A cyclotron (see figure below) designed to accelerate protons has an outer radiu

Question

A cyclotron (see figure below) designed to accelerate protons has an outer radius of 0.330 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 592 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.744 T. Find the cyclotron frequency for the protons in this cyclotron. Find the speed at which protons exit the cyclotron. Find their maximum kinetic energy. How many revolutions does a proton make in the cyclotron? For what time interval does the proton accelerate?

Explanation / Answer

A)

cyclotron frequency is w = (q*B)/m

q is the charge of proton

B is the magnetic field

m is the mass of the proton

then w = (1.6*10^-19*0.744)/(1.67*10^-27) = 71.2*10^6 rad/s

B) r = m*v/(q*B)

speed is v = r*w = 0.33*71.2*10^6 = 23.496 *10^6 m/s

C) maximum KE = Kmax = 0.5*m*v^2 = 0.5*1.67*10^-27*23.496*23.496*10^12 = 461*10^-15 J

D) the KE K is bulit up from 2N passes through V(twice per revolution)

N = K/(2*q*V) = 461*10^-15/(2*1.6*10^-19*592) = 2433 revoltions

E) t = N*2*pi/w = 2433*2*3.142/(71.2*10^6) = 214.77micro Seconds = 241.77*10^-6 S

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