540 likes | 772 Views
M A G N E T IC E F F E C T O F C U R R E N T - I 1 . M a gn e t i c E ff e c t of C u r r e nt – O e r s t e d ’ s E x p e r i m e nt 2 . A m p e r e ’ s S w i m m i ng R u l e 3 . M a x w e l l ’ s C o r k S c r e w R u l e 4 . R i ght H a nd T hu m b R u l e
E N D
MAGNETICEFFECTOFCURRENT-I 1.MagneticEffectofCurrent–Oersted’sExperiment 2.Ampere’sSwimmingRule 3.Maxwell’sCorkScrewRule 4.RightHandThumbRule 5.Biot–Savart’sLaw 6.MagneticFieldduetoInfinitelyLongStraightCurrent– carryingConductor 7.MagneticFieldduetoaCircularLoopcarryingcurrent 8.MagneticFieldduetoaSolenoid
MagneticEffectofCurrent: Anelectriccurrent(i.e.flowofelectriccharge)producesmagneticeffectinthespacearoundtheconductorcalledstrengthofMagneticfieldorsimplyMagneticfield. Oersted’sExperiment: N WhencurrentwasallowedtoflowthroughaE needlekeptdirectlybelowthewire,theneedlewasfoundtodeflectfromitsnormalposition. I K N theneedlewasfoundtodeflectintheE oppositedirectiontotheearliercase. K wireplacedparalleltotheaxisofamagnetic I Whencurrentwasreversedthroughthewire,
Rulestodeterminethedirectionofmagneticfield: Ampere’sSwimmingRule: ImaginingamanwhoswimsinthedirectionofcurrentfromsouthtonorthfacingamagneticneedlekeptunderhimsuchthatcurrententershisfeetthentheNorthpoleoftheneedlewilldeflecttowardshislefthand,i.e. towardsWest.SI Maxwell’sCorkScrewRuleorRightII HandScrewRule: Iftheforwardmotionofanimaginaryrighthandedscrewisinthedirectionofthecurrentthroughalinearconductor,thenthedirectionofrotationofthescrewgivesthedirectionofthemagneticlinesofforcearoundtheconductor. N W N B B
I Ifacurrentcarryingconductorisimaginedtobeheldintherighthandsuchthatthethumbpointsinthedirectionofthecurrent,thenthetipsofthefingersencirclingtheconductorwillgivethedirectionofthemagneticlinesofforce. Biot–Savart’sLaw:BThestrengthofmagneticfielddBduetoasmall currentelementdlcarryingacurrentIatapoint proportionaltoI,dl,sinθandinverselyrP proportionaltothesquareofthedistance(r2)θ whereθistheanglebetweendlandr.dl i)dBαIdBαIdlsinθ ii)dBαdl iii)dBαsinθµ0Idlsinθ iv)dBα1/r24πr2 Right Hand Thumb Rule or Curl Rule: x Pdistantrfromtheelementisdirectly r2 I P’ dB=
Biot–Savart’sLawinvectorform: µ0Idlxr 4πr2 µ0Idlxr 4πr3 Valueofµ0=4πx10-7TmA-1orWbm-1A-1 DirectionofdBissameasthatofdirectionofdlxrwhichcanbedeterminedbyRightHandScrewRule. ItisemergingatP’andenteringxatPintotheplaneofthediagram. Currentelementisavectorquantitywhosemagnitudeisthevector productofcurrentandlengthofsmallelementhavingthedirectionoftheflowofcurrent.(Idl) dB= dB=
MagneticFieldduetoaStraightWirecarryingcurrent: AccordingtoBiot–Savart’slaw µ0Idlsinθ 4πr2 sinθ=a/r=cosФI orr=a/cosФ tanФ=l/aaxB orl=atanФlθФP dl=asec2ФdФdlr SubstitutingforranddlindB, µ0IcosФdФ 4πa Magneticfieldduetowholeconductorisobtainedbyintegratingwithlimits -Ф1toФ2.(Ф1istakennegativesinceitisanticlockwise) Ф2µ0IcosФdФµI(sinФ+sinФ) -Ф14πa4πa dB= Ф2 Ф 1 dB= B=012 B=∫dB =∫
Ifthestraightwireisinfinitelylong,B thenФ1=Ф2=π/2 µ02Iµ0I DirectionofBissameasthatofdirectionofdlxrwhichcanbedeterminedbyRightHandScrewRule. ItisperpendiculartotheplaneofthediagramandenteringintotheplaneatP.MagneticFieldLines: II BB a0a B= 4πa B= 2πa or
MagneticFieldduetoaCircularLoopcarryingcurrent: 1)Atapointontheaxialline: dl XY Ф OФdBsinФ II Ф X’Y’ dl Theplaneofthecoilisconsideredperpendiculartotheplaneofthediagramsuchthatthedirectionofmagneticfieldcanbevisualizedontheplaneofthediagram. AtCandDcurrentelementsXYandX’Y’areconsideredsuchthatcurrentatCemergesoutandatDentersintotheplaneofthediagram. C dBcosФ dB dBco 90°r a xФP dBsinФ dBcosФ dB D
µ0IdlsinθordB=µ0Idl Theangleθbetweendlandris90°becausetheradiusoftheloopisverysmallandsincesin90°=1 Thesemi-verticalanglemadebyrtotheloopisФandtheanglebetweenranddBis90°.Therefore,theanglebetweenverticalaxisanddBisalsoФ. dBisresolvedintocomponentsdBcosФanddBsinФ.Duetodiametricallyoppositecurrentelements,cosФ componentsarealwaysoppositetoeachotherandhencetheycancelouteachother. SinФcomponentsduetoallcurrentelementsdlgetaddedupalongthesamedirection(inthedirectionawayfromtheloop). µ0I(2πa)a 4πr24π(a2+x2)(a2+x2)½ µ0Ia2 dB= r2 4π r2 4π µ0IdlsinФ or B=∫dBsinФ=∫ B= B=2(a2+x2)3/2 (µ0,I,a,sinФareconstants,∫dl=2πaandr&sinФare replacedwithmeasurableandconstantvalues.)
SpecialCases: i)AtthecentreO,x=0.B=2aB ii)Iftheobservationpointisfarawayfromthecoil,thena<<x.So,a2canbeneglectedincomparisonwithx2. B= Differentviewsofdirectionofcurrentandmagneticfieldduetocircularloopofacoil: B BB II I µ0I x0x µ0Ia2 2x3 I
2)Batthecentreoftheloop: dl Theplaneofthecoilislyingontheplanea90°I isclockwisesuchthatthedirectionofxO magneticfieldisperpendicularandintoIdB Theangleθbetweendlandais dB=90°becausetheradiusofthe µ0Idlsin90°=1 4πa2 µ0IB (µ0,I,aareconstantsand∫dl=2πa) 0a ofthediagramandthedirectionofcurrent theplane. µ0Idl µ0Idlsinθ dB= a2 4π a2 4π loopisverysmallandsince B=∫dB=∫ B=2a
MagneticFieldduetoaSolenoid: B xxxxxxx II TIP: Whenwelookatanyendofthecoilcarryingcurrent,ifthecurrentisinanti-clockwisedirectionthenthatendofcoilbehaveslikeNorthPoleandifthecurrentisinclockwisedirectionthenthatendofthecoilbehaveslikeSouthPole.
MAGNETICEFFECTOFCURRENT-II 1.LorentzMagneticForce 2.Fleming’sLeftHandRule 3.ForceonamovingchargeinuniformElectricandMagneticfields 4.ForceonacurrentcarryingconductorinauniformMagneticField 5.Forcebetweentwoinfinitelylongparallelcurrent-carryingconductors 6.Definitionofampere 7.Representationoffieldsduetoparallelcurrents 8.Torqueexperiencedbyacurrent-carryingcoilinauniform MagneticField 9.MovingCoilGalvanometer 10.ConversionofGalvanometerintoAmmeterandVoltmeter 11.DifferencesbetweenAmmeterandVoltmeter
LorentzMagneticForce: Acurrentcarryingconductorplacedinamagneticfieldexperiencesaforcewhichmeansthatamovingchargeinamagneticfieldexperiencesforce. F or q+θB whereθistheanglebetweenvandB SpecialCases: i)Ifthechargeisatrest,i.e.v=0,thenFm=0. notexperienceanyforce.q-θB oranti-paralleltothedirectionofthemagnetic field,thenFm=0.F tothemagneticfield,thentheforceis maximum.Fm(max)=qvB Fm= q(vxB) Fm= (qvBsin θ)n I v I So,a stationarychargeinamagnetic fielddoes ii)Ifθ=0°or180°i.e.ifthechargemovesparallel v iii)Ifθ=90°i.e.ifthechargemovesperpendicular
Fleming’sLeftHandRule: (F)Fieldthumboflefthandarestretchedmutually perpendiculartoeachotherandthecentralfingerpointstocurrent,forefingerpointstomagneticfield,thenthumbpointsinthedirectionofmotion (force)onthecurrentcarryingconductor.Electric Current TIP:(I) Rememberthephrase‘emf’torepresentelectriccurrent,magneticfieldandforceinanticlockwisedirectionofthefingersoflefthand. ForceonamovingchargeinuniformElectricandMagnetic Fields: WhenachargeqmoveswithvelocityvinregioninwhichbothelectricfieldEandmagneticfieldBexist,thentheLorentzforceis F=qE+q(vxB)orF=q(E+vxB) Force Magnetic Ifthecentralfinger,forefingerand (B)
Forceonacurrent-carryingconductorinauniform MagneticField: Forceexperiencedbyeachelectronin F f=-e(vdxB) vd Abetheareaofcrosssectionofthe-B conductor,thenno.ofelectronsintheAl I Forceexperiencedbytheelectronsindlis dF=nAdl[-e(vdxB)]=-neAvd(dlXB) =I(dlxB)whereI=neAvdand-vesignrepresentsthat F=∫dF=∫I(dlxB) F=I(lxB)orF=IlBsinθ I theconductoris Ifnbethenumberdensityofelectrons, θ dl elementdlisnAdl. thedirectionofdlisoppositetothat ofvd)
Forcesbetweentwoparallelinfinitelylongcurrent-carryingconductors:Forcesbetweentwoparallelinfinitelylongcurrent-carryingconductors: MagneticFieldonRSduetocurrentinPQisQS µ0I1(inmagnitude) ForceactingonRSduetocurrentI2throughitisI1I2 F21=2πr2orF21=2πrB2xB1 B1actsperpendicularandintotheplaneofthediagramby RightHandThumbRule.So,theanglebetweenlandB1is90˚.r lislengthoftheconductor. MagneticFieldonPQduetocurrentinRSis µ0I2(inmagnitude)PRForceactingonPQduetocurrentI1throughitis µ0I1I2l(Theanglebetweenland F12=2πr12πremergingout) µ0I1I2l 2πrµII 2πr B1 = 2π r F12 F21 µ0 I1 µ0 I1I2l I l sin90˚ B2= 2πr µ0I2 B2is90˚andB2Is F12 = or Ilsin90˚ F12=F21=F= F/l=012 N/m Forceperunit lengthoftheconductoris
QSQS I1 I1I2 FFFF rrI2 PRPR ByFleming’sLeftHandRule,ByFleming’sLeftHandRule,theconductorsexperiencetheconductorsexperienceforcetowardseachotherandforceawayfromeachotherhenceattracteachother.andhencerepeleachother. x x x
DefinitionofAmpere: ForceperunitlengthoftheF/l=N/m WhenI1=I2=1Ampereandr=1m,thenF=2x10-7N/m. Oneampereisthatcurrentwhich,ifpassedineachoftwoparallelconductorsofinfinitelengthandplaced1mapartinvacuumcauseseachconductortoexperienceaforceof2x10-7Newtonpermetreoflengthoftheconductor. RepresentationofFieldduetoParallelCurrents: I1I2I1I2 BB N µ0I1I2 2πr conductoris
TorqueexperiencedbyaCurrentLoop(Rectangular)inauniformMagneticField:TorqueexperiencedbyaCurrentLoop(Rectangular)inauniformMagneticField: LetθbetheanglebetweentheplaneoftheloopandbS coilisperpendiculartothemagneticfield.θ FRSI |FSP|=IbBsinθxB FQR=I(bxB)l |FQR|=IbBsinθFR θ Moreovertheyactalongthesamelineofaction(axis)Q FPQ=I(lxB) |FPQ|=IlBsin90°=IlBForcesFPQandFRSbeingequalinmagnitudebut oppositeindirectioncancelouteachotheranddonot FRS=I(lxB)produceanytranslationalmotion.Buttheyact |FRs|=IlBsin90°=IlBalongdifferentlinesofactionandhence FSP thedirectionofthemagneticfield.Theaxisofthe P FSP=I(bxB) PQ I Forces FSP and FQR areequalinmagnitudebut oppositeindirectionandtheycancelouteachother. andhencedonotproducetorque. FQR producetorqueabout theaxisofthecoil.
TorqueexperiencedbythecoilisFRS ז=FPQxPN(inmagnitude) ז=IlB(bcosθ)bxS ז=IlbBcosθ ז=IABcosθ(A=lb)PθN ז=NIABcosθ(whereNistheno.ofturns)Fn IfΦistheanglebetweenthenormaltothecoilandthedirectionofthemagneticfield,then Φ+θ=90°i.e.θ=90°-ΦIR So,B ז=IABcos(90°-Φ)Q ז=NIABsinΦ NOTE: Onemustbeverycarefulinusingtheformulaintermsofcosorsinsinceitdependsontheangletakenwhetherwiththeplaneofthecoilorthenormalofthecoil. θ Φ B PQ Φ I n
TorqueinVectorform: ז=NIABsinΦ ז=(NIABsinΦ)n(wherenisunitvectornormaltotheplaneoftheloop) ז=NI(AxB)orז=N(MxB) (sinceM=IAistheMagneticDipoleMoment)Note: 1)Thecoilwillrotateintheanticlockwisedirection(fromthetopview,accordingtothefigure)abouttheaxisofthecoilshownbythedottedline. 2)Thetorqueactsintheupwarddirectionalongthedottedline(accordingtoMaxwell’sScrewRule). 3)IfΦ=0°,thenז=0. 4)IfΦ=90°,thenזismaximum.i.e.זmax=NIAB 5)Units:BinTesla,IinAmpere,Ainm2andזinNm. 6)Theaboveformulaefortorquecanbeusedforanyloopirrespectiveofitsshape.
MovingCoilorSuspendedCoilorD’ArsonvalTypeGalvanometer: TorqueexperiencedbyT ז=NIABsinΦ RestoringtorqueinthePBW coilisEM ז=kα(wherekisFRS restoringtorqueperunitPS angulartwistinthewire) Atequilibrium, B I=kαFPQ ThefactorsinΦcanbeTS eliminatedbychoosing RadialMagneticField. T–TorsionHead,TS–Terminalscrew,M–Mirror,N,S–Polespiecesofamagnet,LS–LevellingScrews,PQRS–Rectangularcoil,PBW–PhosphorBronzeWire thecoilis angulartwist,αisthe NIABsinΦ=kα NABsinΦ LS LS N x SQ R HairSpring
RadialMagneticField: S alongthemagneticlinesofforceinwhicheverNS positionthecoilcomestorestinequilibrium. So,theanglebetweentheplaneofthecoilandB themagneticfieldis0°. ortheanglebetweenthenormaltotheplaneofMirror thecoilandthemagneticfieldis90°. i.e.sinΦ=sin90°=1Lamp2α I=kαorI=GαwhereG=kScale iscalledGalvanometerconstant CurrentSensitivityofGalvanometer: Itisthedefectionofgalvanometerperunitcurrent.I=k VoltageSensitivityofGalvanometer:αNAB Itisthedefectionofgalvanometerperunitvoltage.VkR The(top view PS of) planeof thecoilPQRS lies P NAB NAB αNAB =
ConversionofGalvanometertoAmmeter: GalvanometercanbeconvertedintoammeterIIgG PotentialdifferenceacrossthegalvanometerS andshuntresistanceareequal. IgG I–Ig ConversionofGalvanometertoVoltmeter: Galvanometercanbeconvertedintovoltmeter Potentialdifferenceacrossthegivenloadresistanceisthesumofp.dacrossgalvanometerandp.d.acrossthehigh V Ig byshuntingitwitha verysmallresistance. Is=I- Ig (I–Ig)S= IgG or S= Ig GR byconnecting itwith averyhighresistance. V resistance. -G V=Ig(G+R) or R=
DifferencebetweenAmmeterandVoltmeter: S.No.AmmeterVoltmeter ItisalowresistanceItisahighresistanceinstrument. 2 ResistanceisGS/(G+S)ResistanceisG+R ShuntResistanceisSeriesResistanceis 3(GIg)/(I–Ig)andisverysmall.(V/Ig)-Gandisveryhigh. ItisalwaysconnectedinItisalwaysconnectedinparallel. ResistanceofanidealResistanceofanidealvoltmeter ItsresistanceislessthanthatItsresistanceisgreaterthanthat ItisnotpossibletodecreaseItispossibletodecreasethe 7 therangeofthegivenrangeofthegivenvoltmeter.ammeter. 1 instrument. 4 series. 5 ammeteriszero. isinfinity. 6 ofthegalvanometer. ofthevoltmeter.
MAGNETICEFFECTOFCURRENT-III 1.Cyclotron 2.Ampere’sCircuitalLaw 3.MagneticFieldduetoaStraightSolenoid 4.MagneticFieldduetoaToroidalSolenoid
HFCyclotron: SB D1WD2+ D2 N W W–WindowB-MagneticField Working:ImaginingD1ispositiveandD2isnegative,the+velycharged towardsD2.DuetoperpendicularmagneticfieldandaccordingtoFleming’s WhenitisabouttoleaveD2,D2becomes+veandD1becomes–ve. describethesemi-circularpath.Theprocesscontinuestillthechargetraversesthroughthewholespaceinthedeesandfinallyitcomesoutwithveryhighspeedthroughthewindow. Oscillator B D1 D1,D2–Dees N,S–MagneticPolePieces particlekept atthecentreandinthegapbetween thedeesgetaccelerated LeftHandRulethechargegetsdeflectedanddescribessemi-circularpath. ThereforetheparticleisagainacceleratedintoD1whereitcontinuesto
Theory: Themagneticforceexperiencedbythechargeprovidescentripetalforcerequiredtodescribecircularpath. mv2/r=qvBsin90°(wherem–massofthechargedparticle, Bqrq–charge,v–velocityonthepathof mangleb/nvandB) Iftisthetimetakenbythechargetodescribethesemi-circularpathinsidethedee,then πrπmTimetakeninsidethedeedependsonlyon v Bqthespeedofthechargeortheradiusofthepath. IfTisthetimeperiodofthehighfrequencyoscillator,thenforresonance, 2πm Bq Iffisthefrequencyofthehighfrequencyoscillator(CyclotronFrequency),then Bq 2πm v= radius–r,Bismagneticfieldand90°isthe themagneticfieldandm/qratioandnoton t= or t= T=2t or T= f=
MaximumEnergyoftheParticle: KineticEnergyofthechargedparticleis 222 mm MaximumKineticEnergyofthechargedparticleiswhenr=R(radiusoftheD’s). B2q2R2 m TheexpressionsforTimeperiodandCyclotronfrequencyonlywhenmremainsconstant.(Otherquantitiesarealreadyconstant.) Butmvarieswithvaccordingtom0 Einstein’sRelativisticPrincipleasper[1–(v2/c2)]½ Iffrequencyisvariedinsynchronisationwiththevariationofmassofthechargedparticle(bymaintainingBasconstant)tohaveresonance,thenthecyclotroniscalledsynchro–cyclotron. Ifmagneticfieldisvariedinsynchronisationwiththevariationofmassofthechargedparticle(bymaintainingfasconstant)tohaveresonance,thenthecyclotroniscalledisochronous–cyclotron. NOTE:Cyclotroncannotbeusedforacceleratingneutralparticles.Electronscannotbeacceleratedbecausetheygainspeedveryquicklyduetotheirlightermassandgooutofphasewithalternatinge.m.f.andgetlostwithinthedees. Bqr ( Bqr K.E.=½mv2=½m =½ )2 K.E.max=½ m=
Ampere’sCircuitalLaw: Thelineintegral∫B.dlforaclosedcurveisequaltoµ0timesthenet I BB r Proof:Currentisemerging ∫B.dl=∫B.dlcos0°outandthemagnetic =∫B.dl=B∫dl =B(2πr)=(µ0I/2πr)x2πr ∫B.dl=µ0I currentIthreadingthrough theareabounded bythecurve. ∫ B. dl=µ0I dl I O field isanticlockwise.
MagneticFieldatthecentreofaStraightSolenoid: SaRB PaQ xxxxxxx I(whereIisthenetcurrentI ∫B.dl=∫B.dl+∫B.dl+∫B.dl+∫B.dl =∫B.dlcos0°+∫B.dlcos90°+∫0.dlcos0°+∫B.dlcos90° =B∫dl=B.aandµ0I0=µ0naIB=µ0nI (wherenisno.ofturnsperunitlength,aisthelengthofthepathand Iisthecurrentpassingthroughtheleadofthesolenoid) µ0I0 0 threadingthroughthesolenoid) ∫ B.dl= PQ QR RS SP
MagneticFieldduetoToroidalSolenoid(Toroid): dl B =B∫dl=B(2πr)r AndµI=µn(2πr)IOQ B=µ0nI NOTE:I Themagneticfieldexistsonlyinthe tubularareaboundbythecoilanditdoesnotexistintheareainsideandoutsidethetoroid. i.e.BiszeroatOandQandnon-zeroatP. ∫ ∫ B.dl=µ0I0 P B≠0 ∫ B.dlcos0° B.dl= B=0 00 0 B=0
MAGNETISM 1.BarMagnetanditsproperties 2.CurrentLoopasaMagneticDipoleandDipoleMoment 3.CurrentSolenoidequivalenttoBarMagnet 4.BarMagnetanditDipoleMoment 5.Coulomb’sLawinMagnetism 6.ImportantTermsinMagnetism 7.MagneticFieldduetoaMagneticDipole 8.TorqueandWorkDoneonaMagneticDipole 9.TerrestrialMagnetism 10.ElementsofEarth’sMagneticField 11.TangentLaw 12.PropertiesofDia-,Para-andFerro-magneticsubstances 13.Curie’sLawinMagnetism 14.HysteresisinMagnetism
Magnetism: -Phenomenonofattractingmagneticsubstanceslikeiron,nickel,cobalt,etc. •Abodypossessingthepropertyofmagnetismiscalledamagnet. •Amagneticpoleisapointneartheendofthemagnetwheremagnetismisconcentrated. •Earthisanaturalmagnet. •Theregionaroundamagnetinwhichitexertsforcesonothermagnetsandonobjectsmadeofironisamagneticfield. Propertiesofabarmagnet: 1.AfreelysuspendedmagnetalignsitselfalongNorth–Southdirection. 2.Unlikepolesattractandlikepolesrepeleachother. 3.Magneticpolesalwaysexistinpairs.i.e.Polescannotbeseparated. 4.Amagnetcaninducemagnetisminothermagneticsubstances. 5.Itattractsmagneticsubstances. Repulsionisthesuresttestofmagnetisation:Amagnetattractsironrodaswellasoppositepoleofothermagnet.Thereforeitisnotasuretestofmagnetisation. But,ifarodisrepelledwithstrongforcebyamagnet,thentherodissurelymagnetised.
RepresentationofUniformMagneticField: xxxxx xxxxx xxxxx xxxxx xxxxx UniformfieldUniformfieldperpendicular planeofthediagramplaneofthediagramofthediagram CurrentLoopasaMagneticDipole&DipoleMoment: MagneticDipoleMomentis AM=IAn BSIunitisAm2. TIP: Whenwelookatanyonesideoftheloopcarryingcurrent,ifthecurrent I isinanti-clockwisedirectionthenthatsideoftheloopbehaveslike thatsideoftheloopbehaveslikeMagneticSouthPole. &emergingoutoftheplane Uniformfieldonthe perpendicular&intothe MagneticNorthPoleandifthecurrentisinclockwisedirectionthen
CurrentSolenoidasaMagneticDipoleorBarMagnet: B xxxxxxx II TIP:Playpreviousandnexttounderstandthesimilarityoffieldlines.
BarMagnet: 1.ThelinejoiningthepolesofthemagnetSPMPN MagneticLength magnetiscalledmagneticlengthofthemagnet. 3.Thedistancebetweentheendsofthemagnetiscalledthegeometricallengthofthemagnet. 4.Theratioofmagneticlengthandgeometricallengthisnearly0.84. MagneticDipole&DipoleMoment: Apairofmagneticpolesofequalandoppositestrengthsseparatedbyafinitedistanceiscalledamagneticdipole. Themagnitudeofdipolemomentistheproductofthepolestrengthmandtheseparation2lbetweenthepoles. MagneticDipoleMomentisM=m.2l.lSIunitofpolestrengthisA.m ThedirectionofthedipolemomentisfromSouthpoletoNorthPolealongtheaxisofthemagnet. GeographicLength iscalledmagneticaxis. 2.Thedistancebetweenthepolesofthe
Coulomb’sLawinMagnetism: Theforceofattractionorrepulsionbetweentwomagneticpolesisdirectlyproportionaltotheproductoftheirpolestrengthsandinverselyproportionaltothesquareofthedistancebetweenthem. r αr2 kmmµ0m1m2 F=r24πr2 (wherek=µ0/4πisaconstantandµ0=4πx10-7TmA-1) Invectorformµ0m1m2r µ0m1m2r F αm1m2 m1 m2 F= 12 or F= r2 4π F=4πr3
MagneticIntensityorMagnetisingforce(H): i)MagneticIntensityatapointistheforceexperiencedbyanorthpoleofunitpolestrengthplacedatthatpointduetopolestrengthofthegivenmagnet.H=B/µ ii)Itisalsodefinedasthemagnetomotiveforceperunitlength. iii)Itcanalsobedefinedasthedegreeorextenttowhichamagneticfieldcanmagnetiseasubstance. iv)Itcanalsobedefinedastheforceexperiencedbyaunitpositivechargeflowingwithunitvelocityinadirectionnormaltothemagneticfield. v)ItsSIunitisampere-turnsperlinearmetre.vi)Itscgsunitisoersted. MagneticFieldStrengthorMagneticFieldorMagneticInductionorMagneticFluxDensity(B): i)MagneticFluxDensityisthenumberofmagneticlinesofforcepassingnormallythroughaunitareaofasubstance.B=µH ii)ItsSIunitisweber-m-2orTesla(T). iii)Itscgsunitisgauss.1gauss=10-4Tesla
MagneticFlux(Φ): i)Itisdefinedasthenumberofmagneticlinesofforcepassingnormallythroughasurface. ii)ItsSIunitisweber. RelationbetweenBandH: B=µH(whereµisthepermeabilityofthemedium) MagneticPermeability(µ): Itisthedegreeorextenttowhichmagneticlinesofforcecanpassenterasubstance. ItsSIunitisTmA-1orwbA-1m-1orHm-1 RelativeMagneticPermeability(µr): Itistheratioofmagneticfluxdensityinamaterialtothatinvacuum. Itcanalsobedefinedastheratioofabsolutepermeabilityofthematerialtothatinvacuum. µr=B/B0orµr=µ/µ0
IntensityofMagnetisation:(I): i)Itisthedegreetowhichasubstanceismagnetisedwhenplacedinamagneticfield. ii)Itcanalsobedefinedasthemagneticdipolemoment(M)acquiredperunitvolumeofthesubstance(V). iii)Itcanalsobedefinedasthepolestrength(m)perunitcross-sectionalarea(A)ofthesubstance. iv)I=M/V v)I=m(2l)/A(2l)=m/A vi)SIunitofIntensityofMagnetisationisAm-1. MagneticSusceptibility(cm): i)Itisthepropertyofthesubstancewhichshowshoweasilyasubstancecanbemagnetised. ii)Itcanalsobedefinedastheratioofintensityofmagnetisation(I)inasubstancetothemagneticintensity(H)appliedtothesubstance. iii)cm=I/HSusceptibilityhasnounit. RelationbetweenMagneticPermeability(µr)&Susceptibility(cm): µr=1+cm
MagneticFieldduetoaMagneticDipole(BarMagnet): i)Atapointontheaxiallineofthemagnet: µ02MxBN θ Ifl<<x,thenθ µ02MBS PNS BSBN ofthemagnet:SMNP ll BQ=4π(y2+l2)3/2 Ifl<<y,thenMagneticFieldatapointontheaxiallineacts µ0MMagneticFieldatapointontheequatorialline BQQ yB=B-B θθ x BP=4π(x2–l2)2 BP≈4πx3 ii) Atapointontheequatorialline µ0M alongthedipolemomentvector. BP≈4πy3 actsoppositetothedipolemomentvector.
TorqueonaMagneticDipole(BarMagnet)inUniformMagneticField: TheforcesofmagnitudemBactoppositetoeachotherand hencenetforceactingonthebar2lmB magneticfieldiszero.So,theremBθ magnet. Howevertheforcesarealongdifferentlinesofactionandconstituteacouple.Hencethemagnetwillrotateandexperience torque.M Torque=MagneticForcexdistanceθB t=mB(2lsinθ) =MBsinθt t=MxB DirectionofTorqueisperpendicularandintotheplanecontainingMandB. N M magnetduetoexternaluniform isnotranslationalmotionofthe S B
WorkdoneonaMagneticDipole(BarMagnet)inUniformMagnetic Field: mB dθmB θmB θ1 W=MB(cosθ1-cosθ2) IfPotentialEnergyisarbitrarilytakenzerowhenthedipoleisat90°,thenP.Einrotatingthedipoleandincliningitatanangleθis PotentialEnergy=-MBcosθ Note: PotentialEnergycanbetakenzeroarbitrarilyatanypositionofthedipole. dW=tdθ =MBsinθdθ θ2 θ1 2 W=∫MBsinθdθ B mB
TerrestrialMagnetism: i)GeographicAxisisastraightlinepassingthroughthegeographicalpolesoftheearth.Itistheaxisofrotationoftheearth.Itisalsoknownaspolaraxis. ii)GeographicMeridianatanyplaceisaverticalplanepassingthroughthegeographicnorthandsouthpolesoftheearth. iii)GeographicEquatorisagreatcircleonthesurfaceoftheearth,inaplaneperpendiculartothegeographicaxis.Allthepointsonthegeographicequatorareatequaldistancesfromthegeographicpoles. iv)MagneticAxisisastraightlinepassingthroughthemagneticpolesoftheearth.ItisinclinedtoGeographicAxisnearlyatanangleof17°. v)MagneticMeridianatanyplaceisaverticalplanepassingthroughthemagneticnorthandsouthpolesoftheearth. vi)MagneticEquatorisagreatcircleonthesurfaceoftheearth,inaplaneperpendiculartothemagneticaxis.Allthepointsonthemagneticequatorareatequaldistancesfromthemagneticpoles.
Declination(θ):Geographic TheanglebetweenthemagneticmeridianandMeridian atthatplace.δ BBV Linesshownonthemapthroughtheplacesthathavethesamedeclinationarecalledisogonic line.MagneticMeridian Linedrawnthroughplacesthathavezerodeclinationiscalledanagonicline. DiporInclination(δ): Theanglebetweenthehorizontalcomponentofearth’smagneticfieldandtheearth’sresultantmagneticfieldataplaceisDiporInclinationatthatplace. Itiszeroattheequatorand90°atthepoles. Linesdrawnuponamapthroughplacesthathavethesamediparecalledisocliniclines. Thelinedrawnthroughplacesthathavezerodipisknownasanaclinicline.Itisthemagneticequator. BH thegeographicmeridianataplaceisDeclination θ Itvariesfromplacetoplace.
HorizontalComponentofEarth’sMagneticField(BH): Thetotalintensityoftheearth’smagneticfielddoesnotlieinanyhorizontalplane.Instead,itliesalongthedirectionatanangleofdip(δ)tothehorizontal.Thecomponentoftheearth’smagneticfieldalongthehorizontalatanangleδiscalledHorizontalComponentofEarth’sMagneticField. BH=Bcosδ SimilarlyVerticalComponentisBV=Bsinδ 22 TangentLaw:B2B Ifamagneticneedleissuspendedinaregion wheretwouniformmagneticfieldsare perpendiculartoeachother,theneedlewillN θ thetangentoftheangleistheratioofthetwofields. tanθ=B2/B1 suchthat B=√BH+BV alignitselfalongthedirectionoftheresultant fieldofthetwofieldsatanangleθsuchthat B1
ComparisonofDia,ParaandFerroMagneticmaterials: DIAPARAFERRO 1.DiamagneticParamagneticsubstancesFerromagneticsubstancessubstancesarethosearethosesubstancesarethosesubstancessubstanceswhicharewhicharefeeblyattractedwhicharestrongly feeblyrepelledbyabyamagnet.attractedbyamagnet. magnet.Eg.Aluminium,Eg.Iron,Cobalt,Nickel,Eg.Antimony,Bismuth,Chromium,AlkaliandGadolinium,Dysprosium,Copper,Gold,Silver,Alkalineearthmetals,etc. Quartz,Mercury,Alcohol,Platinum,Oxygen,etc.water,Hydrogen,Air, Argon,etc. 2.WhenplacedinThelinesofforceprefertoThelinesofforcetendtomagneticfield,thelinesofpassthroughthecrowdintothespecimen.forcetendtoavoidthesubstanceratherthanair. substance. NS SNSN
2.Whenplacedinnon-Whenplacedinnon-Whenplacedinnon-uniformmagneticfield,ituniformmagneticfield,ituniformmagneticfield,itmovesfromstrongertomovesfromweakertomovesfromweakertoweakerfield(feeblestrongerfield(feeblestrongerfield(strongrepulsion).attraction).attraction).2.Whenplacedinnon-Whenplacedinnon-Whenplacedinnon-uniformmagneticfield,ituniformmagneticfield,ituniformmagneticfield,itmovesfromstrongertomovesfromweakertomovesfromweakertoweakerfield(feeblestrongerfield(feeblestrongerfield(strongrepulsion).attraction).attraction). 3.WhenadiamagneticWhenaparamagneticrodWhenaparamagneticrodrodisfreelysuspendedinisfreelysuspendedinaisfreelysuspendedina auniformmagneticfield,ituniformmagneticfield,ituniformmagneticfield,italignsitselfinadirectionalignsitselfinadirectionalignsitselfinadirectionperpendiculartothefield.paralleltothefield.paralleltothefieldvery quickly. NSNSNS