ch30-p036 - 36. (a) For path 1, we have d B1 d dB dB 2 = (...

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36. (a) For path 1, we have () 2 23 11 1 1 1 1 3 0.200m 8.50 10 T/s 1.07 10 V B dd B d B d Ed s BA A r dt dt dt dt ππ Φ ⋅= = = = = − × =− × G G v (b) For path 2, the result is 2 3 22 2 2 0.300m 8.50 10 T/s 2.40 10 V B B s r dt dt −− Φ = = = G G v (c) For path 3, we have
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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