College Physics (4th Edition)

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

Chapter 19 - Problems - Page 753: 30


$m = 2.6\times 10^{-25}~kg$

Work Step by Step

We can find an expression for the speed $v$: $\frac{mv^2}{r} = qvB$ $v = \frac{q~B~r}{m}$ The particle has a kinetic energy of $q~\Delta V$. We can find the mass $m$: $K = q~\Delta V$ $\frac{1}{2}mv^2 = q~\Delta V$ $\frac{1}{2}(m)(\frac{q~B~r}{m})^2 = q~\Delta V$ $m = \frac{q~B^2~r^2}{2~\Delta V}$ $m = \frac{(1.6\times 10^{-19}~C)~(1.2~T)^2~(0.125~m)^2}{(2)~(7000~V)}$ $m = 2.6\times 10^{-25}~kg$
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