Explain working of cyclotron.

Explanation

1. Figure shows a schematic view of the cyclotron.

2. Inside the metal boxes the particles is shielded and is not acted on by the electric field. (Electric field is prevailing only around two dees).
3. The magnetic field however acts on the particle and makes it go round in a circular path inside dee.
4. Every time the particle moves from one dee to another it is acted upon by the electric field.
5. The sign of the electric field is changed alternately in tune with circular motion of the particle. This ensures that the particle is always accelerated by the electric field. Each time the acceleration increases the energy of the particle. As energy increases, the radius of the circular path increases. So the path is a spiral one.
6. Charged particles (eg. proton) move in a semicircular path in one of the dees and arrive in the gap between the dees in time interval `T/2`, where T the time period of revolution.
`T=1/V_(C)=(2pim)/(qB)orV_(C)=(qB)/(2pim)" "...(1)`
7. This frequency is called the cyclotron frequency and denoted by `v_(c)`.
8. It is independent from speed of particle, momentum and kinetic energy.
9. The frequency `v_(a)` of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that it takes the ions to complete one half of the revolution. The requirement `v_(a)=v_(c)` is called the resonance condition.
10. The phase of the supply is adjusted so that when the positive ions arrive at the edge of one dees another dees is at a lower potential and the ions are accelerated across the gap.
11. Inside the dees the particles travel in a region free of the electric field. The increase in their kinetic energy is qV each time they cross from one dee to another.
12. For circular motion, centripetal force = magnetic force,
`(mv^(2))/r=qvB`
`thereforer=(mv)/(qB)=p/(qB)" "...(2)` where p is momentum
13. From this equation it is clear that the radius of their path goes on increasing each time their kinetic energy increases.
14. The ions are repeatedly accelerated across the dees until they have the required energy to have a radius approximately that of the dees. They are then deflected by a magnetic field and leave the system via an exit slit.
15. Speed of particle near Dee,
`v=(qBR)/m" "...(3)`
where R is the radius of the trajectory at exit and equals the radius of a dee.
16. Hence, the kinetic energy of the ion is,
`1/2mv^(2)=1/2m((qBR)/m)^(2)`
`therefore1/2mv^(2)=1/2(q^(2)B^(2)R^(2))/m" "...(4)`
17. Equation (1) and (2) indicates that,
(1) (1) The time for one revolution of an ion is independent of its speed or radius of its orbit.
(2) But, its kinetic energy depend on it radius of path. (By increasing speed, radius of the circle increase but frequency `v_(c)` remain constant).