= Solution
Define the positive focusing <frequency>
$$
\boxed{\omega=|\Omega|=\frac{|qB|}{2\gamma_v m}}.
$$
In the <paraxial approximation> $w\simeq v$, traversal time is $T\simeq l/v$. The exit values are
$$
\rho_e=\rho_0\cos(\omega l/v),\qquad
\dot\rho_e=-\omega\rho_0\sin(\omega l/v).
$$
Across the exit fringe, $p_\theta=0$ and $A_\theta\to0$ remove the azimuthal <velocity>. The radial <velocity> remains continuous in the thin-fringe model. Outside, the transverse trajectory is therefore straight along a fixed azimuth, not a continuing helix. Taking the outgoing axial speed also as $v$, the distance to the axis is
$$
f=v\frac{\rho_e}{-\dot\rho_e}
=\boxed{\frac v\omega\cot\left(\frac{\omega l}v\right)}.
$$
It is independent of $\rho_0$, so all equal-speed paraxial rays focus to a common point. The condition $0<\omega l/v<\pi/2$ ensures the first focus lies beyond the exit. For a weak lens, $\boxed{f\simeq4(\gamma_vmv)^2/(q^2B^2l)}$.
The <relativistic solenoid magnetic lens> focuses either charge sign, while its rotation reverses with $qB$. Keeping the exact radius-dependent $w$, or the exit axial-speed change, produces radius-dependent corrections and spoils an exact common focus. Thus the requested focus is appropriately a paraxial result, with the end fields included even when their axial-speed changes are neglected.
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