Download Magnetic Fields due to Currents

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February 20, 2016 CH 29 MagneticFieldsduetoCurrents
I.
CalculatingtheMagneticFieldduetoaCurrent
A. ThemagnitudeofthefielddBproducedatpointPatdistancerby
acurrentlengthelementidsturnsouttobe

whereistheanglebetweenthedirectionsof dS and r̂ ,aunitvector
thatpointsfromdstowardP.Symbol0isaconstant,calledthe
permeabilityconstant,whosevalueis
B. Therefore,invectorform
C. Diagram
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February 20, 2016 II.
MagneticFieldduetoaLongStraightWire:
A. ThemagnitudeofthemagneticfieldataperpendiculardistanceR
fromalong(infinite)straightwirecarryingacurrentiisgivenby
B. Direction:RHRforWires
1.
Visualizationwithironfillings:
a)
Fig.29‐3Ironfilingsthathavebeensprinkledontocardboard
collectinconcentriccircleswhencurrentissentthroughthecentral
wire.Thealignment,whichisalongmagneticfieldlines,iscausedbythe
magneticfieldproducedbythecurrent.(CourtesyEducation
DevelopmentCenter)
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February 20, 2016 2.
RHR:Grasptheelementinyourrighthandwithyourthumbextended
pointinginthedirectionofthecurrentflow.Yourfingerswillthennaturally
curlaroundtheelementinthedirectionofthemagneticfieldlines.
3.
Example:
a)
Fig.29‐4Aright‐handrulegivesthedirectionofthemagneticfielddue
toacurrentinawire.(a)ThemagneticfieldBatanypointtotheleftofthewire
isperpendiculartothedashedradiallineanddirectedintothepage,inthe
directionofthefingertips,asindicatedbythex.(b)Ifthecurrentisreversed,at
anypointtotheleftisstillperpendiculartothedashedradiallinebutnowis
directedoutofthepage,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: Page3
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February 20, 2016 III.
MagneticFieldduetoaCurrentinaCircularArcofWire:
A. StartingwithBiot‐Savartandthebelowdrawing:
B  Note: Page4
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February 20, 2016 B. SampleProblem:
1.
InFig.below,twocirculararcshaveradiia=13.5cmandb=10.7cm,
subtendangleθ=74.0°,carrycurrenti=0.411A,andsharethesamecenterof
curvatureP.Whatarethe(a)magnitudeand(b)direction(intooroutofthe
page)ofthenetmagneticfieldatP?
a)
Solution:
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February 20, 2016 IV.
ForcebetweenTwoParallelWires:
A. Tofindtheforceonacurrent‐carryingwireduetoasecond
current‐carryingwire,firstfindthefieldduetothesecondwireatthe
siteofthefirstwire.Thenfindtheforceonthefirstwireduetothat
field.
B. Parallelcurrentsattracteachother,andantiparallelcurrents
repeleachother.
C. Diagram
D. Page6
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February 20, 2016 E. SampleProblem:
1.
Twoinfinitelylong,parallel,current‐carryingwireswithcurrents
directedoutofthepageareshown.Thewiresareperpendiculartotheplane
ofthepageandareseparatedbyadistanceof0.80m.
a)
Whatisthemagneticfieldvectoratapointmidwaybetweenthe
wires?
b)
Whatistheforceona2.5mlengthofwire1duetowire2?
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February 20, 2016 V.
ForcebetweenTwoParallelWires,RailGun:
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February 20, 2016 VI.
Ampere’sLaw:
A. Remember,whenwewantedtofindqencgivenanE‐field,orvice
versa,wechoseaGaussianSurface,andtooktheclosedloopintegral
ofEdotdA.
B. Now,wewanttoknowiencgivenaB‐field,orviceversa.Sothis
timewechooseanAmperianLoop,andwetaketheclosedloop
integralofBdotdS.
C. TheLaw:
1.
2.
RHR:CurlyourrighthandaroundtheAmperianloop,withthefingers
pointinginthedirectionofintegration.
a)
Acurrentthroughtheloopinthegeneraldirectionofyour
outstretchedthumbisassignedaplussign,
b)
andacurrentgenerallyintheoppositedirectionisassigneda
minussign.
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February 20, 2016 D. Ampere’sLawusedtosolveMagneticFieldOutsideaLongStraight
WireCarryingCurrent:
1.
2.
3.
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February 20, 2016 E. Ampere’sLaw,MagneticFieldInsideaLongStraightWire
CarryingCurrent:
1.
Diagram
2.
3.
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February 20, 2016 VII.
SolenoidsandToroids:
A. ASolenoidisbasicallyacoilwoundintoatightlypackedhelix.In
physics,thetermsolenoidreferstoalong,thinloopofwire,often
wrappedaroundametalliccore,whichproducesamagneticfield
whenanelectriccurrentispassedthroughit.Solenoidsareimportant
becausetheycancreatecontrolledmagneticfieldsandcanbeusedas
electromagnets.
B. Below,averticalcrosssectionthroughthecentralaxisofa
“stretched‐out”solenoid.Thebackportionsoffiveturnsareshown,as
arethemagneticfieldlinesduetoacurrentthroughthesolenoid.
Eachturnproducescircularmagneticfieldlinesnearitself.Nearthe
solenoid’saxis,thefieldlinescombineintoanetmagneticfieldthatis
directedalongtheaxis.Thecloselyspacedfieldlinesthereindicatea
strongmagneticfield.Outsidethesolenoidthefieldlinesarewidely
spaced;thefieldthereisveryweak.
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February 20, 2016 C. ApplicationofAmpere’slawtoasectionofalongidealsolenoid
carryingacurrenti.TheAmperianloopistherectangleabcda.
1.
2.
3.
thesolenoid
Herenbethenumberofturnsperunitlengthof
4.
5.
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February 20, 2016 D. MagneticFieldofaToroid:
1.
Atoroidisadoughnut‐shapedobject.
2.
SothemagneticfieldofaToroidshapedsolenoidis:
3.
whereiisthecurrentinthetoroidwindings
(andispositiveforthosewindingsenclosedbytheAmperianloop)andNis
thetotalnumberofturns.Thisgives
4.
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February 20, 2016 VIII. ACurrentCarryingCoilasaMagneticDipole:
A.
B.
C.
D.
E.
F.
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