Question

A cyclotron uses a magnetic field of 0.575 T to accelerate protons. It has a maximum radius of 0.600 m. (proton mass 1.673x10-27 kg, proton charge 1.602x10-19 C)

a.   Derive an expression for the frequency of the alternating voltage on the dees of the cyclotron.

b.   What is the frequency of the voltage?

c.   What is the maximum kinetic energy of the protons as they emerge from the

instrument?

d.   What is the equivalent accelerating voltage?

Answer 1

a.

The centripetal force of the proton in a cyclotron is

Here, mass of the proton is m, velocity of the proton is v and radius of the circular path is R.

The magnetic force exrted on the proton is

  

Here, charge of the proton is q and magnetic field is B.

The centripetal force of the proton to move in a circular path inside the cyclotron is provided by the magnetic force,

  

F--F 2 mv RaB

Now, the period of the proton is equal to the distance covered by the proton in one cycle divided by the velocity of the proton.

  

Now, freqeuncy of the proton is equal to the reciprocal of the period.

This is the resonant freqeuncy that shoudl matched with the frequency of the alternating voltage on the dees of the cyclotron.

Therefore, the freqeuncy of the alternating voltage on the dees of the cyclotron is

  

b.

Subsitute 0.575 T for B, 1.602 x 10^-19 C for q, 1.673 x 10^-27 kg for m in the frequency above equation,

qB (1.602x 10-19 C(0.575 T) 2 1.673x10-27 kg = 8.77 x 106 Hz

Therefore, the freqeuncy of the volatge is 8.77 x 10⁶ Hz.

c.

The maximum radius of the circular path traced by the proton is

r = 0.600 m

The kinetic energy of the proton emering from the cyclotron is

Therefore, the maximum kinetic nenergy of the emerge proton is 0.0913 x 10^-11 J.

d.

The equivalent accelerating voltage is

  

K.E 0.0913 x10-11 J 1.602x10-19 C =5.69 × 106 V

Therefore, the equivalent accelerating voltage is 5.69 x 10⁶ V.