Answer
$I_2=0.987A$ to the left.
Work Step by Step
We know that
$B_2=\frac{\mu_{\circ}I_2}{2\pi r}$
This can be rearranged as:
$I_2=\frac{2\pi(0.055)(3.59\times 10^{-6})}{4\pi \times 10^{-7}}$
$I_2=0.987A$ to the left.
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 22 - Magnetism - Problems and Conceptual Exercises - Page 799: 114
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