Answer
The maximum current is $~~12.6~mA$
Work Step by Step
In part (a), we found that the total energy in the circuit is $1.98\times 10^{-6}~J$
We can find the maximum current $I$:
$U_B = 1.98\times 10^{-6}~J$
$\frac{L~I^2}{2} = 1.98\times 10^{-6}~J$
$I^2 = \frac{(1.98\times 10^{-6}~J)(2)}{L}$
$I = \sqrt{\frac{(1.98\times 10^{-6}~J)(2)}{25.0\times 10^{-3}~H}}$
$I = 1.26\times 10^{-2}~A$
$I = 12.6~mA$
The maximum current is $~~12.6~mA$
