Suppose a beam of particles, each with momentum $p=\gamma mv_0$ parallel to the axis of a cylinder (of length $L$ and radius $R$ with uniform current $I$) and each with positive charge $q$, impinges on the cylinder's end from the left.
I will neglect the slowing down and scattering of the beam particles by the material of the cylinder, and I will assume the cylinder is much shorter than the focal length (for a ``thin lens'' approximation).
Letting $\hat{z}$ point along the cylinder's axis, the magnetic field produced by this current is $B(\bar{r})=\frac{\mu_0I}{2\pi R^2}(-y\hat{x}+x\hat{y})$. It produces a force $\bar{F}=q\bar{v}\times\bar{B}=\gamma m\dot{\bar{v}}$ on each particle, where $\bar{v}=\dot{x}\hat{x}+\dot{y}\hat{y}+\dot{z}\hat{z}$ has initial condition $v_0\hat{z}$. Note that $v_0=\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}$ because magnetic fields cannot do work. By a thin lens approximation, we can assume $\dot{x},\dot{y}<< v_0$, so that $\dot{z}=v_0\sqrt{1-(\frac{\dot{x}}{v_0})^2-(\frac{\dot{y}}{v_0})^2}\approx v_0$. Taking the $\hat{x}$-component of the force equation gives $\ddot{x}=-\alpha\dot{z}x=-\alpha v_0x$ (where $\alpha\equiv \frac{\mu_0qI}{2\pi\gamma mR^2}$) with oscillatory solution $x(t)=x_0\cos(\sqrt{\alpha v_0}t)$. Now when the particles exit the cylinder, the focal length is $f=v_0t$ where $t$ is the time it takes to reach the focal point. To linear order, $t=\frac{x(\triangle t)}{|\dot{x}(\triangle t)|}$ where $\triangle t=\frac{L}{v_0}$ is the differential time of the focusing. Then $t=\frac{x_0\cos(\sqrt{\alpha v_0}\triangle t)}{\sqrt{\alpha v_0}x_0\sin(\sqrt{\alpha v_0}\triangle t)}\approx \frac{x_0}{\alpha v_0x_0\triangle t}$, giving focal length $f=\frac{2\pi \gamma mv_0R^2}{\mu_0qIL}$. As the $\hat{y}$-component is identical, we see that the particles focus at a point (at the focal length $f$ from the exiting-end of the cylinder).