intermediate physics - · quickcheck 25.10 a bar magnet sits inside a coil of wire that is...

IntermediatePhysics

PHYS102

Dr RichardH.CyburtAssistantProfessorofPhysics

Myoffice:402cintheScienceBuilding

Myphone:(304)384-6006

Myemail:[email protected]

Mywebpage:www.concord.edu/rcyburt

Inpersonoremailisthebestwaytogetaholdofme.

PHYS102

MyOfficeHoursTWR9:30-11:00amW4:00-5:00pm

Meetingsmayalsobearrangedatothertimes,byappointment

PHYS102

ProblemSolvingSectionsIwouldliketohavehour-longsectionsforworkingthroughproblems.

Thiswouldbeanextracomponenttothecourseandcounttowardsextracredit

TR1-2pm

WF10-11am

S308

Ifyoucan’tmakethese,youcanstillpickuptheproblemworksheet.

PHYS102

Midterm2Thursday,March2duringclass8-9:15

Allowedonehalfsheet(8.5x11)pieceofpaperw/notes/formuliCalculatorpencilorblue/blackpen

Review,Wednesday,March17-9pm

Bringquestions!!!

PHYS102

IntermediatePhysics

PHYS102

PHYS102

DouglasAdamsHitchhiker’sGuidetotheGalaxy

Inclass!!

PHYS102

Thislecturewillhelpyouunderstand:Faraday’sLaw

AlternatingCurrent

ACElectricityandTransformers

PHYS102

Lenz’sLaw

©2015PearsonEducation,Inc.

Text:p.813

QuickCheck25.8Thebarmagnetispushedtowardthecenterofawireloop.Whichistrue?

Thereisaclockwiseinducedcurrentintheloop.

Thereisacounterclockwiseinducedcurrentintheloop.

Thereisnoinducedcurrentintheloop.







1. Upward flux from magnet is increasing.2. To oppose the increase, the field of the induced current points down.3. From the right-hand rule, a downward field needs a cw current.






Magnetic flux is zero, so there’s no change of flux.

QuickCheck25.10Abarmagnetsitsinsideacoilofwirethatisconnectedtoameter.Foreachofthefollowingcircumstances1.Thebarmagnetisatrestinthecoil,2.Thebarmagnetispulledoutofthecoil,3.Thebarmagnetiscompletelyoutofthecoilandatrest,4.Thebarmagnetisreinsertedintothecoil,

Whatcanwesayaboutthecurrentinthemeter?A. Thecurrentgoesfromrighttoleft.B. Thecurrentgoesfromlefttoright.C. Thereisnocurrentinthemeter.


QuickCheck25.10Abarmagnetsitsinsideacoilofwirethatisconnectedtoameter.Foreachofthefollowingcircumstances1.Thebarmagnetisatrestinthecoil,C2.Thebarmagnetispulledoutofthecoil,A3.Thebarmagnetiscompletelyoutofthecoilandatrest,C4.Thebarmagnetisreinsertedintothecoil,B

Whatcanwesayaboutthecurrentinthemeter?A. Thecurrentgoesfromrighttoleft.B. Thecurrentgoesfromlefttoright.C. Thereisnocurrentinthemeter.


QuickCheck25.11Amagneticfieldgoesthroughaloopofwire,asatright.Ifthemagnitudeofthemagneticfieldis1.Increasing,2.Decreasing,3.Constant,

Whatcanwesayaboutthecurrentintheloop?Answerforeachofthestatedconditions.A. Theloophasaclockwisecurrent.B. Theloophasacounterclockwisecurrent.C. Theloophasnocurrent.


QuickCheck25.11Amagneticfieldgoesthroughaloopofwire,asatright.Ifthemagnitudeofthemagneticfieldis1.Increasing,B2.Decreasing,A3.Constant,C

Whatcanwesayaboutthecurrentintheloop?Answerforeachofthestatedconditions.A. Theloophasaclockwisecurrent.B. Theloophasacounterclockwisecurrent.C. Theloophasnocurrent.


Themagneticfieldisconfinedtotheregioninsidethedashedlines;itiszerooutside.Themetalloopisbeingpulledoutofthemagneticfield.Whichistrue?

A. Thereisaclockwiseinducedcurrentintheloop.

B. Thereisacounterclockwiseinducedcurrentintheloop.

C. Thereisnoinducedcurrentintheloop.

QuickCheck25.12


Themagneticfieldisconfinedtotheregioninsidethedashedlines;itiszerooutside.Themetalloopisbeingpulledoutofthemagneticfield.Whichistrue?

A. Thereisaclockwiseinducedcurrentintheloop.

B. Thereisacounterclockwiseinducedcurrentintheloop.

C. Thereisnoinducedcurrentintheloop.

1. The flux through the loop is into the screen and decreasing.

2. To oppose the decrease, the field of the induced current must point into the screen.

3. From the right-hand rule, an inward field needs a cwcurrent.

QuickCheck25.12


Section25.4Faraday’sLaw


Faraday’sLawAninducedemf ℇ istheemf associatedwithachangingmagneticflux.

ThedirectionofthecurrentisdeterminedbyLenz’slaw.Thesizeoftheinducedemf isdeterminedbyFaraday’slaw.


Faraday’sLawFaraday’slaw isabasiclawofelectromagneticinduction.Itsaysthatthemagnitudeoftheinducedemf istherateofchange ofthemagneticfluxthroughtheloop:


Faraday’sLawAcoilwireconsistingofN turnsactslikeN batteriesinseries,sotheinducedemf inthecoilis


Faraday’sLawsTherearetwofundamentallydifferentwaystochangethemagneticfluxthroughaconductingloop:1. Theloopcanmoveorexpandorrotate,creatinga

motionalemf.2. Themagneticfieldcanchange.

Theinducedemf istherateofchangeofthemagneticfluxthroughtheloop,regardlessofwhatcausesthefluxtochange.


Faraday’sLaws


Text:p.815

ExampleProblemThefollowingfigureshowsa10-cm-diameterloopinthreedifferentmagneticfields.Theloop’sresistanceis0.1Ohms.Foreachsituation,determinethemagnitudeanddirectionoftheinducedcurrent.


EddyCurrentsTherearetwo“loops”lyingentirelyinametalsheetbetweentwomagnets.

Asthesheetispulled,theloopontherightisleavingthemagneticfield,andthefluxisdecreasing.

AccordingtoFaraday’slaw,thefluxchangeinducesacurrenttoflowaroundtheloop.Lenz’slawsaysthecurrentflowsclockwise.


EddyCurrentsTheloopontheleftsideofthemetalentersthefieldandsothefluxthroughitisincreasing.Lenz’slawrequirestheinduced“whirlpool”currentontheleftlooptobecounterclockwise.


EddyCurrentsEddycurrentsarethespread-outwhirlpoolsofaninducedcurrentinasolidconductor.Bothwhirlpoolsofcurrentaremovinginthesamedirectionastheypassthroughthemagnet.Themagneticfieldexertsaforceonthecurrent,oppositethedirectionofpull,actingasabrakingforce.


EddyCurrentsBecauseofthebrakingforceexertedbythemagneticfield,anexternalforceisrequiredtopullametalthroughamagneticfield.Ifthepullingforceceases,themagneticbrakingforcequicklycausesthemetaltodecelerateuntilitstops.


Section26.1AlternatingCurrent


AlternatingCurrentAbatteryproducesaconstantemf soaflashlighthasaconstantglow.Theelectricitydistributedtohomesisdifferent.Ratherthanaconstantemf,ithasasinusoidalvariation.Alightbulb inyourhomeisonwhentheemf ispositive,butnotwhenitisnegative.Theresultingflickeristoorapidtonoticeundernormalcircumstances.


AlternatingCurrentInChapter25wesawthatanelectricalgeneratorworksbyrotatingacoilofwireinamagneticfield.Thesteadyrotationofthecoilcausestheemf andtheinducedcurrentinthecoiltooscillatesinusoidally,alternatingpositiveandthennegative.Theoscillationforceschargestoflowfirstinonedirectionandthentheotherahalfcyclelater.Thisisanalternatingcurrent:AC.Iftheemf isconstantandthecurrentisalwaysinthesamedirection,theelectricityiscalleddirectcurrent: DC.


AlternatingCurrentTheinstantaneousemf ofanACvoltagesourcecanbewrittenas

ℇ0isthepeakormaximumemf,T istheperiodofoscillation,andf =1/Tisthefrequency.


ResistorCircuitsIncircuitswherethecurrentandvoltageareoscillating,weusei torepresenttheinstantaneous currentandv fortheinstantaneousvoltage.ThepotentialdifferenceacrossaresistorR iscalledtheresistorvoltagevR.ItisrelatedtothecurrentiR throughtheresistorbyOhm’slaw:

vR =iRR


ResistorCircuits


ResistorCircuits

WecananalyzeacircuitwitharesistorconnectedacrossanACemf thesamewaywedidwithaDCresistorcircuit.Kirchhoff’slooplawsaysthatthesumofallthepotentialdifferencesaroundaclosedpathiszero:


ResistorCircuitsTheresistorvoltageisasinusoidalvoltageatfrequencyf:

vR =VR cos(2p ft)VR isthepeakormaximumvoltage,theamplitudeofthesinusoidallyvaryingvoltage.Inthesingleresistorcase,VR =ℇ0,sothecurrentthroughtheresistoris

IR=VR/Risthepeakcurrent.©2015PearsonEducation,Inc.

ResistorCircuitsTheresistor’sinstantaneouscurrentandvoltageoscillateinphase.

Thecurrentisatitsmaximumandminimumvalueswhenthevoltageisatitsmaximumandminimumvalues.


ACPowerinResistorsTheinstantaneouspoweriswrittenas

p =iR2R =[IR cos(2p ft)]2R =IR2R[cos(2p ft)]2


ACPowerinResistorsThepoweroscillatestwiceduringeverycycleoftheemf.


ACPowerinResistorsThecurrentinalightbulbreversesdirection120timespersecond,sothepowerreachesamaximumtwicepersecond.Howeverthebulbglowssteadily,sotheaveragepower ismoreuseful.TheaveragepowerPR is


ACPowerinResistorsRecallthatthepowerinaDCcircuitisPR=I 2R =V 2/R.

Theroot-mean-squarecurrentandtheroot-mean-squarevoltagearemoreusefulexpressionsofpower:

Usingtheroot-mean-squarevalues,wewritetheaveragepowerdissipatedinanACcircuitas:

Aslongasyouworkwithrms voltagesandcurrents,alltheexpressionsforDCpowercarryovertoACpower.


Example26.1TheresistanceandcurrentofatoasterThehotwireinatoasterdissipates580Wwhenpluggedintoa120Voutlet.a. Whatisthewire’sresistance?b. Whataretherms andpeakcurrentsthroughthewire?PREPAREWe’veseenthatthe120Voutletvoltageisanrms value.ThefilamenthasresistanceR.Itdissipates580Wwhenthere’sanrmsvoltageof120Vacrossit.WecansolveEquation26.9forR andthenuseEquations26.9and26.8tofindtherms currentandthepeakcurrent.


Example26.1Theresistanceandcurrentofatoaster(cont.)SOLVE

a. WerearrangeEquation26.9tofindtheresistancefromthermsvoltageandtheaveragepower:


Example26.1Theresistanceandcurrentofatoaster(cont.)b. AsecondrearrangementofEquation26.9allowsustofindthe

currentintermsofthepowerandtheresistance,bothofwhichareknown:

FromEquation26.8,thepeakcurrentis


Example26.1Theresistanceandcurrentofatoaster(cont.)ASSESS Wecandoaquickcheckonourworkbycalculatingthepowerfortheratedvoltageandcomputedcurrent:

PR =Irms Vrms =(4.8A)(120V)=580WThisagreeswiththevaluegivenintheproblemstatement,givingusconfidenceinoursolution.


ExampleProblemA120V(rms)ACpowersupplyisconnectedtoamotor,whichisratedat100W.A. Whatistherms currentinthecircuit?B. Now,supposethatthewiresusedtoconnectthemotortothepowersupply

havearesistanceof7.0Ω.Assumethattherms currentstaysthesame.Whatisthevoltagedropacrosstheresistanceofthewires?Whatisthevoltageatthemotornow?

C. Now,supposethepowersupplyis1200V,andthemotorisratedat100Watthishighervoltage.Whatisthecurrentinthecircuitassumingnoresistanceinthewires?

D. Ifthewireshavearesistanceof7.0Ω,whatisthevoltagedropacrossthewires?Thevoltageatthemotor?


Section26.2ACElectricityandTransformers


ACElectricityandTransformersAtransformerisadevicethattakesanalternating voltageasaninputandproduceseitherahigherorlowervoltageasoutput.Theoperationofatransformerisbasedontheemf producedbychangingmagneticfields,sotheinputmustbeACelectricity.


TransformerOperationAsimplifiedversionofatransformerconsistsoftwocoilswrappedonasingleironcore.


TransformerOperationInasimpletransformer,theprimarycoil isconnectedtoanACvoltage.TheACvoltagecreatesanalternatingcurrent.Thecurrentinthecoilcreatesamagneticfieldthatmagnetizestheironcoretoproduceamuchstrongernetfield.Thestrongfluxfollowstheironcoreandentersthesecondarycoil.


TransformerOperationThecurrentintheprimarycoilofatransformerisanalternatingcurrent;itcreatesanoscillatingmagneticfieldintheironcore.

Thechangingmagneticfieldmeansthereisachangingmagneticfluxthroughthesecondarycoil,whichinducesanemf,anACvoltageV2 inthecoil.

Aresistor,orload, isconnectedtodissipatepower.


TransformerOperation


TransformerOperationThepurposeofatransformeristochangethevoltage,soweneedtounderstandhowthevoltageintheprimarycoilofatransformerrelatestothatinthesecondarycoil.AccordingtoFaraday’slaw,aninstantaneousvoltageacrossN1 turnsduetothemagneticfluxF intheprimarycoilis


TransformerOperationInanidealtransformer, allthefluxis“guided”bytheironcorethroughthesecondarycoil.Thechangingfluxinducesanemf acrossthesecondarycoilgivenby

BecauseΔF/Δt isthesameineachcoil,wecanwritearatioofinstantaneousvoltages:


TransformerOperationWecanalsorelatethepeakandrms voltagesoftheprimaryandsecondarycoils:

Astep-uptransformer, withN2> N1,increasesthevoltage,whileastep-down transformer, withN2 < N1,lowersthevoltage.


Wecanrelatethepeakcurrentsbetweenthecoilsinatransformerwithanexpressionthatmirrorstheexpressionofpeakvoltagesbetweenthecoils:

A step-uptransformerraisesvoltagebutlowerscurrent;astep-downtransformerlowersvoltagebutraisescurrent.

TransformerOperation


QuickCheck26.1Iftheprimarycoilofwireonatransformeriskeptthesameandthenumberofturnsofwireonthesecondaryisincreased,howwillthisaffectthevoltageobservedatthesecondary?

A. Thevoltagewillincrease.B. Thevoltagewillstaythesame.C. Thevoltagewilldecrease.


QuickCheck26.2Supposethatanidealtransformerhas400turnsinitsprimarycoiland100turnsinitssecondarycoil.Theprimarycoilisconnectedtoa120V(rms)electricoutletandcarriesanrms currentof10mA.Whataretherms valuesofthevoltageandcurrentforthesecondary?

A. 480V,40mAB. 480V,2.5mAC. 30V,40mAD. 30V,2.5mA


Example26.2Analyzingastep-downtransformerAbooklighthasa1.4W,4.8Vbulbthatispoweredbyatransformerconnectedtoa120Velectricoutlet.Thesecondarycoilofthetransformerhas20turnsofwire.Howmanyturnsdoestheprimarycoilhave?Whatisthecurrentintheprimarycoil?


Example26.2Analyzingastep-downtransformer(cont.)

Weknowthevoltagesoftheprimaryandthesecondary,sowecancomputetheturnsintheprimarycoilusingEquation26.13.Weknowthevoltageandthepowerofthebulb,sowecanfindthecurrentinthebulb.Thisisthecurrentinthesecondary,whichwecanuseinEquation26.14tofindthecurrentintheprimary,thecurrentprovidedbytheoutlet.


Example26.2Analyzingastep-downtransformer(cont.)SOLVE Thebulbisratedat4.8V;thisistherms voltageatthesecondary,so(V2)rms =4.8V.Thepoweroutlethastheusual(V1)rms =120V,sowecanrearrangeEquation26.13tofind


Example26.2Analyzingastep-downtransformer(cont.)Thebulbconnectedtothesecondarydissipates1.4Wat4.8V;thisisanrms voltage,sotherms currentinthesecondaryis


Example26.2Analyzingastep-downtransformer(cont.)WecanthenrearrangeEquation26.14tofindtherms currentintheprimary:


Example26.2Analyzingastep-downtransformer(cont.)ASSESS Wecancheckourresultsbylookingatthepowersuppliedbythewalloutlet.ThisisP1 =(120V)(0.012A)=1.4W,thesameasthepowerdissipatedbythebulb,asmustbethecasebecausewe’veassumedthetransformerisideal.


PowerTransmission


Example26.3PracticalpowertransmissionToprovidepowertoasmallcity,apowerplantgenerates40MWofACelectricity.Thepowerplantis50kmfromthecity(atypicaldistance),andthe100kmofwireusedinthetransmissionline(tothecityandback)hasaresistanceof7.0Ω.


Example26.3Practicalpowertransmission(cont.)a. Toprovide40MWofpoweratthegeneratorvoltageof25,000V,

whatcurrentisrequired?b. Whatisthepowerdissipatedintheresistanceofthetransmission

lineforthiscurrent?c. Toprovide40MWofpowerat500,000V,whatcurrentis

required?d. Whatisthepowerdissipatedintheresistanceofthetransmission

lineforthishighervoltage?


Example26.3Practicalpowertransmission(cont.)PREPAREWecantreatthecity—andthewiresthattransmitpowertoit—asaload.Allofthevoltagesarerms valuesandthepowerisanaveragepower,sowecanfindthecurrenttoprovideagivenpowerandthepowerdissipatedusingtherelationshipsinEquation26.9.


Example26.3Practicalpowertransmission(cont.)SOLVE

a. Toprovide40MWatthegeneratorvoltageof25,000V,thecurrentis


Example26.3Practicalpowertransmission(cont.)

b. Passingthiscurrentthroughthetransmissionlineswillresultinpowerdissipationinthe7.0Ωresistanceofthewires.Wedon’tknowthevoltagedropacrossthewires,butwedoknowthecurrentandresistance,sowecancompute

Pdissipated inwires =(Irms)2R =(1600A)2 (7.0Ω)=18MW

Thisisnearlyhalfthepowergenerated,clearlyanunacceptableloss.


Example26.3Practicalpowertransmission(cont.)c. Increasingthetransmissionvoltageto500kVreducesthe

necessarycurrent:

Thisisaremarkablysmallcurrenttosupplyacity.Ifyouuseseveralhigh-powerappliancesatonetime,youcouldeasilyusethismuchcurrentinyourhouse.Butthenecessarycurrentforthecitycanbesosmallbecausethevoltageissolarge.


Example26.3Practicalpowertransmission(cont.)d. Therelativelysmallcurrentmeansthatthepowerdissipatedin

theresistanceofthewireswillbesmallaswell:Pdissipated inwires =(Irms)2R =(80A)2 (7.0Ω)=0.045MW

Thisisonlyabout0.1%ofthepowergenerated,whichisquitereasonable.


Example26.3Practicalpowertransmission(cont.)ASSESS Thefinalresult—thepowerdissipatedinthewiresisdramaticallyreducedforanincreasedtransmissionvoltage—isjustwhattheexamplewasdesignedtoillustrate.


PowerTransmissionTransmittingelectricityathighvoltagesmeansasmallercurrentisrequired,andthereforetheresultingpowerlossismoremanageablethanforlowvoltages(andlargercurrents).Thisiswhyelectricaltransmissionlinesrunathighvoltages.Inordertotransmitelectricityathighvoltages,weneedtousetransformerstoincreasethevoltage,whichrequiresACelectricity.ThisiswhyweuseACpowereventhoughitisslightlymoredangerousthanDCpower.


intermediate physics - · quickcheck 25.10 a bar magnet sits inside a coil of wire that is...

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