AQA A2 Physics A chapter 7 textbook answers
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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) Magnetic flux density B: tesla (T) or weber metre −2 (Wb m −2 ) Length l of wire in field: metre (m) 1 It is preferable to give the full name of the units in a question such as this, although the symbols normally used for them (N, A, Tand m) would be accepted. Note that units that are named after eminent physicists do not begin with a capital letter (unlike the name of the persons themselves). The equation F = BIl applies only when the magnetic field is directed at right angles to the direction of the current. 1 When B and I are not at right angles, the force is caused by the component of the magnetic flux density at right angles to the current, B sinθ, where θ is the angle between the current and the field. The more general equation is therefore F = BIl sin θ. It follows that there is no force on a wire when the current is parallel to the magnetic field. (b) (i) Mass m of bar = (25 × 10 −3 ) 2 × 8900 × l = 5.56 l 1 For the bar, mass = volume × density, and volume = cross-sectional area × length. When the bar is supported by the magnetic force acting on it, the upwards magnetic force balances the bar’s weight. The length l of the bar is a common factor, which cancels since it appears on both sides of the equation. Weight mg of bar = 5.56l × 9.81 = 54.6l 1 Magnetic force = weight of bar ∴ mg = BIl 1 BIl = 54.6l gives B = 54.6 ____ I = 54.6 ____ 65 = 0.84 T 1 (ii) Arrow drawn on diagram, labelled M: Into the front surface of the bar, at right angles to it, and in the same horizontal plane as the bar 1 This comes from the application of Fleming’s left-hand rule. The representation may not be easy on the three-dimensional diagram; a correct arrow should point almost ‘north-west’ on the page. 2 (a) (i) Use of F = BIl gives B = F __ Il = 180 _____________ 12 × 10 3 × 0.83 1 The current in X sets up a magnetic field around it. The current in Y experiences a force, because Y is situated in the magnetic field produced by X. This argument applies equally in reverse, so the forces on × and Y are equal and opposite, whether or not the currents in × and Y are the same. = 1.81 × 10 −2 T 1 (ii) Force on X due to Y: • B at X (due to Y) is unchanged because the current in Y is unaltered • F = BIl and the current in X is halved, so the force on X is now 90 N. 2 Originally, when both bus bars carried currents of 180 A, the force between them was 180 N. Halving the current in one of them will reduce this force to 90 N. Chapter 7
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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 norm
Transcript of AQA A2 Physics A chapter 7 textbook answers
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
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
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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.
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(ii) Withinthevelocityselector,magneticforce on ion =electricforceonion
∴BQv = EQ
1 Ionswillpassthroughthevelocityselectorinastraightlineonlywhenthemagneticforceonthemisbalancedbytheelectricforcecausedbytheelectricfield.Onlythoseionswhichsatisfythisconditionwillemergethroughtheslitdirectlyaheadofthem.
InthisequationQcancels,giving
Bv = E,andvelocityofionsv = E __ B ,
meaning that v doesnotdependonQ.
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(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)
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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.
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(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.
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
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