Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle.

#### Solution

When a charged particle with charge *q* moves inside a magnetic field `vecB`with velocity *v*, it experiences a force, which is given by:

`vecF=q(vecvxxvecB)`

Here, `vecv` is perpendicular to `vecB`,`vecF`is the force on the charged particle which acts as the centripetal force and makes it move along a circular path.

Let *m* be the mass of the charged particle and *r* be the radius of the circular path.

`:.q(vecvxxvecB)=(mv^2)/r`

*v* and *B* are at right angles:

`:.qvB=(mv^2)/r`

`r=(mv)/(Bq)`

Time period of circular motion of the charged particle can be calculated as shown below:

`T=(2pir)/v`

`=(2pi)/v(mv)/(Bq)`

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

∴ Angular frequency is

`omega=(2pi)/T`

`:.omega=(Bq)/m` |

Therefore, the frequency of the revolution of the charged particle is independent of the velocity or the energy of the particle.