A cyclotron designed to accelerate protons has an outer radius of 0.290 m. The p

Question

A cyclotron designed to accelerate protons has an outer radius of 0.290 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 600 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field has a magnitude of 0.390 T.
(a) Find the cyclotron frequency.
rad/s
(b) Find the speed at which protons exit the cyclotron.
m/s
(c) Find their maximum kinetic energy.
eV
(d) How many revolutions does a proton make in the cyclotron?

Explanation / Answer

outer radius of the proton r =0.29 m potential difference V =600 V magnitude of magnetic field B =0.39 T mass of the proton m =1.67*10-27 kg charge of the proton q =1.67*10-19 C a) cyclotron frequency f =qB/2m   ........ (1) plug the values in eq (1), we get                   frequency f =0.59*10^7 s-1 b) we know angular speed of the protons is                         =qB/m but angular speed =v/r therefore, the speed at which protons exit the cyclotron is             v =qBr/m    ......... (2)             v =1.08*107 m/s c) maximum kinetic energy K =(1/2)mv2        K =(1/2)(m)(qBr/m)2         K =q2B2r2/2m   .......... (3)         K =0.98*10-13 J             =0.98*10-13(1/1.6*10-19)     (since 1 eV =1.6*10^-19 J)             =6.12*105 eV d) for every revolution, the proton kinetic energy is increased by 2qV. therefore number of revolutions =kinetic energy/2qV             revolutions = (0.98*10-13 J)/(2)(1.6*10-19 C)(600 V)                                =510.4 times              K =q2B2r2/2m   .......... (3)         K =0.98*10-13 J             =0.98*10-13(1/1.6*10-19)     (since 1 eV =1.6*10^-19 J)             =6.12*105 eV d) for every revolution, the proton kinetic energy is increased by 2qV. therefore number of revolutions =kinetic energy/2qV             revolutions = (0.98*10-13 J)/(2)(1.6*10-19 C)(600 V)                                =510.4 times     

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