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ChapterChapter 3030 -- MagneticMagnetic FieldsFields andand TorqueTorque AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, ProfessorProfessorProfessor ofofof PhysicsPhysicsPhysics SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity

© 2007 Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to:

•• DetermineDetermine thethe magnitudemagnitude andand directiondirection ofof thethe forceforce onon aa currentcurrent--carryingcarrying wirewire inin aa BB--fieldfield.. •• CalculateCalculate thethe magneticmagnetic torquetorque onon aa coilcoil oror solenoidsolenoid ofof areaarea AA,, turnsturns NN,, andand currentcurrent II inin aa givengiven BB--fieldfield.. •• CalculateCalculate thethe magneticmagnetic fieldfield inducedinduced atat thethe centercenter ofof aa looploop oror coilcoil oror atat thethe interiorinterior ofof aa solenoidsolenoid.. ForceForce onon aa MovingMoving ChargeCharge

RecallRecall thatthat thethe magneticmagnetic fieldfield BB inin teslasteslas (T)(T) waswas defineddefined inin termsterms ofof thethe forceforce onon aa movingmoving chargecharge::

Magnetic Field FF Intensity B: FF B B vv F vv B  qvsin N S 1 N 1 N 1 T  BB C(m/s) A m ForceForce onon aa ConductorConductor

SinceSince aa currentcurrent IIisis chargecharge qq movingmoving throughthrough aa wire,wire, thethe magneticmagnetic forceforce cancan bebe givengiven inin termsterms ofof current.current.

x x x x x x FF x x Motion Right-hand rule: x x x x x x x x of +q force F is upward. x x x x x x x x I = q/t x x x x L x x x FF == qvBqvB

Since v = L/t, and I = q/t, Lq  F qB  LB we can rearrange to find: tt  The force F on a conductor of length L F = IBL and current I in perpendicular B-field: F = IBL ForceForce DependsDepends onon CurrentCurrent AngleAngle Just as for a moving F charge, the force on a B B wire varies with direction. v sin   I v FF == IBLIBL sinsin  Current I in wire: Length L

Example 1. A wire of length 6 cm makes an angle of 200 with a 3 mT . What current is needed to cause an upward force of 1.5 x 10-4 N?

F 1.5 x 10-4 N I  I = 2.44 A BLsin (3 x 10-3 T)(0.06m)sin20 0 I = 2.44 A ForcesForces onon aa CurrentCurrent LoopLoop

ConsiderConsider aa looploop ofof areaarea AAA === ababab carryingcarrying aa currentcurrent II inin aa constantconstant BBfieldfield asas shownshown below.below.

B b Normal vector x x x x x x x F1 n x x x x x x A x x x x x x x  I a x x x x x x N S x x x x x x x x x x x x x x x x x x x x F2 x x x x x Torque 

TheThe rightright-hand-hand rulerule showsshows thatthat thethe sideside forcesforces cancelcancel each other and the forces F and F cause a torque. each other and the forces F11 and F22 cause a torque. TorqueTorque onon CurrentCurrent LoopLoop Recall that torque is product of force and moment arm.

b The moment arms F a sin x x x x for F and F are: 1 2 x x x x 1 2  n Iout a a x xI x x a sin 2 2  x x x x B x x x x F = F = IBb 1 2 a  2 X I   ()(sin)IBb a  in 1 2 a sin 2 F2 a  2  ()(sin)IBb 2 

a  2(IBb )(2 sin ) IB ( ab )sin   IBAsin

In general, for a loop of N turns carrying a current I, we have:   NIBAsin ExampleExample 2:2: AA 200200--turnturn coilcoil ofof wirewire hashas aa radiusradius ofof 2020 cmcm andand thethe normalnormal toto thethe areaarea makesmakes anan angleangle ofof 30300 withwith aa 33 mTmT BB--field.field. WhatWhat isis thethe torquetorque onon thethe looploop ifif thethe currentcurrent isis 33 AA?? N = 200 turns   NIBAsin n AR22(.2m)  N B S A = 0.126 m2; N = 200 turns

B = 3 mT;  = 300; I = 3 A B = 3 mT;  = 300

NIBAsin (200)(3 A)(0.003T)(0.126 m20 )sin 30

Resultant torque on loop:  = 0.113 Nm

MagneticMagnetic FieldField ofof aa LongLong WireWire

WhenWhen aa currentcurrent IIpassespasses throughthrough aa longlong straightstraight wire,wire, thethe magneticmagnetic fieldfield BBisis circularcircular asas isis shownshown byby thethe patternpattern ofof ironiron filingsfilings belowbelow andand hashas thethe indicatedindicated directiondirection..

Iron TheThe rightright-hand-hand thumbthumb I filings I rule:rule: GraspGrasp wirewire withwith rightright hand;hand; pointpoint thumbthumb inin directiondirection ofof II.. FingersFingers wrapwrap wirewire inin directiondirection ofof B B thethe circularcircular BB-field.-field. CalculatingCalculating BB--fieldfield forfor LongLong WireWire

TheThe magnitudemagnitude ofof thethe magneticmagnetic fieldfield BB atat aa distancedistance rr fromfrom aa wirewire isis proportionalproportional toto currentcurrent II..

 I Circular B Magnitude of B-field for B  0 current I at distance r: 2 r I

The proportionality constant  is called the permeability of free space: r B X -7 Permeability:  = 4 x 10 Tm/A

ExampleExample 3:3: AA longlong straightstraight wirewire carriescarries aa currentcurrent ofof 44 AA toto thethe rightright ofof page.page. FindFind thethe magnitudemagnitude andand directiondirection ofof thethe BB--fieldfield atat aa distancedistance ofof 55 cmcm aboveabove thethe wire.wire. r B=? r = 0.05 m 0 I 5 cm B  I = 4 A I = 4 A 2 r

-7 Tm (4 x 10A )(4 A) B B = 1.60 x 10-5-5 T or 16 T 2 (0.05 m) B = 1.60 x 10 T or 16 T

RightRight-hand-hand thumbthumb B out of paper rule:rule:FingersFingers pointpoint r I = 4 A outout ofof paperpaper inin directiondirection ofof BB-field.-field. ExampleExample 4:4: TwoTwo parallelparallel wireswires areare separatedseparated byby 66 cmcm.. WireWire 11 carriescarries aa currentcurrent ofof 44 AA andand wirewire 22 carriescarries aa currentcurrent ofof 66 AA inin thethe samesame direction.direction. WhatWhat isis thethe resultantresultant BB--fieldfield atat thethe midpointmidpoint betweenbetween thethe wires?wires?

0 I B  2 I2 = 6 A 2 r 6 A 3 cm B into B=? 2 x B is positive paper 3 cm 1 I1 = 4 A B2 is negative B1 out of paper 4 A Resultant is vector 1 sum: BR = B ExampleExample 44 (Cont.):(Cont.): FindFind resultantresultant BB atat midpoint.midpoint.

-7 Tm (4 x 10A )(4 A) B1 26.7 T I = 6 A 2(0.03 m) 3 cm 2 B=? -7 Tm (4 x 10A )(6 A) 3 cm I = 4 A B 40.0 T 1 2 2(0.03 m)

Resultant is vector sum: BR = B  I B  0 2 r BR = 26.7 T – 40 T = -13.3 T

B1 is positive BR is into paper: B = -13.3 T B2 is negative ForceForce BetweenBetween ParallelParallel WiresWires

Recall wire with I1 P creates B1 at P: I  I F2 2 B  01 d d I 1 2 d 1 Out of paper!

Now suppose another wire with current I2 in same direction is parallel to first wire. Wire 2 experiences force F2 due to B1 .

I2 From right-hand rule, ForceForce FF2 isis 2 B what is direction of F2 ? DownwardDownward F2 ParallelParallel WiresWires (Cont.)(Cont.)

Now start with wire 2. B2 into paper I2 creates B2 at P: 2 I  I 2 B  02 d 2 d F1 2 d I1 1 x P INTO paper!

Now the wire with current I1 in same direction is parallel to first wire. Wire 1 experiences force F1 due to B2 .

F From right-hand rule, Force F is 1 Force F11 is B what is direction of F1 ? UpwardUpward I1 ParallelParallel WiresWires (Cont.)(Cont.)

WeWe havehave seenseen thatthat Attraction twotwo parallelparallel wireswires withwith currents in the same F2 I currents in the same d 2 directiondirection areare attractedattracted F1 I1 toto eacheach other.other.

Repulsion F UseUse rightright-hand-hand forceforce 2 rulerule toto showshow thatthat oppositely directed I oppositely directed d I 2 currentscurrents repelrepel eacheach 1 other. other. F1 CalculatingCalculating ForceForce onon WiresWires The field from current in wire 2 is given by: Attraction 2 F 02I 2 I2 B2  d 2 d F1 I1 1 The force F 1 F = I B L L on wire 1 is: F11 = I11 B22 L

02I The same equation results FI11  L 2 d when considering F2 due to B1 F  I I TheThe forceforce perper unitunit lengthlength forfor  012 twotwo wireswires separatedseparated byby dd is:is: L 2 d ExampleExample 5:5: TwoTwo wireswires 55 cmcm apartapart carrycarry currents.currents. TheThe upperupper wirewire hashas 44 AA northnorth andand thethe lowerlower wirewire hashas 66 AA south.south. WhatWhat isis thethe mutualmutual forceforce perper unitunit lengthlength onon thethe wires?wires? F  I I Upper wire  012 2 I = 4 A L 2 d F2 2 d=5 cm F I = 6 A; I = 4 A; d = 0.05 m 1 I = 6 A I1 = 6 A; I2 = 4 A; d = 0.05 m 1 1 L Right-hand rule applied to Lower wire either wire shows repulsion.

F (4 x 10-7 Tm )(6 A)(4 A) F  A  9.60 x 10-5 N/m L 2(0.05 m) L MagneticMagnetic FieldField inin aa CurrentCurrent LoopLoop

RightRight-hand-hand rulerule showsshows BB fieldfield directeddirected outout ofof center.center.

II N B II Out

 I  NI Single B  0 Coil of N B  0 loop: 2R loops: 2R TheThe SolenoidSolenoid

AA solenoidsolenoid consistsconsists ofof manymany Permeability  turnsturns NN ofof aa wirewire inin shapeshape N ofof aa helix.helix. TheThe magneticmagnetic BB-- S fieldfield isis similarsimilar toto thatthat ofof aa barbar .magnet. TheThe corecore cancan bebe airair oror anyany material.material.

If the core is air:   4 x 10-7-7 Tm/A If the core is air:   4 x 10 Tm/A

The relative permeability r uses this value as a comparison.  TheThe relativerelative permeabilitypermeability   or  for a medium (  ): rr 0 for a medium ( r r ): 0 TheThe BB--fieldfield forfor aa SolenoidSolenoid

For a solenoid of length L, Solenoid For a solenoid of length L, L withwith NNturnsturns andand currentcurrent II,, N thethe BB--fieldfield isis givengiven by:by: S

NI B   L

SuchSuch aa BB--fieldfield isis calledcalled thethe magneticmagnetic inductioninduction sincesince itit arisesarises oror isis producedproduced byby thethe current.current. ItIt appliesapplies toto thethe interiorinterior ofof thethe solenoid,solenoid, andand itsits directiondirection isis givengiven byby thethe rightright-hand-hand thumbthumb rulerule appliedapplied toto anyany currentcurrent coil.coil. ExampleExample 6:6: AA solenoidsolenoid ofof lengthlength 2020 cmcm andand 100100 turnsturns carriescarries aa currentcurrent ofof 44 AA.. TheThe relativerelative permeabilitypermeability ofof thethe corecore isis 12,00012,000.. WhatWhat isis thethe magneticmagnetic inductioninduction ofof thethe coil?coil?

I = 4 A; N = 100 turns N = 100 turns 20 cm L = 0.20 m;   r 0 7 Tm  (12000)(4 x10A ) Tm  I = 4 A   0.0151A (0.0151Tm )(100)(4A) B  A B = 30.2 T 0.200 m B = 30.2 T

AA ferromagneticferromagnetic corecore cancan significantlysignificantly increaseincrease thethe B-fieldB-field !! SummarySummary ofof FormulasFormulas

The force F on a wire The force F on a wire F carryingcarrying currentcurrent II inin B B aa givengiven BB-field.-field. I sin   I v FF == IBLIBL sinsin  Current I in wire: Length L

F1 n TheThe torquetorque onon aa looploop oror coilcoil ofof NN turnsturns andand currentcurrent II inin A  aa BB-field-field atat knownknown angleangle .. N S   NIBAsin B F2 SummarySummary (Continued)(Continued)

AA circularcircular magneticmagnetic fieldfield BBisis inducedinduced Circular B byby aa currentcurrent inin aa wire.wire. TheThe directiondirection isis givengiven byby thethe rightright-hand-hand thumbthumb rulerule.. I

The magnitude depends 0 I on the current I and the B  r B distance r from the wire. 2 r X

I Permeability:  = 4 x 10-7-7 Tm/A Permeability:  = 4 x 10 Tm/A

SummarySummary (Continued)(Continued)

The force per unit length for F  I I  012 two wires separated by d is: L 2 d

 I  NI Single B  0 Coil of N B  0 loop: 2R loops: 2R

For a solenoid of length L, NI with N turns and current I, B  L the B-field is given by: CONCLUSION:CONCLUSION: ChapterChapter 3030 TorqueTorque andand MagneticMagnetic FieldsFields