A cyclotron is designed to accelerate protons (mass 1.67 $\times$ 10^-27 kg) up to a kinetic energy of 2.5 $\times$ 10^-13 J.If the radius of the dipole magnets is 96 cm, what is the magnetic field used in the cyclotron? A) 35 mT B) 0.13 T C) 0.19 T D) 0.21 T

Question

Quizplus · Accepted Answer

The cyclotron frequency, f, is given by

f = qB/2*pi*m

where q is the charge of the proton (1.6 x 10^-19 C) and m is its mass (1.67 x 10^-27 kg).

The cyclotron period, T, is given by

T = 1/f.

The kinetic energy of the proton, E, is related to its speed, v, by

E = (1/2) m v^2.

Using these equations, we can solve for B:

v = sqrt(2E/m) = 2.44 x 10^7 m/s

f = qB/2*pi*m = v/(2*pi*r)

B = 2*pi*m*v/(q*r) = 0.19 T (approximately)

Therefore, the correct answer is C.

A Cyclotron Is Designed to Accelerate Protons (Mass 1 ×\times× 10-27 Kg) Up to a Kinetic Energy of 2

A Cyclotron Is Designed to Accelerate Protons (Mass 1 $\times$ 10^-27 Kg) Up to a Kinetic Energy of 2