Question

If an alpha particle enters perpendicularly in a uniform magnetic field with velocity $v$, then the time period of revolution of particle is $T$. If the same particle enters in the same field with velocity $2v$, then time period of revolution will be:

• A. T/2
• B. T
• C. 2T
• D. 4T

Answer 1

Answer: (B)

T

Answer 2

A charged particle moving perpendicularly in a uniform magnetic field experiences a magnetic force that acts as the centripetal force, causing circular motion. The magnetic force is given by $F_m = qvB$, and the centripetal force is $F_c = \frac{mv^2}{r}$. Equating these forces, we get $qvB = \frac{mv^2}{r}$. This leads to the radius of the circular path $r = \frac{mv}{qB}$. The angular velocity is $\omega = \frac{v}{r} = \frac{v}{(mv/qB)} = \frac{qB}{m}$. The time period of revolution is $T = \frac{2\pi}{\omega} = \frac{2\pi m}{qB}$. This formula for the time period is independent of the particle's velocity $v$. Therefore, if the velocity changes from $v$ to $2v$, the time period of revolution remains the same, $T$.