Answer to Question #104714 in Astronomy | Astrophysics for Nikhil

Question

An electron is moving with a speed of 0.97c in a magnetic field of strength 10^8
G.
Calculate the peak frequency at which the electron will radiate.

Answer 1

Let us solve this problem applying classical mechanics principles.

First, convert 100 000 000 Gauss to a SI unit of Tesla by dividing this by 10 000. So, it will be 10 000 T. Now apply Newton's second law to find the radius of electron's orbit:

$m_e\frac{v^2}{R}=evB,\\ \space\\ R=\frac{m_ev}{eB}.$

Of course, the electron must move perpendicularly to the field.

How much time will it take for the electron to make one lap? The answer is

$T=\frac{2\pi R}{v}.$

The frequency, as we know, is

$f=\frac{1}{T}=\frac{v}{2\pi R}=\frac{eB}{2\pi m_e}=2.8\cdot10^{14}\text{ Hz}.$

This frequency is called cyclotron frequency or gyrofrequency.