Biot-Savart Law & Circular Coils question from JEE Main 2021 (26 Aug, Shift 1)

Answer: (d)

1. $B_{axis} = B_c\left(1+\dfrac{r^2}{a^2}\right)^{-3/2} \approx B_c\left(1-\dfrac{3}{2}\dfrac{r^2}{a^2}\right)$ for $r \ll a$.
2. Fractional change $\dfrac{B_c - B_{axis}}{B_c} = \dfrac{3}{2}\dfrac{r^2}{a^2}$
3. So the answer is (d).