Question

A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.30 T. The field is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron?

Answer 1

When a charged particle enters a perperdicular magnetic field, it executes circular motion. Here the centripetal force for the circular motion is provided by the magnetic force,

The velocity and radius are same for both proton and electron.

For the proton,

For the electron,

Dividing both equations,

Given B(p) = 0.30T. Also m(p) = 1.67 x 10^-27kg, q(p) = 1.6 x 10^-19C, m(e) = 9.1 x 10^-31kg, q(e) = 1.6 x 10^-19C. Therefore,

So the magnitude of the field used for the electron 1.63 x 10^-4T.