1. **State the problem:** We need to find the self-inductance $L$ of a coil with 500 turns, where each turn has a magnetic flux $\phi = 0.10$ Wb when a current $I = 2.0$ A flows through it. 2. **Formula used:** The self-inductance $L$ is defined by the relation between the total magnetic flux linkage $N\phi$ and the current $I$: $$L = \frac{N\phi}{I}$$ where $N$ is the number of turns, $\phi$ is the flux per turn, and $I$ is the current. 3. **Calculate total flux linkage:** $$N\phi = 500 \times 0.10 = 50 \text{ Wb-turns}$$ 4. **Calculate self-inductance:** $$L = \frac{50}{2.0} = 25 \text{ H}$$ 5. **Explanation:** The self-inductance measures how much magnetic flux is linked per unit current. Here, the coil's 500 turns each contribute 0.10 Wb flux, so total flux linkage is 50 Wb-turns. Dividing by the current 2.0 A gives the inductance 25 H.