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 magnetic field. 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 AR22(.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 Nm 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 Tm/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.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.