Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2024 (8 Apr, Shift 2)

Answer: (c)

1. Inside: $B = \dfrac{\mu_0 I r}{2\pi a^2}$; outside: $B = \dfrac{\mu_0 I}{2\pi r}$.
2. At $a/2$: $B_1 = \dfrac{\mu_0 I (a/2)}{2\pi a^2} = \dfrac{\mu_0 I}{4\pi a}$.
3. At $2a$: $B_2 = \dfrac{\mu_0 I}{2\pi (2a)} = \dfrac{\mu_0 I}{4\pi a}$.
4. So the answer is (c) $1 : 1$.