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 looses one half of its kinetic energy without change in its direction. How will

1. the radius of its path
2. its time period of revolution change?

Answer 1

1. The radius of a charged particle moving in a uniform magnetic field is given by:

r = `(mv)/(qB)`

The initial kinetic energy of the particle:

K_i = `1/2 mv_i^2`

After losing half of its kinetic energy:

`K_f  (1/2 mv_f^2) = 1/2 K_i = 1/2(1/2 mv_i^2)`

v_f = `v_i/sqrt2`

r = `(mv)/(qB)`

r_f = `(mv_f)/(qB)`

= `(mv_i)/(qB sqrt2)`

= `1/sqrt2 r_i`

Hence, the radius decreases by a factor of `1/sqrt2`.

2. The time period of revolution for a charged particle in a magnetic field is given by:

T = `(2 pi m)/(qB)`

Even when the kinetic energy is decreased, the time period does not change because it is independent of velocity.