Question

If a charged particle enters a uniform magnetic field normally with certain velocity, then the time period of revolution of the particle:

• A. Decreases with increase of velocity of the particle
• B. Increases with increase of radius of the orbit
• C. Increases with increase of magnetic field
• D. Decreases with increase of specific charge of the particle

Hint:

For uniform circular motion in a magnetic field, the time period of revolution is independent of velocity and only depends on the specific charge \( \frac{q}{m} \).

Answer 1

The Correct Option is D

Step 1: Time Period in a Magnetic Field A charged particle moving perpendicular to a uniform magnetic field undergoes circular motion. The time period \( T \) of revolution is given by: \[ T = \frac{2\pi m}{qB} \] where: - \( m \) = mass of the particle, - \( q \) = charge of the particle, - \( B \) = magnetic field strength. Step 2: Effect of Specific Charge The specific charge is defined as: \[ \frac{q}{m} \] From the equation: \[ T \propto \frac{m}{q} \] Thus, when the specific charge \( \frac{q}{m} \) increases, \( T \) decreases. Conclusion Thus, the correct answer is: \[ \text{Decreases with increase of specific charge of the particle} \]