Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 20 - Magnetism - Problems - Page 585: 43

Answer

$4.66\times10^{-5}T$.

Work Step by Step

At the point indicated, the magnetic field due to the top wire will be toward the right. The magnetic field due to the bottom wire points out of the page. Because these fields are at right angles, the net field there can be found using the Pythagorean Theorem with the two individual fields. The distance, r, from the field point to each wire is the same. $$B_{net}=\sqrt{(\frac{\mu_o I_{top}}{2\pi r})^2+(\frac{\mu_o I_{bot}}{2\pi r})^2}$$ $$ =\frac{\mu_o }{2\pi r}\sqrt{ I_{top}^2+ I_{bot}^2}$$ $$ =\frac{4\pi\times10^{-7} }{2\pi (0.100m)}\sqrt{ (20.0A)^2+ (12.0A)^2}=4.66\times10^{-5}T $$
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