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: 5



Work Step by Step

Given: q= 0.050 C B= 0.080 T $\theta =45^{\circ}$ F= 10 N Recall: $F= qvBsin\theta$ or $v=\frac{F}{qBsin\theta}$ Substituting the values, we get $v=\frac{10N}{0.050C\times0.080T\times sin45^{\circ}}= 3.5\times10^{3}m/s$
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