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 763: 39


$0.104$ Volts/meter

Work Step by Step

Recall that $J = \sigma\vec{E}, \quad\rightarrow\quad I = A\sigma\vec{E}$ If the current is the same in both wires, $I_{nichrome} = I_{aluminum}$, then we have the equation: $ A_n\sigma_n\vec{E}_n = A_A\sigma_A\vec{E}_A$ Table 27.2 gives the conductivity for nichrome as $\sigma_n = 6.7\times10^5\Omega$m and for aluminum as $\sigma_A = 3.5\times10^7\Omega$ m Solving for $\vec{E}_n$... $\displaystyle \vec{E}_n = \frac{A_A\sigma_A\vec{E}_A}{A_n\sigma_n} = \frac{(\pi(0.0005)^2)(3.5\times10^7)(0.008)}{(\pi(0.001)^2)(6.7\times10^5)} = 0.104$ Volts/meter
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.