A cyclotron (figure) designed to accelerate protons has an outer radius of 0.321

Question

A cyclotron (figure) designed to accelerate protons has an outer radius of 0.321 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 630 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.856 T The black, dashed, curved lines Alternating AV represent the path of the particles. After being accelerated, the particles exit here North pole of magnet (a) Find the cyclotron frequency for the protons in this cyclotron. rad/s (b) Find the speed at which protons exit the cyclotron. m/s (c) Find their maximum kinetic energy. eV (d) How many revolutions does a proton make in the cyclotron revolutions (e) For what time interval does the proton accelerate?

Explanation / Answer

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

mv^2/r = qvB =====> v = qrB/m

Their velocity can also be related to their period T = 1/ f by

v = dr/dt = 2pi*r/T = 2pi*rf

f = qB/2pi*m = 13.05 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

ve = qRB/m = (1.6*10^-19*0.321*0.856)/1.67*10^-27 = 26.33 Mm/s

c) K = 0.5mv^2 = (qrB)^2/2m = 5.8*10^-13 J

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

N = K/(2*q*V) = (5.8*10^-13)/(2*1.6*10^-19*630) = 2870.52

e) t = TN = N/f = 219.96 µs

  

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