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 27 - Current and Resistance - Exercises and Problems - Page 762: 18


1.7 A

Work Step by Step

Given: Electric field E= 0.012 V/m, Area=$ 2.0 mm\times2.0mm= 4.0\times10^{-6}m^{2}$ For aluminium, Conductivity $\sigma= 3.5\times10^{7}\Omega^{-1}m^{-1}$ Recall: I= JA where J is the current density equal to $\sigma E$. Then, $I= \sigma EA$. Substitute the values of $\sigma$,E and A: Current I= $3.5\times10^{7}\Omega^{-1}m^{-1}\times0.012 \frac{V}{m}\times4.0\times10^{-6}m^{2}\approx 1.7 A$
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