To solve this problem, we will use Ampère's Law to find the magnetic field $B(r)$ inside and outside the cable.$\\$Given:$\\$• Radius of the cable $R \\$• Total current $I$ uniformly distributed across the cross-section$\\$1. Inside the cable $(r < R): \\$Consider a circular path of radius $r$ inside the cable.$\\$The current enclosed by this path is proportional to the area:$\\ I_{enclosed} = I \cdot \frac{\pi r^2}{\pi R^2} = I \cdot \frac{r^2}{R^2} \\$Using Ampère's Law:$\\ \oint B \cdot dl = \mu_0 I_{enclosed} \\ B \cdot 2\pi r = \mu_0 \cdot I \cdot \frac{r^2}{R^2} \\ B = \frac{\mu_0 I r}{2\pi R^2} \\$This shows that $B$ is directly proportional to $r$ inside the cable.$\\$2. Outside the cable $(r \geq R): \\$Consider a circular path of radius $r$ outside the cable.$\\$The entire current $I$ is enclosed by this path.$\\$Using Ampère's Law:$\\ \oint B \cdot dl = \mu_0 I \\ B \cdot 2\pi r = \mu_0 I \\ B = \frac{\mu_0 I}{2\pi r} \\$This shows that $B$ is inversely proportional to $r$ outside the cable.$\\$Combining both regions, the graph of $B(r)$ is:$\\$• Linear increase inside the cable ($r < R$).$\\$• Hyperbolic decrease outside the cable ($r \geq R$).$\\$The correct graph is represented by Option 3.