Question

A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha, is:

a.0.5
b.1
c.2
d.4
e.8

Answer 1

The ratio of the speeds of a proton and an alpha particle, both experiencing the same magnetic force while traveling perpendicular to a magnetic field, is 2. This is due to the relationship F = qvBsin(θ), considering the charges of the proton (e) and alpha particle (2e).

Step-by-step explanation:

The question concerns the comparison of the speeds of a proton and an alpha particle when they are both subjected to the same magnetic force in a perpendicular magnetic field. The force exerted on a charged particle by a magnetic field is given by the equation F = qvBsin(θ), where F is the force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.

Since both particles are traveling perpendicular to the magnetic field, θ is 90 degrees, and sin(θ) is 1. For the forces to be equal, and given that the proton's charge is e and the alpha particle's charge is 2e, the equation must hold for each particle: Fproton = evprotonB = Falpha = 2evalphaB. To find the ratio of their speeds, we divide the two equations and have vproton/valpha = 2. Therefore, the correct answer is c.2.