A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons? If the radius of its ‘dees’ is 60 cm, calculate the kinetic energy (in MeV) of the proton beam produced by the accelerator.
Oscillator frequency, f = 10 MHz = 107 Hz = Be/2πm
Where e = charge on a proton = 1.6 × 10-19 C
And m = mass of a proton = 1.67 × 10-27 kg
The applied magnetic field, B = 2mf/e
= ![]()
= 0.66 T
Radius of dees, R = 0.6 m
Kinetic energy of the proton beam = B2e2R2/2m
= ![]()
= 1.2 × 10-12 J
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.
