Biot-Savart Law & Circular Coils question from JEE Main 2022 (30 Jun, Shift 1)

Answer: (d)

1. Turns per unit radial width: $\dfrac{n}{r_2-r_1}$; a ring of radius $r$, width $dr$ gives $dB = \dfrac{\mu_0 I}{2r}\cdot\dfrac{n\,dr}{r_2-r_1}$.
2. $B = \dfrac{\mu_0 nI}{2(r_2-r_1)}\displaystyle\int_{r_1}^{r_2}\dfrac{dr}{r} = \dfrac{\mu_0 nI}{2(r_2-r_1)}\ln\dfrac{r_2}{r_1}$
3. So the answer is (d).