Answer
(a) $0.025A$
(b) $9.7\mu J$
Work Step by Step
(a) We know that
$I=(\frac{\epsilon}{R})(1-e^{\frac{-tR}{L}})$
We plug in the known values to obtain:
$I=(\frac{9.0}{180})(1-e^{-\frac{(0.120\times 10^{-3})(180)}{0.0031}})$
$I=0.025A$
(b) We can find the energy stored as
$U=(\frac{1}{2})LI^2$
We plug in the known values to obtain:
$U=(\frac{1}{2})(0.0031)(0.025)^2$
$U=9.7\mu J$
