problem29_56

problem29_56 - capacitor b 2 s m 2 3 N 88 2 A 4 2 = = = = =...

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29.56: The bar will experience a magnetic force due to the induced current in the loop. According to Example 29.6, the induced voltage in the loop has a magnitude , BLv which opposes the voltage of the battery, . ε Thus, the net current in the loop is . R BLv I - = The acceleration of the bar is . ) ( ) sin(90 mR LB BLv m ILB m F a - ° = = = a) To find mR LB BLv dt dv a t v ) ( set ), ( - = = and solve for v using the method of separation of variables: - - - = - = = - v t t e e BL v dt mR LB BLv dv mR L 0 0 ). (1 ) s m 10 ( ) 1 ( ) ( s 3.1 t 2 2 B Note that the graph of this function is similar in appearance to that of a charging
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Unformatted text preview: capacitor. b) 2 s m 2 . 3 N; 88 . 2 A; 4 . 2 = = = = = = m F a ILB F R ε I c) When 2 s m 6 . 2 ) (5.0 kg) (0.90 T) 1.5 ( m) (0.8 )] s m (2.0 m) (0.8 T) (1.5 V 12 [ , s m . 2 = Ω-= = a v d) Note that as the velocity increases, the acceleration decreases. The velocity will asymptotically approach the terminal velocity , s m 10 m) (0.8 T) (1.5 V 12 = = ε BL which makes the acceleration zero....
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This homework help was uploaded on 02/03/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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