College Physics (4th Edition)

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

Chapter 19 - Problems - Page 753: 28


The minimum radius is 21 cm

Work Step by Step

$F = \frac{mv^2}{r}$ $F = qvB$ We can equate the two expressions for $F$ to find the minimum required value for $r$: $\frac{mv^2}{r} = qvB$ $r = \frac{mv}{qB}$ $r = \frac{(1.67\times 10^{-27}~kg)(1.0\times 10^7~m/s)}{(1.6\times 10^{-19}~C)(0.50~T)}$ $r = 0.21~m$ The minimum radius is 21 cm
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