induction and inductance - usna 30... · [shivok sp212] february 20, 2016 page 1 ch 30 induction...
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[SHIVOK SP212] February 20, 2016
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CH 30
InductionandInductance
I. Faraday’sExperiments
A. Let’sexaminetwosimpleexperimentstoprepareforourdiscussionaboutFaraday’sLaw:
B. FirstExperiment:
1. MoveamagnetthroughaloopinducinganEMFintothatloop.
2. Acurrentappearsonlyifthereisrelativemotionbetweentheloopandthemagnet(onemustmoverelativetotheother);thecurrentdisappearswhentherelativemotionbetweenthemceases.
3. Fastermotionproducesagreatercurrent.
4. Ifmovingthemagnet’snorthpoletowardtheloopcauses,say,clockwisecurrent,thenmovingthenorthpoleawaycausescounterclockwisecurrent.Movingthesouthpoletowardorawayfromtheloopalsocausescurrents,butinthereverseddirections.
5. Thecurrentthusproducedintheloopiscalledinducedcurrent.
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C. SecondExperiment:
1. ForthisexperimentweusetheapparatusofFig.30‐2,withthetwoconductingloopsclosetoeachotherbutnottouching.
2. IfwecloseswitchS,toturnonacurrentintheright‐handloop,themetersuddenlyandbrieflyregistersacurrent—aninducedcurrent—intheleft‐handloop.
3. Ifwethenopentheswitch,anothersuddenandbriefinducedcurrentappearsinthelefthandloop,butintheoppositedirection.
4. Wegetaninducedcurrent(andthusaninducedemf)onlywhenthecurrentintheright‐handloopischanging(eitherturningonorturningoff)andnotwhenitisconstant(evenifitislarge).
II. Faraday’sLawofInduction:
A. AnEMFisinducedintheloopattheleftinFigures30‐1and30‐2whenthenumberofmagneticfieldlinesthatpassthroughtheloopischanging.
B. ThemagnitudeoftheEMF(ScriptE)inducedinaconductingloopisequaltotherateatwhichthemagneticfluxBthroughthatloopchangeswithtime.
C. SupposealoopenclosinganareaAisplacedinamagneticfieldB.Thenthemagneticfluxthroughtheloopis
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D. Iftheloopliesinaplaneandthemagneticfieldisperpendiculartotheplaneoftheloop,andIfthemagneticfieldisconstant,then
E. TheSIunitformagneticfluxisthetesla–squaremeter,whichiscalledtheweber(abbreviatedWb):
F. Finally,Faraday’sLawofInduction:
1. IfwechangethemagneticfluxthroughacoilofNturns,aninducedemfappearsineveryturnandthetotalemfinducedinthecoilisthesumoftheseindividualinducedemfs.Ifthecoilistightlywound(closelypacked),sothatthesamemagneticfluxBpassesthroughalltheturns,thetotalemfinducedinthecoilis
2. Herearethegeneralmeansbywhichwecanchangethemagneticfluxthroughacoil:
a) ChangethemagnitudeBofthemagneticfieldwithinthecoil.
b) Changeeitherthetotalareaofthecoilortheportionofthatareathatlieswithinthemagneticfield(forexample,byexpandingthecoilorslidingitintooroutofthefield).
c) ChangetheanglebetweenthedirectionofthemagneticfieldBandtheplaneofthecoil(forexample,byrotatingthecoilsothatthefieldBisfirstperpendiculartotheplaneofthecoil,andthenalongthatplane).
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G. SampleProblem:
1. InFig.below,a120‐turncoilofradius1.8cmandresistance5.3Ωiscoaxialwithasolenoidof220turns/cmanddiameter3.2cm.Thesolenoidcurrentdropsfrom1.5AtozerointimeintervalΔt=25ms.WhatcurrentisinducedinthecoilduringΔt?
a) Solution:
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III. Lenz’sLaw:
A. Aninducedcurrenthasadirectionsuchthatthemagneticfielddudetothecurrentopposesthechangeinthemagneticfluxthatinducesthecurrent.
B. OppositiontoPoleMovement.Theapproachofthemagnet’snorthpoleinFig.30‐4increasesthemagneticfluxthroughtheloop,inducingacurrentintheloop.Toopposethemagneticfluxincreasebeingcausedbytheapproachingmagnet,theloop’snorthpole(andthemagneticmoment)mustfacetowardtheapproachingnorthpolesoastorepelit.ThecurrentinducedintheloopmustbecounterclockwiseinFig.30‐4.Ifwenextpullthemagnetawayfromtheloop,acurrentwillagainbeinducedintheloop.Now,theloopwillhaveasouthpolefacingtheretreatingnorthpoleofthemagnet,soastoopposetheretreat.Thus,theinducedcurrentwillbeclockwise.
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C. Fig.30‐5Thedirectionofthecurrentiinducedinaloopissuchthatthecurrent’smagneticfieldBindopposesthechangeinthe
magneticfieldinducingi.Thefieldisalwaysdirectedoppositeanincreasingfield(a)andinthesamedirection(b)asadecreasingfieldB.Thecurled–straightright‐handrulegivesthedirectionoftheinducedcurrentbasedonthedirectionoftheinducedfield.
D. Ifthenorthpoleofamagnetnearsaclosedconductingloopwithitsmagneticfielddirecteddownward,thefluxthroughtheloopincreases.Toopposethisincreaseinflux,theinducedcurrentimustsetupitsownfieldBinddirectedupwardinsidetheloop,asshownin
Fig.30‐5a;thentheupwardfluxofthefieldBindopposestheincreasing
downwardfluxoffield.Thecurled–straightright‐handrulethentellsusthatimustbecounterclockwiseinFig.30‐5a.
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E. SampleProblem(Lenz’sLaw):
1. Thefollowingtwosituationsareseparate.Ontheleft,asquareloopofwireispenetratedbyamagneticfieldoutofthepagethatisincreasinginstrength.Ontheright,thenorthpoleofamagnetismovingawayfromacoilofwire.Therowthatcorrectlygivesthedirectionoftheinducedcurrentthrougheachresistoris
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IV. InductionandEnergyTransfers:
A. Considerthepullingaconductingloopoutofamagneticfieldasshownbelow:
B. Iftheloopispulledataconstantvelocityv,onemustapplyaconstantforceFtotheloopsinceanequalandoppositemagneticforceactsonthelooptoopposeit.ThepowerisP=Fv.
C. Astheloopispulled,theportionofitsareawithinthemagneticfield,andthereforethemagneticflux,decrease.AccordingtoFaraday’slaw,acurrentisproducedintheloop.Themagnitudeofthe
fluxthroughtheloopisB=BA=BLx.
D. Therefore,
E. Theinducedcurrentistherefore
F. Thenetdeflectingforceis:
G. Thepoweristherefore
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H. SampleProblems:
1. InFig.belowametalrodisforcedtomovewithconstantvelocityalongtwoparallelmetalrails,connectedwithastripofmetalatoneend.AmagneticfieldofmagnitudeB=0.350Tpointsoutofthepage.(a)IftherailsareseparatedbyL=25.0cmandthespeedoftherodis55.0cm/s,whatemfisgenerated?(b)Iftherodhasaresistanceof18.0Ωandtherailsandconnectorhavenegligibleresistance,whatisthecurrentintherod?(c)Atwhatrateisenergybeingtransferredtothermalenergy?
a) Solution:
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2. InFig.below,alongrectangularconductingloop,ofwidthL,resistance
R,andmassm,ishunginahorizontal,uniformmagneticfield thatisdirectedintothepageandthatexistsonlyabovelineaa.Theloopisthendropped;duringitsfall,itacceleratesuntilitreachesacertainterminalspeedvt.Ignoringairdrag,findanexpressionforvt.
a) Solution:
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I. EddyCurrents
V. InducedElectricField:
A. Achangingmagneticfieldproducesanelectricfield.
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B. InducedElectricFields,ReformulationofFaraday’sLaw:
1. Consideraparticleofchargeq0movingaroundthecircularpath.The
workWdoneonitinonerevolutionbytheinducedelectricfieldisW=Eq0,
whereEistheinducedemf.
2. Fromanotherpointofview,theworkis
3. Herewhereq0Eisthemagnitudeoftheforceactingonthetestcharge
and2risthedistanceoverwhichthatforceacts.
4. Ingeneral,
C. InducedElectricFields,ANewLookatElectricPotential:
1. ElectricPotentialhasmeaningonlyforelectricfieldsthatareproducedbystaticcharges;ithasnomeaningforelectricfieldsthatareproducedbyinduction.
2. Whenachangingmagneticfluxispresent,theintegralisnotzerobutisd
B/dt.
3. Thus,assigningelectricpotentialtoaninducedelectricfieldleadsustoconcludethatelectricpotentialhasnomeaningforelectricfieldsassociatedwithinduction.
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D. SampleProblem:
1. Alongsolenoidhasadiameterof12.0cm.Whenacurrentiexistsinitswindings,auniformmagneticfieldofmagnitudeB=30.0mTisproducedinitsinterior.Bydecreasingi,thefieldiscausedtodecreaseattherateof6.50mT/s.Calculatethemagnitudeoftheinducedelectricfield(a)2.20cmand(b)8.20cmfromtheaxisofthesolenoid.
a) Solution:
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VI. InductorsandInductance:
A. Aninductor(symbol)canbeusedtoproduceadesiredmagneticfield.
B. Ifweestablishacurrentiinthewindings(turns)ofthesolenoidwhichcanbetreatedasourinductor,thecurrentproducesamagneticflux
Bthroughthecentralregionoftheinductor.
C. Theinductanceoftheinductoristhen
D. TheSIunitofinductanceisthetesla–squaremeterperampere(T
m2/A).
Wecallthisthehenry(H),afterAmericanphysicistJosephHenry.
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E. InductanceofaSolenoid:
1. Consideralongsolenoidofcross‐sectionalareaA,withnumberofturnsN,andoflengthl.Thefluxis
Herenisthenumberofturnsperunitlength.
2. ThemagnitudeofBisgivenby:
3. Therefore,
4. Theinductanceperunitlengthnearthecenteristherefore:
Here,
F. Self‐Induction:
1. Aninducedemfappearsinanycoilinwhichcurrentischanging.
2. Thisprocess(seefigurebelow)iscalledself‐induction,andtheemfthatappearsiscalledself‐inducedemf.ItobeysFaraday’slawofinductionjustasotheremfsdo.
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3. ,butremember
4.
G. Sampleproblems:
1. Theinductanceofacloselypackedcoilof400turnsis8.0mH.Calculatethemagneticfluxthroughthecoilwhenthecurrentis5.0mA.
a) Solution:
2. Atagiveninstantthecurrentandself‐inducedemfinaninductoraredirectedasindicatedinFig.below.(a)Isthecurrentincreasingordecreasing?(b)Theinducedemfis17V,andtherateofchangeofthecurrentis25kA/s;findtheinductance.
a) Solution:
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VII. RLCircuits:
A. Initially,aninductoractstoopposechangesinthecurrentthroughit.Alongtimelater,itactslikeanordinaryconnectingwire.
B. Thecircuit:
1. Currenteqn:
2. Riseofcurrent:
3. Thusthetimeconstant:
4. Ifwesuddenlyremovetheemffromthissamecircuit,thechargedoesnotimmediatelyfalltozerobutapproacheszeroinanexponentialfashion:
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5. Graphs:
C. Sampleproblem:
1. Asolenoidhavinganinductanceof6.30μHisconnectedinserieswitha1.20kΩresistor.(a)Ifa14.0Vbatteryisconnectedacrossthepair,howlongwillittakeforthecurrentthroughtheresistortoreach80.0%ofitsfinalvalue?(b)Whatisthecurrentthroughtheresistorattimet=1.0τL?
a) Solution:
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VIII. EnergyStoredinaMagneticField:
A. Thecircuit:
B. KVL:
C. Power:
D. ThisistherateatwhichmagneticpotentialenergyU
Bisstoredinthe
magneticfield.
E. Thus
F. Finally,thetotalenergystoredbyaninductorLcarryingacurrentiis:
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G. Sampleproblem:
1. ForthecircuitofFig.below,assumethat =10.0V,R=6.70Ω,andL=5.50H.Theidealbatteryisconnectedattimet=0.(a)Howmuchenergyisdeliveredbythebatteryduringthefirst2.00s?(b)Howmuchofthisenergyisstoredinthemagneticfieldoftheinductor?(c)Howmuchofthisenergyisdissipatedintheresistor?
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H. EnergyDensityofaMagneticField:
1. Consideralengthlnearthemiddleofalongsolenoidofcross‐sectionalareaAcarryingcurrenti;thevolumeassociatedwiththislengthisAl.
2. TheenergyUBstoredbythelengthlofthesolenoidmustlieentirely
withinthisvolumebecausethemagneticfieldoutsidesuchasolenoidisapproximatelyzero.Also,thestoredenergymustbeuniformlydistributedwithinthesolenoidbecausethemagneticfieldis(approximately)uniformeverywhereinside.
3. Thus,theenergystoredperunitvolumeofthefieldis
4.
I. Sampleproblem:
1. Asolenoidthatis85.0cmlonghasacross‐sectionalareaof17.0cm2.Thereare950turnsofwirecarryingacurrentof6.60A.(a)Calculatetheenergydensityofthemagneticfieldinsidethesolenoid.(b)Findthetotalenergystoredinthemagneticfieldthere(neglectendeffects).
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IX. MutualInduction:
A. Diagram
B. ThemutualinductanceM21ofcoil2withrespecttocoil1is
definedas
1.
2. Therightsideofthisequationis,accordingtoFaraday’slaw,justthemagnitudeoftheemfE
2appearingincoil2duetothechangingcurrentincoil
1.
3. Similarly,
C. Sampleproblem:Twocoilsareatfixedlocations.Whencoil1hasnocurrentandthecurrentincoil2increasesattherate15.0A/s,theemfincoil1is25.0mV.(a)Whatistheirmutualinductance?(b)Whencoil2hasnocurrentandcoil1hasacurrentof3.60A,whatisthefluxlinkageincoil2?