Q 12-04-057JEE MainJEE Main 2025 (24 Jan, Shift 2)Easy
A long straight wire of a circular cross-section with radius $a$ carries a steady current $I$. The current $I$ is uniformly distributed across this cross-section. The plot of magnitude of magnetic field $B$ with distance $r$ from the centre of the wire is given by
Answer: (D) see figure
Ampère's law:
- Inside ($r < a$): the enclosed current is $I\dfrac{r^2}{a^2}$, so $B = \dfrac{\mu_0Ir}{2\pi a^2} \propto r$ (a straight line from the origin).
- Outside ($r \ge a$): $B = \dfrac{\mu_0I}{2\pi r} \propto \dfrac{1}{r}$.
$B$ rises linearly to a maximum at $r = a$ and then falls off as $1/r$: graph (4).
Solution by Sreeraj P, M.Sc Physics