= Ideal magnetic response to a cylindrical cyclonic event
{title2=$A=B_0r\sin(\phi-ft),\quad B_z=B_0tg'\cos(\phi-ft)$}
For an axisymmetric velocity $\mathbf u=(0,rf(r),g(r))$ independent of $z$, write $\mathbf B=B_z\widehat{\mathbf z}+\nabla\times(A\widehat{\mathbf z})$. Ideal induction becomes
$$
(\partial_t+f\partial_\phi)A=0,\qquad(\partial_t+f\partial_\phi)B_z=\frac{g'}r\partial_\phi A.
$$
The first equation expresses <magnetic flux freezing> of the transverse flux function, while the second is axial stretching $B_rg'$. Starting from a uniform field $B_0\widehat{\mathbf x}$ gives $A(r,\phi,t)=B_0r\sin(\phi-f(r)t)$ and $B_z(r,\phi,t)=B_0t g'(r)\cos(\phi-f(r)t)$. In complex azimuthal-mode notation, $\widehat A=B_0r e^{-ift}$ and $\widehat B_z=iB_0t g'e^{-ift}$, obtained by elementary integrating factors. The transverse field has $B_r=r^{-1}A_\phi$ and $B_\phi=-A_r$, preserving zero <divergence>.
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