In the given figure of a cyclotron, showing the particle source S and the dees.A uniform magnetic field is directed up form the plane of the page.Circulating protons spiral outward within the hollow dees gaining energy every time they cross the gap between the dees.Suppose that a proton, injected by source S at the centre of the centre of the cyclotron in figure initially moves toward a negatively charged dee.It will accelerate toward this dee and enter it.Once inside, it is shielded from electric field by the copper walls of the dee,that is the electric field does not enter the dee.The magnetic field, however, is not screened by the (nonmagnetic) copper dee, so the proton moves in circular path whose radius, which depends on its speed, is given by r=(mv)/(qB) ...(1) Let us assume that at the instant the proton emerges into the center gap from the first dee, the potential difference between the dees is reversed.Thus, the proton again faces a negatively charged dee and is again accelerated.Thus,the proton again faces a negatively charged dee and is again accelerated.This process continues, the circulating proton always being in step.with the oscillations of the dee potential, until the proton has spiraled out to the edge of the dee system.There a deflector plate sends it out through a portal.The key to the operation of the cyclotron is that the frequency f at which the proton circulates in the field (and that does not depend on its speed) must be equal to the fixed frequency f_(osc) of the electrical oscilliator, or f=f_(osc)("resonance condition") ...(2) This resonance condition says that, if the energy of the circulating proton is to increase, energy must be fed to it at a frequency f_(osc) that is equal to the natural frequency f at which the proton circulates in the magnetic field. Combining equation 1 and 2 allows us to write the resonance condition as qB=2pimf_(osc) ..(3) For the proton, q and m are fixed.The oscillator (we assume) is designed to work at a single fixed frequency f_(osc) We then "tune" the cyclotron by barying B until eq. 3 is satisfied and then many protons circulate through the magnetic field, to emerge as a beam. If q//m for a charge particle is 10^(6) ,frequency of applied AC is 10^(6)Hz .Then applied magnetic field is
