College Physics (4th Edition)

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

Chapter 18 - Problems - Page 699: 20


$v_d = 1.14\times 10^{-4}~m/s$

Work Step by Step

We can find the number $N$ of copper atoms per $m^3$: $N = \frac{9.0~g/cm^3}{64~g/mol}$ $N = 0.140625~mol/cm^3$ $N = (0.140625~mol/cm^3)(6.02\times 10^{23}~atoms/mol)(10^6~cm^3/m^3)$ $N = 8.4656\times 10^{28}~atoms/m^3$ We can find the number $n$ of conduction electrons per $m^3$: $n = 1.3\times (8.4656\times 10^{28})$ $n = 1.1\times 10^{29}$ We can write an expression for the drift speed: $v_d = \frac{I}{n~q~A}$ $I$ is the current $n$ is the number of conduction electrons per $m^3$ $q$ is the charge of one electron $A$ is the cross-sectional area We can find the drift speed $v_d$: $v_d = \frac{I}{n~q~A}$ $v_d = \frac{2.0~A}{(1.1\times 10^{29}~m^{-3})(1.6\times 10^{-19}~C)(1.00\times 10^{-6}~m^2)}$ $v_d = 1.14\times 10^{-4}~m/s$
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