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 34 - Electromagnetic Fields and Waves - Exercises and Problems - Page 1030: 30

Answer

$({\bf 2\times 10^5}\;\hat j ) \;\rm V/m$

Work Step by Step

From the given graph, we can see that: - $B= -0.010\;\hat z\;\rm T$ - $v=2\times 10^7\;\hat i\;\rm m/s$ - $\sum F=0\;\rm N$ Since the electron will move without deflecting, the net force exerted on it is zero, as mentioned above. Recalling that the net force exerted on a moving charge in an electromagnetic field is given by $$\sum F=q(\vec E+\vec v\times \vec B)$$ $$0=q(\vec E+\vec v\times \vec B)$$ Thus, $$\vec E=-\vec v\times \vec B$$ Plug the known; $$\vec E=-(2\times 10^7\;\hat i)\times(-0.010\;\hat z)$$ $$\vec E=-(2\times 10^5\;\hat j) $$ But this is the direction of the electric field if the charge is positive, so when the charge is negative (as our electron), then the direction of the electric field must be opposite. Hence, $$\vec E =(\color{red}{\bf 2\times 10^5}\;\hat j ) \;\rm V/m$$ It is obvious now that the direction of the electric field is upward, in the positive $y$-axis.
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