College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 19 - Magnetism - Learning Path Questions and Exercises - Exercises - Page 691: 9


a). $3.84\times10^{-18}N$, b). $2.715\times10^{-18}N$, c). 0, d). 0.

Work Step by Step

$F=qvBsin\theta$ a). $F=1.6\times10^{-19}\times2\times10^{4}\times1.2\times10^{-3}\times sin90^{\circ}=3.84\times10^{-18}N$ b). $F=1.6\times10^{-19}\times2\times10^{4}\times1.2\times10^{-3}\times sin45^{\circ}=2.715\times10^{-18}N$ c). Since, $\theta=0^{\circ}, sin\theta=0$, So, $F=0$ d). Since, $\theta=180^{\circ}, sin\theta=0$, So, $F=0$
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