Answer to Question #104750 in Astronomy | Astrophysics for rupal

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

Interestingly, the peak frequency of radiation of an electron that makes circles in a magnetic field that is at 90-degree angle to the velocity vector does not depend on the speed and is sometimes called gyrofrequency or cyclotron frequency. For a non-relativistic particle that frequency is

$$f=\frac{qB}{2\pi m}.$$

For a relativistic particle that value decreases:

$$f_\text{rel}=f\sqrt{1-\frac{v^2}{c^2}}=\frac{qB}{2\pi m}\sqrt{1-\frac{v^2}{c^2}}.$$

Calculations give us the values of

$$f=2.8\cdot10^{14}\text{ Hz}$$

if we omit the fact that the particle moves at such a high speed and

$$f_\text{rel}=6.8\cdot10^{13}\text{ Hz}$$

if we calculate it according to the second equation.

Don't forget that G (Gauss) is not a SI unit! 10 000 Gauss is 1 T (Tesla), which is a SI unit.