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 $B\left(r\right)$ with the distance $r$ from the axis of a thick cable of radius $R$ carrying current $I$, we can apply Ampere's Circuital Law.

Ampere's Circuital Law is given by:

$\oint B·dl={\mu }_0{I}_{\mathrm{enclosed}}$

Let us analyze the two regions:

1. Inside the cable ($r<R$):
Since the current $I$ 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:

$I_{\mathrm{enclosed}}=J·\left(\pi r^2\right)=I\frac{r^2}{R^2}$

Applying Ampere's Law around the loop of radius $r$:

$B·\left(2\pi r\right)={\mu }_0·(I\frac{r^2}{R^2}$

Solving for $B$:

$B=\frac{{\mu }_{0}Ir}{2\pi R^2}$

Therefore, inside the cable, the magnetic field is directly proportional to $r$:

$B\propto r$

This gives a linear relationship starting from zero at the center ($r=0$) and reaching a maximum value at the surface ($r=R$).

2. Outside the cable ($r>R$):
For any point outside the cable, the entire current $I$ is enclosed by the Amperean loop of radius $r$:

$I_{\mathrm{enclosed}}=I$

Applying Ampere's Law:

$B·\left(2\pi r\right)={\mu }_{0}I$

Solving for $B$:

$B=\frac{{\mu }_{0}I}{2\pi r}$

Therefore, outside the cable, the magnetic field is inversely proportional to $r$:

$B\propto \frac{1}{r}$

This represents a rectangular hyperbola decreasing towards zero as $r$ approaches infinity.

Conclusion:
The graph increases linearly from $r=0$ to $r=R$, and then decreases as $1/r$ for $r>R$. This variation corresponds to the graph shown in Option B.