= Solution
Write $D_t=\partial_t+\mathbf u\cdot\nabla$ for the <material derivative>. The <resistive induction equation>, together with the <solenoidal> conditions on both fields, is
$$
D_t\mathbf B=(\mathbf B\cdot\nabla)\mathbf u+\eta\Delta\mathbf B,\qquad \nabla\cdot\mathbf B=\nabla\cdot\mathbf u=0.
$$
Since $D_t\mathbf x=\mathbf u$, the <product rule> gives
$$
D_tP=\mathbf x\cdot D_t\mathbf B+\mathbf B\cdot\mathbf u,\qquad
\mathbf B\cdot\nabla Q=\mathbf x\cdot[(\mathbf B\cdot\nabla)\mathbf u]+\mathbf B\cdot\mathbf u.
$$
Also $\Delta P=\mathbf x\cdot\Delta\mathbf B+2\nabla\cdot\mathbf B=\mathbf x\cdot\Delta\mathbf B$. Substitution proves the <radial magnetic induction scalar> equation:
$$
\boxed{\partial_tP+\mathbf u\cdot\nabla P=\mathbf B\cdot\nabla Q+\eta\Delta P\quad(r<a).}
$$
In the insulating exterior the quasistatic <magnetic field> is current-free, so $\nabla\times\mathbf B=0$ and $\nabla\cdot\mathbf B=0$. Thus $\Delta\mathbf B=0$ and \b[$\Delta P=0$ for $r>a$], with decay at infinity for an isolated field. This is an instantaneous exterior <Laplace equation>, rather than a diffusion equation with the interior value of $\eta$.
Use the usual nonmagnetic-interface assumptions: equal <magnetic permeability> on both sides and no surface current. The <electromagnetic boundary conditions> then make $B_r$ and the tangential <magnetic field> continuous. Hence $P=rB_r$ is continuous. In spherical coordinates, the <solenoidal magnetic-field constraint> reads
$$
\partial_rP=-B_r-\operatorname{div}_{S^2}\mathbf B_{\mathrm{tan}},
$$
where the last term is the angular <divergence> on the unit sphere. Its right-hand side is also continuous, giving the <insulating boundary condition for the radial magnetic scalar>:
$$
\boxed{[P]_{r=a}=0,\qquad[\partial_rP]_{r=a}=0.}
$$
Equivalently, the decaying exterior degree-$l$ <spherical harmonic> is proportional to $r^{-(l+1)}$, so its boundary amplitude satisfies $\partial_rP_{lm}(a)=-(l+1)P_{lm}(a)/a$.
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