A cyclotron of radius 0.25 m accelerates nonrelativistic protons in a 1.7-T magnetic field.What is the kinetic energy of the protons when they leave the cyclotron? , ,m_proton = 1.67 × 10^-27 kg) A) 3.7 MeV B) 7.8 keV C) 8.7 keV D) 7.8 MeV E) 8.7 MeV

Question

Quizplus · Accepted Answer

The cyclotron frequency is given by ω = qB/m where q is the charge of the proton, B is the magnetic field, and m is the mass of the proton.
Using the radius of the cyclotron, we can find the speed of the protons: v = rω = qBr/m
The kinetic energy of the proton is then K = (1/2)mv^2 = (1/2)m(qBr/m)^2 = (1/2)q^2B^2r^2/m
Plugging in the given values, we get K = 8.7 MeV, which corresponds to choice E.

A Cyclotron of Radius 0