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 999: 47

Answer

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Work Step by Step

$$\color{blue}{\bf [a]}$$ The induced emf for a wire that moves in a magnetic field is given by $$\varepsilon=vlB$$ Plug the known and referring to Figure 33.26. $$\varepsilon=(100)(0.04)(1.0)=\color{red}{\bf 4}\;\rm V$$ $$\color{blue}{\bf [b]}$$ The induced current is given by $$I=\dfrac{\varepsilon}{R}$$ where $R_{net}=4R=4(0.01)=0.04\;\Omega$; $$I=\dfrac{4}{0.04}=\color{red}{\bf 100}\;\rm A$$ $$\color{blue}{\bf [c]}$$ The moving wire works as a battery. It has an internal resistance of 0.01$\Omega$, so the potential difference between its ends is then $$\Delta V=\varepsilon-IR_{\rm wire}=4-(100)(0.01)$$ $$\Delta V=\color{red}{\bf 3}\;\rm V$$
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