A cyclotron (figure) designed to accelerate protons has an outer radius of 0.324

Question

A cyclotron (figure) designed to accelerate protons has an outer radius of 0.324 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 598 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.848 T. The black, dashed, curved lines represent the path of the particles. Alternating DI D: After being accelerated, the particles exit here. North pole of magnet (a) Find the cyclotron frequency for the protons in this cyclotron 8.12E7 rad/s (b) Find the speed at which protons exit the cyclotron. 2.7475E7 m/s (c) Find their maximum kinetic energy 7.926E6 Your response differs from the correct answer by more than 100%. ev (d) How many revolutions does a proton make in the cyclotron? revolutions (e) For what time interval does the proton accelerate?

Explanation / Answer

part c )  

KEmax = mv^2/2

m = mass of proton = 1.67 x 10^-27 kg

v = 2.7475 x 1067 m/s

KEmax = 6.3 x 10^-13 J

in ev = 6.3 x 10^-13 / ( 1.6 x 10^-19) = 3.94 x 10^6 eV

part d )

kinectic energy of proton change by 598 eV twice during each revolution

number of revolution = 3.94 x 10^6 eV/ ( 2 *598 eV)

= 3293.9 = 3293

part e )

theta = w*dt

dt = theta/w

theta = 3293 revolution

w = 8.12 x 10^7 rad/s

dt = (3293 rev/w ) * ( 2pi rad/1rev) = 2.55 x 10^-4 s

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