Answer to Question #159802 in Electric Circuits for zain ul abdeen

Question

A cyclotron designed to accelerate protons has an outer radius of 0.350 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 600 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.800 T. (a) Find the cyclotron frequency for the protons in this cyclotron. (b) Find the speed at which protons exit the cyclotron and (c) their maximum kinetic energy. (d) How many revolutions does a proton make in the cyclotron? (e) For what time interval does one proton accelerate? 

Answer 1

(a) Protons with velocities v in a constant magnetic field will move in circles of radius r in the plane perpendicular to the magnetic field. The center ward acceleration is given by

                     $F_c = m\large\frac{v^2}{r}$ $= F_B = qvB$

                                             $v =\large\frac{qrB}{m}$

Their velocity can also be related to their period $T = \large\frac{1}{f}$ by

                             $v = \large\frac{dr}{dt}$ $= \large\frac{2\pi r}{T}$ $=2\pi rf$

So

                                 $2\pi rf =$ $\large\frac{qrB}{m}$   $\to f = \large\frac{qB}{2\pi m}$ $= 12.2 [MHz]$

This is the frequency of revolution for a proton anywhere inside the cyclotron.

(b) Using our v(r) equation from (a) when r = R, we have

   $v_e = \large\frac{qrB}{m}$ $= \large\frac{(1.6*10^{-19}C)*(0.350m)*(0.800T)}{1.69*10^{-27}kg}$ = $26.8[Mm/s]$

(c)       $E_k = \frac{1}{2}mv^2 = \frac{(qrB)^2}{2m} = 6.01*10^{-13} [J]$

(d)   The kinetic energy K is built up from 2N passes through V (twice per revolution).

                            $N = \frac{E_k}{2qv}$ $= \large\frac{6.01*10^{-13}J}{2*(1.6*10^{-19}C)*(600V)}$ $= 3130$

(e)           $\delta t = TN = \large\frac{N}{f}$ $257[\mu s]$