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

Question

A uniform magnetic field \(\overrightarrow{B}\) is set up along the positive x-axis. A particle of charge ‘q’ and mass ‘m’ moving with a velocity \(\overrightarrow{V}\) enters the field at the origin in X-Y plane such that it has velocity components both along and perpendicular to the magnetic field \(\overrightarrow{B}\). 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 fig.

If the velocity component v_y is perpendicular to the magnetic field B, the magnetic force acts like a centripetal force qv_y B.

Since tangent velocity v_y = rω

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 

\(x=v_x.\,T=v_x.\frac{2\pi m}{qB}\)