A cyclotron (figure) designed to accelerate protons has an outer radius of 0.354

Question

A cyclotron (figure) designed to accelerate protons has an outer radius of 0.354 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 614 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.736 T. Find the cyclotron frequency for the protons in this cyclotron. I 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) f = B*q/(2*pi*m)

= 0.736*1.6*10^-19/(2*pi*1.67*10^-27)

= 1.12*10^7 hz

w = 2*pi*f

= 1.12*10^7*2*pi

= 7.05*10^7 rad/s

b) apply, r = m*v/(B*q)

==> v = B*q*r/m

= 0.736*1.6*10^-19*0.354/(1.67*10^-27)

= 2.5*10^7 m/s

c) KE = (1/2)*m*v^2

= (1/2)*1.67*10^-27*(2.5*10^7)^2

= 5.22*10^-13 J

= 5.22*10^-13/(1.6*10^-19)

= 3.26*10^6 eV

d) Let N is the no of revolutions made.

use work-energy throrem,

Workdone on proton = gain in kinetic energy

2*N*q*delta_V = KE

==> N = KE/(2*q*delta_V)

= 5.22*10^-13/(2*1.6*10^-19*614)

= 2657

e) total time taken = N*time periode

= N*(2*pi*m)/(B*q)

= 2657*2*pi*1.67*10^-27/(0.736*1.6*10^-19)

= 2.37*10^-3 s

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