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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Document preview page 1

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

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

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

Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition offers a comprehensive guide to solving every question in your textbook, helping you master the material.

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Page 1 of 16
Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 1 preview imageSolutionManualforFundamentalsofElectromagneticsforElectricalandComputerEngineeringNannapaneniNarayanaRaoEdwardC.JordanProfessorEmeritusofElectricalandComputerEngineeringUniversityofIllinoisatUrbana-Champaign,USADistinguishedAmritaProfessorofEngineeringAmritaVishwaVidyapeetham(AmritaUniversity),India
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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 2 preview imageDownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com
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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 3 preview imageCONTENTSCRaPLEr1covvniiieee]CRAPLET2Levieseen1SChapter3...eee29)Chapter4Lo...eeeAChapter5.......oiiiiiiiiieDO)CRAPLEr6Leveeen82Chapter7........oiiiiiiiiniin104CHAPLET8vivitar.134CRAPLEL9viseeeeeccrine153Chapter10......ouviiiiiiiiiiiiiieceeLTOAPPENAIXALouieseee183)AppendixBo.oo.UBT
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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 4 preview imageCHAPTER11.1(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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 5 preview imagea,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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 6 preview image1.7.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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 7 preview image1.12.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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 8 preview image1.16.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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 9 preview image1.19.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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 10 preview image1.21.(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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 11 preview image1.26.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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 12 preview image129.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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 13 preview image1.32.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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 14 preview imageNe?.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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 15 preview image-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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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition - Page 16 preview image11.40.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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