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 998: 38

Answer

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Work Step by Step

We have here a loop and a changing flux due to the changing in the current in the solenoid, so we must have an induced current and an induced emf. We know that the induced current is given by $$I_{\rm loop}=\dfrac{\varepsilon_{\rm loop}}{R}\tag 1$$ We know that the induced emf $\varepsilon$ is given by $$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm solenoid}}{dt}\right|\tag 2$$ where $ \Phi =\vec A\cdot \vec B =AB\cos\theta=AB $ since $\theta=0^\circ$, so $$d\Phi =A dB$$ where $B_{\rm solenoid}=\mu_0 NI/l$ $$d\Phi =A_{\rm loop}\dfrac{\mu_0NdI}{l} =\pi r^2_{\rm loop}\dfrac{\mu_0NdI}{l} $$ Plug into (2), $$\varepsilon_{\rm loop}=\left|\pi r^2_{\rm loop}\dfrac{\mu_0N}{l} \dfrac{dI }{dt}\right| $$ $$\varepsilon_{\rm loop}=\pi r^2_{\rm loop}\dfrac{\mu_0N}{l} \left|\dfrac{dI }{dt}\right| $$ Plug into (1), $$I_{\rm loop}=\dfrac{\pi r^2_{\rm loop}\dfrac{\mu_0N}{l} \left|\dfrac{dI }{dt}\right|}{R} $$ During $0\lt t\lt 0.02$, $|dI/dt|=50$, At $t=0.01$ s, $$I_{\rm loop}=\dfrac{\pi (0.005)^2 \dfrac{(4\pi\times 10^{-7})(120)}{(0.08)} \left|-50\right|}{(0.50)} $$ $$I_{\rm loop}=\color{red}{\bf 1.48\times 10^{-5}}\;\rm A$$
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