College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 20 - Problems - Page 793: 7

Answer

$B = 3.3~T$

Work Step by Step

We can convert the angular speed to units of rad/s: $\omega = 350~rpm \times \frac{1~min}{60~s} \times \frac{2\pi~rad}{1~rev} = 36.65~rad/s$ We can find the strength of the magnetic field: $emf = \omega~NBA$ $B = \frac{emf}{\omega~NA}$ $B = \frac{emf}{\omega~N~\pi~r^2}$ $B = \frac{17.0~V}{(36.65~rad/s)(50)(\pi)(0.030~m)^2}$ $B = 3.3~T$
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