An ideal inductor of inductance 40 mH is connected in series with a...

In a DC circuit at steady state, an ideal inductor acts as a short circuit (zero resistance). The steady current $i$ in the circuit is determined entirely by the battery voltage and the resistance.

Using Ohm's law:
$i = \dfrac{V}{R} = \dfrac{10}{5} = 2\text{ A}$

The magnetic energy $U$ stored in the inductor is:
$U = \dfrac{1}{2} L i^2$

Substitute $L = 40 \times 10^{-3}\text{ H}$ and $i = 2\text{ A}$:
$U = \dfrac{1}{2} \times (40 \times 10^{-3}) \times (2)^2$
$U = 20 \times 10^{-3} \times 4$
$U = 80 \times 10^{-3}\text{ J} = 80\text{ mJ}$

Assuming no energy is stored because the current is constant (and thus induced emf is zero) incorrectly yields $0\text{ mJ}$. Forgetting the factor of $\dfrac{1}{2}$ gives $160\text{ mJ}$, and forgetting to square the current gives $40\text{ mJ}$.

Answer: $80\text{ mJ}$