高中物理解題思路全解全析(Highschoolphysicsproblemsolvingthinkingtotalsolutionanalysis)

高中物理解題思路全解全析（Highschoolphysics,problemsolving,thinking,totalsolution,analysis）Highschoolphysics,problemsolving,thinking,totalsolution,analysisI.ideasandmethodsofsolvingstaticsproblems1.determinetheobjectofstudy:isolatetheobject--.Whennecessary,thesubjectsshouldbeconverted.Inthistypeoftransformation,onecaseischangedtoanotherobject,oneistoincludetheoriginalobject,buttoenlargetherangeandincludeanotherobject.2.,analyzetheexternalforcesof"object",andanalyze"primitiveforce",nottoanalyzeanddealwiththeforces.Representedbyforceofeffort.3.,accordingtothesituation,orwiththeparallelogramrule,orwiththetrianglerule,orbytheorthogonaldecompositionrule,improvethepurposeofforcesynthesisanddecomposition,andreduceblindness.forequilibriumproblems,theequilibriumconditionsareF=0,sigmaM=0,thecolumnequationsaresolved,andthendiscussed.forthechangeofequilibriumstate,theproblemofforcechangecanbestudiedbyanalyticmethodorgraphicmethod.Staticlearningproblemscanbedividedintothreecategories:Theuseofforcesynthesisandthelawofdecomposition.Thebalanceandthechangeoftheconcurrentforce.Balanceandchangeofthefixedaxis.UnderstandingthebalanceandbalanceconditionsofobjectsForaparticle,iftheparticleremainsstationaryoruniformlylinearattheactionoftheforce,i.e.,theaccelerationiszero,itiscalledabalance.Tobalancethemass,theremustbeasigmaF=0.Iftheorthogonaldecompositionofeachforce,thereare:SigmaFX=0,sigmaFY=0.Forarigidbody,equilibriummeansthatthereisnotranslationalacceleration,thatis=0,andthereisnorotationalacceleration,=0(stationaryorevenspinning),atthistime:SigmaF=0,sigmaM=0.Itshouldbepointedoutthatwhentheobjectisbalancedunderthreeforces(nonparallelforces),thefollowingconclusionscanbedrawnaccordingtosigmaF=0:Thethreeforceswillbeconcurrent.Thesethreeforcevectorsformaclosedtriangle.Theresultantforceofanytwoforcesmustbeequivalenttotheinverseofthethirdforces.Analysisandprocedureofforceonanobjectstressanalysispoints:1,cleartheobjectofstudy2.Analyzethenumberanddirectionofforceactingonanobjectoranode.Ifitisaconnectingbodyoranoverlappingbody,usetheisolationmethod"3,whendrawingalargeforcelinecorrespondinglylonger4,eachforcemarksthecorresponding

symbol(powerful

mustbefamous),inEnglishletters5objectsornodes:SolvethedynamicequationsofthecolumnbyorthogonaldecompositionWhentheforceisbalancedWhentheforceisunbalanced7,theforcecharacteristicsofsomeobjects:8,thesameropeonthesmoothpulleyorsmoothhook,theropeonbothsidesoftheforcesizeisequal,whenthemorethanthreesectionsofropeknotattheintersection,eachroperopeforcesizeisgenerallynotequal.(two)stressanalysisprocedure:1,todeterminethenumberofobjects,contactforce(elasticityandfriction);fieldforce,magneticforce)

andmapping:(gravity);fieldforce(electric2,thedirectionofjudgment:Judgeaccordingtothenatureoftheforceandthecause;Accordingtothemovementoftheobjecttojudge;AisjudgedbyNewton'sthirdlaw;BisjudgedbyNewton'ssecondlaw(inthedirectionofacceleration,theobjectisforced).Two.BasicmethodsandstepsofsolvingkinematicsproblemsThebasicconceptsofkinematics(displacement,velocity,acceleration,etc.)andthebasiclawsarethebasisforsolvingproblems.Theyaretheweaponsthatweunderstandproblems,analyzeproblemsandseekwaystosolvethem.Onlywhenwehaveaprofoundunderstandingofconceptsandlawscanwesolveallkindsofproblemsflexibly,butsolvingproblemsisanecessarylinktodeeplyunderstandtheconceptsandlaws.Accordingtothebasicconceptsandlawsofkinematics,thebasicmethodandprocedureforsolvingkinematicsproblemsareshown2.Youunderstand,sketch,cleartheknownquantity,unknownunknowns.makecleartheobjectofstudy.Selectthereferenceframeandcoordinatesystem.analyzethetime,displacement,initialvelocity,accelerationandsoon.(4)applythelawofmotionandgeometryrelationtoestablishtheequationofsolution.(5)solvingequations.Three.ThebasicmethodofsolvingdynamicsproblemsWeanalyzeandsolvedynamicsexerciseswiththebasicconceptsandbasiclawsofdynamics.Becauseofthecomplexdynamics,weclassifytheexercisesaccordingtodifferentdynamics.1,theapplicationofNewton'slawtosolvetheproblem,Therearetwobasictypesofthisproblem:(1)theknownobjectissubjecttothemotionoftheobject;(2)theforceoftheobjectisdeterminedbythemotionoftheknownobject;therearemanyproblemsinthetwobasicproblems.Fromtheobjectofstudy,therearemultipleobjectswithasingleobject.(1)basicmethodsofsolvingproblemsAccordingtoNewton'slaw,thebasicmethodofsolvingexercisesisAccordingtotheselectedresearchobject,determinethem.Analysetheforceoftheobjectanddeterminetheforceofthepainting.Analyzethemovementoftheobjectanddeterminethea.AccordingtoNewton'slaw,theconceptofforce,thelawandtheformulaofkinematics,theequationofsolutionissetup.Solvingequation.Thechecking,discussion.Theabove,thesecondandthethirdarethebasisofsolvingproblems.Theyareofteninterrelatedandcannotbeseparated.TheproblemofsolvingbyusingkineticenergytheoremThetheoremofkineticenergyisthattheenergy,force,displacement,kineticenergy,velocity,massandsooncanbecalculatedaccordingtokineticenergytheorem.Whatisthebasicmethodtosolvetheproblembyapplyingkineticenergytheorem?Adisplacementofselectedobjectsandobjectstospecifymands.Analysetheforceoftheobjectandcombinethedisplacementtomakeclear.Analyzetheinitialvelocityoftheobjecttodeterminethekineticenergyatthebeginning.Then,accordingtokineticenergytheoremandequalcolumnequation,thesolutionequationischeckedanddiscussed.(example)asshowninFigure4-5,thequalityoftheboardis3meterslong.Objectmass.Thecoefficientoffrictionbetweenanobjectandaboard,thecoefficientoffrictionbetweenaboardandahorizontalgroundTherightendoftheboardisinastationarystate.Thehorizontalhorizontalpullofthecowisnowused.Theobjectwillslideonthewoodenboard.After2seconds(1),howmuchwilltheforceFdo?(2)whatisthekineticenergyofanobject?(M/S2)TheproblemofsolvingbymomentumtheoremFromthetheoremofmomentum,thistheoremcanbeappliedtoimpulse,force,time,momentum,velocity,mass,etc..ThebasicmethodofsolvingmomentumtheoremisSelecttheobjectandasectionoftheprocesstodefinemandt.Analysetheforceoftheobjecttodefinitetheimpulse.Theanalysisofobjectsatthebeginningandtheendofthespeedtoclearatthebeginningandtheendofmomentum.Then,accordingtothemomentumtheorem,theequationsareestablished,theequationsaresolved,andthecomputationsarediscussed.[example8]theweightof10kilogramsofheavyhammerfrom3.2metershighfreefallhittheworkpiece,hammerhittheworkpieceafterjumping0.2meters,hittingtimeis0.01seconds.Calculatetheaveragestrikingforceofthehammerontheworkpiece.TheapplicationofthelawofconservationofmechanicalenergyTheformulaofconservationofmechanicalenergyisknown.Itcanbeusedtocalculatekineticenergy,velocity,mass,potentialenergy,height,displacement,etc..ThebasicmethodofapplyingthelawofconservationofmechanicalenergyisTheselectedsystemandasectionofdisplacementareselected.Analyzingtheexternalforce,internalforceandtheworkdonebythesystemtodeterminewhetherthemechanicalenergyofthesystemisconserved.Theanalysissystemofobjectsintheinitialfinalstatepositionandvelocitytodeterminetheinitialfinalstatemachinery.Then,accordingtotheequationsofconservationofmechanicalenergy,theequationsaresolvedandchecked.Four.BasicmethodsofsolvingelectricfieldproblemsThemainproblemofthischapteristhedescriptionofelectricfieldpropertyandtheeffectofelectricfieldoncharge.Whensolvingaproblem,itisnecessarytofindoutseveralphysicalquantitiesdescribingtheelectricfieldpropertyandstudythelawofelectricfield.1,howtoanalyzeelectricfield,electricpotential,electricfieldforceandelectricpotentialenergyinelectricfieldfirst,theelectricfieldstudiedistheelectricfieldformedbythosefieldcharges.findoutthemeaningofthesymbolsofthephysicalquantitiesintheelectricfield.(3)correctapplicationofsuperpositionprinciple(vectorsumorscalarsum).Themeaningsofthequantitysymbolsarebrieflydescribedbelow:Thepositiveandnegativeoftheelectricityonlyshowthedifferenceoftheelectricity,butnotthesizeoftheelectricity.Theelectricfieldintensityandelectricfieldforcearevectors.WhenapplyingCoulomblawandfieldintensityformula,donotsubstitutethesignofelectricenergy,Thesizeanddirectionshouldbedeterminedbycalculation.Inspace,thedirectionoffieldandelectricforcecannotbesimplyexpressedby'+','-'Thepotentialandelectricpotentialenergyarescalar,andpositiveandnegativerepresentthesize.Whentheyareusedtocalculate,theycansubstitutetheirsymbols,suchasUispositive,qisnegative,andsoisnegative,suchasU1>U2>0andQarenegative.Whentheelectricfieldworkispositiveornegative,theelectricpotentialenergyoftheelectricchargeiscorrespondinglychanged.WhentheWABispositive(i.e.,theelectricfieldforceispositivework),theelectricpotentialenergyofthechargedecreases;whentheWABisnegative,theelectricpotentialenergyofthechargeincreases.Therefore,whenapplied,thesymboloftheelectricfieldforcecanbedeterminedbyreplacingthesymbolsofeachquantity.Ofcourse,youcanalsousethesizeofthework,andthentheelectricfieldforceanddirectionofmovementtodeterminethepositiveandnegativework.Buttheformercanbedirectlycomparedwithothers.2,howtoanalyzethebalanceandmotionofelectricchargeinelectricfieldBalanceandmovingchargeintheelectricfieldisacomprehensiveknowledgeaboutthemechanicsoftheelectricfield;SichuanXi?Comprehensiveproblemcanbesolved,todeepenunderstandingofrelevantconceptsandrules,improvetheanalysis,haveasignificantroleincomprehensiveabilityproblems.Theanalysismethodofthiskindofproblemisthesameasthatofmechanicalanalysis:determinetheobjectofstudy(achargedbody).analyzetheexternalforceofchargedbody.accordingtotheanalysisofthephysicalprocess,shouldpayattentiontodiscussvariousconditions,analysisoftheimpliedconditionsinquestion,thisisthekeytosolvingtheproblems.accordingtothephysicalprocessandtheknownphysicalquantities,thepropermechanicslawischosen.discusstheresults.exampleinFigure7-3shows,if(triton)and(helium)verticalelectricfielddirectionintothesamedeflectionfield,andinthecasewhenthelateraldisplacementoftheirsizeratio.(1)enteratthesameinitialvelocity;(2)enteratthesameinitialkineticenergy;(3)enteratthesameinitialmomentum;(4)enterthesameacceleratedelectricfieldbeforeentering.Analysisandsolvingofchargedparticlesintheelectricfieldelectricforceisfargreaterthanthegravity,sogravitycanbeignored.Thechargedparticlesinthedeflectedelectricfieldaresubjectedtotheactionoftheelectricfieldforce,whichissimilartothemotionofaflatthrow,whichmovesuniformlyintheoriginalvelocitydirectionandacceleratesuniformlyatthetransversevelocity.ByusingNewton'ssecondlawanduniformlyacceleratedmotionformulaentertheelectricfieldatthesameinitialspeedV0,becauseE,landV0arethesame,so(2)entertheelectricfieldatthesameinitialkineticenergy,Ek0,becauseE,landMV2arethesame,so...entertheelectricfieldatthesameinitialmomentumasP0,becauseE,landMV0arethesametheelectricfieldoftheparticleisacceleratedafterthesameacceleratingelectricfieldisacceleratedandthenenterstheelectricfield(U1foracceleratingvoltage)fromBecauseE,l,andU1arethesame,thesizeoftheYisindependentofparticlemassandpower,so:NotethatintheratioforlateraldisplacementoftheY,shouldfirstfindtheexpressionofY,thecondition,findouttheYandm,theparticlemasspowerratioofQ,thenlistthanthesolutionthisisthegeneralmethodofsolvingratio.3,howtoanalyzetheproblemofparallelplatecapacitor?Attentionshouldbepaidtotheanalysisofsuchproblems(1)theparallelplatecapacitorisopencircuitinthedirectcurrentcircuit,andthevoltagebetweenthetwoplatesisthesameasthatoftheappliance(orbranch)connectedinparallel.(2)ifthecapacitorisconnectedwiththepowersupplyandtheswitchisclosed,whenthetwoplatedistanceischangedorthetwoplateisfacingthearea,thetwoplateelectricityisinvariable,andthechargedquantityofthepolarplateischanged.Iftheswitchisswitchedoff,thedistancebetweenthetwopolesischangedandthevoltagewillbechangedaccordinglywhenthedistancebetweenthepolesischangedorthetwoplateisfacingthearea.(3)theparallelplatecapacitorisauniformelectricfield,Theelectricfieldintensitybetweenthetwoplatescanbediscussed,andfurtherdiscussionismadeonthebalanceandtransportofthechargesbetweenthetwoplates.4,theelectricfieldandpotentialareanalyzedbyusingthecharacteristicsofthepowerlineandequipotentialsurfacePowerlinesandequipotentialsurfacescanvisuallydescribefieldstrengthsandpotentials.Thenumberofelectriclinesdrawnaroundthechargeisproportionaltothechargeofthecharge.Thedensityanddirectionofthepowerlineindicatethemagnitudeanddirectionoftheelectricfieldstrength,thedecreaseofthepotentialofthepowerline,theequipotentialsurface,theverticalpowerlineandsoon......CanhelpusanalyzethefieldstrengthandEMF[example]thereisasphericallychargedcavityconductorthatputsanegativecharge,Q,intothecavity,asshown.Ask:(1)duetoelectrostaticinduction,whatistheelectricityintheinnerandouterwallsofthecavityconductor?Whatarethecharacteristicsofelectricfieldintensityinthecavity,conductorandconductor,andwhatisthedirectionoftheelectricfieldstrength?iftheinnerwallofthecavityconductorisgrounded,whatelectricpowerdoestheinnerandouterwallsofthecavityconductorhave?Whatistheelectricpotentialratiointhecavity,insidetheconductorandoutsidetheconductor?(3)removetheearthwireandthentakethefieldfromthecavityconductor.Whatistheelectricity

chargeQawayintheinnerandouterwallsofthecavityconductor?Whatisthefieldstrengthandelectricpotentialinthecavity,insidetheconductorandoutsidetheconductor?Analysisandreconciliationofthesubject,usingpowerlinestoanalyzemoreclearlythenegativechargeplacedinthecavity,thenegativechargearoundwillgenerateelectricfield(drawpowerlinewhosedirectionispointingtothenegativecharge)freeelectronpotentialfromlowtohighpotential(inverseelectronpowerlinemotion)electrostaticinduction,theconductorwallwithapositivechargequantityisQ,theguidewallwithinvitroelectricityisanegativechargeofQ,asshowninfigureHollowconductorinsideandoutsidepowerlinenumber(forpowerlinenumberisproportionaltotheelectricpowerlinetoempty)outsidetheplasticmetalconductor(powerlineendsatnegativecharge).Thecloserthecavityconductor,thegreaterthefieldstrength.Thereisnopowerlineintheconductor,theelectricfieldintensityiszero,andthecloserthenegativechargeisinthecavity,themorecloselytheQpowerlineis,andthegreatertheelectricfieldintensityis.Thedecreaseofthepowerlinepotential,suchastheprescribedinfinity,thepotentialiszero,andtheclosertothecavityconductor,thelowerthepotential,theinternalpotentialoftheconductorisequal,andthecloserthenegativechargeinsidethecavity,thelowerthepotentialoftheQ.Theelectricpotentialeverywhereislessthanzero.(2)iftheinnerwallofthecavityconductorisgroundedandtheelectronsarefromlowpotentialtohighpotential,thefreeelectronsontheconductorwillenterthegroundthroughtheearthwire.Aftertheelectrostaticbalance,theinnerwalloftheconductorisstillpositivelycharged,andtheouterwalloftheconductorisnotelectrified.Sincethepowerlinenumberisproportionaltothefieldcharge,thefieldchargeQisnotchanged,sothedistributionofthepowerlineinthecavityisunchanged,andtheelectricfieldintensityinthecavityisunchanged.Thefieldstrengthinsidetheconductorisstillzero.Sincetheouterwalloftheconductorisnotcharged,thereisnopowerlineoutsidetheconductor,andtheexternalfieldstrengthoftheconductoriszero.TheconductormustbeearthedtomaketheouterspaceoftheconductorfreefromtheinfluenceoftheinnerfieldchargeofthecavityAftertheelectrostaticbalance,theconductorandthegroundpotentialareequaltozero.Thepotentialintheinnercavityoftheconductorisstillnegative.Theclosertothefieldchargepotential,thelowerthepotentialishigherthantheconductorbeforetheground.suchasremovingtheground,thentakeawayfromthecavityfieldchargeconductor,becauseofelectrostaticinduction,outersurfaceoftheconductorsurface.Thefreeelectroninwardelectrostaticequilibriumconductor,innersurfaceoftheoutersurfaceisnotcharged,positivelycharged,thechargedamountisQ.Atthistime,thereisnopowerlineinsidetheconductorandthecavity,thefieldstrengthiszero,thesurfacestrengthoftheconductorsurfaceisperpendicular,theconductorsurfacepointstotheconductor,thefartherawayfromtheconductor,themoreloosethepowerline,thesmallerthefieldstrength.Thepotentialofthepowerlinedecreases,thepotentialatinfinityiszero,andtheclosertotheconductor,thehigherthepotential.Theelectricpotentialintheconductorisequaltothatinthecavity,andthepotentialofeachpointisgreaterthanzero.Whentheconductorisgrounded,theoutersurfaceoftheconductorisnotcharged,andthepowerlinecanalsobeusedforanalysis.Iftheoutersurfaceisnegativelycharged,thereisapowerlinepointingfromtheinfinitetotheconductor,andtheconductorwillhaveapotentialoflessthanzero,whichisincontradictionwiththeconductorpotential.Iftheconductorsurfaceispositivelychargedatlast,powerlinefromtheoutersurfaceoftheconductortoinfinity,whilethepotentialoftheconductorwillbegreaterthanzero,andalsotopotentialconflicts.So,thiswillbethegroundedconductor,theconductorsurfaceisnolongercharged.Thecurrentandvoltageofthecircuitwillnotbeaffected.Thewireisremoved,astheammetervoltmetercircuitcalculation.Butasarealtable,theyallhaveresistors,whichshowboththecurrentandthevoltageofthecircuitandshowitsowncurrentorvoltage.Ifthetrueammeterisasmallresistance,ahighresistancevoltmeteristrue,theywillaffecttheaccesscircuitofcurrentandvoltagevaluesofcircuit.Therefore,whensolvingproblems,weshouldmakesurethattheelectricmeterinthecircuitisregardedasanidealtable.Two,basicmethodsofsolvingproblems、determinationofmagneticfieldandmagneticforcedirectiondeterminationofthedirectionofcurrentandmagneticfield--correctapplicationofAmpere'sruleForthelinecurrent,loopcurrentandelectrifiedsolenoidspacearoundthemagneticfielddistribution,tobeskilledtocorrectgraphicallyrepresentedbyfieldlines,drawthedistribution--includingtherightdirectionandthedensityofthegeneralsituationofthemagneticline,butalsocansolveproblemsaccordingtoneedtochoosedifferenticons(suchasthree-dimensionalmap,profileorcrosssectionetc.).Amongthem,thedeterminationofthedirectionofthemagneticfielddirection,accordingtothedirectionofthecurrenttocorrectlygraspthetwokindsofAmpererule,thatis:Foralinearcurrent,holdthewire(current)withtherighthand,sothatthedirectionofthepointingthumbisthesameasthedirectionofthecurrent,andthedirectionofthefourpointedfingeristhedirectionofthemagneticlineofforcesurroundingthecurrent.Fortheringcurrentandelectrifiedsolenoid,shouldmaketherighthandfourfingerswillbenddirectionandcurrentdirection,thenstraightthumbreferstoisthedirectiontothecenterringcurrentaxismagneticfield,orsolenoidinternalmagneticfield(i.e.thethumbpointingthroughthesolenoidforceline,inchoation,Arctic).Fortheenergizedsolenoid,itsinternalmagneticfielddirectionisdirectedfromtheNpoletotheSpole,whiletheinternalmagneticfielddirectionisdirectedfromtheSpoletotheNpole.Thusformingaclosedcurve.thedeterminationofAmpereforceandLorentzforcedirection--correctapplicationofthelefthandruleUsingthelefthandruletodeterminethedirectionofAmpereforcedependsonthedirectionofthemagneticfieldBanddirectionofthecurrentI.AslongasthedirectionoftheB

theisnotparalleltothatoftheIL,theampereforceispresentandisperpendiculartotheplanedeterminedbyBandIL.ForthegeneralcaseofBandILisnotvertical,youmustfirstuseBvectorisdecomposedintotwocomponents:oneisperpendiculartotheIL,anotherisparalleltotheIL,asshowninFigure9-2,accordingtothecurrentdirectionandthendeterminethedirectionofampI.Underthepremisethattheanglebetweenthemagneticfieldandtheconductingwireisgiven,iftheampereforce,Fmagneticfield,BandthedirectionofanytwoquantitiesoftheconductingwireILaredetermined,thedirectionofthethirdquantitiescanbedeterminedaccordingtothelefthandrule.UsingthelefthanddingzejudgeLorentzforceaccordingtothesamedirection,thedirectionofBandthecurrentdirectionofthemotionofchargedparticlesformed(current,positivelychargedparticlemovementinthedirectionandspeedofV,current,negativelychargedparticlemotiondirectionandthevelocityofVformationintheoppositedirection).AslongasthedirectionsofBandVarenotparallel,theLorentzforceexistsandisperpendiculartotheplanedeterminedbyBandV.ForthegeneralcaseofBandVisnotvertical,itstillneedstheBvectorisdecomposedintotwocomponents:oneisperpendiculartotheV,anotherisparalleltotheV,asshowninFigure9-3,(ortheUvectorisdecomposedintotwocomponents:oneisperpendiculartotheB,anotherisparalleltotheB,asshowninFigure9-3).ThendeterminethedirectionoftheLorentzforceaccordingtothedirectionandthedirectionoftheV(orthedirectionanddirectionoftheB).InthepremiseofmagneticfieldBandtheknownelectricparticlevelocityVdirectiongiven,ifitisdeterminedinarbitraryLorentzforceF,Bmagneticfieldandparticlevelocityinthetwovolumeofthedirection,alsocanaccordingtothethirdvolumeofthelefthanddingzejudgedirection.2.CalculationofmagneticfieldforceanditseffectcalculationofAmpereforcesize,theangleofwhichisBandIL(seeFigure9-2),wecanseefromtheangleofthedifferentvalues,valuesofAmpereforcechange,which,whenthevalueofFiszero,whenthemaximumF-measure.Applicableconditionsofthisformula,Generallyspeaking,shouldbethegeneralelectricstraightwireILinthemagneticfieldofB,butthereareexceptions,suchastheBvectorinthenonuniformmagneticfieldaslongastheelectricitystraightguidelineILlocationalongthewireofthemostequal(Bvaluesareequalandthesamedirection),theampereforcecanalsousetheformula.AsfortheeffectofAmpereforce,theusualproblemsencounteredinsolvingproblemsareillustratedbelow:Theinteractionbetweenparallelthroughelectricalconductors;thephasecurrentsinthesamedirection,reversecurrentrepulsion.Thisisanimportantexampleofthecurrentmagneticinteraction.UnderthecombinedactionofAmpereforceandotherforces,theconductingconductorisinabalancestate,soastodeterminethewaitingvalueofBorI.Ifthemagneticinductionintensityismeasuredbycurrentbalance,thecurrentintensityismeasuredbymagnetoelectricammeter.[Fig.2)Fig.9-5showsacurrentbalanceformeasuringthemagneticinductionintensityofauniformmagneticfield.Arectangularcoilisplacedatoneendofthebalance,thebottomofwhichisplacedintheuniformmagneticfieldBtobemeasured,andthedirectionoftheBisperpendiculartothesurfaceofthepaper.Knownasncoilturns,thebottomedgeofLwhenthecoilinacounterclockwisedirection,thecurrentintensityisI,thebalan