The figure shows a long straight wire of circular cross-section (radius $a$) carrying a steady current $I$. The current $I$ is uniformly distributed across this cross-section. Calculate the magnetic field in the region $r < a$ and $r > a$.

(N/A) Consider the case $r > a$. The Amperian loop,labelled $2$,is a circle concentric with the cross-section. For this loop,the path length is $L = 2 \pi r$.
The current enclosed by the loop is $I_e = I$.
Using Ampere's circuital law,$\oint B \cdot dl = \mu_0 I_e$,we get:
$B(2 \pi r) = \mu_0 I$
$B = \frac{\mu_0 I}{2 \pi r}$
Thus,$B \propto \frac{1}{r}$ for $r > a$.
$(b)$ Consider the case $r < a$. The Amperian loop is a circle labelled $1$ with radius $r$.
The path length is $L = 2 \pi r$.
Since the current is uniformly distributed,the current enclosed $I_e$ is proportional to the area of the loop:
$I_e = I \left( \frac{\pi r^2}{\pi a^2} \right) = I \frac{r^2}{a^2}$.
Using Ampere's circuital law:
$B(2 \pi r) = \mu_0 \left( I \frac{r^2}{a^2} \right)$
$B = \left( \frac{\mu_0 I}{2 \pi a^2} \right) r$
Thus,$B \propto r$ for $r < a$.