ch 28 structured notes 2015 - usna 28 structured notes with...electrically charged particles, such...
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CH 28
MagneticFields
I. WhatProducesMagneticField?
A. Onewaythatmagneticfieldsareproducedistousemovingelectricallychargedparticles,suchasacurrentinawire,tomakeanelectromagnet.Thecurrentproducesamagneticfieldthatisutilizable.
B. Theotherwaytoproduceamagneticfieldisbymeansofelementaryparticlessuchaselectrons,becausetheseparticleshaveanintrinsicmagneticfieldaroundthem.
1. Themagneticfieldsoftheelectronsincertainmaterialsaddtogethertogiveanetmagneticfieldaroundthematerial.Suchadditionisthereasonwhyapermanentmagnet,hasapermanentmagneticfield.
2. Inothermaterials,themagneticfieldsoftheelectronscancelout,givingnonetmagneticfieldsurroundingthematerial.
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II. TheDefinitionofBfield:
A. Wecandefineamagneticfield,B,byfiringachargedparticlethroughthepointatwhichistobedefined,usingvariousdirectionsandspeedsfortheparticleanddeterminingtheforcethatactsontheparticleatthatpoint.Bisthendefinedtobeavectorquantitythatisdirectedalongthezero‐forceaxis.
B. Themagneticforceonthechargedparticle,FB,isdefinedtobe:
Hereqisthechargeoftheparticle,visitsvelocity,andBthemagneticfieldintheregion.
C. Themagnitudeofthisforceisthen:
HereisthesmallestanglebetweenvectorsvandB.
D. FindingtheMagneticForceonaParticle:
1. TheforceFBactingonachargedparticlemovingwithvelocityvthroughamagneticFieldBisALWAYSperpendiculartobothvandB.
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E. TheSIunitforBthatfollowsisnewtonpercoulomb‐meterpersecond.Forconvenience,thisiscalledthetesla(T):
1. Anearlier(non‐SI)unitforBisthegauss(G),and
2. Table
F. Sampleproblem:
1. Analphaparticletravelsatavelocity ofmagnitude550m/sthrough
auniformmagneticfield ofmagnitude0.045T.(Analphaparticlehasa
chargeof+3.2×10‐19Candamassof6.6×10‐27kg.)Theanglebetween
and is52°.Whatisthemagnitudeof(a)theforce actingontheparticle
duetothefieldand(b)theaccelerationoftheparticledueto ?(c)Doesthespeedoftheparticleincrease,decrease,orremainthesame?
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III. MagneticFieldLines:
A. ThedirectionofthetangenttoamagneticfieldlineatanypointgivesthedirectionofBatthatpoint.
B. ThespacingofthelinesrepresentsthemagnitudeofB—themagneticfieldisstrongerwherethelinesareclosertogether,andconversely.
C. Oppositemagneticpolesattracteachother,andlikemagneticpolesrepeleachother.
D. NosuchthingasaMagneticMonopoleEveryNorthPolehasanassociatedSouthPole!
E. FieldlinesemanatefromtheNorthPoleandterminateintheSouthPole.
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IV. CrossedFields,DiscoveryofanElectron:
A. Figure
B. WhenthetwofieldsinFig.28‐7areadjustedsothatthetwodeflectingforcesactingonthechargedparticlecancel,wehave
C. Thus,thecrossedfieldsallowustomeasurethespeedofthechargedparticlespassingthroughthem.
D. Thedeflectionofachargedparticle,movingthroughanelectricfield,E,betweentwoplates,atthefarendoftheplates(inthepreviousproblem)is
Here,vistheparticle’sspeed,mitsmass,qitscharge,andListhelengthoftheplates.
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V. CrossedFields,TheHallEffect:
A. AHallpotentialdifferenceVisassociatedwiththeelectricfieldacrossstripwidthd,andthemagnitudeofthatpotentialdifferenceisV=Ed.Whentheelectricandmagneticforcesareinbalance(Fig.28‐
8b), wherevdisthedriftspeed.But,
1. WhereJisthecurrentdensity,Athecross‐sectionalarea,etheelectroniccharge,andnthenumberofchargesperunitvolume.
B. Therefore, here,l=(A/d),thethicknessofthestrip.
1. Fig.28‐8Astripofcoppercarryingacurrentiisimmersedinamagneticfield.(a)Thesituationimmediatelyafterthemagneticfieldisturnedon.Thecurvedpaththatwillthenbetakenbyanelectronisshown.(b)Thesituationatequilibrium,whichquicklyfollows.Notethatnegativechargespileupontherightsideofthestrip,leavinguncompensatedpositivechargesontheleft.Thus,theleftsideisatahigherpotentialthantherightside.(c)Forthesamecurrentdirection,ifthechargecarrierswerepositivelycharged,theywouldpileupontherightside,andtherightsidewouldbeatthehigherpotential.
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VI. ACirculatingChargedParticle:
A. Consideraparticleofchargemagnitude|q|andmassmmovingperpendiculartoauniformmagneticfieldB,atspeedv.
B. Themagneticforcecontinuouslydeflectstheparticle,andsinceBandvarealwaysperpendiculartoeachother,thisdeflectioncausestheparticletofollowacircularpath.
C. Themagneticforceactingontheparticlehasamagnitudeof|q|vB.
D. Foruniformcircularmotion
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E. Diagram
1. Fig.28‐10Electronscirculatinginachambercontaininggasatlowpressure(theirpathistheglowingcircle).Auniformmagneticfield,B,pointingdirectlyoutoftheplaneofthepage,fillsthechamber.NotetheradiallydirectedmagneticforceF
B;forcircularmotiontooccur,F
Bmustpoint
towardthecenterofthecircle,(CourtesyJohnLeP.Webb,SussexUniversity,England)
VII. HelicalPaths:
A. Fig.28‐11(a)Achargedparticlemovesinauniformmagneticfield,theparticle’svelocityvmakingananglefwiththefielddirection.(b)Theparticlefollowsahelicalpathofradiusrandpitchp.(c)Achargedparticlespiralinginanonuniformmagneticfield.(Theparticlecanbecometrapped,spiralingbackandforthbetweenthestrongfieldregionsateitherend.)Notethatthemagneticforcevectorsattheleftandrightsideshaveacomponentpointingtowardthecenterofthefigure.
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B. Thevelocityvector,v,ofsuchaparticleresolvedintotwocomponents,oneparalleltoandoneperpendiculartoit:
C. Theparallelcomponentdeterminesthepitchpofthehelix(thedistancebetweenadjacentturns(Fig.28‐11b)).Theperpendicularcomponentdeterminestheradiusofthehelix.
Pitch = and Radius =
D. Themorecloselyspacedfieldlinesattheleftandrightsidesindicatethatthemagneticfieldisstrongerthere.Whenthefieldatanendisstrongenough,theparticle“reflects”fromthatend.Iftheparticlereflectsfrombothends,itissaidtobetrappedinamagneticbottle.
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VIII. SampleproblemsofChargedParticlesinMagneticFields
A. CrossedFields:
1. Anelectricfieldof1.50kV/mandaperpendicularmagneticfieldof0.400Tactonamovingelectrontoproducenonetforce.Whatistheelectron'sspeed?
B. TheHallEffect:
1. Astripofcopper150μmthickand4.5mmwideisplacedinauniform
magneticfield ofmagnitude0.65T,with perpendiculartothestrip.Acurrenti=23AisthensentthroughthestripsuchthataHallpotentialdifferenceVappearsacrossthewidthofthestrip.CalculateV.(Thenumberofchargecarriersperunitvolumeforcopperis8.47×1028electrons/m3.)
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C. ACirculatingChargedParticle:
1. AnelectronisacceleratedfromrestthroughpotentialdifferenceVandthenentersaregionofuniformmagneticfield,whereitundergoesuniformcircularmotion.Figure28‐37givestheradiusrofthatmotionversusV1/2.Theverticalaxis
scaleissetbyrs=3.0mm,andthehorizontalaxisscaleissetby .Whatisthemagnitudeofthemagneticfield?
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IX. MagneticForceonaCurrent‐CarryingWire:
A. AforceactsonacurrentcarryingwireinaBfield:
B. ConsideralengthLofthewireinthefigure.Alltheconductionelectronsinthissectionofwirewilldriftpastplanexxinatimet=L/vd.
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C. Thus,inthattimeachargewillpassthroughthatplanethatisgivenby
HereLisalengthvectorthathasmagnitudeLandisdirectedalongthewiresegmentinthedirectionofthe(conventional)current.
D. Ifawireisnotstraightorthefieldisnotuniform,wecanimaginethewirebrokenupintosmallstraightsegments.Theforceonthewireasawholeisthenthevectorsumofalltheforcesonthesegmentsthatmakeitup.Inthedifferentiallimit,wecanwrite
andwecanfindtheresultantforceonanygivenarrangementofcurrentsbyintegratingEq.28‐28overthatarrangement.
E. Sampleproblem:
1. Awire1.80mlongcarriesacurrentof13.0Aandmakesanangleof35.0°withauniformmagneticfieldofmagnitudeB=1.50T.Calculatethemagneticforceonthewire.
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X. TorqueonaCurrentLoop:
A. Considernow,aloopofwirewithcurrentflowingthroughit:
B. ThetwomagneticforcesFand–Fproduceatorqueontheloop,tendingtorotateitaboutitscentralaxis.
C. Todefinetheorientationoftheloopinthemagneticfield,weuseanormalvectornthatisperpendiculartotheplaneoftheloop.Figure28‐19bshowsaright‐handruleforfindingthedirectionofn.InFig.28‐19c,thenormalvectoroftheloopisshownatanarbitraryangletothedirectionofthemagneticfield.
RHR for µ
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D. Forside2themagnitudeoftheforceactingonthissideisF2=ibB
sin(90°‐)=ibBcos=F4.F
2andF
4canceloutexactly.
E. ForcesF1andF
3havethecommonmagnitudeiaB.AsFig.28‐19c
shows,thesetwoforcesdonotsharethesamelineofaction;sotheyproduceanettorque.
F. ForNloops,whenA=ab,theareaoftheloop,thetotaltorqueis:
G. SampleProblem
1. Anelectronmovesinacircleofradiusr=5.29×10‐11mwithspeed2.19×106m/s.Treatthecircularpathasacurrentloopwithaconstantcurrentequaltotheratiooftheelectron'schargemagnitudetotheperiodofthemotion.IfthecircleliesinauniformmagneticfieldofmagnitudeB=7.10mT,whatisthemaximumpossiblemagnitudeofthetorqueproducedontheloopbythefield?
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XI. TheMagneticDipoleMoment,:
A. Definition:
Here,Nisthenumberofturnsinthecoil,iisthecurrentthroughthecoil,andAistheareaenclosedbyeachturnofthecoil.
B. Direction:Thedirectionofisthatofthenormalvectortotheplaneofthecoil.
C. Thedefinitionoftorquecanberewrittenas:
D. Justasintheelectriccase,themagneticdipoleinanexternalmagneticfieldhasanenergythatdependsonthedipole’sorientationinthefield:
E. Amagneticdipolehasitslowestenergy(‐Bcos0=‐B)whenitsdipolemomentislinedupwiththemagneticfield.Ithasitshighestenergy(‐Bcos180°=+B)whenisdirectedoppositethefield.
F. Fromthebelowequations,onecanseethattheunitofcanbethejoulepertesla(J/T),ortheampere–squaremeter.
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G. SampleProblem:
1. Amagneticdipolewithadipolemomentofmagnitude0.020J/Tisreleasedfromrestinauniformmagneticfieldofmagnitude52mT.Therotationofthedipoleduetothemagneticforceonitisunimpeded.Whenthedipolerotatesthroughtheorientationwhereitsdipolemomentisalignedwiththemagneticfield,itskineticenergyis0.80mJ.(a)Whatistheinitialanglebetweenthedipolemomentandthemagneticfield?(b)Whatistheanglewhenthedipoleisnext(momentarily)atrest?