= Solution
In the inertial frame, rigid corotation gives $\mathbf u=\boldsymbol\Omega\mathbin\times\mathbf r$. The <ideal magnetohydrodynamics> condition is the <motional electric field>
$$
\mathbf E=-\frac{\mathbf u\mathbin\times\mathbf B_0}{c}.
$$
Apply <Gauss's law> in Gaussian units and use the <divergence and curl of a cross product>:
$$
4\pi n_q=\nabla\mathbin\cdot\mathbf E
=-\frac1c\left[\mathbf B_0\mathbin\cdot(\nabla\mathbin\times\mathbf u)-\mathbf u\mathbin\cdot(\nabla\mathbin\times\mathbf B_0)\right].
$$
The exterior field is produced by currents inside the star, so $\nabla\mathbin\times\mathbf B_0=0$ there, while rotation with constant <angular velocity> has $\nabla\mathbin\times\mathbf u=2\boldsymbol\Omega$. Hence the required <Goldreich-Julian charge density> is
$$
\boxed{n_q(\mathbf r)=-\frac{\boldsymbol\Omega\mathbin\cdot\mathbf B_0(\mathbf r)}{2\pi c}}.
$$
Here $n_q$ denotes electric charge per unit volume, as in the question. If each carrier has charge $q$, its signed number density is $n_q/q$.
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