The figure given below shows a long straight solid wire of circular cross-section of radius '$a$' carrying steady current $I$. The current $I$ is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field $(B)$ with distance $(r)$ from the axis of the conductor in the region is :

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786187069/questions/yxkl9cjyz915g8sabh8q.png)

Question

The figure given below shows a long straight solid wire of circular cross-section of radius ' $a$ ' carrying steady current $I$ . The current $I$ is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field $(B)$ with distance $(r)$ from the axis of the conductor in the region is :

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786187069/questions/yxkl9cjyz915g8sabh8q.png)

Accepted Answer

A. 1. Since the current is uniformly distributed over the cross-section, we use **Ampère's circuital law**, $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$, taking a circular Amperian loop of radius $r$ coaxial with the wire.

2. For a point inside the wire ($r < a$), only the fraction of current enclosed within radius $r$ contributes. Since current density is uniform, the enclosed current is $I \cdot \dfrac{r^2}{a^2}$.

Applying Ampère's law:
$$
B(2\pi r) = \mu_0 I \frac{r^2}{a^2}
$$
Solving for $B$:
$$
B = \frac{\mu_0 I r}{2\pi a^2}
$$
This shows that inside the wire, $B$ increases linearly with $r$.

3. For a point outside the wire ($r > a$), the entire current $I$ is enclosed by the Amperian loop.

So,
$$
B(2\pi r) = \mu_0 I
$$
$$
B = \frac{\mu_0 I}{2\pi r}
$$
This shows that outside the wire, $B$ decreases as $\dfrac{1}{r}$.

4. Combining both results: $B$ rises linearly from zero at the centre to a maximum at $r = a$, and then falls off as $\dfrac{1}{r}$ for $r > a$.

This behaviour is correctly shown in **Option A**.