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For the ideal magnetic response to a cylindrical cyclonic event, the plane-integrated $x$-component of $\mathbf{u}\times \mathbf{B}$ at duration $t$ isThis follows from $(\mathbf{u}\times \mathbf{B}{)}_x=(rf{B}_z+g{A}_r)\cos \varphi -(g/r)A_{\varphi }\sin \varphi$ and azimuthal integration. If boundary terms vanish, integration by parts gives the equivalent form $\pi B_0{\int }_0^{\infty }{r}^2{g}^{\prime }[\sin (ft)+tf\cos (ft)]dr$. For $f=g=a^2-r^2$ on $r<a$, zero outside, the response isThe value at zero is its continuous limit zero. It is initially $-\pi B_0{a}^{6}t/3$, but changes sign repeatedly at the nonzero roots of $\tan x=x$, with oscillatory envelope of order $t^{-1}$. The flow's kinetic helicity density is $2f^2\ge 0$, showing that the sign of a finite-duration ideal response is not fixed by the instantaneous helicity sign.

The ansatz gives $B_r=A_{\varphi }/r$, $B_{\varphi }=-A_r$, and $B_z=B$. Initially $A=B_{0}r\sin \varphi$ and $B=0$, since $B_r=B_0\cos \varphi$ and $B_{\varphi }=-B_0\sin \varphi$ are a uniform Cartesian field. In ideal magnetohydrodynamics, the transverse flux function is advected, while the axial field is stretched by the axial velocity gradient. More explicitly, the transverse induction equations are equivalent to $({\partial }_t+f{\partial }_{\varphi })A=0$; the axial component of ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}$ is $({\partial }_t+f{\partial }_{\varphi })B=B_r{g}^{\prime }$. Substituting the azimuthal mode therefore givesThe initial coefficients are $\hat{A}(r,0)=B_{0}r$ and $\hat{B}(r,0)=0$. Multiplication by $e^{ift}$ integrates both equations, giving $\hat{A}=B_{0}r{e}^{-ift}$ and $\hat{B}=i{B}_{0}t{g}^{\prime }{e}^{-ift}$. Thus the ideal magnetic response to a cylindrical cyclonic event at its end isWith ideal induction these fields remain frozen after the flow is switched off. The EMF calculated below is the active-flow value at $T^{-}$; at $T^{+}$ the velocity itself is zero.

In cylindrical components, $\mathbf{u}\times \mathbf{B}=(rfB+g{A}_r)\hat{\mathbf{r}}+(g{A}_{\varphi }/r)\hat{\mathbf{\varphi }}-f{A}_{\varphi }\hat{\mathbf{z}}$. Put $\psi =\varphi -fT$. Since $A_r=B_0(\sin \psi -rT{f}^{\prime }\cos \psi )$, its Cartesian $x$-component reduces toUse ${\int }_0^{2\pi }\cos \varphi \cos (\varphi -fT)d\varphi =\pi \cos (fT)$ and the area element $rdrd\varphi$. The integrated electromotive response of a finite cyclonic event isWhen the boundary term vanishes, the last two terms combine as $-g{\partial }_r[r^2\sin (fT)]$, giving the useful equivalent integralFor the compactly supported profile, $f{g}^{\prime }-g{f}^{\prime }=0$ on $0<r<a$. Substitution of $v=a^2-r^2$ into the first integral givesAt zero the continuous limiting value is zero, and $\mathcal{E}(T)=-\pi B_0{a}^{6}T/3+O(T^3)$. For $B_0>0$ it begins negative; the nonzero zeros satisfy $\tan (a^{2}T)=a^{2}T$, the first at $a^{2}T\simeq 4.4934$. It then alternates sign, with an asymptotic $1/T$ envelope. The requested sketch uses $x=a^{2}T$ and scales the EMF by $\pi B_0{a}^4$:

This sign alternation is possible even though the flow has fixed-sign kinetic helicity. Indeed, its vorticity components are $(0,-g^{\prime },2f+r{f}^{\prime })$, so $\mathbf{u}\cdot \mathbf{\omega }=2fg+r(g{f}^{\prime }-f{g}^{\prime })$, which reduces to $2f^2\ge 0$ for this profile. The finite-duration response retains winding and phase information; it cannot be assigned a universal sign from the instantaneous helicity alone.

A small magnetic diffusivity replaces the ideal mode equations byDifferential rotation creates radial phase gradients of order $T{f}^{\prime }$, so the resistive damping rate grows like $\eta T^2{f}^{\prime 2}$. Locally, magnetic phase mixing under differential rotation gives an accumulated damping factor of order $e^{-\eta f^{\prime 2}{T}^3/3}$. Thus the ideal result is a good early-time approximation but diffusion eventually limits winding and stretching, smooths the sharp field gradients at $r=a$, and modifies the later EMF oscillations. Small diffusion does not immediately remove all finite-time sign reversals, and a long-time resistive response requires the diffusive boundary-value problem rather than the ideal formula. After the event, the remaining distorted field relaxes by diffusion toward the imposed uniform background.