An electromagnet has an inductance of 2.8 H and carries a current of 0.40 A…

Step 1. The emf. $|\varepsilon| = L\dfrac{\Delta I}{\Delta t} = 2.8\ \text{H} \times \dfrac{0.40\ \text{A}}{1.4 \times 10^{-3}\ \text{s}} = 800\ \text{V}$.

Step 2. The energy. $U = \dfrac{1}{2}LI^2 = \dfrac{1}{2} \times 2.8\ \text{H} \times (0.40\ \text{A})^2 = 0.224\ \text{J}$.

Step 3. The power. $P = \dfrac{U}{\Delta t} = \dfrac{0.224\ \text{J}}{1.4 \times 10^{-3}\ \text{s}} = 160\ \text{W}$.

Step 4. The spark. The coil pushes current through the circuit with an emf of $800$ V. The circuit is open, so that voltage stands across the gap of the switch. The air in the gap breaks down, and a spark carries the current. The wire has no gap, so there is no spark in it.