A certain cyclotron that has a magnetic field whose magnitude is 1.8 T is design

Question

A certain cyclotron that has a magnetic field whose magnitude is 1.8 T is designed to accelerate protons to a kinetic energy of 25 Mev. (a) What is the cyclotron frequency for this cyclotron? (b) What must the minimum radius of the magnet be to achieve this energy? (c) If the alternating potential difference applied to the dees has a maximum value of 50 kV, how many revolutions must the protons make before emerging with kinetic energies of 25MeV? Please explain math steps and theory. Also, what is a "dees"? What is a cyclotron?

Explanation / Answer

a)f=qB/2mp =1.6*10-19*1.8/2*(1.67*10-27) =27.45KHZ

b)r =(2Km)/qB =(2*25*106*1.6*10-19*1.67*10-27)/(1.6*10-19*1.8) =0.4m

c)Energy gain/revolution T=4qVo =4*1.6*10-19*(50KV/2) =100KV*(1.6*10-19) =100KeV

Number of revolutions =25*106/100*103 =250 revolutions

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