College Physics (4th Edition)

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

Chapter 19 - Problems - Page 753: 27


$v = 2.8\times 10^7~m/s$

Work Step by Step

$F = \frac{mv^2}{r}$ $F = qvB$ We can equate the two expressions for $F$ to find the maximum possible value for $v$: $\frac{mv^2}{r} = qvB$ $v = \frac{q~B~r}{m}$ $v = \frac{(1.6\times 10^{-19}~C)(0.360~T)(0.820~m)}{1.67\times 10^{-27}~kg}$ $v = 2.8\times 10^7~m/s$
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