A cyclotron (figure) designed to accelerate protons has an outer radius of 0.363

Question

A cyclotron (figure) designed to accelerate protons has an outer radius of 0.363 m. The protons are emitted nearly at rest from a source at the center and are accelerated through632 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.784 T.

(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?
s

Explanation / Answer

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

Fc=mv^2/r =Fb=qvB

v=qrB/m

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

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

2rf=qrBm

so f=qB/2m = 11.96 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

= (1.60 · 1019 C)*(0.363 m)*(0.784 T)/ 1.69 * 1027 kg = 27.2 Mm/s

(c) K = 1/2 mv^2 = (qrB)^2 /2m = 6.20 * 10^13 J

d)

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

N = K/2qV = 6.20 * 10^-13 J/ 2 *(1.60 * 10^19)*(632 V) = 3065

e) t = TN = N/ f = 256.2 µs  

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