Question

An electron enters a region of space containing a uniform 0.0000281-T magnetic field. Its speed is 121 m/s and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion Find the radius r of the electron's path, and the frequency fof the motion. Number r- Number Hz f-

Answer 1

The force experienced by the particle when it enters perpendicular to the field F = qvB

where q is the charge of the particle( here, charge of electron), v is the velocity of the particle and B is magnetic field or magnetic flux density.

When in circular motion, acceleration = $v^{2}/R$ where v is the velocity and R is the radius of circular motion.

We know that F = ma where m is the mass of particle and a is the accelartion(mass of electron is 9.1 * 10^-31 kg).

Thus, =

R = m * v²/F = 24.4 * 10^-18 m = 24.4 attometer

The frequency of particle in circular motion due to magnetic field is called cyclotron freuqency.

This frequency = $f = qB/ (2\pi m)$ = $1.6 * 10^{-19}*0.0000281/ (2\pi * 9.1 * 10^{-31})$

= 786330.35 Hz