Question

In a certain cyclotron a proton moves in a circle of radius 0.380 m. The magnitude of the magnetic field is 1.80 T. (a) What is the oscillator frequency (in Hz)? (b) What is the kinetic energy (in eV) of the proton?

Answer 1

Given

   proton charge of proton is q = 1.6*10^-19

   mass of proton is m = 1.67262*10^-27 kg

   radius of the circle is r= 0.380 m

   magnetic field B = 1.80T

we know that the frequency of the cyclotron f = Bq/(2pim)

       f = 1.80*1.6*10^-19 /(2pi*1.67262*10^-27) Hz

       f = 27404087 Hz

Bqv = mv^2/r => v = Bqr/m ==> v = 1.80*1.6*10^-19*0.380 /(1.67262*10^-27) m/s

               V = 65430283 m/s

   kinetic energy is k.e = 0.5*mv^2

   k.e = 0.5*m*(Bqr/m)^2

   = B^2*q^2*r^2/2m

   = 1.80^2*(1.6*10^-19)^2*0.380^2/(2*1.67262*10^-27) J

   = 3.580*10^-12 J

kinetic energy of the proton is k.e = 3.580*10^-12 J