Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 44

Answer

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

Work Step by Step

The force $F$ provides the centripetal force to keep the particle moving in a circle. We can find an expression for the speed; $F = \frac{m~v^2}{r}$ $q~v~B = \frac{m~v^2}{r}$ $v = \frac{q~r~B}{m}$ We then find the period $T$ of the circular motion: $T = \frac{distance}{speed}$ $T = \frac{2\pi~r}{v}$ $T = \frac{2\pi~r}{(q~r~B)/m}$ $T = \frac{2\pi~m}{q~B}$
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