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

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