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 1000: 56

Answer

${\bf 3.93 }\;\rm V$

Work Step by Step

We know that moving the coil away from the flux zone, to a zero flux zone, will create an induced current. This induced current is given by $$I_{\rm coil}=\dfrac{\varepsilon_{\rm coil}}{R}\tag 1$$ So the induced emf is given by Faraday’s law, $$\varepsilon_{\rm coil}=N\left| \dfrac{d\Phi}{dt}\right|$$ Plug into (1); $$I_{\rm coil}=\dfrac{N}{R}\left| \dfrac{d\Phi}{dt}\right|$$ Recalling that $I=dq/dt$, the author put it as a hint. $$\dfrac{dq}{dt}=\dfrac{N}{R}\left| \dfrac{d\Phi}{dt}\right|$$ Hence, $$dq=\dfrac{N}{R}\left| d\Phi \right|$$ Replacing $d$ by $\Delta $; $$\Delta q=\dfrac{N}{R}\left| \Delta\Phi \right|$$ $$dq=\dfrac{N}{R}\left| \Phi_f-\Phi_i \right|$$ where $\Phi=AB=\pi r^2 B$ Noting that the coil will be exposed to zero flux after pulling it out of the flux region, so $\Phi_f=0$, $$\Delta q=\dfrac{N}{R}\left| 0-\pi r^2 B \right|$$ $$\Delta q=\dfrac{N}{R}\left| -\pi r^2 B \right|\tag 2$$ We need to find the final voltage across the capacitor due to the induced current after pulling out the coil. Recalling that the voltage across the capacitor is given by $$\Delta V_C=\dfrac{\Delta q}{C}$$ Plug $\Delta q$ from (2), $$\Delta V_C=\dfrac{N\left| -\pi r^2 B \right|}{RC}$$ Plug the known; $$\Delta V_C=\dfrac{(10)\left| -\pi (0.005)^2 (1\times 10^{-3}) \right|}{(0.20)(1\times 10^{-6})}$$ $$\Delta V_C=\color{red}{\bf 3.93 }\;\rm V$$
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