Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2025 (24 Jan, Shift 2)

Answer: (b)

1. Inside ($r < a$): enclosed current $I r^2/a^2$, so $B = \dfrac{\mu_0 I r}{2\pi a^2} \propto r$ (straight line from origin).
2. Outside ($r > a$): $B = \dfrac{\mu_0 I}{2\pi r} \propto \dfrac{1}{r}$ (falling curve).
3. Both give $B = \dfrac{\mu_0 I}{2\pi a}$ at $r = a$, the peak.
4. So the answer is (b): linear rise to $r = a$, then $1/r$ fall.