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CHAPTER55.1.GiventhecurrentdensityJ=−104[sin(2x)e−2yax+cos(2x)e−2yay]kA/m2:a)Findthetotalcurrentcrossingtheplaney=1intheaydirectionintheregion0x1,0z2:ThisisfoundthroughI=ZZSJ·nØØØSda=Z20Z10J·ayØØØy=1dxdz=Z20Z10−104cos(2x)e−2dxdz=−104(2)12sin(2x)ØØØ10e−2=−1.23MAb)Findthetotalcurrentleavingtheregion0x,x1,2z3byintegratingJ·dSoverthesurfaceofthecube:Notefirstthatcurrentthroughthetopandbottomsurfaceswillnotexist,sinceJhasnozcomponent.Alsonotethattherewillbenocurrentthroughthex=0plane,sinceJx=0there.Currentwillpassthroughthethreeremainingsurfaces,andwillbefoundthroughI=Z32Z10J·(−ay)ØØØy=0dxdz+Z32Z10J·(ay)ØØØy=1dxdz+Z32Z10J·(ax)ØØØx=1dydz=104Z32Z10£cos(2x)e−0−cos(2x)e−2§dxdz−104Z32Z10sin(2)e−2ydydz=104µ12∂sin(2x)ØØØ10(3−2)£1−e−2§+104µ12∂sin(2)e−2yØØØ10(3−2)=0c)Repeatpartb,butusethedivergencetheorem:WefindthenetoutwardcurrentthroughthesurfaceofthecubebyintegratingthedivergenceofJoverthecubevolume.Wehave∇·J=∂Jx∂x+∂Jy∂y=−10−4£2cos(2x)e−2y−2cos(2x)e−2y§=0asexpected5.2.GivenJ=−10−4(yax+xay)A/m2,findthecurrentcrossingthey=0planeinthe−aydirectionbetweenz=0and1,andx=0and2.Aty=0,J(x,0)=−104xay,sothatthecurrentthroughtheplanebecomesI=ZJ·dS=Z10Z20−104xay·(−ay)dxdz=2×10−4A585.3.LetJ=400sinθr2+4arA/m2a)Findthetotalcurrentflowingthroughthatportionofthesphericalsurfacer=0.8,boundedby0.1πθ0.3π,0φ2π:ThiswillbeI=ZZJ·nØØØSda=Z2π0Z.3π.1π400sinθ(.8)2+4(.8)2sinθdθdφ=400(.8)22π4.64Z.3π.1πsin2dθ=346.5Z.3π.1π12[1−cos(2θ)]dθ=77.4Ab)FindtheaveragevalueofJoverthedefinedarea.TheareaisArea=Z2π0Z.3π.1π(.8)2sinθdθdφ=1.46m2TheaveragecurrentdensityisthusJavg=(77.4/1.46)ar=53.0arA/m2.5.4.Ifvolumechargedensityisgivenasρv=(cosωt)/r2C/m3insphericalcoordinates,findJ.ItisreasonabletoassumethatJisnotafunctionofθorφ.Weusethecontinuityequation(5),alongwiththeassumptionofnoangularvariationtowrite∇·J=1r2∂∂r°r2Jr¢=−∂ρv∂t=−∂∂tµcosωtr2∂=ωsinωtr2Sowemaynowsolve∂∂r°r2Jr¢=ωsinωtbydirectintegrationtoobtain:J=Jrar=ωsinωtrarA/m2wheretheintegrationconstantissettozerobecauseasteadycurrentwillnotbecreatedbyatime-varyingchargedensity.595.5.LetJ=25ρaρ−20ρ2+0.01azA/m2a)Findthetotalcurrentcrossingtheplanez=0.2intheazdirectionforρ0.4:UseI=ZZSJ·nØØØz=.2da=Z2π0Z.40−20ρ2+.01ρdρdφ=−µ12∂20ln(.01+ρ2)ØØØ.40(2π)=−20πln(17)=−178.0Ab)Calculate∂ρv/∂t:Thisisfoundusingtheequationofcontinuity:∂ρv∂t=−∇·J=1ρ∂∂ρ(ρJρ)+∂Jz∂z=1ρ∂∂ρ(25)+∂∂zµ−20ρ2+.01∂=0c)Findtheoutwardcurrentcrossingtheclosedsurfacedefinedbyρ=0.01,ρ=0.4,z=0,andz=0.2:ThiswillbeI=Z.20Z2π025.01aρ·(−aρ)(.01)dφdz+Z.20Z2π025.4aρ·(aρ)(.4)dφdz+Z2π0Z.40−20ρ2+.01az·(−az)ρdρdφ+Z2π0Z.40−20ρ2+.01az·(az)ρdρdφ=0sincetheintegralswillcanceleachother.d)ShowthatthedivergencetheoremissatisfiedforJandthesurfacespecifiedinpartb.Inpartc,thenetoutwardfluxwasfoundtobezero,andinpartb,thedivergenceofJwasfoundtobezero(aswillbeitsvolumeintegral).Therefore,thedivergencetheoremissatisfied.5.6.Insphericalcoordinates,acurrentdensityJ=−k/(rsinθ)aθA/m2existsinaconductingmedium,wherekisaconstant.DeterminethetotalcurrentintheazdirectionthatcrossesacirculardiskofradiusR,centeredonthezaxisandlocatedata)z=0;b)z=h.Integrationoveradiskmeansthatweusecylindricalcoordinates.Thegeneralfluxintegralassumestheform:I=ZsJ·dS=Z2π0ZR0−krsinθaθ·az|{z}−sinθρdρdφThen,usingr=pρ2+z2,thisbecomesI=Z2π0ZR0kρpρ2+z2=2πkpρ2+z2ØØØR0=2πkhpR2+z2−ziAtz=0(parta),wehaveI(0)=2πkR,andatz=h(partb):I(h)=2πk£√R2+h2−h§.605.7.Assumingthatthereisnotransformationofmasstoenergyorvice-versa,itispossibletowriteacontinuityequationformass.a)Ifweusethecontinuityequationforchargeasourmodel,whatquantitiescorrespondtoJandρv?Thesewouldbe,respectively,massfluxdensityin(kg/m2−s)andmassdensityin(kg/m3).b)Givenacube1cmonaside,experimentaldatashowthattheratesatwhichmassisleavingeachofthesixfacesare10.25,-9.85,1.75,-2.00,-4.05,and4.45mg/s.Ifweassumethatthecubeisanincrementalvolumeelement,determineanapproximatevalueforthetimerateofchangeofdensityatitscenter.Wemaywritethecontinuityequationformassasfollows,alsoinvokingthedivergencetheorem:Zv∂ρm∂tdv=−Zv∇·Jmdv=−IsJm·dSwhereIsJm·dS=10.25−9.85+1.75−2.00−4.05+4.45=0.550mg/sTreatingour1cm3volumeasdifferential,wefind∂ρm∂t.=−0.550×10−3g/s10−6m3=−550g/m3−s5.8.Atruncatedconehasaheightof16cm.Thecircularfacesonthetopandbottomhaveradiiof2mmand0.1mm,respectively.Ifthematerialfromwhichthissolidconeisconstructedhasaconductivityof2×106S/m,usesomegoodapproximationstodeterminetheresistancebetweenthetwocircularfaces.Considertheconeupsidedownandcenteredonthepositivezaxis.The1-mmradiusendisatdistancez=`fromthex-yplane;thewideend(2-mmradius)liesatz=`+16cm.`ischosensuchthatiftheconewerenottruncated,itsvertexwouldoccurattheorigin.Theconesurfacesubtendsangleθcfromthezaxis(insphericalcoordinates).Therefore,wemaywrite`=0.1mmtanθcandtanθc=2mm160+`Solvingthese,wefind`=8.4mm,tanθc=1.19×10−2,andsoθc=0.68◦,whichgivesusaverythincone!Withthisunderstanding,wecanassumethatthecurrentdensityisuniformwithθandφandwillvaryonlywithsphericalradius,r.Sothecurrentdensitywillbeconstantoverasphericalcap(ofconstantr)anywherewithinthecone.Astheconeisthin,wecanalsoassumeconstantcurrentdensityoveranyflatsurfacewithintheconeat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