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

A charged particle's motion in a magnetic field is analyzed by decomposing its velocity into two components: 1. A component \( v_{\perp} \) perpendicular to the magnetic field, which induces circular motion. 2. A component \( v_{\parallel} \) parallel to the magnetic field, which results in linear motion along the field's direction. The total velocity \( v \) is the vector sum of these components, where \( v_{\perp} = v \sin \theta \) and \( v_{\parallel} = v \cos \theta \). The combination of circular and linear motion creates a helical path. The magnetic force acts as the centripetal force for the circular motion, determining the radius \( r \) of the circular path according to the equation \( \frac{mv_{\perp}^2}{r} = q v_{\perp} B \). Solving for \( r \) yields \( r = \frac{mv_{\perp}}{qB} \). The time period \( T \) for one full revolution is calculated as the circumference divided by the perpendicular velocity: \( T = \frac{2\pi r}{v_{\perp}} = \frac{2\pi m}{qB} \). The frequency of revolution \( u \) is the inverse of the time period: \( u = \frac{1}{T} = \frac{qB}{2\pi m} \). Consequently, the particle traverses a helical path with a revolution frequency of \( u = \frac{qB}{2\pi m} \).