Answer
$I=13.3mA$
Work Step by Step
We know that:
$U_{max}=\frac{1}{2}LI^2$
This can be rearranged as;
$I=\sqrt\frac{2U_{max}}{L}$
We plug in the known values to obtain:
$I=\sqrt\frac{2(10\times 10^{-6})}{1.50\times 10^{-3}}=13.3\times 10^{-3}=13.3mA$
