = V. +Vti I = I. + It where V. and I. are as given in Prob. 14.5.3 and with (from 14.3.18-19) V+ = ~o sin[,B(z + l)]; in regions I and III, and in regions II and IV. 14.6 REFLECTION COEFFICIENT REPRESENTATION OF TRANSMISSION LINES 14.6.1 The Smith chart solution is like the case of the Quarter-Wave Section exem­ plified using Fig. 14.6.3. The load is at r = 2, x = 2 on the chart. The impedance a quarater-wave toward the generator amounts to a constant radius clockwise rota­ tion of 1800 to the point where r = 0.25 and x = -jO.25. Evaluation of (14.6.20) checks this result, because it shows that .I 1 1 2-j2 r+ JX = . %=-1 rL +JXL 2+j2 8 14.6.2 From (14.6.3), f = 0.538 + jO.308 and IfI= 0.620. It follows from (14.6.10) that the VSWR is 4.26. These values also follow from drawing a circle through r+ jx = 2+ j2, using the radius of the circle to obtain IfIand the construction of Fig. 14.6.4a to evaluate (14.6.10). 14-14 Solutions to Chapter 14 14.6.3 The angular distance on the Smith charge from the point y = 2+ iO to the circle where y has a real part of 1 is I = 0.0975~. To cancel the reactance, where y =1 + iO.7 at this point, the distance from the shorted end of the stub to the point where it is attached to the line must be I. = 0.347~. 14.6.4 Adjustment of the length of the first stub makes it possible to be anywhere on the circle 9 = 2 of the admittance chart at the terminals of the parallel stub and load. IT this admittance can be transferred onto the circle 9 = 1 by moving a distance I toward the generator (clockwise), the second stub can be used to match the line by compensating for the reactive part of the impedance. Thus, determination of the stub lengths amounts to finding a pair of points on these circles that are at the same radius and separated by the angle 0.042~. This then gives both the combined stub (1) and load impedance (for the case given, y = 2+ i1.3) and combined stub (2) and line impedance at z = -I (for the case given, y = 1+ i1.16). To create the needed susceptance at the load, 11 = 0.04~. To cancel the resulting susceptance at the second stub, h = 0.38~. 14.6.5 The impedance at the left end of the quarter wave section is 0.5. Thus, normal­ ized to the impedance of the line to the left, the impedance there is Z/Z: = 0.25. It follows from the Smith chart and (14.6.10) that the VSWR = 4.0. 14.7 DISTRIBUTED PARAMETER EQUIVALENTS AND MODELS WITH DISSIPATION 14.1.1 The currents must sum to zero at the node. With those through the conduc­ tance and capacitance on the right, avI(z) - I(z + ~z) = G~zV + C~z­at The voltage drop around a loop comprised of the terminals and the series resistance and inductance must sum of zero. With the voltage drops across the resistor and inductor on the right, aIV(z) - V(z + ~z) = R~zI + L~z­at In the limit where ~z -0, these expressions become the transmission line equa­ tions, (14.7.1) and (14.7.2). 14.1.2 (a) IT the voltage is given by (14.7.12), as a special case of (14.7.9), then it follows that I(z,t) is the special case of (14.7.10) ~ (e-ifJ-_ eifJ-) . 1-R g :Jwt -e Zo (ej{J1 +e-i{Jl) e 14-15 Solutions to Chapter 14 (b) The desired impedance is the ratio of the voltage, (14.7.12), to this cUlTent, evaluated at z = -I. (eiJJI + e-iJJ1 ) Z = Zo (eiJJI _ e-iJJ1 ) (c) In the long-wave limit, 1,811 <: 1, exp(i,81) -+ 1 +i,81 and this expression becomes Z = Zo = (R + iwL) = 1 i,81 -,821 [G + iwCl1 where (14.7.8) and (14.7.11) have been used to write the latter equality. (Note that (14.7.8) is best left in the form suggested by (14.7.7) to obtain this result.) The circuit having this impedance is a conductance lG shunted by a capacitance lC. 14.1.3 The short requires that V(O, t) = °gives V+ = V_. With the magnitude ad­ justed to match the condition that V(-l, t) = Vg(t), (14.7.9) and (14.7.10) become Thus, the impedance at z = -I is Z = Zo(eiJJI -e-iJJI)j(eiJJI +e-iJJ1 ) In the limit where 1,811 <: 1, it follows from this expression and (14.7.8) and (14.7.11) that because expi,81 -+ 1 +i,81 Z -+ Zoi,81 = I(R + iwL) which is the impedance of a resistance lR in series with an inductor lL. 14.1.4 (a) The theorem is obtained by adding the negative of V times (1) to the negative of 1 times (2). (b) The identity follows from (c) Each of the quadratic terms in the power theorem take the form of (1), a time independent part and a part that varies sinusoidally at twice the driving frequency. The periodic part time-averages to zero in the power flux term on the left and in the dissipation terms (the last two terms) on the right. The only contribution to the energy storage term is due to the second harmonic, and 14-16 Solutions to Chapter 14 that time-averages to zero. Thus, on the time-average there is no contribution from the energy storage terms. The integral theorem, (d) follows from the integration of (c) over the length of the system. Integration of the derivative on the left results in the integrand evaluated at the end points. Because the current is zero where z = 0, the only contribution is the time-average input power on the left in (d). (d) The left hand side is evaluated using (14.7.12) and (14.7.6). First, using (14.7.11), (14.7.6) becomes 1= -yotTg tan f3z (2) Thus, (3) That the right hand side must give the same thing follows from using (14.7.3) and (14.7.4) to write GtT* = jwctT* _ di* (4)dz Rl =-dt' -jwLl (5)dz Thus, dtT/0.!:Re [1* RI+ tTtT*G]dz = /0 .!:Re [_l * _ jwLl1* -12 -12 dz ~~ ~ di*]+ jwCVV* -V ---;J; dz 0 1 dt'* dl* = - -Re [1- + tT-]dz (6) / _12 dz dz = -.!:Re/O d(iV*) dz 2 _I dz = iRe ItT*IZ=-I which is the same as (3). 14.8 UNIFORM AND TEM WAVES IN OHMIC CONDUCTORS 14-17 Solutions to Chapter 14 14.8.1 In Ampere's law, represented by (12.1.4), Ju = uE. Hence, (12.1.6) becomes oil) 2 oA 02AV(V .A + Jjuil) + JjE-) -VA = -JjU- -JjE- (1)ot ot ot2 Hence, the gauge condition, (14.8.3), becomes oA. oil)V .A =--=-Jjuil) -JjE- (2)OZ ot Evaluation of this expression on the conductor surface with (14.8.9) and (14.8.11) gives 01 oVL- = -JjuV -LC- (3)OZ ot From (8.6.14) and (7.6.4) (4) Thus, 01 = -GV _coV (5)OZ ot This and (14.8.12) are the desired transmission line equations including the losses represented by the shunt conductance G. Note that, provided the conductors are "perfect", the TEM wave represented by these equations is exact and not quasi­ one-dimensional. 14.8.2 The transverse dependence of the electric and magnetic fields are respectively the same as for the two-dimensional EQS capacitor-resistor and MQS inductor. The axial dependence of the fields is as given by (14.8.10) and (14.8.11). Thus, with (Prob. 14.2.1) uC = 21rE/ln(a/b); L = Jjo In(a/b); G = -C = 21rU /In(a/b)21r E and hence {j and Zo given by (14.7.8) and (14.7.11) with R = 0, the desired fields are E-R ~ vg (e-;fJ. +e;fJ.) ;wt. - e rln(a/b)(e;fJ 1 + e-;fJ 1) e lr ~ (-;fJ lIS ;fJ lIS )H R vg e -e ;wt. = e 21rrZ (e;fJ1 + e-;fJ1) e l<ll o 14-18 Solutions toChapter14 14.8.3 Thetransverse dependence ofthepotential followsfrom(4.6.18)-(4.6.19), (4.6.25)and(4.6.27). Thus,withtheaxialdependence givenby(14.8.10), E=-Bq>ix_Bq>i BxBy'Y where [Vl~-:t)2+y, ] VgInVlv'P-R'+:t)'+Y' (e-i~z+ei~z) 'wt q>--Re- e3 - 2In[:k+V(l/R)2_1] (ei~l+ei~l) Using(14.2.2)Azfollowsfromthispotential. where IIV:J(v'l2-R2-x)2+y2(e-i~z_ei~z). A--R!:!!.......J!..l 3wt z-en ('~l '~l)e21rZoy(v'l2_R2+x)2+y2e3+e3 Intheseexpressions, f3andZoareevaluated from(14.7.8)and(14.7.11) usingthe valuesofCandLgivenby(4.6.27)and(4.6.12)withR=0andG=(u/€)C. 14.8.4 (a)Theintegral ofEaroundthegivencontour isequaltothenegative rateof changeofthemagnetic fluxlinked.Thus, andinthelimit~z-+0, BEaBEb BHa--+b--=-J1.o(a+b)--YBzBz Bt Because €aEa=€bEb,thisexpression becomes €aBEa BH (a+-b)-B=-J1.o(a+b)-BY4z t(2) (3) IfEaandHyweretoberespectively writtenintermsofVandI,thiswould bethetransmission lineequation representing thelawofinduction (seeProb. 14.1.1). (b)Asimilarderivation usingthecontour closingattheinterface gives (4) 14-19 Solutions to Chapter 14 and in the limit ti.z -+ 0, aH" aEa E. = -aJ.'0ljt -a az (5) With the use of (3), this expression becomes HE. =-[aJ.'o(Ea -l)b/(a + Ea b)] aa " (6) Eb Eb t Finally, for a wave having a z dependence exp(-j,8z), the desired ratio follows from (6) and (3). IE.I = b(,8a) 11-Ea I (7)lEal a+b Eb Thus, the approximation is good provided the wavelength is large compared toa and b and is exact in the limit where the dielectric is uniform. 14.9 QUASI-ONE-DIMENSIONAL MODELS 14.9.1 From (14.9.11) 2R=-­ 1rWU while, from (4.7.2) and (8.6.12) respectively c- 211"E • L = e'n[('/a) + Y(I/a)2 -1]-In[(l/a) + y(l/a) -II' 11" To make the skin depth small compared to the wire radius 6=V2 :> R => w <: 2/a2J.W WJ.&U For the frequency to be high enough that the inductive reactance dominates 2 wL = wJ.'ua In [(l/a) + Y(I/a)2 _ 1]R 2 Thus, the frequency range over which the inductive reactance dominates but the constant resistance model is still appropriate is 2 2 -==-:-:-::-:--:---;:;:;;::;::::;::===: < w <-­J.'uR2Inl(l/a) + y(l/a)2 -11 a2J.'u For this range to exist, the conductor spacing must be large enough compared to their radii that 1 <: In[( i) + Y(I/a)2 -1]a Because of the logarithmic dependence, the quantity on the right is not likely to be very large. 14-20 Solutions to Chapter 14 14.9.2 From (14.9.11), 1 1 1 1 1 R = 0'27l'a~ + 0'7l'b2 = 'frO' [a~ + b2 ] while, from Prob. 14.2.1 L = ~;ln(a/b)j C= 2'frE In(a/b) For the skin depth to be large compared to the transverse dimensions of the con­ ductors 0== V2 ::> ~ or b => W -< 2/b21J0' and 2/~21J0' WIJO' This puts an upper limit on the frequency for which the model is valid. To be useful, the model should be valid at sufficiently high frequencies that the inductive reactance can dominate the resistance. Thus, it should extend to wL = WlJoO' In(a/b)/[..!.... + -!.] >1 R 2 a~ b2 For the frequency range to include this value but not exceed the skin depth limit, 2[~+;,\] 2 d 2 1J00In(a/b) -< W -< b21J0' an ~21J0' which is possible only if 1 In(a/b) -< [a~ + ;,\](b2and~2) Because of the logarithms dependence of L, this is not a very large range. 14.9.3 Comparison of (14.9.18) and (10.6.1) shows the mathematical analogy between the charge diffusion line and one-dimensional magnetic diffusion. The analogous electric and magnetic variables and parameters are H;s +-+ V, Kp +-+ Vp , IJO' +-+ RC, b +-+ I, :z: +-+ % Because the boundary condition on V at % =0 is the same as that on H;s at :z: =0, the solution is found by following the steps of Example 10.6.1. From 10.6.21, it follows that the desired distribution of V is % co (-l)n. n1r% V = -v. --'"2V. --Sin (_)e-t /.,.....P, LJ P n7l' I ' n=l This transient response is represented by Fig. 10.6.3a where H;s/Kp -V /Vp and z/b -%/1. 14.9.4 See solution to Prob. 10.6.2 using analogy described in solution to Prob. 14.9.3. SOLUTIONS TOCHAPTER 15 15.1SOURCE ANDMATERIAL CONFIGURATIONS 15.1.1 TABLEPI5.l.1. ModalFieldRepresentation Cartesian Laplace's Eq. Poisson's Eq. Polar Laplace's Eq. InitialValue Helmholtz Eq.Physical Constraints EQSPotential EQS3-Dimensional Polarization Conduction ChargeRelax. MQS,Equi-A Magnetization MQSE MQSEddyCurrent EQSPotential MQSEqua-A EQSPotential Conduction MQS,Constrained Current MQS,Equi-A MQSE Diffusion Eq. TMModes 3-Dimensional TEModesExample/Prob. Sec.5.5,Demo.5.5.1 Probs.5.5.1-7 Examp. 5.10.1 Probs.5.10.1,3 Examp. 6.6.3,6.7.1 Prob.6.3.10,6.6.9 Prob.6.7.1 Examp. 7.4.1 Prob.7.9.12 Examp. 8.6.3 Demo.8.6.2 Prob.8.6.10 Prob.9.6.9 Prob.10.1.2 Prob.10.1.5 Probs.5.6.7-9,13 Prob.8.6.7 Examp. 5.8.2-3 Probs.5.8.3-9 Prob.7.4.4,7.5.6 Prob.8.5.2 Prob.8.6.5 Examp. 10.12 Prob.10.1.3 Examp. 10.6 Prob.10.6.1-2 Examp. 13.3.1 Prob.13.3.1-6 Probs.13.4.3-4 Demo.13.3.1 Examp. 13.3.2 Demo.13.3.2 1 Solutions to Chapter 15 15-2 15.2 MACROSCOPIC MEDIA ~ 15.2.1 In each case, the excitation is an imposed uniform field at infinity. For (a), the field is tangential to the spherical surface everywhere except at the singular points at the poles. Thus, i) the system could be EQS with the regions insulating dielectrics and fa :> fb, ii) the system could be a stationary conductor with the field lines either J orE and O'a :> O'b, iii) it could be MQS with the lines B or H, the materials insulating and /Sa :> /Sb and iv) it could be a perfectly conducting sphere in an insulating media with the lines either B or H changing in time rapidly enough to induce the currents in the sphere required to exclude the field. For (b), the field is perpendicular to the surface. Thus, i) it could be EQS and a perfect conductor in an insulating medium with the lines representing E, ii) it could be EQS E with the materials perfect insulators (the field changing rapidly compared to the charge relaxation time in either material) with fb :> fa, iii) it could be J or E in stationary conduction with 0'1> :> O'a, iv) and it could be MQS H or B with the materials insulating and /Sb:> /Sa• .j 15.2.2 The excitation is inside the sphere. In (a), the field in that region is perpendic­ ular to the interface. Thus, i) the lines could be EQS E with the inside an insulator and the outside a perfect conductor, ii) the system could again be EQS and the lines could be E with both materials perfect insulators and fa :> fb, iii) it could be stationary conduction with the lines either E or J and a dipole current source with O'a :> O'b and iv) the lines could be MQS H or B with a magnetic dipole and the regions magnetizable insulators with /Sa :> /Sb. In (b), the interior field lines are tangential to the surface. Thus, i) the dipole could be electric and the materials perfect insula.tors ha.ving fb :> fa, ii) the dipole could be a current source for stationary conduction with the lines E orJ and O'b :> O'a, iii) the system could be MQS with the dipole magnetic and the materials magnetizable insulators having /Sb :> /Sa, and iv) the system could be MQS with a magnetic dipole varying rapidly enough with time to make the outer material a perfect conductor while the interior one remains a perfect insulator. 15.3 CHARACTERIC TIMES, PHYSICAL PROCESSES, AND APPROXIMATIONS 15.3.1 Because it does not involve O',W is normalized to rem. Thus, the horizontal axis is log(wr em) =log(wlyPE) Then Wf f 0' wre =-= wrem -- = 1 => wrem = --==-­ 0' O'l.,jiif (Vi7/J/l) Thus, with the characteristic conductivity defined as 0'* == v;[;/l 15-3 Solutions to Chapter 15 the critical line indicating charge relaxation, W'I"e = 1, is written in terms of the independent variables of normalized frequency and conductivity as logw'I"em = log (uU .. ) Similarly, 'l"em l{iii ( U )-1 ( U )W'I"m =1 => W'I"em =-= 1=- => logw'I"em = -log ­'l"m P.U 2 u.. u.. log("./".·) " " " MQS"""" "" -1 " " -- QSC ---+----:~------ - / / / / / / / EQS / / Figure S15.3.1 Thus, the plot is as shown in Fig. S15.3.1. H U > u.., raising the frequency results in a transition from stationary conduction to the MQS regime while if u < u.. , the transition is to the EQS regime. 15.3.2 (a) In the limit of zero frequency, the electric and magnetic fields are as summa­ rized by (7.5.7) and (7.5.11) and by (11.3.10) and (11.2.12). With (a) and (b) respectively designating the nonconducting annulus and the rod, fl.Eb = -1. (1)L EG_ fI [Z• In(r/a).] (2)--In(a/b) rL II' + L I. D b uflr.=--141 (3)L2 15-4 Solutions toChapter15 UO=~b21<l> L2r Themagnetic fieldisinducedbytheuniformcurrentdensity(4) O<r<b (5) whichisreturned asthesurfacecurrentdensityK.=-IO'ub2/2La]inthe perfectly conducting wall.Thereisnovolumechargedensityintheinteriorof therod.Onitssurfaceandontheinnersurfaceoftheouterwall,thesurface chargedensities are ( )fotl Z 0'.r=a=In(a/b)aLi( )fotl 11 0'.r=b=-In(a/b)Lb(6) Thesefieldsandsourcesaresketched inFig.815.3.2a. 0! J0 0/(:) 0,r-'+0-+0 0-+0+0..... 0 0 0 0 0 0 - -+ (a) Fleur.SlI.S.2_,b (b)Withalldimensions onthesameorder,theargument isasgiveninthissection. Anyone ofthedimensions, a,borListhetypicaldimension. Theratioof thatdimension toeitheroftheothertwoispresumed tobeperhaps2or3. 15-5 Solutions to Chapter 15 The permittivity and permeability can similarly be taken as that of either region with the respective ratios of these quantities again presumed to be less than an order of magnitude. Thus, the system is first EQS as the frequency is raised if the characteristic dimension, a, b or L, is small compared to 1*, where the latter is based on the conductivity of the rod and the permittivity and permeability of either region. In the case where the charge relaxation time is the longest of the characteristic times, the EQS case, the magnetic induction is not important as the frequency is raised to the point where the sources begin to alter their distribution. In this case, the dominant source is the charge density, specifically the surface charge density. With each half­ cycle, the surface charge density on the surface of the rod undergoes a sign reversal. To change this charge, the current density of (5) must be revised so that there is a component normal to the interface. In the "distributed circuit" picture of Fig. PI5.3.2a, this is the current required to charge the capacitors. (In the next problem, the energy stored in the capacitors is used as a means of establishing the equivalent capacitance needed to account for the charging of the surface.) In the case where the characteristic length is large compared to 1*, the system is MQS. The displacement current is negligible. This is equivalent to saying that the accumulation of charge has essentially no effect on the current density, which is itself solenoidal Thus, the conductivity of the rod is large enough that the current that enters at one end is negligibly diverted by supplying surface charge, essentially all reaching the far end. However, because the magnetic induction is important, these currents try to link as little magnetic flux as possible. As suggested by the distributed circuit picture of Fig. PI5.3.2b, the current distribution tends to crowd to the outer surface of the rod. The inductive reactance for a current circulating through the interior of the rod is less than that of a current nearer the surface. Thus, as the frequency is raised, the dominant field source, the current density, displays skin effect. In Cartesian rather than cylindrical geometry, Example 10.7.1 illustrates the distribution of magnetic field and current density. The radial direction in this problem plays the role ofthe z direction in the example. In both cases, the field and current density are independent of the axial direction (y in the example and z in this problem). One dimensional magnetic diffusion was pictured in Sec. 14.8 in terms of an L-G transmission line (negligible capacitance). Note that this is equivalent to the R -L distributed circuit used to schematically portray the MQS behavior in Fig. PI5.3.2b. The transmission line would be an exact representation if the rod were replaced by a "slab" conductor and the return conductors were planar rather than circular cylindrical. Such a configuration is shown in Fig. SI5.3.2b. Demonsatration 10.7.1 makes use of a transformer rather than a current source to drive the currents through the conductor. In the limit where the probed conductor is very long compared to its depth, it gives rise to the same current distribution as obtained in the slab conductor of Fig. SI5.3.2b. In the problem, the current distribution is somewhat different from that in the slab when the skin depth is on the order of the rod radius because of the cylindrical geometry. 15-6 Solutions toChapter15 (c)Theconditions areasdiscussed inSec.14.9.Sothattheskindepthislarge compared totherodradius,thefrequency mustbelowenoughthatthecurrent distribution inthecenterconductor isessentially uniform. Theinductance willnevertheless beself-consistently retained inthemodelprovided thatthe conditions foundinProb.14.9.2aresatisfied. 1<:In(a/b) (1) (Here,theouterconductor hasbeeneffectively madetohaveaninfinitecon­ ductivity bysettingli.-00inthesolutiontoProb.14.9.2.). Oncewehave decidedtoconsider systemsthatarelongintheaxialdirection, z,compared tothetransverse dimensions andtakenthequasi-one-dimensional modelas representing thedynamics, itisinteresting toseehowthelength,I,inthez direction determines theorderofthecharacteristic times L "M=-jRI "em=-=hlLOjc"s=z2RO (2) •log(l/lO) -WTM=1 .----------==~~------..., ..~log(WTM) WTE=1 • (c) FigureS15.3.Jc Inthelimitwheretheinductance isnotimportant, thesystemisachargediffusion lineasdiscussed inSec.14.9.Interestingly, thecharacteristic timeassociated with thisEQSlimitingmodeldepends onthesquareofthelength.Again,bycontrast withasystemhavingasingletypicallength,theinteraction betweentheinductance andtheresistance isindependent oflength(magnetic relaxation ratherthandiffu­ sion).Thus,inconstructing alength-frequency planeforsortingoutthephysical possibilities, itisthetimeL/Rthatcanbeselectedfornormalizing thefrequency. Thus,inthisplanethecriticallinesare I I WTM=IjWTem=1=>1*==(WTM)-ljW"s=1=>1*=(WTM)-1/2(3) Solutions to Chapter 15 15-1 and it follows (see Fig. S15.3.2c) that for the system to first be EQS as the frequency is raised, I> 1* == JL/C/R. 15.4 ENERGY, POWER, AND FORCE 15.4.1 The electric field intensity in the three regions follows from Example 7.5.2. Feom (7.5.7) and (7.5.11), respectively, (1) EG= tI [z.+,n(r/a).] (2)In(a/b) rLII' L I. The magnetic field intensity is summarized in Example 11.3.1. Feom (11.3.10) and (11.2.12), respectively, Uti. ()Ub = -rio#> 32L Uti b2• UG= --141 (4)L 2r The required electric energy, magnetic energy, and dissipation follow by carrying out the piece-wise volume integrations. (5) (6) and 0 b Pd =1r uEb . Eb21rrdrdz (7) -L 10 Note that this last integral is essentially one of the two carried out in (5). Evaluation of these expressions, using (1)-(6), gives (8) (9) (10) 15-8 Solutions to Chapter 15 Written with the voltage replaced by the total current, 2 . (U'lrb)\=1) -- (11)L the magnetic energy, (9), becomes _! [paL1n(a/b) PbL] .2 (12)Wm -2 2'1r + 'irS \ Feom a comparison of (S), (12), and (10), respectively, to (13) it follows that the quasi-stationary parameters that model the system at frequencies that are low compared to either R/L or llRO, whichever is the lower, are L -![ Lln(a/b) PbL] (15) -2 Pa 2'1r + S'Ir 7rb2 G = Ub (16) L (Note that L on the right is the length L of the device, to be distinguished from the inductance L on the left in (15).) Written in the form of (15.2.S), the ratio of the total magnetic to the total electric energy is, from (9) and (S) Wm = K(~)2. (17) We 1* ' where _( Pb) {4(alb)2 [1 2K = In(alb) + 4pa / ln2(a/b) ilL/a) In(a/b) 11 1]+ -[-+ (b/a)2[ln(b/a) -ln2(b/a) --I] (17)222 2Eb}+­Ea Provided the ratio of all dimension.s and of the permittivities and permeabilities are on the same order, the coefficient K is "of the order of unity." | | PRENTICE HALL, ENGLEWOOD CLIFFS, NEW JERSEY 07632 ISBN 0-13-248980-5