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Question
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.
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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.
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