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 - General Problems - Page 587: 67

Answer

$0.3\;\rm N$

Work Step by Step

We know that the magnetic force exerted on a wire that carries a current is given by $$F_{\text{on wire}}=B_{Earth}I_{wire}l_{wire}\sin\theta$$ In our case, the direction of the current is toward the east while we know that the direction of Earth's magnetic field is toward the north which means that the angle between them is $90^\circ$. Thus, $$F_{\text{on wire}}=B_{Earth}I_{wire}l_{wire}\sin90^\circ= B_{Earth}I_{wire}l_{wire}$$ Plugging the known; $$F_{\text{on wire}}= 5\times10^{-5}\cdot 330\cdot 18=0.397\;\rm T$$ $$F_{\text{on wire}}\approx \color{red}{\bf 0.3}\;\rm N$$ According to the right-hand rule, the direction of this force must be upward on the wire in the direction that is perpendicular to the plane of the current and the magnetic field. And since Earth's magnetic field dip at an angle of $22^\circ$, the direction of the force is $68^\circ$ above the horizontal.
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