We apply Ampere's Law in integral form:
$\begin{align*}\oint \mathbf{B} \cdot d \mathbf {l} = \mu_0 I_{\text{enc}}\end{align*}$
By symmetry, this can simplified to
$\begin{align}2 \pi r B = \mu_0 I_{\text{enc}} \label{eqn:1}\end{align}$
Because the wire is hollow, equation (1) implies that $B = 0$ when $r < R$ .
Let $J$ be the constant volume current density in the region $R < r < 2 R$ . Then,
$\begin{align*}I_{\text{enc}} = \left(\pi r^2 - \pi a^2\right) J\end{align*}$
So,
$\begin{align*}B = \frac{\mu_0 J}{2 r}\left(r^2 - a^2 \right)\end{align*}$
When $r > 2R$ , $I_{\text{enc}}$ is constant and equal to $\pi\left(b^2 - a^2 \right)J$ , so
$\begin{align*}B = \frac{\mu_0 J}{2 r}\left(b^2 - a^2 \right)\end{align*}$
Only answer (E) shows the correct behavior for $r < R$ and $r > 2R$ , therefore this answer is correct.

Solution to 2008 Problem 92