= Solution
For an <axisymmetric vector field>, write $\mathbf B=B_R\mathbf e_R+B_\phi\mathbf e_\phi+B_z\mathbf e_z$. Pure <differential rotation> gives
$$
\mathbf u\times\mathbf B=R\Omega B_z\mathbf e_R-R\Omega B_R\mathbf e_z.
$$
This vector has no azimuthal component and is independent of $\phi$, so its <curl> has zero radial and vertical components. Its azimuthal component is
$$
\{\nabla\times(\mathbf u\times\mathbf B)\}_\phi=\partial_z(R\Omega B_z)+\partial_R(R\Omega B_R)
=R(B_z\partial_z\Omega+B_R\partial_R\Omega)+\Omega\{R\partial_zB_z+\partial_R(RB_R)\}.
$$
The final braces are $R\nabla\cdot\mathbf B=0$ by <Gauss's law for magnetism>. The <ideal magnetohydrodynamic induction equation> therefore yields <axisymmetric magnetic winding>:
$$
\boxed{\partial_t\mathbf B=R(\mathbf B_p\cdot\nabla\Omega)\mathbf e_\phi,\qquad\partial_t\mathbf B_p=0.}
$$
<Differential rotation> creates a <toroidal magnetic field> from the fixed <poloidal magnetic field>. A stationary field under the same purely rotational assumptions requires <Ferraro's law of isorotation>, $\mathbf B_p\cdot\nabla\Omega=0$.
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