Certainly! Here’s how to respond to the question by explaining the magnetic field variation in a long straight solid wire carrying a steady current.

To find the variation of the magnetic field \( B \) with distance \( r \) from the axis of a solid wire carrying a steady current \( I \), we can use Ampère's Law and the concept of current distribution.

The given wire has a circular cross-section with radius \( a \), and the current is uniformly distributed across this cross-section. Let's consider two cases based on the distance \( r \) from the axis:

For \( r \leq a \) (inside the wire):

By symmetry, the magnetic field \( B \) at a distance \( r \) from the axis will be in the form of concentric circles. Using Ampère's Law:

Here, \( I_{\text{enclosed}} = I \frac{\pi r^2}{\pi a^2} = I \left( \frac{r^2}{a^2} \right) \), since the current is uniformly distributed.

Therefore, \(B(2\pi r) = \mu_0 I \frac{r^2}{a^2}\).

Solving for \( B \):

\(B = \frac{\mu_0 I r}{2\pi a^2}\), which shows that \( B \) increases linearly with \( r \).

For \( r > a \) (outside the wire):

For any point outside the conductor, the magnetic field is determined by the total current \( I \) enclosed, irrespective of how it is distributed inside.

Thus, \(B (2\pi r) = \mu_0 I\).

Solving for \( B \):

\(B = \frac{\mu_0 I}{2\pi r}\), which shows that \( B \) decreases with \( \frac{1}{r} \).

The graph of \( B \) with respect to \( r \) will initially increase linearly inside the conductor (for \( r \leq a \)) and then decrease inversely outside the conductor (for \( r > a \)).

This corresponds to the correct representation of the magnetic field variation. Therefore, the correct option is represented by the image above.