Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2022 (28 Jun, Shift 1)

Answer: (d)

1. Inside the hollow cylinder ($r < R$) an Amperian loop encloses no current, so $B = 0$.
2. Outside ($r > R$): $B = \dfrac{\mu_0 I}{2\pi r}$, maximum at $r = R$ and falling as $1/r$.
3. So the graph is zero up to $R$, jumps to $\dfrac{\mu_0 I}{2\pi R}$, then decreases: answer (d).