College Physics (4th Edition)

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

Chapter 18 - Problems - Page 698: 4

Answer

$I = 9.6\times 10^{-6}~A$

Work Step by Step

We can find the current: $I = (3.0\times 10^{13}~s^{-1})(2)(1.6\times 10^{-19}~C)$ $I = 9.6\times 10^{-6}~A$
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