# s06x2aan - NAME PHYS-1200 PHYSICS II QUIZ 2 SOLUTIONS PART...

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NAME PHYS-1200 PHYSICS II QUIZ 2 MARCH 29, 2006 SOLUTIONS PART A. 1. D 6. A 2. B 7. B 3. D 8. B 4. B 5. C PART B . 1. TO THE BOTTOM OF THE PAGE The current in the wire produces a decreasing magnetic flux within the wire loop. According to Lenz’s law, the induced electric field would be in a direction to oppose that decrease. It must be counterclockwise about the loop, so at P , it is toward the bottom of the page. 2. a) TO THE LEFT The same direction as the current in the wires. b) OUT OF THE PAGE Based on the direction of the current and the right hand rule. 3. a) s t x v 04 . 0 m 0 . 1 = = v = 25 m/s b) The wavelength can be read directly from the graph. λ = 4.0 m c) m 0 . 4 m/s 25 = = λ v f f = 6.2 Hz 1

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NAME PART C . 1. a) F) 10 0 . 4 H)( 10 0 . 2 ( 2 1 2 1 7 3 - - × × = = π π LC f f = 5.6 ×10 3 Hz = 5600 Hz b) dt di L = E . The easy way to get this is 0 = + C q dt di L , so C q dt di L - = . Then, just after the switch is closed, C Q C q dt di L = = = E , so F 10 0 4 C 10 5 . 1 7 6 - . × × = - E V 8 . 3 = E Another approach is q = Q cos ϖ t , so t Q dt dq i ϖ ϖ sin - = = , and t Q dt di ϖ ϖ cos 2 - = .
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