College Physics (4th Edition)

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

Chapter 19 - Problems - Page 754: 32


$B = 0.17~T$

Work Step by Step

$F = \frac{mv^2}{r}$ $F = qvB$ We can equate the two expressions for $F$ to find an expression for the speed $v$: $\frac{mv^2}{r} = qvB$ $v = \frac{q~B~r}{m}$ The ion has a kinetic energy of $q~\Delta V$. We can find the magnitude of the field: $K = q~\Delta V$ $\frac{1}{2}mv^2 = q~\Delta V$ $\frac{1}{2}(m)(\frac{q~B~r}{m})^2 = q~\Delta V$ $B^2 = \frac{2~m~\Delta V}{q~r^2}$ $B = \sqrt{\frac{2~m~\Delta V}{q~r^2}}$ $B = \sqrt{\frac{(2)~(12)(1.66\times 10^{-27}~kg)~(5000~V)}{(1.6\times 10^{-19}~C)~(0.21~m)^2}}$ $B = 0.17~T$
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