Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 29 - The Magnetic Field - Exercises and Problems - Page 831: 28


$B = 1.6\times 10^{-3}~T$

Work Step by Step

We can find the magnetic field: $f = \frac{qB}{2\pi~m}$ $B = \frac{2\pi~m~f}{q}$ $B = \frac{(2\pi)~(9.1\times 10^{-31}~kg)~(45\times 10^6~Hz)}{1.6\times 10^{-19}~C}$ $B = 1.6\times 10^{-3}~T$
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