Induction and Inductance - USNA 30... · [SHIVOK SP212] February 20, 2016 Page 1 CH 30 Induction...

[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?

Induction and Inductance - USNA 30... · [SHIVOK SP212] February 20, 2016 Page 1 CH 30 Induction...

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