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# Chapter-32 - Chapter 32 1 We use 6 n =1 Bn = 0 to obtain 5...

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1253 Chapter 32 1. We use 6 1 0 Bn n = Φ = to obtain ( 29 5 6 1 1Wb 2Wb 3Wb 4Wb 5Wb 3Wb . B Bn n = Φ = - Φ = - - + - + - = + 2. (a) The flux through the top is + (0.30 T) π r 2 where r = 0.020 m. The flux through the bottom is + 0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the total of the previously mentioned fluxes. Thus (in magnitude) the flux though the sides is 1.1 mWb. (b) The fact that it is negative means it is inward. 3. (a) We use Gauss’ law for magnetism: z = r r B dA 0 . Now, z = + + r r B dA C Φ Φ Φ 1 2 , where Φ 1 is the magnetic flux through the first end mentioned, Φ 2 is the magnetic flux through the second end mentioned, and Φ C is the magnetic flux through the curved surface. Over the first end the magnetic field is inward, so the flux is Φ 1 = –25.0 μ Wb. Over the second end the magnetic field is uniform, normal to the surface, and outward, so the flux is Φ 2 = AB = π r 2 B , where A is the area of the end and r is the radius of the cylinder. It value is Φ 2 2 3 5 0120 160 10 7 24 10 72 4 = × = + × = + - - π . . . . . m T Wb Wb b gc h μ Since the three fluxes must sum to zero, Φ Φ Φ C = - - = - = - 1 2 250 72 4 47 4 . . . . μ μ μ Wb Wb Wb Thus, the magnitude is | | 47.4 Wb. C μ Φ = (b) The minus sign in c Φ indicates that the flux is inward through the curved surface. 4. From Gauss’ law for magnetism, the flux through S 1 is equal to that through S 2 , the portion of the xz plane that lies within the cylinder. Here the normal direction of S 2 is + y . Therefore,

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CHAPTER 32 1254 0 0 1 2 left 1 ( ) ( ) ( ) 2 ( ) 2 ln3 . 2 2 r r r B B r r r i iL S S B x L dx B x L dx L dx r x μ μ - - - Φ = Φ = = = = π - π 5. We use the result of part (b) in Sample Problem 32-1: ( 29 2 0 0 , 2 R dE B r R r dt μ ε = to solve for dE / dt : ( ( ( 29( 29( 29 7 3 13 2 2 12 2 2 3 0 0 2 2.0 10 T 6.0 10 m 2 V 2.4 10 . m s 4 T m A 8.85 10 C /N m 3.0 10 m dE Br dt R μ ε - - -7 - - × × = = = × π×10 × × 6. The integral of the field along the indicated path is, by Eq. 32-18 and Eq. 32-19, equal to 0 0 2 enclosed area (4.0 cm)(2.0 cm) (0.75 A) 52 nT m total area 12 cm d i μ μ = = . 7. (a) Noting that the magnitude of the electric field (assumed uniform) is given by E = V / d (where d = 5.0 mm), we use the result of part (a) in Sample Problem 32-1 ( 29 0 0 0 0 . 2 2 r r dE dV B r R dt d dt μ ε μ ε = = We also use the fact that the time derivative of sin ( ϖ t ) (where ϖ = 2 π f = 2 π (60) 377/s in this problem) is ϖ cos( ϖ t ). Thus, we find the magnetic field as a function of r (for r R ; note that this neglects “fringing” and related effects at the edges): ( 29 0 0 0 0 max max max cos 2 2 r rV B V t B d d μ ε μ ε ϖ ϖ ϖ = = where V max = 150 V. This grows with r until reaching its highest value at r = R = 30 mm: ( ( ( ( ( ( 29 12 3 0 0 max max 3 12 4 H m 8.85 10 F m 30 10 m 150V 377 s 2 2 5.0 10 m 1.9 10 T. r R RV B d μ ε ϖ -7 - - = - - π×10 × × = = × = × (b) For r 0.03 m, we use the expression max 0 0 max / 2 B rV d μ ε ϖ = found in part (a) (note the B r dependence), and for r 0.03 m we perform a similar calculation starting with the result of part (b) in Sample Problem 32-1:
1255 ( 29 ( 29 2 2 2 0 0 0 0 0 0 max max max max max 2 0 0 max cos 2 2 2 for 2 R R R dE dV B V t r dt rd dt rd R V r R rd μ ε

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Chapter-32 - Chapter 32 1 We use 6 n =1 Bn = 0 to obtain 5...

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