Question

A beam of protons moves in a circle of radius 0.53 m. The protons move perpendicular to a 0.21 T magnetic field. What is the speed of each proton?

Answer 1

Magnetic force on a charged particle moving perpendicular to the magnetic field = qvB, where q is charge, v is velocity and B is magnetic field.

Also, for circular motion, centripetal force = mv^2 /r , where m is mass, v is velocity and r is radius of the circular trajectory.

Centripetal force is provided by the magnetic force. So, they must be equal.

So, qvB=mv^2/r

=>v=qBr/m

Here,q=charge on one proton = 1.6*10^-19 C, B=0.21 T, r=0.53 meters, m=mass of proton = 1.672*10^-27 kg.

So, v=(1.6*10^-19)*(0.21)*(0.53) / (1.672*10^-27) = 10650717.7 m/s.