An inductor of 0.40 H carries a current of 3.0 A. Compute the stored magnetic energy in joules, to one decimal place.

1.8 (answers within ±0.05 count)

$U = \tfrac{1}{2}LI^2 = 0.5 \times 0.40 \times 9.0 = 1.8$ J, the exact magnetic counterpart of $\tfrac{1}{2}CV^2$ for a capacitor.