Question

Cyclotron derivation frequency angular frequency limitations

Answer 1

THEORY :
Let a particle of charge qq and mass mm enter a region of magnetic field B with a velocity v normal to the field B.The particle follows a circular path, the necessary centripetal force begin provided by the magnetic field. Therefore,

Magnetic force on charge q = Centripetal force on charge q
or
qvBsin90
=mv2rqvBsin⁡90
=mv2r or
r=mvqBr=mvqB

Period of revolution of the charged particle is given by

T=2πrv
=2πv.mvqB
=2πmqBT
=2πrv
=2πv.mvqB
=2πmqB

Hence frequency of revolution of the particle will be


fc=1T
=qB2πmfc
=1T=qB2πm

Clearly,this frequency is independent of both the velocity of the particle and the radius of the orbit and is called cyclotron frequency or magnetic resonance frequency.This is the key fact which is made use of in the operation of a cyclotron.





LIMITATIONS OF CYCLOTRON:
1.According to the Einstein’s special theory of relativity,the mass of a particle increases with the increase in its velocity as
m=m01−v2/c2√m=m01−v2/c2
Where momo is the rest mass of the particle. At high velocities,the cyclotron frequency (fc=qB/2πm)(fc=qB/2πm)will decrease due to increase in mass. This will throw the particles out of resonance with the oscillating field. That is,as the ions reach the gap between the dees, the polarity of the dees is not reversed at that instant. Consequently the ions are not accelerated further.
The above drawback is overcome either by increasing magnetic field as in a synchrotron or by decreasing the frequency of the alternating electric field as in a synchro-cyclotron.



2. Electrons cannot be accelerated in a cyclotron.A large increase in their energy increases their velocity to a very large extent.This throws the electrons out of step with the oscillating field.

3.Neutron,being electrically neutral,cannot be accelerated in a cyclotron.