= Solution
Set $t=0$ just inside the entrance, $z=0$, $\rho=\rho_0$ and $\theta=\theta_0$. The stated continuity gives $\dot\rho(0)=0$. Until the first axis crossing,
$$
\boxed{\rho(t)=\rho_0\cos(\Omega t),\quad\theta(t)=\theta_0-\Omega t,\quad z(t)=wt},\qquad
w=\sqrt{v^2-\Omega^2\rho_0^2}.
$$
For the full orbit, including any crossings, use the transverse complex coordinate $u=x+iy$:
$$
\boxed{u(t)=\rho_0e^{i\theta_0}\cos(\Omega t)e^{-i\Omega t}
=\frac{\rho_0e^{i\theta_0}}2(1+e^{-2i\Omega t})},\qquad 0<t<l/w.
$$
This <helical solenoid trajectory with zero incident angular momentum> is a circle of radius $\rho_0/2$ about transverse guiding centre $u_0/2$, combined with uniform axial motion. Its cyclotron <angular frequency> is $|qB|/(\gamma m)=2|\Omega|$, and its axial pitch per turn is $\pi w/|\Omega|$. The speed check is $|\dot u|^2+w^2=\Omega^2\rho_0^2+w^2=v^2$.
The physical radius is $|u|=\rho_0|\cos\Omega t|$. When the cosine changes sign, a nonnegative-radius coordinate adds $\pi$ to $\theta$; this is a coordinate change, not a discontinuous turn of the particle. For $q=0$ or $B=0$, the limit is simply $u=u_0$, $z=vt$.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2017-q2-solenoid-lens.png]
{title=Original solenoid-lens trajectories: transverse circular motion, helical drift and paraxial focusing after the exit fringe}
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