JEE Main 2025 Jan 24 Shift 2, Physics Q5: Ampere's Circuital Law

Apply Ampere's law to a circular loop of radius $r$ coaxial with the wire.

Inside the wire ($r<a$): the enclosed current is the fraction of $I$ within radius $r$, $I_{\text{enc}}=I\frac{r^2}{a^2}$. So

$B(2\pi r)={\mu }_{0}I\frac{r^2}{a^2}\Rightarrow B=\frac{{\mu }_{0}Ir}{2\pi a^2}$

$B\propto r$ — linear rise from $0$ to a maximum $B_{\max }=\frac{{\mu }_{0}I}{2\pi a}$ at $r=a$.

Outside the wire ($r>a$): the full current $I$ is enclosed, so

$B=\frac{{\mu }_{0}I}{2\pi r}\Rightarrow B\propto \frac{1}{r}$

$B$ decays as $1/r$ from the peak at $r=a$.

The plot showing a linear rise inside and a $1/r$ decay outside, continuous at $r=a$, matches option (1).

Correct option: (1)