Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition

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SolutionManualforFundamentalsofElectromagneticsforElectricalandComputerEngineeringNannapaneniNarayanaRaoEdwardC.JordanProfessorEmeritusofElectricalandComputerEngineeringUniversityofIllinoisatUrbana-Champaign,USADistinguishedAmritaProfessorofEngineeringAmritaVishwaVidyapeetham(AmritaUniversity),India

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DownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com

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CONTENTSCRaPLEr1covvniiieee]CRAPLET2Levieseen1SChapter3...eee29)Chapter4Lo...eeeAChapter5.......oiiiiiiiiieDO)CRAPLEr6Leveeen82Chapter7........oiiiiiiiiniin104CHAPLET8vivitar.134CRAPLEL9viseeeeeccrine153Chapter10......ouviiiiiiiiiiiiiieceeLTOAPPENAIXALouieseee183)AppendixBo.oo.UBT

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CHAPTER11.1(a)Totaldistance=1+++Lod=omwrT2727:1-45"(b)Distancenorth=[—++=—=—L_=0.8mTae1+yA1111,1.1Distanceeast=a(ae=04m.~.Finalpositionis(0.8,0.4)(c)Straightlinedistance=+/(0.8)2+(0.4)?=0.8944m1.2.A+B+C=2a;+3a+2a3—(1)2A+B-C=a;+3a,—(2)A-2B+3C=4a+50,+a;—(3)(1)+2)—3A+2B=3a;+16a,+2a;—(4)2)x3+(3)—TA+B=Ta,+lday+a;—(5)[G)X2-(@)]+11>A=a,+2a,—(6)5)-(6)x7—B=a;—(H=-©)-N—C=a+a+a;—(8)13.(A+B)*(A-B)=A*A-A+*B+B*A-B+B=A*—B(A+B)Xx(A-B)=AXA-AXB+BXA-BXxB=2BxAForA=3a;—5a;+4a;andB=a;+a,—2as,A+B=4a;—4a;+2a;,A—B=2a;—6a;+6as,A’=9+25+16=50,andB*=1+1+4=6A+B)»(A-B)=8+24+12=44=4>-pB1—StudyXY

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a,a,a,(A+B)x(A-B)=|4-42|=-12a,-20a,-16a,266a,a,a,=2|11-2=2BxA35414.BxC=-4a,+2a,+8a,Ax(BxC)=8a,+16a,CxA=-a,-2a,+7a,Bx(CxA)=-12a,-8a,—4a,AxB=a,+2a,+3a,,Cx(AxB)=4a,—8a,+4a,AxBXC)+Bx(CxA)+Cx(AxB)=0Infact,thisquantityiszeroforanyA,B,andC.15.Arca=SABsinc=LiaXB|2"2Forthepoints(1,2,1),(3,4,5),.8and(2,-1,-3),7BsindA=4a,+6a,—4a,B=5a,+3a,—8a,A)AxB=-36a,+12a,-18a,‘2Area=ze?+(12)*+(~18)*=21units.1.6.Areaofthebase=[BXx|;Heightofparallelepiped=ProjectionEs=ofAontothenormaltothebaseA[E]ZA.BxCB[BxC]|+.Volumeofparallelepiped=Areaofbasexheight=A«BxCForA=4a,,B=2a,+a,+3a,andC=2a,+6a,A*BxC=0.Hence,volumeoftheparallelepipediszero.Thethreevectorslieinaplane.2—StudyXY

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1.7.ThevectorAmustbeperpendiculartoboth(a,+2a,)and(a,—2a,).HenceA=C(-a,+2a,)X(a,—2a;)=C(2a,+2a,+a,)whereCisaconstant.TofindC,wenotethata,XxA=a,XC(2a,+2a,+a,)=2a,—a, C=1andA=2a,+2a,+a,Verification:a,xA=a,X(2a,+2a;+a)=a,—2a.1.8.VectorfromA(S,0,3)toB(3,3,2)=2a,+3a,—a,VectorfromC(6,2,4)toD(3,3,6)=3a,+a,+2a,CD6+3-2ComponentofABalongCD=AB+——=————==1.8708P€CDJ9+1+41.9.Writingtheequationfortheplaneas=5tog=1,wefindtheinterceptsonthex,y,andz-axestobeat15,-12,and20,respectively.Thus3Rup=-15a,—12a,:fooRac=-15a,+20a,)RacxRap=240a,—300a,+180a,<2,a=RyoxRyp_da5,+3a,AsRacXRgp572x.Distancefromorigintotheplane=15a,*a,=2.1.10.Fory=2x,z=4y,wehavedy=2dx,dz=4dy=8dx.sod=dea+dya,+dza;=dra,+2dxa,+8dxa,=(a,+2a,+8a,)dx,independentofthepoint.1.11.Forx=y=7%,wehavedx=dy=27dz.Atthepoint(4,4,2),dx=dy=4dzsod=dva,+dyay+dza;=4dza,+4dzay+dza,=(4a,+4a,+a,)dz3—StudyXY

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1.12.Differentiallengthvectorhavingprojectiondya,=dya,dy...i.Differentiallengthvectorhavingxeyer/ik3x:projectiondxa,isJSaipat1[dx%[07dxay+dza;=doac—5da,tLreya=(a“3aJdx,4sinceforx+2z=2,dz=—:dx,independentofthepoint.._11sds=a,~5a;dxXdya,=Fata,dxdy.1.13.Onevectortangentialtothe¥surfaceisdza,.Anothertangentialvectorisgivenbydl=dxa,+dya,142,4,=dra,+2xdxa,@*)=(a,+4a,)dxld.~.Vectornormaltotheplane=(a,+4a,)dxXdza,=(4a,—a,)dxdzUnitvect1totheplfa,nitvectornormaltotheplane=——==.?71.14.DenotingK(x,y)tobetheheightfield,wehavePry+r=4,7+)<4or,h=\J4=x2~y2FP+y<4.1.15.Thenumberfieldisx+y+z..~.Constantmagnitudesurfacesaretheplanesx+y+z=constant.4—StudyXY

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1.16.d(x,y,2)=xa,+ya,+za,Constantmagnitudesurfacesarex”+y*+z*=constant,andhencearesphericalsurfacescenteredatthecorner.Directionlinesareradiallinesemanatingfromthecorner.1.17.v=—rwsinga,+rwcosga,=aX-ya,+xa,)YN91.18.f(z,1)=10cos(27x10't~0.177)1CUNYaro!t=0kzLix0E:idt=Fat=gxto(®:—SarVeSSCSAAVANaSCO>320322-53=5 3=1S310’wiosHZWTSON.4Te>><>f(z,1)representsatravelingwaveprogressingwithtimeinthepositivez-direction.5+StudyXY

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1.19.f(z,1)=10cos(27x10"+0.172)111-1felixk=2x6blyEzixkGt=0§z3€3©~RYMNVO)LlBRNSN.CONN-ARRETOS<<>>:zlo27.6=522.53=0w3PVNVLTYOYTIONTSSINCOONB=<>-f(z,©)representsatravelingwaveprogressingwithtimeinthenegativez-direction.1.20.f(z,£)=10cos272x10'¢cos0.172t=0=:,exe”“fiSEIN0Bs——>0Thm=7.01Tez2xd”|“teSeasxe.5-®Cem7.01PaSeTN+%-1.07{yztbm“10KaziomAf(z,1)representsastandingwave.6

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1.21.(a)Thetwocomponentsareinphase;hence,linearpolarization.(b)Thetwocomponentsareperpendicularindirection,differinphaseby90°andequalinamplitude;hence,circularpolarization.(c)Thetwocomponentsareperpendicularindirection,differinphaseby90°butunequalinamplitude;henceellipticalpolarization.1.22.F;andF;differinphaseby90°.1.3..|=+/3+1coswt=2cosar;[Fy=FREEsinwt=2sinwt.i..FyandF;areequalinamplitude.FF,=BLL=0...FyisperpendiculartoFs.ThusFy+F;iscircularlypolarized.1.23.\[~|HENHLySECoThepolarizationisellipticalwithmajoraxisinthey-direction,minoraxisinthex-direction,andeccentricityequaltoV2.1.24.10cos(ax—30°)+10cos(ax+210°)Iw|l107”+10M”=1030°500”Re.~.Thesumis10cos(ax—90°)=10sinar.~~.ad1.25.3cos(ax+60°)—4cos(ax+150°)YorbeImJ3e=deSEANogJ60°-53.13)_£,j68T36060°s=5e=5eFaRe—5cos(ax+6.87°).£8037—StudyXY

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1.26.Replacingzby710°7ibyI,and13cos10%by13¢/,wehave51070xj10°T+127=13¢/"GT=13,T=01Bmeror,(12+j5)I=13,1=25WET=leThusi=1cos(10%-22.62°)=1cos(10%—0.1267)1.27.Fromtheconstructionshown,0—=mg=tan45°=1[3©Elo&2oN—FR,or,Q=187£91mg470,20VoVELQm31.28.(a)Atthepoint(0,0,100),Ix476499)47g,(101)Q.1__0107-99"iN747g)99%1012°x0(100+1)%-(100-1)?Q400YTarrveyseppeeyLlpspL475(100-1)%x(100+1Eo(100%—1)400EELBy,47€)100100%78,(b)Atthepoint(100,0,0)201E=————————>a478,(1007+12)¥2°FA2760(100%)8—StudyXY

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129.E=Q|a,+2a,+2a,LAA,+2,LAAs,ta,47g,93/2932932a+a,+2a,2a,+a,+a,a,+2a,+a,tearTarTTenLaat,2a,Lata,ta,1232332=<S42Lo(a,+a,+a;)dm3(orWhyWBC——£(0.18519+0.27217+0.04811+0.19245)(a+a,+a,)XxyZz47g,0.0555Q=ata+a,)N/C.1.30.Fortheithsegment,.-3kdz=2-1andcharge=EUC.if10050__LFR"50Cpe2N071¥-=a2%PREYyEh(2+1)50107°a.aTZ-“Ta2.10(2i-1+1]a,iyy_2i-1os2i-11.31.Fortheithsegment,z=100,chargedensity=10100C/m,.6andcharge=107°KamiBX=10@i-yC.506.Cpl210°°2i-1)1SB)52372i=1(2+1)750107’.LyaTY?=2,010@i-1)+1]a,[=9—StudyXY

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1.32.Dividingthecircularringintonsegmentsandusingthesymmetryofthefieldaboutthez-axis,weobtain_¥x22(2)%107°1B=)imap(2+1)1_4mx107008944x107Came&‘4=1.012x10"a,NIC.Xx1.33.Forthe(i/)tharea,C2i-1_2j-1¥*=Tao7100Mcharge=——x10=4x10°77CJ10000WaITLL:y4S&axa”ii-EeNNEa,[ATT77TTAELedLed(i4241)Y?3i=lj=lx_75050_4x10”RTAIi“mm2210@i-D*+107*@2j-1)+1]a,i=lj=2i-12j—-1.34.Forthe(i/)tharea,x=Soy=us,2-1)2j-1Ychargedensity=107(G35)=107°(2i-1)2j-1)*¢/m?_4TTI“130:yin1\2charge=JoesX10”2i-1)(2j1)?=4x107°@i-D2)-1)CpedShyerner,47gpg+?+1)?:i.5050_4x107?Qi-12j-1)?ATpe[10Qi-12+1072)-1)+1]10—StudyXY

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Ne?.1.35.(a)J=Nev=—Eysinaxa,maw122.10738=MOTXLOREXI010sin2mx10’a,9.1083x1077"x27x10=0.4485%x10sin27x107a,A/m?(b)AI=J+AS=0.4485x10sin27x10"a,+0.01(a,+a)=0.4485x108sin27x107A1.36.Denotingxtobethedisplacement,wewritetheequationofmotiontobe2mE=mgkx+gEcosaxldrym2E,coswtdxonm=z+kx=mg+qEocosat9tThesteadystatesolutionconsistsoftwoparts.Oneisx==duetomg.Tofindthesecondpartx,wewrite(je)mx,+ky=qEge’®,or,(k—a&’m)%,=gE,==15mg4kX)=——5—.Thusx=x+x,=—>+—"—>—coswx?k-a’mTRk-a’m.dxqEywVelocity=—=—————sinwrYdk-a'mI,dya,X(-a1.37.dF,=I;dxa,xHelada,x(Ca,)47(l)¥=0I,dysyIdxa_xa,.1:Fy=I,ya,x|LL1xEy47(l)*-HoHo=1,dya,xdxa,=zladxdya,11—StudyXY

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-a,+a,+a1.38.(2)For0,1,1),ag=——2—"%R=4/3,and3pote!dx(a,+2a,t2)(a,+a,+a,)ar3NetoIdx=——=—(-a,+a,)ofCh(b)For(2,2,2)acta,+24,R=3,andor(2,2,2),ag=——7=——,R=3,an*Ndpot!dx@,+2a,+2,)(a,+2,+2%,)493=01.39.(a)At(0,0,1),thecomponentsofBMperpendiculartothez-axiscancel,1whereasthezcomponentsadd.ThusBe1Hg0012,X(0.00%,+a)aTY4(140.0052)?a“5%107-[2000005,£0012,a,“22H,(b)Atthepoint(0,1,0),poto|__001a+0.01a,47](1-0.005)(140.005)N0.0la,X(=0.005a,+a)B0.0la,x(0.005a,+a.)(1+0.005%)?(1+0.005%)7?Ho|0.01x4x1x0.0052x0.01x0.0057EPISaSnoITEAz(1-0.005%)(1+0.0052)107%,TTT12—StudyXY

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11.40.Fortheithelement,dl=30%N.\2kd2i-1R=(5%)a"~R_12i-1iag=|a,~|TogJaz|»anItl1(EA),|LxAB=r505%TooJ|7Al2007R50io]50J"=—_—4[—B=2)"dB20>[1410@i-1)]a,i=li=11.41.Dividingtheloopintonsegmentsandusingthesymmetryofthefield3aboutthez-axis,weobtainol222)2gB=——a:32“2Dirn(2422)y_8mul_*=Ins=0.179la,1.42.Equatingthemagneticforcetothecentripetalforce,wehave2.mymyevBy=—onr=ByPReroomv._oo_eBOrbitalfrequency=5=omForBy=5x107,13—StudyXY

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