Physics, Astronomy & Cosmology / Re: What is the ratio between electric and magnetic field in electromagnetic waves?« on: 01/03/2018 12:56:03 »
Things are different in electrodynamics because the magnetic field B is defined differently, it's not always a mere conversion from a value in a unit of measure to the other value in the other unit, sometimes even the equations are different. But your is a good question. Let's see the effect of this, that is the magnetic force on a moving charge q at velocity v in a field B which, for simplicity, is orthogonal to v.How can merely change the unit from SI to cgsWhat is the ratio between amplitude of electric and magnetic field in electromagnetic waves?In which unit of measure? In SI is c, as already said: ||E|| = c||B||; in other systems, the rate may not be c; in CGS - GAUSS the rate is 1: ||E|| = ||B||.
And this in the "far field" of the wave, that is far enough from the sources, or things become very complicated
1) In SI: F = q*v x B where "x" here stays for vectorial product. So, for an em wave in the void in the far field: ||F|| = q||v||*||B|| = q||v||*||E||/c.
2) In CGS: F = q*(v/c) x B. So, for an em wave in the far field:
||F|| = q(||v||/c)*||B|| = q(||v||/c)*||E|| = q||v||*||E||/c
and the two are the same
In the case of slowly rotating magnet, we can get amplitude of B much larger than amplitude of E when measured nearby.
But if we measure from adequately far distance, amplitude of B diminish quicker than E until their ratio approach a constant?Yes. Do you remember what I wrote? Compute the wavelenght λ; how far from the source, let's say at distance r, you should evaluate the fields to be sure that r >> λ?
To compute the wavelenght remember that λ = c/f where f is the frequency. It's a trivial exercise.