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: 19

Answer

(a) $E = 10~V/m$ (b) $J = 9.1\times 10^6~A/m^2$ (c) The diameter of the wire is $0.53~mm$

Work Step by Step

(a) We can find the electric field: $E = \frac{\Delta V}{d}$ $E = \frac{1.5~V}{0.15~m}$ $E = 10~V/m$ (b) We can find the current density: $J = \sigma~E$ $J = (9.1\times 10^5~S/m)(10~V/m)$ $J = 9.1\times 10^6~A/m^2$ (c) We can find the radius: $I = J~A$ $I = J~\pi~r^2$ $r^2 = \frac{I}{J~\pi}$ $r = \sqrt{\frac{I}{J~\pi}}$ $r = \sqrt{\frac{2.0~A}{(9.1\times 10^6~A/m^2)~(\pi)}}$ $r = 0.264\times 10^{-3}~m$ $r = 0.264~mm$ Since the diameter is double the radius, the diameter of the wire is $0.53~mm$
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