The figure given below shows a long straight solid wire of circular...

For a long straight solid wire of radius $a$ carrying a steady current $I$ uniformly distributed across its cross-section, we can find the magnetic field $B$ at a distance $r$ from the axis using Ampere's circuital law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$.

Case 1: Inside the wire ($r < a$)
The current enclosed by an Amperian loop of radius $r$ is $I_{enclosed} = I \left(\dfrac{\pi r^2}{\pi a^2}\right) = I \dfrac{r^2}{a^2}$.
Applying Ampere's law:
$B(2\pi r) = \mu_0 \left(I \dfrac{r^2}{a^2}\right)$
$B = \left(\dfrac{\mu_0 I}{2\pi a^2}\right) r$
Thus, $B \propto r$. The graph is a straight line passing through the origin.

Case 2: Outside the wire ($r \ge a$)
The current enclosed by an Amperian loop of radius $r$ is the total current $I$.
Applying Ampere's law:
$B(2\pi r) = \mu_0 I$
$B = \dfrac{\mu_0 I}{2\pi r}$
Thus, $B \propto \dfrac{1}{r}$. The graph is a rectangular hyperbola.

At the surface ($r = a$), the magnetic field is maximum: $B = \dfrac{\mu_0 I}{2\pi a}$.

The plot that correctly represents this variation shows a linear increase from the origin up to $r = a$, followed by a $1/r$ decrease for $r > a$. This matches the first graph.

Answer: