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: 9


(a) $I = 0.80~A$ (b) $J = 7.1\times 10^7~A/m^2$

Work Step by Step

(a) We can find the current $I$: $I = J~A$ $I = J~\pi~r^2$ $I = (4.5\times 10^5~A/m^2)~(\pi)~(0.75\times 10^{-3}~m)^2$ $I = 0.80~A$ (b) We can find the current density in the filament: $J = \frac{I}{A}$ $J = \frac{I}{\pi~r^2}$ $J = \frac{0.80~A}{(\pi)~(0.06\times 10^{-3}~m)^2}$ $J = 7.1\times 10^7~A/m^2$
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