= Solution
With $s=r\sin\theta$ and $z=r\cos\theta$, radial <differential rotation> gives $\partial_z\omega=\cos\theta\,\omega'(r)$. Thus the source is $B_0r\omega'(r)\sin\theta\cos\theta$. Put $B=f(r)g(\theta)$ with $g=\sin\theta\cos\theta$. The toroidal diffusion operator is
$$
\left(\nabla^2-\frac1{s^2}\right)(fg)=\left(f''+\frac2rf'\right)g+\frac f{r^2}\left[\frac1{\sin\theta}\frac d{d\theta}\left(\sin\theta\frac{dg}{d\theta}\right)-\frac g{\sin^2\theta}\right].
$$
Direct differentiation of $g$ makes the bracket $-6g$. Therefore the steady radial equation is
$$
\eta\left(f''+\frac2rf'-\frac6{r^2}f\right)=-B_0r\omega'(r).
$$
Define $I(r)=\int_0^r x^4\omega(x)\,dx$ and $h=I/r^3$. Since $I'=r^4\omega$,
$$
h'=r\omega-\frac{3I}{r^4},\qquad h''=r\omega'-2\omega+\frac{12I}{r^5},
$$
so $h''+2h'/r-6h/r^2=r\omega'$. Taking $f=-B_0h/\eta$ proves the <steady toroidal induction by spherical differential rotation>:
$$
\boxed{B(r,\theta)=-\frac{B_0}{\eta}\frac{\sin\theta\cos\theta}{r^3}\int_0^rx^4\omega(x)\,dx.}
$$
Near zero, $I(r)=\omega(0)r^5/5+O(r^6)$, hence $B=O(r^2)\sin\theta\cos\theta$ and is regular. Its leading <vector field> is proportional to $z(-y,x,0)$, so the apparent azimuthal-basis singularity on the axis is harmless. The rapid decay of $\omega$ makes $I_\infty=\int_0^\infty x^4\omega(x)\,dx$ finite, giving $B\sim-(B_0I_\infty/\eta)r^{-3}\sin\theta\cos\theta$ at large radius. The homogeneous radial solutions are $r^2$ and $r^{-3}$; regularity at zero excludes the latter, and decay at infinity excludes the former. Thus these boundary requirements also single out the displayed solution in this angular sector.
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