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CH 29 

MagneticFieldsduetoCurrents

I. CalculatingtheMagneticFieldduetoaCurrent

A. ThemagnitudeofthefielddBproducedatpointPatdistancerbyacurrentlengthelementidsturnsouttobe

 

 

 

whereistheanglebetweenthedirectionsofdSand r̂ ,aunitvector

thatpointsfromdstowardP.Symbol0isaconstant,calledthe

permeabilityconstant,whosevalueis

 

B. Therefore,invectorform 

 

 

C. Diagram

 

 

 

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II. MagneticFieldduetoaLongStraightWire:

A. ThemagnitudeofthemagneticfieldataperpendiculardistanceRfromalong(infinite)straightwirecarryingacurrentiisgivenby

 

 

 

B. Direction:RHRforWires

 

1. Visualizationwithironfillings:

a) Fig.29‐3Ironfilingsthathavebeensprinkledontocardboardcollectinconcentriccircleswhencurrentissentthroughthecentralwire.Thealignment,whichisalongmagneticfieldlines,iscausedbythemagneticfieldproducedbythecurrent.(CourtesyEducationDevelopmentCenter)

 

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2. RHR:Grasptheelementinyourrighthandwithyourthumbextendedpointinginthedirectionofthecurrentflow.Yourfingerswillthennaturallycurlaroundtheelementinthedirectionofthemagneticfieldlines.

3. Example:

 

 

a) Fig.29‐4Aright‐handrulegivesthedirectionofthemagneticfieldduetoacurrentinawire.(a)ThemagneticfieldBatanypointtotheleftofthewireisperpendiculartothedashedradiallineanddirectedintothepage,inthedirectionofthefingertips,asindicatedbythex.(b)Ifthecurrentisreversed,atanypointtotheleftisstillperpendiculartothedashedradiallinebutnowisdirectedoutofthepage,asindicatedbythedot.

 

C. ProofforEquationforSolvingforBmagnitude: 

 

Equation if semi‐infinite straight wire: Note that the Magnetic field due to either the lower half or the 

upper half of the infinite wire in figure 29‐5, is half that value; this is: 

 

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III. MagneticFieldduetoaCurrentinaCircularArcofWire:

A. StartingwithBiot‐Savartandthebelowdrawing:

 

 

 

 

Note: 

 

 

 

 

 

 

 

B  

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B. SampleProblem:

1. InFig.below,twocirculararcshaveradiia=13.5cmandb=10.7cm,subtendangleθ=74.0°,carrycurrenti=0.411A,andsharethesamecenterofcurvatureP.Whatarethe(a)magnitudeand(b)direction(intooroutofthepage)ofthenetmagneticfieldatP?

 

 

 

  

a) Solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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IV. ForcebetweenTwoParallelWires:

A. Tofindtheforceonacurrent‐carryingwireduetoasecondcurrent‐carryingwire,firstfindthefieldduetothesecondwireatthesiteofthefirstwire.Thenfindtheforceonthefirstwireduetothatfield.

B. Parallelcurrentsattracteachother,andantiparallelcurrentsrepeleachother.

C. Diagram

 

D.

 

 

 

 

 

 

 

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E. SampleProblem:

1. Twoinfinitelylong,parallel,current‐carryingwireswithcurrentsdirectedoutofthepageareshown.Thewiresareperpendiculartotheplaneofthepageandareseparatedbyadistanceof0.80m.

 

a) Whatisthemagneticfieldvectoratapointmidwaybetweenthewires?

 

 

 

 

 

 

 

b) Whatistheforceona2.5mlengthofwire1duetowire2? 

 

 

 

 

 

 

 

 

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V. ForcebetweenTwoParallelWires,RailGun:

 

 

 

 

 

 

 

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VI. Ampere’sLaw:

A. Remember,whenwewantedtofindqencgivenanE‐field,orviceversa,wechoseaGaussianSurface,andtooktheclosedloopintegralofEdotdA.

B. Now,wewanttoknowiencgivenaB‐field,orviceversa.SothistimewechooseanAmperianLoop,andwetaketheclosedloopintegralofBdotdS.

 

C. TheLaw:

1.

2. RHR:CurlyourrighthandaroundtheAmperianloop,withthefingerspointinginthedirectionofintegration.

a) Acurrentthroughtheloopinthegeneraldirectionofyouroutstretchedthumbisassignedaplussign,

b) andacurrentgenerallyintheoppositedirectionisassignedaminussign.

 

 

 

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D. Ampere’sLawusedtosolveMagneticFieldOutsideaLongStraightWireCarryingCurrent:

1.

2.

3.  

 

 

 

 

 

 

 

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E. Ampere’sLaw,MagneticFieldInsideaLongStraightWireCarryingCurrent:

1. Diagram

2.

3.

 

 

  

 

 

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VII. SolenoidsandToroids:

A. ASolenoidisbasicallyacoilwoundintoatightlypackedhelix.Inphysics,thetermsolenoidreferstoalong,thinloopofwire,oftenwrappedaroundametalliccore,whichproducesamagneticfieldwhenanelectriccurrentispassedthroughit.Solenoidsareimportantbecausetheycancreatecontrolledmagneticfieldsandcanbeusedaselectromagnets.

 

B. Below,averticalcrosssectionthroughthecentralaxisofa“stretched‐out”solenoid.Thebackportionsoffiveturnsareshown,asarethemagneticfieldlinesduetoacurrentthroughthesolenoid.Eachturnproducescircularmagneticfieldlinesnearitself.Nearthesolenoid’saxis,thefieldlinescombineintoanetmagneticfieldthatisdirectedalongtheaxis.Thecloselyspacedfieldlinesthereindicateastrongmagneticfield.Outsidethesolenoidthefieldlinesarewidelyspaced;thefieldthereisveryweak.

 

 

 

 

 

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C. ApplicationofAmpere’slawtoasectionofalongidealsolenoidcarryingacurrenti.TheAmperianloopistherectangleabcda.

 

1.

2.  

 

 

3. Herenbethenumberofturnsperunitlengthofthesolenoid

4.

5.  

 

 

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D. MagneticFieldofaToroid:

1. Atoroidisadoughnut‐shapedobject.

2. SothemagneticfieldofaToroidshapedsolenoidis:

3.

whereiisthecurrentinthetoroidwindings(andispositiveforthosewindingsenclosedbytheAmperianloop)andNisthetotalnumberofturns.Thisgives

4.  

 

 

 

 

 

 

 

 

 

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VIII. ACurrentCarryingCoilasaMagneticDipole:

A.

B.

C.

D.

E.

F.  

 

 

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