Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 21 - Magnetic Forces and Magnetic Fields - Problems - Page 608: 1

Answer

$5.9\times 10^{12}$ $m/s^{2}$

Work Step by Step

The magnetic fore acting on a particle having charge $q$ and moving with a velocity $(\vec v)$ in a uniform magnetic field $(\vec B)$ is given by $ \vec F=q(\vec v\times \vec B)$ ............. $(1)$ Here, $\vec v \perp \vec B$ and $ |\vec B|=1.6\times10^{-5}$ $T$, $|\vec v|=2.1\times10^{6}$ $m/s $, $q=1.6\times10^{19}C$ Putting the above values in equation $(1)$, we get, $|\vec F|=qvB$ $|\vec F|=1.6\times10^{-19}\times2.1\times10^{6}\times1.6\times10^{-5}$ Now, $|\vec F|=m_{e}a$, where $m_{e}$ is the mass of electron and $a$ is the magnitude of the acceleration caused by the magnetic force. Thus, $a=\frac{|\vec F|}{m_{e}}$ or, $a=\frac{1.6\times10^{-19}\times2.1\times10^{6}\times1.6\times10^{-5}}{9.11\times10^{-31}}$ $m/s^{2}$ or, $a=5.9\times 10^{12}$ $m/s^{2}$ Therefore, the magnitude of the acceleration of the electron caused by the magnetic force is $5.9\times 10^{12}$ $m/s^{2}$.
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