AQA A2 Physics A chapter 8 textbook answers

Answers to examination-style questions

AQA Physics A A2 Level © Nelson Thornes Ltd 2009 1

Answers Marks Examiner’s tips

1 (a) Ammeterdeflectsinonedirection 1 Themagneticfluxthroughthecoilincreasesandinonedirection,andthendecreases,astheN-polepassesthrough.ThisprocessisthenrepeatedintheoppositedirectionastheS-polepassesthrough.Theammeterdeflectsonlywhenthereisrelativemovementbetweenthemagnetandcoil.

andthenintheoppositedirection. 1

(b) (i) Theaccelerationofthemagnetwouldbereduced . . .

1 AccordingtoLenz’s law,thedirectionofaninducedemfisalwayssuchastooppose the change of fluxthatproducesit.Inthisexamplethechangeoffluxcanbeopposedbytryingtopreventthemagnetfromdroppingintothecoil.

becausethemagneticfieldproducedbythecurrentinducedinthecoilgivesanupwardsforceontheN-poleofthemagnetthatopposesthemovementoftheN-poletowardsthecoil.

1

(ii) Theaccelerationofthemagnetwouldagainbereduced . . .

1 Asthemagnetleaves,thechangeoffluxcanbeopposedbytryingtokeepthemagnetinthecoil.TheinducedmagneticfieldthereforecausesaforceofattractionontheS-poleofthemagnet.

becausethemagneticfieldproducedbythecurrentinducedinthecoilgivesanupwardsforceontheS-poleofthemagnetthatopposesthemovementoftheS-poleawayfromthecoil.

1

(c) Relevant points include: • Themagnetwouldnowfallwithan

accelerationofg • Anemfwouldstillbeinducedinthecoil • Nocurrentcouldbeproducedbythe

inducedemf • Noinducedmagneticfieldwouldbe

produced • Therewouldbenoopposingforceson

themagnetduetoaninducedcurrent.

any 3 Therecanbeacurrentonlyifacircuitiscomplete.Withthemeterdisconnected,theemfcannotgiveacurrent,andthecoilwillthereforenotproduceanopposingmagneticfield.An alternative explanationwouldbeintermsofenergy:noenergycouldbedissipatedfromthecircuitiftherewerenocurrent,andthemagnetwouldlosenoneofitskineticenergy.

2 (a) (i) frepresentsmagneticflux 1 Donotconfusethiswithflux linkage,whichis(numberofturnsoncoil)× f.

(ii) Theunitoffisweber(Wb),orteslametre2(Tm2).

1 f =(fluxdensity)× (areaatrightanglestoflux) = BATheunitofBistesla,sotheunitoffisT m2.1Tm2 =1Wb

(b) (i) ThemaximumemfoccurswhenΔB ___ Δt

isamaximumi.e.wherethegradientoftheB–tgraphisgreatest,

∴drawtangenttolineatt =0.5or1.0s (t =0isalsoacceptable).

1 Themagnitudeoftheinducedemfisgivenby

ε = N Δf

 ___ Δt = N Δ(BA)

______ Δt .

InthisexampleN andAareconstant,

henceε ∝ ΔB ___ Δt

Chapter 8

Chapter 8Answers to examination-style questions



Fromtangent,ΔB ___ Δt ≈ 6 ×10−3Ts−1

∴ maximumemf= N A ΔB ___ Δt

1 Usearulertodrawthelongeststraightlineyoucanthatisatangenttothegraphlineatitsmaximumslope.Takereadingsfromtheendsofyourline.Sometolerance(say20%)wouldhavetobeallowedintheanswer:perhaps(2.9to4.3)×10−4V.

=240×2.5×10−4 × 6 ×10−3

=3.6×10−4 V1

(ii) Graph drawn to show: • Both+and−values,withzeroemf

at0.25and0.75s • Peakemfat0.5and1.0sonagraph

ofthecorrect(cosine)shape

2 AccordingtoLenz’slaw,apositiverateofchangeofflux(gradient)willproduceanegativeemf.Technically,yourgraphshouldbea−(cosine)curve.

(iii)Pendulumwouldhavetobeshorter; 1 ForapendulumT ∝ √______

length,and

frequencyf = 1 __ T . 1 _ 4 oftheoriginallength. 1

(iv) Themaximumspeedofthemagnetisincreased

1 Insimpleharmonicmotion,vmax = 2πfA,implyingvmax∝ fwhenAisunchanged. ∴ thecoilcutsthefluxatahigherrate

orrateofchangeoffluxincreases.1

(v) Possiblealternativeways: • useastrongermagnet • useacoilwithmoreturns • useacoilofgreaterarea • placeasoftironcoreinthecoil • usealargeramplitudeofoscillation

ofthemagnet.

any 2 Thereareplentyofotherwaysofincreasingthemaximumemf.Anythingthatincreasestherateofchangeoffluxwillsuffice.Takecarewithyourwords,however: not a biggermagnet,andnot a magnetofgreaterarea.

3 (a) Distancebetweenpulses=7.2cm ∴timebetweenpulses=72ms

1 EachsquareinFig.5shouldhaveasideof1cm;thediagramisnottoscale.Sometolerancewouldhavetobeallowedintheanswer,saybetween71and73ms.Thereisa+and−voltagepulseeverytimeoneofthemagnetspassesthecoil.Thishappensfourtimesperrevolutionoftheaxle.Theoscilloscopetimebaseisseton10 mscm−1,fromwhichthetimebetweenpulsescanbefound.

Timefor1revolutionofaxle=4×72 =288ms

1

Numberofrevolutionsperminute

= ( 1 _________ 288 ×10−3 ) × 60 = 208

or205−211withanextendedrangeofvaluesofthetimebetweenpulses.

1

(b) (i) Relevant points are: • Movementofamagnetchangesthe

magneticfluxthroughthecoil(orchangesthefluxlinkedwiththecoil)

• Theinducedemfisproportionaltotherateofchangeoffluxlinkedwiththecoil(oranemfisproducedbecausethemagneticfluxcutsthroughthecoil).

2 AstheN-poleofamagnetapproachesthecoil,theincreaseofmagneticfluxproducesapositivevoltagepulse;asitleavesthedecreaseinfluxproducesanegativepulse.Faraday’slawrelatesinducedemftotherateofchangeoffluxlinkingthecoil.




(ii) Peakinducedemf=1.5×5=7.5mV 1 ReadingshavetobetakenfromthecoarsescaleonthescreeninFigure2,sotherewouldhavetobesometoleranceintheacceptedanswers.Readingthepeakemfas7.6mVwouldgive2.17Wbs−1asthefinalanswerhere.

inducedemf= N Δf

 ___ Δt 1

∴ Δf

 ___ Δt = inducedemf ___________ N = 7.5×10−3 _________ 350 1

=2.14×10−5Wbs−1

or2.1 ×10−5Wbs−1 orTm2s−1

1

(c) Relevant points include: • Directionofinducedemf(orcurrent)

opposesthechangethatproducesit • Apulseisproducedasamagnet

approachesand leavesacoil • Forcesonamagnetrepelitasit

approachesandattractitasitleaves • Currentthereforeflowsonewayasthe

magnetapproachesandtheoppositewayasitleaves.

any 3 Theemfinducedinacoilcausesacurrentinit.Thiscurrentproducesamagneticfieldthatopposesthemovementofthemagnet,repellinganapproachingmagnetandattractingamagnetthatismovingaway.Thecurrentflowsinoppositedirectionstoprovidereversalofthefielddirection.

(d) WaveformdrawnonFigure2toshow: • largerpeakvoltages • +and−peaksclosertogether • narrower(sharper)peaks • setsofpeaksclosertogether.

any 3 Fasterrotationwouldproduceagreaterrateofchangeofflux(thereforelargervoltages),andquickereventsthatoccurmorefrequently.

4 (a) Diagramcompletedtoshowaworkablearrangement,with:

• ahorizontalcurrent-carryingconductoratrightanglestothefluxlines,apivotandabalancingweight

• magneticforceshownbyalabelledarrowdirected,forexample,downwards

• currentdirectionlabelledconsistentwithdirectionofforce:forexample,intotheplaneofthepage(towardsthebackofthearrangementofmountedmagnets).

3 Acurrentbalanceisanapparatuswhichgivesthepossibilityofmeasuringamagneticfieldby‘weighing’theforcethatactsonacurrent-carryingconductorinamagneticfield.Asimplearrangementconsistsofawireframethatispivoted,withweightsononesidetobalancethemagneticforceontheotherside.TheforceandcurrentdirectionsshowninyourdiagrammustbeconsistentwithFleming’sleft-handrule.

(b) (i) Forceononewireinthecoil = BIl =40×10−3 ×0.55×0.15 =3.3×10−3N

1 Part (b)revisesworkcoveredinChapter7.Thecoilissquare,soitslengthisindistinguishablefromitswidth.Eachsideisoflength0.15m.Donotoverlookthefactthatthecoilhas20turns.

Forceononesideofthecoil = 20 ×3.3×10−3 =6.6×10−2N

1

(ii) Anemfisinducedoncethecoilisrotating,becausethecoilcutsthemagneticfluxcontinuously.

1 TheinducedemfgeneratedinarotatingmotorcoilisknownasabackemfandisanexampleofLenz’slaw.Itseemsstrangethatamotorcarriesmaximumcurrentwhenitfirststartsandminimumcurrentwhenitrotatesfastest.

Thisinducedemfopposestheemfthatiscausingthecoiltorotate,therebyreducingthecurrent.

1




(iii)Appliedpd/voltagefromsupply= I0R =0.55×0.50=0.275V netpd/voltagewhencoilrotates= IR =0.14×0.50=0.070V

1 ThevoltagefromthesupplyconnectedtothemotorcanbefoundusingV = IR whenthecoilisnotturningsincethereisnobackemf.Whenthecoilstartstoturnthereisasmallervoltageacrosstheresistanceofthecoil–thedifferencesinthesetwovoltagesistheinducedemfinthecoil.Thisinducedemfisequaltotherateofchangeofflux linkage,nottherateofchangeofflux.Onceagain,donotforgetthatthecoilhas20 turns.

∴backemf=0.275−0.070=0.205V 1

Inducedbackemf= N Δf

 ___ Δt gives

Δf

 ___ Δt = inducedemf ___________ N = 0.205 _____ 20

1

∴maximumrateatwhichfluxiscutbycoil=1.03(or1.02) ×10−2Wbs−1

1

5 (a) Magneticfluxlinkageistheproductofmagneticfluxandthenumberofturnsthatcuttheflux

or = Nf( = NBA,whereBisnormaltoA),withtermsdefined.

1 Magneticfluxlinkageisausefulideabecauseaninducedemfisproportionaltotherateofchangeoffluxlinkage.ItsunitiswrittenbyconventionasWbturn,eventhoughthenumberofturnsisjustanumber.NfisnotusuallyexpressedinWb,althoughthatisdimensionallycorrect.

TheunitofNfisweberturn(Wbturn) 1

(b) Relevant points include: • Thealternatingvoltage,Vin,suppliedto

theinputcausesanalternatingcurrentintheprimarycoil

• Theacintheprimarycoilproducesanalternatingmagneticfluxinthecore

• Thisalternatingfluxpassesthroughthesecondarycoil

• Theinducedemf,Vout,inthesecondarycoilisproportionaltotherateofchangeoffluxlinkage

• Therearefewerturnsonthesecondarythanontheprimary,soVout < Vin

any 4 A step-down transformerisonethatisusedtoreducethevoltageofanalternatingsupply.Themagneticfieldproducedbythecurrentintheprimaryisconcentratedinthemagneticallysoftcore.Anemfisinducedinthesecondarycoilbecauseitisinacontinuouslychanging,alternatingmagneticfield.AnidealtransformerwithequalnumbersofturnsonthetwocoilswouldgiveVout = Vin.Fewerturnsonthesecondaryreducesthefluxlinkagewiththesecondary,reducingVout.

(c) (i) UseofVs __ Vp

= Ns ___ Npgives

Vs ____ 230 = 1 ___ 15 1 The‘turnsratio’,whichisgiveninthequestionas15:1,isNp:Ns,i.e.primaryturns:secondaryturns.Theoutputvoltageisthevoltageacrossthesecondarycoil.

fromwhichoutputvoltage Vout = Vs =15.3 V

1




(ii) Reasons for imperfect efficiency: • Thermalenergylossescausedbythe

currentsinthecoils,whichhaveresistance

• Energyisrequiredtomagnetise,demagnetiseandre-magnetisethecore

• Eddycurrentsinducedinthecorecausethermalenergylossesinthecore

• Imperfectfluxlinkage:notallthemagneticfluxproducedbytheprimarycoilpassesthroughthesecondarycoil.

any 2 Norealtransformercaneverbe100%efficient,butpracticaltransformersdohaveveryhighefficiencies(approaching,andsometimesexceeding,99%).Donotjustwrite“heating”or“heatloss”,trytogivesomedetailofwhereandhowtheheatlossoccurs.

6 (a) (i) UseofNs ___ Np

= Vs __ Vpgives

Ns _____ 3000 = 12 ____ 230 1 Correctuseoftheturnsratioequation(whichisgivenintheDataBooklet)shouldleadtotwoeasymarks.Theanswerworksouttobe156.52,butyoumuststateitasawhole number.Transformerscannothavefractionalnumbersofturns!

fromwhichnumberofturnsonsecondaryNs =157or156

1

(ii) ApplyingP = I Vforonelampgives

currentineachlampI = P __ V = 30 ___ 12 =2.5A

1 Part (a)(ii)dependsonyourfamiliaritywithcurrentelectricity,coveredinAS Physics AUnit1.

Alternative solutions are possibleForexampletotalpowersupplied = 8 × 30 =240W

andusingP = V2 ___ R gives

R = V2 ___ P = 12

2 ____ 240 =0.60Ω

Totalcurrentsuppliedbytransformer Itot = 8 ×2.5= 20 A

1

∴ Totalresistanceofthelamps

R = V ___ Itot = 12 ___ 20 =0.60Ω

1

(iii)Iftransformerisperfectlyefficient,primarypower=secondarypower

(orIp Vp = IsVs)

1 Ifadeviceisperfectlyefficient,theenergysuppliedtoitpersecondisequaltotheenergyitsuppliespersecond.Alternative solution:Pout = IsVs = 20 ×12=240WSincethetransformerisperfectlyefficient, Pin = Pout =240W

Powerinput=primarypower =secondarypower =poweroutput= 8 × 30

1

=240W 1




(b) Relevant points include: • Thependulumbobcutsthroughthe

magneticfluxofthefieldofthemagnet • Thereisaninducedemfinthebob • Thisemfcausescirculatingcurrents

(eddycurrents)inthebrassbob • Thesecurrentscausethermalenergy

lossesinthebob • Theenergyconvertedinthiswaycomes

fromthekineticenergyofthebob,causingittoslowdownrapidlythatis,heavydamping

any 4 Electromagneticdampingcanbeappliedverysuccessfullyinpracticaldevices,suchastheelectromagneticretardersthatarefittedtobusesandcoachestoenhancetheirbrakingsystems.AllrelyonLenz’slaw:theinducedemfsactsoastoopposethechangeoffluxproducingthem.Alternative approach insteadofthelasttwopoints:• thecurrentsinthebobproducemagneticfields . . .

• whichinteractwiththeexistingmagneticfieldtoopposethemotionofthebob

7 (a) (i) PowerP = I rmsVrmsgives

Irms = P ____ Vrms = 960×10

3 ________ 11×103

1 Electricalpowerdistributionreliesonbasiccurrentelectricitytheory,whichwascoveredinAS Physics A Unit1.P = I Vappliestoac,justasitdoestodc,providedrmsvaluesareused.Thepeakcurrentistheinstantaneousvalueofac,whereasthermscurrentisamoremeaningfulaveragevalue.

=87.3A 1 PeakcurrentI0 = Irms √

__ 2 =87.3√

__ 2

=123Aor120A1

(ii) Powerlostfromcables= Irms2 R =87.32 ×1.8

1 NotethatthemeanpowerdissipatedisproportionaltoIrms2,andnottoI0

2.(Part(a)(i)ofthisquestioncouldmisleadyou!)11kVisastandarddistributionvoltageusedonlocaldistributionnetworks.At11kVthepowerlossisslight.Ifyourepeatthiscalculationforthesamepowerlineoperatedinthesamewaybutat3kV,youwillfindthattheoutputpowerisonly81%oftheinputpower.Thepowerlossincreasesastheoperatingvoltagedecreases,becausethecurrentincreases.

=13.7×103W 1 ∴ poweravailableatoutputend =960−14=946kW

1

Percentageofinputpoweravailableat

output= 946 ____ 960 ×100=98.5%or98.6%

or 99%.

1




(b) Relevant points include: • Powerislostfromthecables(orpdis

droppedalongthem) • Longercableshavegreaterresistance • Powerlostfromcables= irms2r • Lesspoweriswastedifahighvoltageis

used • Becauseathighvoltagethecurrent

neededisless(forthesamepower) • acvoltagescanbechangedusing

transformers • Transformerscannotbeoperatedby

steadydc.

any 5 Themainobjectofmodernpowerdistributionnetworks,liketheNationalGrid,istominimisethepowerwastedasaresultoftheheatingofthecables.Thisinvolvestakingthepowerforasmuchofthedistanceaspossibleatthehighestacceptablevoltage.Atthesametime,thesystemmustbesafe–andhighvoltageelectricityisinherentlydangerous.Thereductioninvoltageisthereforecarriedoutinstagesbytransformersattheconsumers’endofthesystem:fromtheNationalGridat450kV→ 132kV→ 33kV→11kV→ 230 V at the consumer.

Nelson Thornes is responsible for the solution(s) given and they may not constitute the only possible solution(s).

AQA A2 Physics A chapter 8 textbook answers

Documents

Transcript of AQA A2 Physics A chapter 8 textbook answers