A thick current carrying cable of radius ‘R’ carries current ‘I’ uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance ‘r’ from the axis of the cable is represented by :

To find the variation of the magnetic field $B\left(r\right)$ due to a thick current-carrying cable of radius $R$ carrying a uniformly distributed current $I$ at a distance $r$ from the axis, we can apply Ampere's Circuital Law.

Ampere's Circuital Law states that the line integral of the magnetic field $B$ around a closed loop is equal to ${\mu }_0$ times the total current enclosed by the loop:
$\oint B·dl={\mu }_0{I}_{\text{enclosed}}$

We consider two cases based on the distance $r$ from the axis of the cable:

Case 1: Inside the cable ($r<R$)
Since the current is uniformly distributed across the cross-sectional area $\pi R^2$, the current density $J$ is:
$J=\frac{I}{\pi R^2}$

The current enclosed within an Amperean loop of radius $r$ is:
$I_{\text{enclosed}}=J·\left(\pi r^2\right)=I\frac{r^2}{R^2}$

Applying Ampere's Law for the circular loop of radius $r$:
$B·\left(2\pi r\right)={\mu }_{0}I\frac{r^2}{R^2}$

Solving for $B$ gives:
$B=\left(\frac{{\mu }_{0}I}{2\pi R^2}\right)r$

Therefore, inside the cable, the magnetic field is directly proportional to $r$:
$B\propto r$ (which represents a straight line starting from the origin).

At the surface of the cable where $r=R$, the magnetic field reaches its maximum value:
$B_{\text{max}}=\frac{{\mu }_{0}I}{2\pi R}$

Case 2: Outside the cable ($r>R$)
For an Amperean loop of radius $r$ outside the cable, the total enclosed current is the entire current flowing through the cable:
$I_{\text{enclosed}}=I$

Applying Ampere's Law:
$B·\left(2\pi r\right)={\mu }_{0}I$

Solving for $B$ gives:
$B=\frac{{\mu }_{0}I}{2\pi r}$

Therefore, outside the cable, the magnetic field is inversely proportional to $r$:
$B\propto \frac{1}{r}$ (which represents a rectangular hyperbola).

Conclusion:
The variation is represented by a linear increase from $r=0$ to $r=R$, followed by a hyperbolic decrease for $r>R$. Looking at the provided options, Option 3 (shown in the third image) correctly depicts this variation with a dashed vertical line marking the boundary at $r=R$.