A proton that is being accelerated travels along arcs of a circle through the magnetic fields of a cyclotron. Near the beginning of its path through the cyclotron, the speed of the proton is slow and the radii of the arcs are small. Near the end of its path, the radii are large and the speed is c/3, so the motion of the particle must be treated relativistically. What must the ratio of the average magnetic field at the edge of the cyclotron to that at the center be to compensate for the relativistic momentum of the proton? HINT: The cyclotron frequency f = qB/2πm must remain constant for the proton throughout its path.
A) 0.94
B) 0.33
C) 1.3
D) 1.06
E) 1.7

Question

Quizplus · Accepted Answer

Since the cyclotron frequency is constant, we have:

qB1/2πm = qB2/2πm

where B1 is the average magnetic field at the edge and B2 is the average magnetic field at the center.

We also know that the kinetic energy of the proton must be:

K = (γ - 1)mc^2

where γ is the Lorentz factor, given by:

γ = 1/√(1 - v^2/c^2)

Near the end of its path, the speed is c/3, so γ = 4/3.

Using the equation for the relativistic momentum, p = γmv, we can solve for the ratio of the average magnetic fields:

p1/p2 = B1/B2 = (γ1/γ2)(v2/v1) = (4/3)/(1/√8) = 1.06

Therefore, the ratio of the average magnetic field at the edge of the cyclotron to that at the center must be 1.06 to compensate for the relativistic momentum of the proton. The answer is D.

A Proton That Is Being Accelerated Travels Along Arcs of a Circle