Time Period of Charged Particle in a Magnetic Field - Detailed Solution

Time Period of a Charged Particle in a Magnetic Field

Question: Under the influence of a uniform magnetic field, a charged particle moves with a constant speed \( v \) in a circle of radius \( R \). What is the time period of rotation of the particle? (2009, 2007)

(a) Depends on \( R \) and not on \( v \)

(b) Is independent of both \( v \) and \( R \)

(c) Depends on both \( v \) and \( R \)

(d) Depends on \( v \) and not on \( R \)

Detailed Step-by-Step Explanation

Step 1: Magnetic Force and Circular Motion
When a charged particle of charge \( q \) and mass \( m \) moves perpendicular to a uniform magnetic field \( B \), it experiences a magnetic force given by: \[ F = qvB. \] This force acts as the centripetal force that keeps the particle in circular motion.

Step 2: Equating Centripetal and Magnetic Forces
For circular motion, the centripetal force is: \[ F_c = \frac{mv^2}{R}. \] Equate this with the magnetic force: \[ qvB = \frac{mv^2}{R}. \]

Step 3: Solve for the Radius
Rearranging, we get: \[ R = \frac{mv}{qB}. \] Notice that the radius \( R \) depends on the speed \( v \) and other quantities.

Step 4: Time Period of Circular Motion
The time period \( T \) for one complete revolution is: \[ T = \frac{2\pi R}{v}. \] Substitute \( R \) from the previous step: \[ T = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB}. \]

Step 5: Conclusion
The expression for \( T \) shows that the time period depends only on the mass \( m \), the charge \( q \), and the magnetic field \( B \). It is independent of both the speed \( v \) and the radius \( R \).

Final Answer: The time period is independent of both \( v \) and \( R \), which corresponds to option (b).