Question

A uniform magnetic field $B$ is set up along the positive $x$ -axis. A particle of charge $q$ and mass $m$ moving with 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

Hint: Helical would be the path of the charged particle. In this way, the charge is pushed linearly into the magnetic field, velocity $v\cos \theta $ as the round path due to the speed $v\sin \theta $ . The perpendicular velocity component to the magnetic field induces circular motion while the velocity component parallel to the field pushes the particle down a straight line. Two consecutive circles are horizontally separated. This is a helical movement.

Complete step by step solution:
Let $\theta $ be the magnetic field angle of the particle velocity.
Therefore, Velocity component perpendicular to the magnetic field $B$ will be,
${v_p} = v\sin \theta $
Where the velocity of particle is $v$
Velocity component parallel to the magnetic field $B$ will be,
${v_{||}} = v\cos \theta $

The magnetic field is perpendicular to both the magnetic field and ${v_p} $ which allows the electron to travel about in a circular fashion. Which implies a centripetal force. The field does not impact ${v_l} $ however, since this is a constant component.

The particle's trajectory is helical in the field.
The formula for centripetal force is,
$F = \dfrac{{m{v_p}^2}}{R} = e {v_p} B$
Where the radius is $R$, $m$ is mass and $B$ is the magnetic field.
$R = \dfrac{{m{v_p}}}{{eB}}$
The formula for time period is given by,
$T = \dfrac{{2\pi R}}{{{v_p}}}$ Or
$T = \dfrac{{2\pi m}} {{eB}} $
The pitch is the direction the magnetic field travels by the particle in a single amount of time.
$ = {v_{||}} T = \dfrac{{2\pi mv\sin \theta}}{{eB}}$

Therefore, distance moved by the particle along the magnetic field in one rotation is $\dfrac {{2\pi mv\sin \theta}}{{eB}}.$

Note: It is said that a force acting on the particle conducts work while a portion of it occurs in the direction of the particle's motion. When we consider charges in a magnetic field of uniform magnitude $B$ , where we have a charged particle holding a charge $q$ , the magnetic force acts perpendicular to the particle's speed.