A coil has an inductance of 2 H and resistance of 4 $$\Omega$$. A 10 V is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be _______ $$\times 10^{-2}$$ J

At equilibrium, the current through the coil has reached its steady-state value where the inductor acts as a short circuit.

$$I_0 = \frac{V}{R} = \frac{10}{4} = 2.5 \text{ A}$$

The energy stored in the magnetic field:

$$U = \frac{1}{2}LI_0^2 = \frac{1}{2} \times 2 \times (2.5)^2 = \frac{1}{2} \times 2 \times 6.25 = 6.25 \text{ J}$$

$$= 625 \times 10^{-2} \text{ J}$$