College Physics (4th Edition)

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

Chapter 18 - Problems - Page 698: 15


$v_d = 5.87\times 10^{-5}~m/s$

Work Step by Step

We can write an expression for the drift speed: $v_d = \frac{I}{n~q~A} = \frac{I}{n~q~\pi~r^2}$ $I$ is the current $n$ is the number of electrons per unit of volume $q$ is the charge of one electron $A$ is the cross-sectional area $r$ is the cross-sectional radius We can find the drift speed $v_d$: $v_d = \frac{I}{n~q~\pi~r^2}$ $v_d = \frac{2.50~A}{(8.47\times 10^{28}~m^{-3})(1.6\times 10^{-19}~C)~(\pi)~(1.00\times 10^{-3}~m)^2}$ $v_d = 5.87\times 10^{-5}~m/s$
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