Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2021 (27 Jul, Shift 2)

Answer: (c)

1. Inside a wire of radius $R$ (uniform current): $B = \frac{\mu_0 I r}{2\pi R^2}$, a straight line through O with slope $\propto \frac{1}{R^2}$.
2. Outside: $B = \frac{\mu_0 I}{2\pi r}$, the same curve for both wires since $I$ is the same.
3. Peak value $\frac{\mu_0 I}{2\pi R}$ occurs at $r = R$, so wire a ($a < b$) has a steeper line and a higher peak at a smaller $r$.
4. Wire b peaks lower and later at $r = b$, then merges with the same $1/r$ fall.
5. So the answer is (c).