College Physics (4th Edition)

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

Chapter 19 - Problems - Page 754: 36


$T = \frac{2\pi~m}{q~B}$

Work Step by Step

$F = \frac{mv^2}{r}$ $F = qvB$ We can equate the two expressions for $F$ to find an expression for the speed $v$: $\frac{mv^2}{r} = qvB$ $v = \frac{q~B~r}{m}$ We can find an expression for the period: $T = \frac{distance}{speed}$ $T = \frac{2\pi~r}{v}$ $T = \frac{2\pi~r}{(\frac{q~B~r}{m})}$ $T = \frac{2\pi~m}{q~B}$ We can see that the period is independent of the particle's speed.
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