The distance moved by a charged particle along the magnetic field (the component of velocity is parallel to…
The distance moved by a charged particle along the magnetic field (the component of velocity is parallel to the magnetic field) in one rotation is given by ( $\mathrm{m}$ - mass of the particle, $\mathrm{v}$ - velocity of the particle, $\mathrm{q}$ charge of the particle, B - magnetic field)
$\frac{2 \pi m v}{q B}$
$\frac{\pi \mathrm{mv}}{\mathrm{qB}}$
$\frac{4 \pi \mathrm{mv}}{\mathrm{qB}}$
$\frac{2 \pi m v}{q B^2}$
Solution
If the velocity component $\mathrm{v}_{\mathrm{y}}$ is perpendicular to the magnetic field $\mathrm{B}$, the magnetic force acts like a centripetal force $\mathrm{qv}_{\mathrm{y}} \mathrm{B}$.
$\begin{aligned} & q v_y B=\frac{m_y^2}{r} \\ & v_y=\frac{q B r}{m} \\ & v_y=r \omega \\ & \omega=\frac{q B}{m}\end{aligned}$
Time taken for one revolution
$\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi \mathrm{m}}{\mathrm{qB}}$
The distance moved by a changed particle along the magnetic field.
$\begin{aligned} & x=V t \\ & =\frac{2 \pi \mathrm{mv}}{q B}\end{aligned}$