Answer
$I=5.58mA$
Work Step by Step
We know that
$U=\frac{1}{2}LI^2$
We rearrange the formula as:
$I=\sqrt\frac{2U}{L}$
We then plug in the known values to obtain:
$I=\sqrt\frac{2(1.1681\times 10^{-6})}{0.075}=5.58\times 10^{-3}=5.58mA$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 31 - Electromagnetic Oscillations and Alternating Current - Problems - Page 935: 1b
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