University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 30 - Inductance - Problems - Exercises - Page 1013: 30.14

Answer

6.92 mA/s.

Work Step by Step

By definition, $L=\frac{N\Phi_B}{i}$. $$ L=\frac{800(3.25\times10^{-3}Wb) }{2.90A}=0.8966H$$ $$L = \frac{\epsilon}{|di/dt|}$$ Solve for the rate of change of the current. $$|\frac{di}{dt}|=\frac{|\epsilon|}{L}=\frac{6.20\times10^{-3}V}{0.8966H}=6.92 mA/s$$
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