aqa a2 physics a chapter 7 textbook answers
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Chapter 7Answers to examination-style questionsAnswers1 (a) the units of the quantities are: Force F: newton (N) Current I: ampere (A) Magnetic flux density B: tesla (T) or weber metre−2 (Wb m−2) Length l of wire in field: metre (m) T he equation F = BIl applies only when the magnetic field is directed at right angles to the direction of the current.Marks Examiner’s tips1 It is preferable to give the full name of the units in a question such as this, although the symbols normTRANSCRIPT
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Answers to examination-style questions
AQA Physics A A2 Level © Nelson Thornes Ltd 2009 1
Answers Marks Examiner’s tips
1 (a) the units of the quantities are: Force F: newton (N) Current I: ampere (A) MagneticfluxdensityB:tesla(T) or weber metre−2 (Wb m−2) Length lofwireinfield:metre(m)
1 Itispreferabletogivethefullnameoftheunitsinaquestionsuchasthis,althoughthesymbolsnormallyusedforthem(N,A,Tandm)wouldbeaccepted.Notethatunitsthatarenamedaftereminentphysicistsdonotbeginwithacapitalletter(unlikethenameofthepersonsthemselves).
TheequationF = BIlappliesonlywhenthemagneticfieldisdirectedatrightanglestothedirectionofthecurrent.
1 When BandIarenotatrightangles,theforceiscausedbythecomponentofthemagneticfluxdensityatrightanglestothecurrent,Bsinθ,whereθistheanglebetweenthecurrentandthefield.ThemoregeneralequationisthereforeF = BIlsinθ.Itfollowsthatthereisnoforceonawirewhenthecurrentisparalleltothemagneticfield.
(b) (i) Massm of bar = (25 × 10−3)2 × 8900 × l =5.56l
1 Forthebar,mass=volume×density,andvolume=cross-sectionalarea× length.Whenthebarissupportedbythemagneticforceactingonit,theupwardsmagneticforcebalancesthebar’sweight.Thelengthlofthebarisacommonfactor,whichcancelssinceitappearsonbothsidesoftheequation.
Weight mg of bar =5.56l ×9.81=54.6l 1 Magnetic force = weight of bar ∴ mg = BIl
1
BIl =54.6lgives
B = 54.6 ____ I = 54.6 ____ 65 =0.84T
1
(ii) Arrow drawn on diagram, labelled M: Intothefrontsurfaceofthebar,at
rightanglestoit,andinthesamehorizontalplaneasthebar
1 ThiscomesfromtheapplicationofFleming’sleft-handrule.Therepresentationmaynotbeeasyonthethree-dimensionaldiagram;acorrectarrowshouldpointalmost‘north-west’onthepage.
2 (a) (i) UseofF = BIlgives
B = F __ Il = 180 _____________ 12 × 103 ×0.83
1 ThecurrentinX setsupamagneticfieldaroundit.ThecurrentinYexperiencesaforce,becauseYissituatedinthemagneticfieldproducedbyX.Thisargumentappliesequallyinreverse,sotheforceson× andYareequalandopposite,whetherornotthecurrentsin× andYarethesame.
=1.81× 10−2T 1
(ii) Force on XduetoY: • B at X(duetoY)isunchanged
becausethecurrentinYisunaltered • F = BIlandthecurrentinX is
halved,sotheforceonXisnow90N.
2 Originally,whenbothbusbarscarriedcurrentsof180A,theforcebetweenthemwas180N.Halvingthecurrentinoneofthemwillreducethisforceto90N.
Chapter 7
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Chapter 7Answers to examination-style questions
AQA Physics A A2 Level © Nelson Thornes Ltd 2009 2
Answers Marks Examiner’s tips
Force on YduetoX: • B at Y(duetoX)ishalvedbecause
the current in Xishalved • F = BIlandthecurrentinYis
unchanged,sotheforceonYisalsonow90N.
2 Onceyouhaveworkedouttheforceononebusbar,itwouldbeacceptabletodeducethatonthesecondbypointingoutthattheyexperienceequalandoppositeforces.ThisisreallyanexampleofNewton’s3rdlawofmotion.
(b) (i) Relevant points include: • Themagneticfieldduetothecurrent
ineachbusbaralternateswiththecurrent in the bar
• Theforcewillbezerowhenthecurrentineitherwireiszero
• Bothcurrentandfieldreversetogether,sotheforcedoesnotreversedirectionwhenthecurrentreverses
• Theforceoneachwirevariesperiodically,makingthewiresvibrate
• Thefrequencyofvibrationistwicetheacfrequency.
any 3 Thesetwobusbarswill,infact,attracteachother–becausetheycarrycurrentsinthesamedirection.(Ifonecurrentalonewerereversed,theywouldrepel.)Withac,themagnitudeoftheforcevariesbecausethecurrentsvarywithtime.Butthebusbarsstillattracteachotherwhenthecurrentsarereversed.Ineachhalfcycleoftheactheforceswillrisefrom0toamaximumandthenfallto0again.Hencetherearetwoattractioncyclesforeverycycleofthealternatingcurrents.
(ii) Theamplitudeofvibrationcouldbereducedby:
• clampingeachbaralongitlength • attachingadampingdevicetoeach
bar • movingthebarsfurtherapart.
any 1 Eitherthephysicalmovementhastoberestricted(byclampingordamping)orthewireshavetobeinaweakerfield;thelattercouldbeachievedbyincreasingtheirseparation.
3 (a) (i) Theforceontheionisdirectedoutoftheplaneofthepage.
1 ApplyFleming’sleft-handrule;positiveionsmoveinthesamedirectionastheconventionalelectriccurrent.
(ii) Inthemagneticfieldtheionmovesinacircularpath...
1 Theforceexperiencedbymovingchargesinamagneticfieldisalwaysatrightanglestotheirvelocity.Likeamassoftheendofastring,theymoveinacircularpath.Youarerequiredtodescribethepathoftheioninwords,andtocarryoutacalculation.Sincethepathiscircular,theobviousfeaturetocalculateistheradiusofthepath.Themagnetic force (BQv)isthecentripetalforceonthemovingcharge.Thechargeoftheionis+2e,becauseitisdoubly-charged.ThevalueofeisgivenintheDataBooklet.
inahorizontalplane(oroutoftheplaneofthepage).
1
theforceequationforthecircular
motionisBQv = mv2 ____ r
∴radiusofpathr = mv ___ BQ
1
= 1.05× 10−25 ×7.8× 105 ___________________ 0.28× 2 ×1.60× 10−19 =0.914m
or0.91m
1
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Chapter 7Answers to examination-style questions
AQA Physics A A2 Level © Nelson Thornes Ltd 2009 3
Answers Marks Examiner’s tips
(b) Relevant points are: Whenmagneticfieldstrengthisdoubled • theradiusofthepathdecreases • tohalftheoriginalvalue. Whenasinglechargedionisused • theradiusofthepathincreases • todoubletheoriginalvalue.
any 3 Fromtheequationgivingtheradiusofthe path in part (a)(ii),itfollowsthat
r ∝ 1 __ B whentheotherfactorsremain
constant,andthatr ∝ 1 __ Q forionsofthe
sameisotopethattravelinthesamefieldatthesamevelocity.ThereforewhenQishalved,risdoubled.
4 (a) (i) Thefluxdensityofamagneticfieldis1teslaifa1metrelengthofwirecarryingacurrentof1ampereexperiencesamagneticforceof1newtonwhenplacedinthefield...
1 ByrearrangingF = BIl,themagneticflux
densityofthefieldisgivenbyB = F __ Il
fromwhich1T= N A−1 m−1.SinceF = BIl isonlyvalidwhenBand I are perpendicular,thisconditionisanessentialpartofthedefinition.
insuchawaythatthemagneticfieldisatrightanglestothecurrent.
1
(ii) Withinthevelocityselector,magneticforce on ion =electricforceonion
∴BQv = EQ
1 Ionswillpassthroughthevelocityselectorinastraightlineonlywhenthemagneticforceonthemisbalancedbytheelectricforcecausedbytheelectricfield.Onlythoseionswhichsatisfythisconditionwillemergethroughtheslitdirectlyaheadofthem.
InthisequationQcancels,giving
Bv = E,andvelocityofionsv = E __ B ,
meaning that v doesnotdependonQ.
1
(iii)Velocityselected
v = E __ B = 2.0× 104 ________ 0.14 =1.43× 105ms−1
(whichisabout140kms−1)
1 ThevaluesofEandBaregivenatthestartofpart(a);directsubstitutioninto
v = E __ B givestherequiredresult.
(b) (i) Radiusofcircularpathwhilstinion
separatorr = mv ___ BQ
1 Seepart(a)(ii)ofQuestion3forthetheoryunderlyingthisequation.
Theioncontains58nucleons,eachofmassaround1u.Theradiusr of the path isonehalfofitsdiameter(0.28m),whichisgivenasthedistancefromP to thepointwheretheplateisstruck.
Massof58 28 Ni ion = 58 ×1.661×10−27
=9.64× 10−26kg1
∴ B = mv ___ rQ = 9.64× 10−26 ×1.43× 105 ____________________ 0.14×1.6× 10−19
=0.615Tor0.62T
1
(ii) Radiusr of path ∝massm of ion
∴newradius= ( 60 ___ 58 ) ×0.140=0.145m =0.145m
1 Thisproportionalityfollowsfromtheequationinpart(b)(i),whenv,BandQ areallunchanged(aconditionthatissatisfiedhere).Alternatively,youcouldcalculatethe
newradiusbysubstitutinginto r = mv ___ BQ
butitwouldtakelonger.
newdiameter=0.290m,soseparationofisotopesonplate=0.290−0.280 =0.010m(10mm)
1
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Chapter 7Answers to examination-style questions
AQA Physics A A2 Level © Nelson Thornes Ltd 2009 4
Answers Marks Examiner’s tips
5 (a) Theforceequationforthecircularmotion
isBQv = mv2 ____ r
∴speedofparticlev = BQr
____ m
1 Themagneticforceactsasthecentripetal
force.Whenthisisequatedwithmv2 ____ r ,
one v cancels,andrearrangementgivesthisresult.
(b) Sincetheparticlesfollowthesamepath,theradiusofcurvatureristhesame.
1 Theradiusofcurvatureisr = mv ___ BQ .An
antiprotonhasthesamecharge(−e)asanegativepionandbothtypesofparticlearetravellingthroughthesamemagneticfield,meaningthatbothBandQ are the sameforthetwoparticles.Youshouldremember from AS Physics Unit 1 that theparticleshavedifferentmasses.
Theyhavethesamemomentumbecauser ∝ mv when BandQarethesame.
1
mvisthesame,butmisdifferentfortheantiprotonandthenegativepion,
∴ v mustbedifferent.
1
(c) Identifycorrectformatforbothparticles: antiproton:3antiquarks negative pion:quark+antiquark
1 Abaryon,suchasaproton,consistsof3quarksandanantibaryonof3antiquarks.Amesonisaquark–antiquarkcombination.
antiproton:2upantiquarks+1downantiquark(
_ u,_ u,_ d)
1 3antiquarksareneededtogiveatotalcharge of −1e:− 2_3 e − 2_3 e + 1_3 e = −1e
negative pion:1upantiquark+1downquark(
_ u,_ d)
1 Thetotalchargehasagaintobe−1e:− 2_3 e − 1_3 e = −1e
6 (a) (i) Themagneticfieldisdirectedintotheplaneofthepage
1 ApplyFleming’sleft-handrule:theforceactstowardsthecentreofthesemicircleandthearrowonthepositiveiongivesthedirectionoftheconventionalcurrent.
(ii) Relevant points include: • Themagneticfieldisdirectedat
rightanglestothevelocityoftheions.
• Themagneticforceactsatrightanglestoboththemagneticfieldandthevelocity.
• Hencethemagneticforceactsatrightangletothevelocityoftheions.
• Thisforcechangesthedirectionofthevelocityoftheionsbutnotitsmagnitude.
• Theforceremainsperpendiculartothevelocityasthedirectionofmovementoftheionschanges...
• sotheforceactsasacentripetalforce.
any 4 Thispartofthequestionasksforanexplanationinwordsofthesemicircularpathoftheions,ratherthanamathematicalderivationofthecentripetalforceequation.MostoftheargumentfollowsfromtheapplicationofFleming’sleft-handrule,byshowingthatFisperpendiculartov inthiscase.Theforcethereforeactstowardsthecentreofthesemicircleandiscentripetal,likethetensioninastringpullingonamassthatiswhirledaround.
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Chapter 7Answers to examination-style questions
AQA Physics A A2 Level © Nelson Thornes Ltd 2009 5
Answers Marks Examiner’s tips
(iii)Theforceequationforthecircular
motionisB Q v mv2 ____ r
1 Theforceequationforchargesmovinginamagneticfieldisaregulartopicinpastquestions.Herethediameter of the path isrequired,notitsradius. ∴diameterofpathd = 2r = 2mv ____ BQ 1
(b) Chargetomassratioofions
Q
__ m = 2v ___ Bd = 2 ×7.5× 104 _______________ 0.34× 110 × 10−3
1 ‘Chargetomassratio’ofionsshouldbefamiliarfromAS Physics AUnit1.Rearrangementoftheresultfrompart(a)(iii),togetherwithsubstitutionofthegivenvalues,leadstothisresult.
=4.01× 106Ckg−1 1
(c) Relevant points include: (i) Ionshaveadifferentmass Diameter d of path ∝massm of ion Duetoisotopesofthesameelement (ii) Ionsaredoublyionised
Diameter d of path ∝ 1 __ Q ;ifQis
doubledthendishalved
any 3 Part (c)becomesatestofwhetheryoucanunderstandtheimplicationsoftheequationinpart(a)(iii).Aslightly differentradiusindicateseitheraslight differenceincharge(whichisnotpossible),oraslightdifferenceinmass(whichis).
Possible alternativeanswerfor(i):Mutualrepulsionofions(allchargedpositively)causessmearingofthespotaroundR.
7 (a) B Q v mv2 ____ r 1 Inthisquestiontheequationforthe
circularmotionofchargedparticleshastobeusedtofindavalueforB.Thediameterofthecircleismarkedonthediagramas240m.Thismagneticfieldmustactupwards,outoftheplaneofthepage(Fleming’sleft-handrule).
givesmagneticfluxdensity
B = 1.67× 10−27 ×1.2× 105 ___________________ 1.60× 10−19 × 120
1
=1.04× 10−5T 1
(b) Asthekineticenergyoftheprotonsincreases,themagneticfluxdensityhastoincrease...
1 In a synchrotrontheradiusofthepathtravelledbythechargedparticleshastoremainconstantastheyareprogressivelyaccelerated.Forthistobeachieved,theincreaseinmagneticfluxdensityhastobe synchronisedwiththeincreaseinvelocity.
becauseinapathofconstantradius(usingthesamechargedparticles)magneticfluxdensityB ∝ v
1
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