= Solution
For the aligned <magnetic dipole field>,
$$
B_r=\frac{2\mu\cos\theta}{r^3},
\qquad
B_\theta=\frac{\mu\sin\theta}{r^3}.
$$
Since $\mathbf e_z=\cos\theta\,\mathbf e_r-\sin\theta\,\mathbf e_\theta$,
$$
\boldsymbol\Omega\mathbin\cdot\mathbf B_0
=\frac{\Omega\mu}{r^3}(3\cos^2\theta-1).
$$
The corotation velocity is $\mathbf u=\Omega r\sin\theta\,\mathbf e_\phi$. The <current density> is therefore the advected <charge density>,
$$
\boxed{
\mathbf j=n_q\mathbf u
=-\frac{\Omega^2\mu}{2\pi c r^2}
\sin\theta(3\cos^2\theta-1)\,\mathbf e_\phi}.
$$
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