Question

A particle of mass \( m \) and charge \( q \) is moving with a velocity \( \vec{v} \) in a magnetic field \( \vec{B} \) as shown in the figure. Show that it follows a helical path. Hence, obtain its frequency of revolution.

 

Hint:

A charged particle moving in a magnetic field will follow a helical path due to the combination of circular motion caused by the perpendicular velocity component and linear motion caused by the parallel velocity component.

Answer 1

The motion of a charged particle in a magnetic field can be analyzed into two components of velocity: 1. A component \( v_{\perp} \) perpendicular to the magnetic field, causing circular motion. 2. A component \( v_{\parallel} \) parallel to the magnetic field, causing linear motion along the direction of the field. The total velocity \( v \) is the vector sum of these two components: \[ v_{\perp} = v \sin \theta, \quad v_{\parallel} = v \cos \theta. \] The perpendicular component causes circular motion, and the parallel component causes linear motion along the magnetic field direction, resulting in a helical path. The magnetic force provides the centripetal force for the circular motion, which gives the radius \( r \) of the circular path: \[ \frac{mv_{\perp}^2}{r} = q v_{\perp} B. \] Solving for \( r \): \[ r = \frac{mv_{\perp}}{qB}. \] The time period \( T \) for one complete revolution is the circumference of the circle divided by the velocity: \[ T = \frac{2\pi r}{v_{\perp}} = \frac{2\pi m}{qB}. \] The frequency of revolution \( \nu \) is the reciprocal of the time period: \[ \nu = \frac{1}{T} = \frac{qB}{2\pi m}. \] Thus, the particle follows a helical path with a frequency of revolution \( \nu = \frac{qB}{2\pi m} \).