Question

A uniform magnetic field B is set up along the positive X-axis. A particle of charge `q' and mass 'm' moving with a velocity v enters the field at the origin in X-Y plane such that it has velocity components both along and perpendicular to the magnetic field B. Trace, giving reason, the trajectory followed by the particle. Find out the expression for the distance moved by the particle along the magnetic field in one rotation.

Answer 1

If component $v_x$ of the velocity vector is along the magnetic field, and remain constant, the charge particle will follow a helical trajectory; as shown in figure below.
If the velocity component $v_y$ is perpendicular to the magnetic field B, the magnetic force acts like a centripetal force $q{v}_{y}B$ .



So, $q{v}_{y}B=\frac{m{v}_y^2}{r}$

$v_y=\frac{qBr}{m}$

Since tangent velocity $v_y=r\omega$

$\Rightarrow r\omega =\frac{qBr}{m}$

$\Rightarrow \omega =\frac{qB}{m}$

Time taken for one revolution,

$T=\frac{2\pi }{\omega }=\frac{2\pi m}{qB}$

and the distance moved along the magnetic field in the helical path is

$x=v_{x}T$

$=v_x.\frac{2\pi m}{qB}$