2009S2UnitExam4ASolutions

2009S2UnitExam4ASolutions - ID A UCSD Physics 2B Answer...

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ID: A 1 UCSD Physics 2B Unit Exam 4A Magnetism & Induction Answer Section MULTIPLE CHOICE 1. ANS: B r c = mv qB v = qBr c m = 1.60 × 10 19 C Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3.22 × 3 T Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 28.1 × 3 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 1.67 × 27 kg Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 8.67 × 3 m s TOP: CYCLOTRON RADIUS 2. ANS: D For the trajectory to be undisturbed, the combined net force must be zero: F = qE + vB () F = 0 E =− vB = 7.33 × 6 m s Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 8.31 × 5 T Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 609 V m TOP: CYCLOTRON RADIUS 3. ANS: B B = μ 0 NIa 2 2 z 2 + a 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3/2 = 1.26 × 6 H / m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 150 125 × 3 A Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3.0 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 2 23 . 0 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 2 + 3.0 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 2 È Î Í Í Í Í Í Í Í Í ˘ ˚ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ = 1.39 × 4 T = 139 T TOP: LOOP FIELD 4. ANS: D The magnitude is F ||= I l × B || = I lB sin θ = 1.5 A 3.0 m 1.0 T sin 58 = 3.8 N TOP: FORCE | WIRE 5. ANS: C The magnetic moment is the number of turns * current * area = NIA = NI π R 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 100 5.20 A 3.14 12.0 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 2 = 23.5 A m 2 TOP: DIPOLE MOMENT
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ID: A 2 6. ANS: A B = μ 0 In = 0 I N L L = 0 IN B = 1.26 × 10 6 H / m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3.42 A () 169 375 × 6 T = 1.94 m TOP: SOLENOID 7. 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.
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This note was uploaded on 04/04/2010 for the course PHYSICS 2B taught by Professor Lever during the Summer '09 term at UCSD.

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2009S2UnitExam4ASolutions - ID A UCSD Physics 2B Answer...

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