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Question
Obtain the expression for the cyclotron frequency.
Solution
When a charged particle having charge q moves inside a magnetic field `vecB`with velocity v, it experiences a force `vecF=q(vecvxxvecB)`.
When `vecv` is perpendicular to`vecB`, the force `vecF` on the charged particle acts as the centripetal force and makes it move along a circular path.
Let m be the mass of charged particle and r be the radius of the circular path.
Then `q(vecv xx vecB )=(mv^2)/r`
∵ v and B are at right angles
`∴ qvB=(mv^2)/r`
`⇒r=(mv)/(Bq)`
Time period of the circular motion of a charged particle is given by
`T=(2pir)/v`
`T=(2pi)/v (mv)/(Bq)`
`therefore T=(2pim)/(Bq)`
Angular frequency,`omega=(2pi)/T`
`therefore omega=(Bq)/m`
This is often called cyclotron frequency.
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