College Physics (4th Edition)

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

Chapter 19 - Problems - Page 753: 29


$B = 2.9~T$

Work Step by Step

$F = \frac{mv^2}{r}$ $F = qvB$ We can equate the two expressions for $F$ to find the value for $B$: $\frac{mv^2}{r} = qvB$ $B = \frac{mv}{qr}$ $B = \frac{(4.003)(1.67\times 10^{-27}~kg)(0.458)(3.0\times 10^8~m/s)}{(2)(1.6\times 10^{-19}~C)(1.00~m)}$ $B = 2.9~T$
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