= Solution
Use the <resistive induction equation> with constant <magnetic diffusivity>,
$$
\partial_t\mathbf B_{\mathrm{tot}}=\nabla\times(\mathbf u\times\mathbf B_{\mathrm{tot}})+\eta\nabla^2\mathbf B_{\mathrm{tot}},\qquad\nabla\cdot\mathbf B_{\mathrm{tot}}=0.
$$
Seek $\mathbf B_{\mathrm{tot}}=B_0\mathbf e_z+B(s,z,t)\mathbf e_\phi$. The axisymmetric velocity $\mathbf u=s\omega(s,z)\mathbf e_\phi$ is incompressible, and the proposed field is <solenoidal>. The <toroidal magnetic field> is parallel to the velocity, so it does not contribute to their <cross product>. The remaining term is
$$
\mathbf u\times\mathbf B_{\mathrm{tot}}=s\omega B_0\mathbf e_s,\qquad\nabla\times(s\omega B_0\mathbf e_s)=sB_0\partial_z\omega\,\mathbf e_\phi.
$$
The uniform axial field has zero vector <Laplacian>. For the azimuthal part, differentiation of the cylindrical basis gives $\partial_\phi^2\mathbf e_\phi=-\mathbf e_\phi$. Consequently
$$
\nabla^2(B\mathbf e_\phi)=\left(\partial_s^2B+\frac1s\partial_sB+\partial_z^2B-\frac B{s^2}\right)\mathbf e_\phi.
$$
There is no source for a radial or axial perturbation, so zero initial values of those components remain zero. Equating the toroidal components proves
$$
\boxed{\partial_tB=s(\mathbf B_0\cdot\nabla)\omega+\eta\left(\nabla^2-\frac1{s^2}\right)B.}
$$
The last Laplacian is the scalar axisymmetric <Laplacian>. The additional $-B/s^2$ term is essential for a vector component in a rotating basis. The source is the <Omega effect>: rotation that varies along an axial field line winds it into a <toroidal magnetic field>.
Back to article page
