# f01x2aan - q is a maximum so the current is zero i = 0 A c...

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NAME PHYS-1200 PHYSICS II QUIZ 2 OCTOBER 24, 2001 SOLUTIONS PART A. 1. E 6. D 2. D 7. B 3. B 8. D 4. B 5. C PART B . 1. TO THE TOP 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 top of the page. 2. a) TO THE LEFT The same direction as the current in the wires. b) INTO THE PAGE Based on the direction of the current and the right hand rule. 3. a) NO DIRECTION, IT IS ZERO That is the equilibrium position. b) TO THE RIGHT Back toward the equilibrium position. c) NO DIRECTION, IT IS ZERO That is the equilibrium position.

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NAME PART C . 1. a) rad/s 10 16 . 3 F) 10 H)(2.0 10 5 . 0 ( 1 1 and , 2 4 6 3 × = × × = = = - - LC T ϖ ϖ π Then, rad/s 10 16 . 3 2 4 × = π T T = 1.99 ×10 -4 s = 2.0 ×10 -4 s = 0.20 ms b) The capacitor is fully charged,
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Unformatted text preview: q is a maximum, so the current is zero. i = 0 A c) After one period, everything is the same as at t = 0. i = 0 A g) The induced emf will oppose the increase of current. 2. a) 1 m 0.020 2 2 so , 2-= = = λ k k = 100 m b) 2 s 10 6 2 1 6-× = = f f = 3.0 ×10 6 Hz = 3.0 MHz c) Negative x direction The + sign before the t term means it is going in the – direction. d) B must be along the y direction, perpendicular to the direction of propagation, and E . To move in the – x direction it must be of the form B = + B m sin ( kx + t ) y ˆ k and are the same as for the electric field, so only B m is needed. T 10 . 1 m/s 10 3.0 V/m . 3 8 8-× = × = = c E B m m . Then, B = (1.0 ×10-8 T) sin [(0.020 π m-1 ) x + (6.0 π ×10 6 s-1 ) t ] y ˆ...
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