Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 30 - Current and Resistance - Exercises and Problems - Page 887: 22

Answer

nichrome

Work Step by Step

We can find the resistivity of the wire: $\rho = \frac{E}{J}$ $\rho = \frac{E}{I/A}$ $\rho = \frac{E~A}{I}$ $\rho = \frac{E~\pi~r^2}{I}$ $\rho = \frac{(0.0075~V/m)~(\pi)~(0.50\times 10^{-3}~m)^2}{3.9\times 10^{-3}~A}$ $\rho = 1.5\times 10^{-6}\Omega~m$ This value for the resistivity is closest to the resistivity of nichrome. We can conclude that the wire is made of nichrome.
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