College Physics (4th Edition)

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

Chapter 19 - Problems - Page 753: 22

Answer

The electron is moving horizontally at an angle of $20.9^{\circ}$ east of south or $20.9^{\circ}$ west of north.

Work Step by Step

We can find the angle $\theta$ between the velocity vector and the magnetic field: $F = q~v ~B~sin~\theta$ $sin~\theta = \frac{F}{q~v~B}$ $\theta = arcsin~(\frac{F}{q~v~B})$ $\theta = arcsin~\left(\frac{1.6 \times 10^{-14}~N}{(1.6\times 10^{-19}~C)(2.0\times 10^5~m/s)(1.4~T)}~\right)$ $\theta = 20.9^{\circ}$ Since the charge of the electron is negative, by the right hand rule, the velocity vector must be directed horizontally at an angle of $20.9^{\circ}$ east of south or $20.9^{\circ}$ west of north.
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