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: 39

Answer

$ 14.65 \;\rm A $

Work Step by Step

We can assume that Earth's magnetic field is the same in both cases. And since the compass only shows the net direction of the magnetic field in both cases, it shows Earth's direction magnetic field only [in the first case (outside)] and the direction of the net magnetic field of both the Earth and the wire [in the second case (inside)]. In the figure below, it it obvious that the direction of the electric field due to the wire is downward. Now we need to find the magnitude of the electricfiled due to the current in the wire. We have a triangle below and we found its angles. Using the sines law; $$\dfrac{B_E}{\sin 148^\circ }=\dfrac{B_{w}}{\sin 15^\circ }$$ Solving for $B_w$; $$B_w=\dfrac{B_E}{\sin 148^\circ }\cdot \sin 15^\circ\tag 1$$ $$B_w=\dfrac{\mu I}{2\pi r}$$ Solving for $I$; $$I=\dfrac{2\pi r B_w}{\mu} $$ Plugging from (1); $$I=\dfrac{2\pi r }{\mu} \cdot \dfrac{B_E}{\sin 148^\circ }\cdot \sin 15^\circ$$ Plugging the known; $$I=\dfrac{2\pi \cdot 0.12 }{4\pi\times 10^{-7}} \cdot \dfrac{0.5\times 10^{-4}}{\sin 148^\circ }\cdot \sin 15^\circ$$ According to right-hand rule, the direction of the current in the wire must be upward since the compass is to the east from the wire. $$I=\color{red}{\bf 14.65}\;\rm A$$
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