Other: Induction & Displacement Current question from NEET 2015 (AIPMT, Cancelled Paper)

Answer: (b)

1. Field of the wire $B = \frac{\mu_0 I}{2\pi r}$. The left side is at $x - \frac{a}{2}$, the right side at $x + \frac{a}{2}$.
2. Only the vertical sides give motional emf, and they oppose: $e = (B_1 - B_2) a V$.
3. $e = \frac{\mu_0 I a V}{2\pi}\left(\frac{1}{x - a/2} - \frac{1}{x + a/2}\right) = \frac{\mu_0 I a V}{2\pi} \cdot \frac{4a}{(2x-a)(2x+a)}$
4. So $e \propto \frac{1}{(2x-a)(2x+a)}$: answer (b).