A non-relativistic electron originates at a point $A$ lying on the axis of a straight solenoid and moves with velocity $v$ at an angle $\alpha$ to the axis. The magnetic induction of the field is equal to $B$. Find the distance $r$ from the axis to the point on the screen into which the electron strikes. The screen is oriented at right angles In the axis and is located at a distance $\ell$ from the point $\mathrm{A}$.

Let the axis of the solenoid is along the $z-$axis. Magnetic field of the solenoid is along the $z-$axis. The charged particle enters into the magnetic field at an angle $\alpha$ with the axis of the solenoid.
The magnetic force $(q\overrightarrow{v}\times \overrightarrow{B})$ on the charged is in $\mathrm{x}-\mathrm{y}$ plane perpendicular to the magnetic field.
The component of velocity in the $x-y$ plane is $v_0=v\sin \alpha$. The path of the particle in the $x-y$ plane is circular as shown in Fig.

The radius of the circular path is

$R=\frac{m{v}_0}{qB}=\frac{mv\sin \alpha }{qB}$

and           $\omega =\frac{qB}{m}$

$x$ and $y$ coordinates are respectively

$x=-(R-R\cos \omega t)=R(\cos \omega \text{t}-1)$

$y=-R\sin \omega \text{t}$

$v_y=\frac{dy}{dt}=-R\omega \cos \omega \text{t}$

and  $v_z=v_0\cos \alpha$

$z=v_{z}t=vt\cos \theta$

Screen is perpendicular to z-axis (i.e., in $x-y$ plane), and its position is $z=\ell$

$\ell =v{t}_0\cos \alpha$

$t_0=\frac{\ell }{v\cos \alpha }$

Let the particle strikes on the screen at point $\text{p}\left({\text{x}}_0,{\text{y}}_0,\ell \right)$

$r=\sqrt{x_0^2+y_0^2}=\sqrt{{\left[R\left(\cos \omega t_0-1\right)\right]}^2+{\left(-R\sin \omega t_0\right)}^2}$

    $=R\sqrt{1+{\cos }^2\omega t_0-2\cos \omega t_0+{\sin }^2\omega t_0}$

     $=R\sqrt{2-2\cos \omega t_0}=2R\sin \frac{\omega t_0}{2}$

Here           $R=\frac{mv\sin \alpha }{\text{qB}}$

                    $\omega =\frac{qB}{m},t_0=\frac{\ell }{v\cos \alpha }\text{ and }q=e$

Thus,           $r=2\frac{mv\sin \alpha }{eB}\sin \frac{eB}{m}·\frac{\ell }{v\cos \alpha }$

                     $r=2\rho \sin \frac{\varphi }{2}$

where,           $\rho =\frac{mv}{eB}\sin \alpha$

and                $\varphi =\frac{\ell eB}{mv\cos \alpha }$