A particle of charge 2.0 × 10^{−8} C and mass 2.0 × 10^{−10} g is projected with a speed of 2.0 × 10^{3} m s^{−1} in a region with a uniform magnetic field of 0.10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.

#### Solution

Given:

Charge of the particle, *q* = 2.0 × 10^{−8} C

Mass of the particle, *m* = 2.0 × 10^{−10} g

Projected speed of the particle*, v* = 2.0 × 10^{3} m s^{−1}

Uniform magnetic field, *B* = 0.10 T.

As per the question, the velocity is perpendicular to the field.

So, for the particle to move in a circle,the centrifugal force to the particle will be provided by the magnetic force acting on it.

Using *qvB* =`(mv^2)/(r)` , where *r* is the radius of the circle formed,

`r = (mv)/(qB)`

= `(2xx10^-13xx2xx10^3)/(2xx10^-8xx0.10)`

= 20 cm

Time period,

`T = (2pim)/(qB)`

=` (2xx3.14xx2xx10^-13}/(2xx10^-8xx0.10)`

= 6.28 × 10^{-4 }s