A pure inductor of 40 mH carries the current i = 2.5 sin(400t), with i in A and t in s…

Step 1. The reactance. $X_L = \omega L = 400\ \text{rad/s} \times 0.040\ \text{H} = 16\ \Omega$.

Step 2. The peak voltage. $V_m = I_m X_L = 2.5\ \text{A} \times 16\ \Omega = 40$ V.

Step 3. The voltage as a sine. The voltage leads the current by $90^\circ$, so $v = 40 \sin(400t + \pi/2)\ \text{V}$, which is the same as $v = 40 \cos(400t)\ \text{V}$. (Check with $v = L\,di/dt = 0.040 \times 2.5 \times 400 \cos 400t = 40 \cos 400t$.)

Step 4. The greatest energy. The energy of the coil is $\tfrac{1}{2} L i^2$. It is greatest when $i$ is at its peak. $U = \tfrac{1}{2} \times 0.040\ \text{H} \times (2.5\ \text{A})^2 = 0.125$ J. At that instant the voltage is zero.