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Homework 12 Solutions

# Homework 12 Solutions - Chapter 32 1 We use 6 n =1 Bn = 0...

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1243 Chapter 32 1. We use 6 1 0 Bn n = Φ= to obtain () 5 6 1 1Wb 2Wb 3Wb 4Wb 5Wb 3Wb . BB n n = −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 ⋅= G G BdA 0 . Now, z ⋅=+ + G G C ΦΦΦ 12 , 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 = r 2 B , where A is the area of the end and r is the radius of the cylinder. Its value is Φ 2 2 35 0120 160 10 7 24 10 72 4 = + × = + −− π .. . . . m T Wb b g ch Since the three fluxes must sum to zero, ΦΦ Φ C =− = = − 250 72 4 47 4 . . Wb Thus, the magnitude is | | 47.4 Wb. C (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 1244 00 12 l e f t 1 () 2 2 l n 3 . 22 rr r BB r ii L SS B x L d x B x L d x L d x rx μ ππ −− Φ= = = = ∫∫ 5. We use the result of part (b) in Sample Problem — “Magnetic field induced by changing electric field,” 2 , 2 R dE B rR rd t με =≥ to solve for dE / dt : ( ) ( ) ( ) 73 13 2 2 12 2 2 3 2 2.0 10 T 6.0 10 m
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Homework 12 Solutions - Chapter 32 1 We use 6 n =1 Bn = 0...

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