College Physics (4th Edition)

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

Chapter 18 - Problems - Page 698: 18


The thickness of the strip is $8.1\times 10^{-7}~m$

Work Step by Step

We can write an expression for the drift speed: $v_d = \frac{I}{n~q~A} = \frac{I}{n~q~W~T}$ $I$ 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 $W$ is the cross-sectional width $T$ is the cross-sectional thickness We can find the thickness $T$: $v_d = \frac{I}{n~q~W~T}$ $T = \frac{I}{n~q~W~v_d}$ $T = \frac{130 \times 10^{-6}~A}{(8.8\times 10^{22}~m^{-3})(1.6\times 10^{-19}~C)~(260\times 10^{-6}~m)(44~m/s)}$ $T = 8.1\times 10^{-7}~m$ The thickness of the strip is $8.1\times 10^{-7}~m$
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