Question

A charged particle is moving in a circular path with velocity \( \vec{V} \) in a uniform magnetic field \( \vec{B} \). It is made to pass through a sheet of lead and as a consequence, it loses 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:

When the kinetic energy of a charged particle decreases, its velocity decreases, which in turn reduces the radius of its circular path in a magnetic field.

Answer 1

The radius \( r \) of the circular path of a charged particle moving in a magnetic field is given by: \[ r = \frac{mv}{qB} \] where \( m \) is the mass of the particle, \( v \) is its velocity, \( q \) is the charge, and \( B \) is the magnetic field. When the particle loses half of its kinetic energy, its velocity \( v \) changes because kinetic energy \( K.E. = \frac{1}{2} mv^2 \). If the kinetic energy is halved, then the new velocity \( v' \) will be: \[ v' = \frac{v}{\sqrt{2}} \] Since the radius is directly proportional to the velocity, the new radius \( r' \) is: \[ r' = \frac{m v'}{qB} = \frac{m}{qB} \cdot \frac{v}{\sqrt{2}} = \frac{r}{\sqrt{2}} \] Thus, the radius of the path decreases by a factor of \( \sqrt{2} \). The time period \( T \) of revolution is given by: \[ T = \frac{2 \pi r}{v} \] Since the radius decreases by a factor of \( \sqrt{2} \) and the velocity also decreases by a factor of \( \sqrt{2} \), the time period will not change because both the radius and the velocity decrease by the same factor. Therefore, the time period remains the same. Thus: 1. The radius of the path decreases by a factor of \( \sqrt{2} \). 2. The time period of revolution remains unchanged.