An electron and a proton have the same radius of curvature i

An electron and a proton have the same radius of curvature in a uniform magnetic field, and the electron speed is twice that of the proton. What is the momentum of each particle?

Solution

It is not quite possible to find the magnitude of the momentum of each particle with only the ratio of speeds given. I will assume that the ratio of the angular momentum for the two particles was sought and not the invidual magnitudes.

In a magnetic field of magnitude B, for a charge q moving with velocity v, the net force acting on it is given as:

F = qvB now as the charges in the given case are moving around a circle hence the angular momentum of the particles would be given as: mvR

That is, angular momentum of electron = Me2VR

Angular momentum of proton = MpVR

Therefore the ratio of the two angular momentums would be given as: M2e / Mp = 2 x 9.10938356 × 10^-31 / 1.6726219 × 10^-27

That is the required ratio would be: 10.89 x 10^-4