A non-conducting cylindrical shell of radius R is placed in uniform magnetic field. The axis of the cylinder is parallel to the direction of magnetic field. A hole is drilled in cylinder and a particle having charge q and mass m is projected with velocity ${\text{v}}_{\text{0}}$  perpendicular to magnetic field and directed towards the axis of cylinder as shown in diagram. The particle collides elastically with inner wall of the cylinder and rebounds.

From fig the minimum number of collisions are two

$\theta =60°\text{Tan}\theta =\frac{\text{r}}{\text{R}}=\sqrt{3}$

$\frac{{\text{mV}}_{\text{0}}}{\text{BqR}}=\sqrt{3}$

$\Rightarrow {\text{V}}_{\text{0}}=\left(\frac{\text{BqR}}{\text{m}}\right)\sqrt{3}$

For $360°$ turn in a magnetic field

$\text{T}=\frac{2\pi \text{m}}{\text{Bq}}$

$\text{for}60°\text{turn}$

$\text{Time}\text{is}\left(\frac{\pi \text{m}}{3\text{Bq}}\right)$

Required time for 3 turns

$\text{t}=\frac{\pi \text{m}}{\text{Bq}}$

$i\text{f}\text{V}=\frac{{\text{V}}_{\text{0}}}{\text{3}}\text{Tan}\theta =\frac{\text{V}}{\text{R}}=\frac{\text{mv}}{\text{BqR}}$

$\text{Tan}\theta =\frac{\text{m}}{\text{BqR}}.\frac{{\text{V}}_{\text{0}}}{3}=\frac{\text{m}}{\text{3BqR}}\left[\frac{\text{BqR}}{\text{m}}\right]\sqrt{3}$

$\text{Tan}\theta =\frac{1}{\sqrt{3}}\Rightarrow \theta =30°$

$\therefore$ At centre are path of particle makes$60°$ at the centre of cylinder. The particle makes five collisions and came out from the field$\left(6\times 60°=360°\right)$