Question

A charged particle is moving in a circular path with velocity \( V \) in a uniform magnetic field \( \vec{B} \). It is made to pass through a sheet of lead and as a consequence, it looses one half of its kinetic energy without change in its direction. How will (1) the radius of its path change? (2) its time period of revolution change?

Hint:

A reduction in kinetic energy affects the velocity and radius of a charged particle's circular path but does not change its time period of revolution.

Answer 1

(1) The radius of the path will decrease. Since the kinetic energy is reduced by half, the velocity decreases. The radius of a charged particle's path in a magnetic field is given by: \[ r = \frac{mv}{qB} \] Since the velocity \( v \) is halved, the radius will also be halved.

(2) The time period of revolution will remain unchanged. The time period of revolution \( T \) for a charged particle moving in a magnetic field is given by: \[ T = \frac{2\pi m}{qB} \] Since the magnetic field \( B \) and the charge \( q \) are constant, the time period is independent of the kinetic energy, and hence remains the same.