Answer
$i=5.58A$
Work Step by Step
Energy is given as $U=uV$.
We plug in the known values to obtain:
$U=(70)(0.0200)=1.4J$
Inductor energy is given as:
$U=\frac{1}{2}Li^2$
This can be rearranged as:
$i=\sqrt\frac{2U}{L}$
We plug in the known values to obtain:
$i=\sqrt\frac{2(1.4)}{90\times 10^{-3}}=5.58A$
