Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 37 - The Foundation of Modern Physics - Exercises and Problems - Page 1083: 35


$\frac{q}{m} = 9.58\times 10^7~C/kg$ This particle is a proton.

Work Step by Step

If the particle passes through without deflection, then $v = \frac{E}{B}$ We can find the charge-to-mass ratio: $F_B = qvB = \frac{mv^2}{r}$ $\frac{q}{m} = \frac{v}{B~r}$ $\frac{q}{m} = \frac{E}{B^2~r}$ $\frac{q}{m} = \frac{187,500~V/m}{(0.1250~T)^2~(0.12525~m)}$ $\frac{q}{m} = 9.58\times 10^7~C/kg$ We could guess that the charge of the particle is $1.6\times 10^{-19}~C$ We can find the mass: $\frac{q}{m} = 9.58\times 10^7~C/kg$ $m = \frac{q}{9.58\times 10^7~C/kg}$ $m = \frac{1.6\times 10^{-19}~C}{9.58\times 10^7~C/kg}$ $m = 1.67\times 10^{-27}~kg$ This particle is a proton.
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