Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$I_{max} = 0.050~A$
We can find the maximum current: $\frac{1}{2}LI^2 = \frac{1}{2}C(\Delta V)^2$ $I = (\Delta V)~\sqrt{\frac{C}{L}}$ $I = (5.0~V)~\sqrt{\frac{0.10\times 10^{-6}~F}{1.0\times 10^{-3}~H}}$ $I = 0.050~A$