A cyclotron (figure) designed to accelerate protons has an outer radius of 0.357

Question

A cyclotron (figure) designed to accelerate protons has an outer radius of 0.357 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 596 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.792 T. (a) Find the cyclotron frequency for the protons in this cyclotron. (b) Find the speed at which protons exit the cyclotron. (c) Find their maximum kinetic energy. Your response differs from the correct answer by more than 10%. Double check your calculations. (d) How many revolutions does a proton make in the cyclotron? (e) For what time interval does the proton accelerate?

Explanation / Answer

a) cyclotron frequency is given by = qB/m = (1.6X10-19C)(0.792T)/[6.28X1.67X10-27kg] = 7.58X107rad/s

b) let the speed at which the proton exits be v

then v = qBr/m = (7.58X107rad/s)(0.357m) = 2.71X107m/s

c) Maximum KE = 1/2mv2 = 1/2(1.67X10-27kg)(2.71X107m/s)2 = 6.13X10-13J

Now the enrgy expressed in electron volt = (6.13X10-13J/1.6X10-19J)eV = 3.83X106eV.

d) let us suppose that the proton makes n revolutions and comes out with speed v = 2.71X107m/s

If it starts at zero speed, each revolution gives it energy = qV = 1.6X10-19CX596V = 953.6X10-19J

it's final energy is 6.13X10-13J which is achieves after n revolutions

then nqV = n(953.6X10-19J) = 6.13X10-13J

or n = 6.4X103

This concludes the answers. If you find anything lacking please let me know.. I will resolve your query without delay....

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