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2009S2UnitExam4BSolutions

2009S2UnitExam4BSolutions - ID B UCSD Physics 2B Answer...

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ID: B 2 UCSD Physics 2B Unit Exam 4B Magnetism & Induction Answer Section MULTIPLE CHOICE 1. ANS: B Two ways to view this problem: 1. By Faraday's Law, the induced emf = − d Φ dt Ê Ë Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ so curl the finger of the right hand in the direction of the current ( emf ) and the thumb points to the right. Hence the flux is increasing to the left (decreasing to the right) and so the magnet (South pole to the left) must be moving to the right. 2. Lenz's Law says the current tries to oppose the change in flux. Since the current loop makes magnetic North pole pointing to the right, the loop is trying to attract the bar magnet so it must be moving to the right. TOP: LENZ 2. ANS: C Field at the center of a full circle is B = μ 0 I 2 R so the field for a semicircle must be half that. The two semicircular currents flow in opposite directions, so the fields subtract : B = μ 0 I 4 1 R 1 1 R 2 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ = 1.26 × 10 6 N / m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 595 A ( ) 4 1 0.023 m 1 0.032 m Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ = 2.29 × 10 3 T = 2.29m T TOP: FIELD SUPERPOSITION 3. ANS: B The magnetic flux is the product of the coil area A , the number of turns N , the magnetic field strength B and the cosine of the angle at which the field lines penetrate the coil: Φ = N B A = NBA cos θ N = Φ BA cos θ = 0.75 Wb 0.5 T ( ) 10 × 10 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 2 cos60 ( ) = 300 TOP: FLUX 4. ANS: B B = μ 0 In = μ 0 I N L L = μ 0 IN B = 1.26 × 10 6 H / m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3.42 A ( ) 169 ( ) 375 × 10 6 T = 1.94 m TOP: SOLENOID
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ID: B 2 5. ANS: A First we use Faraday’s Law to get the EMF. The flux through the loop is the field B times the area A = h × x . The change in the flux is only due to the penetration depth x since the field B and height h are fixed. The power dissipated by the resistor is P =
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