Ch30-p008 - This shows that the induced emf is induced = − 4.0 µ V Now we use Faraday’s law = − d Φ B dt = − A dB dt = − A a Plugging

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8. From the datum at t = 0 in Fig. 30-41(b) we see 0.0015 A = V battery /R , which implies that the resistance is R = (6.00 µ V)/(0.0015 A) = 0.0040 . Now, the value of the current during 10 s < t < 20 s leads us to equate ( V battery + ε induced ) /R = 0.00050 A.
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Unformatted text preview: This shows that the induced emf is induced = − 4.0 µ V. Now we use Faraday’s law: = − d Φ B dt = − A dB dt = − A a . Plugging in = − 4.0 × 10 − 6 V and A = 5.0 × 10 − 4 m 2 , we obtain a = 0.0080 T/s....
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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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