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1 Faraday's Law 203 *Q31.13 (i) Answer (b). The battery makes counterclockwise current 1, in the primary coil, so its magnetic field El is to the right and increasing just after the switch is closed. The secondary coil will oppose the change with a leftward field E2, which comes from an induced clockwise current 12 that goes to the right in the resistor. The upper pair of hands in the diagram represent this effect. \ \. _ mcreasing ~Bl',' (ii) Answer (d). At steady state the primary magnetic field is unchanging, so no emf is induced in the secondary. \SB ~ --J decreasing (iii) Answer (a). The primary's field is to the right and decreasing as the switch is opened. The secondary coil opposes this FG. Q31.13 decrease by making its own field to the right, carrying counterclockwise current to the left in the resistor. The lower pair of hands diagrammed represent this chain of events. *Q31.14 A positive electric charge carried around a circular electric field line in the direction of the field gains energy from the field every step of the way. t can be a test charge imagined to exist in vacuum or it can be an actual free charge participating in a current driven by an induced emf. By doing net work on an object carried around a closed path to its starting point, the magneticallyinduced electric field exerts by definition a nonconservative force. We can get a larger and larger voltage just by looping a wire around into a coil with more and more turns. SOLUT!6NS'TO:;~ROBl:EMS":':"",,:~' e~ '.,,'~;-". c_ ". ", Section 31.1 Section 31.3 Faraday's Law of nduction Lenz's Law *P31.1 (a) (b) =-N MA/).t cose =-lnr2 (B /).t - Bj }ose =_[ n( m)2 J( s T - O} =-(8.04xlO-6m2)12.5 T/s = 1101 flv tending to produce clockwise current as seen from above The rate of change of magnetic field in this case is (-0.5 T T)/0.08 s = 25 T/s. t is twice as large in magnitude and in the opposite sense from the rate of change in case (a), so the emf is also twice as large in magnitude and in the opposite sense. =1/),<1>81=/).(E.A) /).t /).t = (2.50T-0.500T)(8.00xlO st m2)( N s C m )c1 N v.c) m 1.60 mv 1 1 = 1.60 mv and 1,oop = R = ,Q-=_0_.80_0_mA_

2 204 Chapter 31 P31.3 =_NM3Acose!!.t =_NBnr2(coSef-cose;)!!.t = (50.0 x 10-6 T)[n(0.500 m)2 J(cos cos s 00) = mv *P31.4 Ca) We evaluate the average emf: l>-n MJAco,e!!.t!!.t =-Nn,'( Rt- B, }o,e = -l2[ n(o.021o m)' J(O-O.ll st }o,oo = V The average induced current will then be V/ == 4.42 ma. f the meter has a sufficiently short response time, it will register the current. The average current may even run the meter offscale by a factor of 4.42, so you might wish to slow down the motion of the coil. Cb) Positive. The coil sees decreasing external magnetic flux toward you, so it makes some flux of its own in this direction by carrying counterc1ockwise current, that enters the red terminal of the ammeter. (P31. \..- " Ca) Cb) =_ de>b=-a db = (0.160 m2)(0.350 T) e-4 OOj2 00 =13.79 mv = 2.00 s Cc) At t = 0 = \28.0 mv *P31.6 de> d =_N_B =-N-(BAcose) =-NBcose (/la) -M =-200(50.0 X 10-6 T)(cos62.00) ('90XJO~ s- m') = -1_0.2_1_V_ " P31.7 Noting unit conversions from F = qv x Band U = qv, the induced voltage is d(b.a) =-N-;jf = V [=-- R _ V A

3 -----~----~ 206 Chapter 31 c9<1>, =(!l.nl)a~,"" =-N-=-Nl1on 7rrsolenoid - d<p B (2) d =-15.0( 47rX 10-7 T mj A)(1.00x 103 m-) 7r( m)2 (600 Ajs )cos(120t) =-14.2cos(120t) mv *P31.12 (a) =-20(130)47rX 10-7(0-6[1]/13 X 10-6)7r(0.03)21/0.8 = vi (b) A flat compact circular coil with 130 turns and radius 40.0 cm carries current 3.00 A counterclockwise. The current is smoothly reversed to become 3.00 A clockwise after 13.0,lis. At the center of this primary coil is a secondary coil in the same plane, with 20 turns and radius 3.00 cm. Find the emf induced in the secondary. "! i.,j j, P31.13 For a counterclockwise trip around the left-hand loop, with B=At ~[ At (2a2 )cos 0 J- 11 (5R) - pqr = 0 and for the right-hand loop, a p B0 t PQ ,'!: " ji where PQ = is the upward current in QP. 2a FG. P31.13 Q a 2 5( 2 ) 2Aa - 6R PQ-"3 Aa + pqr = 0 Aa2 PQ = -23-R upward, and since R = (0.100 Q/m)(0.650 m) = Q PQ = (1.00 X 10-3 Tjs) (0.650 m)2 = 283 1A upward.. 23( Q) -----,.,.1 P = ~:B = N( ~} = N ( Ot)A At t = 5.00 s, lel= 30.0 (0.410 T/S)[ 7r( m)2 J= mv 1 P31.15 B = 10n = l1on(30.0 A)(1- e-1.60t) n turns/m ~ <PB = f BdA = l1on(30.0 A)(1- e-.60)f da N turns FG. P31.15 = (68.2 mv) e counterclockwise 1

4 208 Chapter 31 P31.19 The upper loop has area n:(0.05 m)2 = 7.85 x 10-3 m2. The induced emf in it is d db _ =-N-BAcos8=-lAcosoo-=-7.85xlO-3 m2(2 T/s)=-1.57XlO 2 V The minus sign indicates that it tends to produce counterclockwise current, to make its own magnetic field out of the page. Similarly, the induced emf in the lower loop is =-NAcos8dB =-n:(0.09 m)22 T/s=-5.09xlO-2 V=+5.09xlO-2 V to produce '! counterclockwise current in the lower loop, which becomes clockwise current in the upper loop, The net emf for current in this sense around the figure 8 is 5.09 X 10-2 V X 10-2 V = 3.52 X 10-2 V. t pushes current in this sense through series resistance [2n:(0.05 m) + 2n:(0.09 m)]3 Q/m = 2.64 Q X 10-2 V The current S 1= R = 2.64 Q -_1_3_.3_m_A_. Section 31.2 Motional emf Section 31.3 Lenz's Law P31.20 (a) For maximum induced emf, with positive charge at the top of the antenna, F+ = q+ (v X:B), so the auto must move east. 1i:i'j, ". 'i, *P31.21 (a) =B v=(1.2xlo-6 T)(14.0m)(70m/s)=11.18XO-3 V. A free positive test charge in the wing feels a magnetic force in direction v x :B= north x down = west. Then it migrates west to make the left-hand wingtip positive. (b) No change. The charges in the horizontally-moving wing respond only to the vertical component of the Earth's field. (c) No. f we tried to connect the wings into a circuit with the light bulb, we would run an extra insulated wire along the wing. With the wing it would form a one-turn coil, in which the i' " -;i (' B v i P31.22 ~ 1=- emf = - i's zero = 2.5(1.2)v/6 as the coil= moves 0.5 in a uniform field. l R R \ / \~ v = 1.00 m/s R 1) FG. P31.22

5 Faraday's Law 215 P31.43 (a) )\ (b) The emf is directly proportional to vt' but the current is inversely proportional to R. A large R means a small current at a given speed, so the loop must travel faster to get FB = mg. FG. P31.43 (c) At a given speed, the current is directly proportional to the magnetic field. But the force is proportional to the product of the current and the field. For a small B, the speed must increase to compensate for both the small B and also the current, so vt cc 11B2 Additional Problems *P31.44 (a) The circuit encloses increasing flux of magnetic field into the page, so it tries to make its own field out of the page, by carrying counterclockwise current. (b) 1= R = mvlr = (0.4 T 0.8 m S m1s)/48 Q = 1100 ma (c) The magnetic field exerts a backward magnetic force on the induced current. With the values in (b), this force is B = 0.1 A 0.8 m 0.4 T = N, much less than 0.6 N. The speed of the bar increases until the backward magnetic force exerted on the current in the bar is equal to 0.6 N. The terminal speed is given by 0.6 N = B = (B )lvr. Then v = 0.6 N R(B )2 = 0.6 N 48 Q/(0.4 T 0.8 m)2 = 1281 m/s. (d) (e) (f) Fv = 0.6 N 281 m/s = 1169 W The terminal speed becomes larger. The bar must move faster to generate a larger emf to produce enough current in the larger resistance to feel the 0.6-N magnetic force. The power delivered to the circuit by the agent moving the bar, and then converted into internal energy by the resistor, is described by P = Fv = F2 R B2 2. Thus the power is directly proportional to the resistance and becomes larger as the bulb heats up. E=-N -(BA cos e) = -N (nr2) cos 0 - d (db ) E=-(30.0{n(2.70XlO-3 mfj(1) :r[so.o mt+(3.20 mt)sin(2n[s23t S-J)] E=-(30.0)[n(2.70XlO-3 mf](3.20xlo-3 T)[2n(S23 s-1)cos(2n[s23t S-lJ)J E=-(7.22XlO-3 V)cos[2n(S23t S-l)]

6 n. Faraday's Law 217 &Th, 'mf induced betweeu tb, end< of b, moving b", ;, C = B v = (2.50 T)(0.350 m)(8.00 m/s) = 7.00 V The left-hand loop contains decreasing flux away from you, so the induced current in it will be ~ clockwise, to produce its own field directed away from you. Let 1[ represent the current flowing upward through the 2.00-Q resistor. The right-hand loop will carry counterc1ockwise current. Let 13 be the upward current in the 5.00-Q resistor. (a) Kirchhoff's loop rule then gives: V - 11(2.00 Q) = 0 and V - 13 (5.00 Q) = 0 (b) The total power converted in the resistors of the circuit is 11= 3.50 A 13 = A (c) P = = CU + 13) = (7.00 V)(3.50 A A) = W Method 1: The current in the sliding conductor is downward with value 12= 3.50 A + 1AO A = 4.90 A. The magnetic field exerts a force of Fm = B = (4.90 A)(0.350 m)(2.50 T) = 4.29 N directed ~ toward the right on this conductor. An outside agent must then exert a force of N to the left to keep the bar moving. Method 2: The agent moving the bar must supply the power according to P = F.v = Fv cos 0. The force required is then: P W = N F = -;; m/s P = C +Cjnduced R and C' =-~(BA) Coinduced F = m dv = 1Bd - D 11 i:l FG. P ,- (out of paper) -- dv _ 1Bd = Bd (C '" - m mr dv = Bd ce-bvd) mr +Cinduced) To solve the differential equation, let u = C - Bvd du =-Bd dv 1 du Bd ---=-u Bd mr so Jdu =_f(bd? "0 u 0 mr ntegrating from t = 0 to t = t, or Since v = 0 when t = 0, and n~=_(bd)2 Uo mr ~ = e-s'd't/mr Uo Uo =C u = C - Bvd t Therefore, v = :d (1- e-s'd't/mr)

21 MAGNETIC FORCES AND MAGNETIC FIELDS

21 MAGNETIC FORCES AND MAGNETIC FIELDS CHAPTER 1 MAGNETIC FORCES AND MAGNETIC FIELDS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1 (d) Right-Hand Rule No 1 gives the direction of the magnetic force as x for both drawings A and B In drawing C, the