JEE Physics 2025 Question with Solution

Given: A long straight wire of radius $a$ carries a steady current $I$ uniformly distributed over its cross-section.

Find: The correct plot of magnetic field magnitude $B$ versus distance $r$ from the centre.

For a uniformly distributed current, the magnetic field inside the wire increases linearly with radius:

$$B(r) = \frac{\mu_0 I r}{2\pi a^2}, \qquad r<a$$

So, for $r<a$, we have $B \propto r$.

For points outside the wire, the entire current is enclosed, so:

$$B(r) = \frac{\mu_0 I}{2\pi r}, \qquad r>a$$

Thus, for $r>a$, we have $B \propto \frac{1}{r}$.

Therefore, the graph must rise linearly from the centre up to $r=a$ and then fall as $\frac{1}{r}$ beyond $a$.

The solution states that the correct option is C, which corresponds to Plot 3 in the source labeling. However, from the given options shown, the graph with linear rise up to $a$ followed by inverse fall is Plot 2, which is option B. This is the most defensible match to the working.

Therefore, the correct option is B.

Given: A long straight current-carrying wire with uniform current density.

Find: Which qualitative graph matches $B$ as a function of $r$.

By Ampere's law, for a circular Amperian loop of radius $r$,

$$B(2\pi r) = \mu_0 I_{\text{enc}}$$

Inside the wire, current enclosed is proportional to area:

$$I_{\text{enc}} = I\frac{\pi r^2}{\pi a^2} = I\frac{r^2}{a^2}$$

Substituting,

$$B(2\pi r) = \mu_0 I\frac{r^2}{a^2}$$ $$B = \frac{\mu_0 I r}{2\pi a^2}$$

Hence, $B$ increases linearly with $r$ inside the wire.

Outside the wire, $I_{\text{enc}}=I$, so

$$B(2\pi r) = \mu_0 I$$ $$B = \frac{\mu_0 I}{2\pi r}$$

Hence, $B$ decreases as $\frac{1}{r}$ outside the wire.

So the correct plot is the one with a straight-line increase up to $r=a$ and a decreasing curve afterward. That corresponds to option B among the displayed plots.