Solution to 2008 Problem 92

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*}
\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.

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