Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 33 - Electromagnetic Induction - Exercises and Problems - Page 997: 21

Answer

${\bf 9.5\times 10^{-5}}\;\rm J$

Work Step by Step

We know that this energy depends on the inductance and the current. So the energy stored in the solenoid is given by $$U_{\rm L}=\frac{1}{2}LI^2\tag 1$$ The inductance of the solenoid is given by $$L=\dfrac{\mu_0N^2 A}{l}$$ Plug into (1), $$U_{\rm L}=\frac{1}{2}\left[ \dfrac{\mu_0N^2 A}{l}\right]I^2 $$ $$U_{\rm L}=\frac{1}{2}\left[ \dfrac{\mu_0N^2 \pi r^2 }{l}\right]I^2 $$ Plug the known; $$U_{\rm L}=\frac{1}{2}\left[\pi \dfrac{(4\pi \times 10^{-7})(200)^2 (1.5\times 10^{-2})^2}{(0.12)}\right](0.80)^2 $$ $$U_{\rm L}=\color{red}{\bf 9.5\times 10^{-5}}\;\rm J$$
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