# Optics | Electromagnetic Waves | Electric/Magnetic Fields

heycoa

## Homework Statement

An isotropic quasimonochromatic point source radiates at a rate of 100 W. What is the flux density at a distance of 1 m? What are the amplitudes of the E- and B- fields at that point?

## Homework Equations

I think (but not at all sure) that the equation for flux density is power/(4*pi*meters^2)

## The Attempt at a Solution

I plugged my info into that equation and came up with 78.54. I don't know the units and I don't know how to do the last part and get the amplitudes of electric and magnetic field. Please help me learn this!! I appreciate it very much!

## Answers and Replies

dydxforsn
The flux density at r = 1m (which you have correctly calculated) is the time averaged Poynting vector at r = 1m. So how is the amplitudes of E and B related to <S> (the time averaged Poynting vector). Also recall that the amplitudes of E and B themselves are related to each other, so you can represent <S> as being either a function of E^2 or B^2.

Also the units for <S> are just W/m^2 if you used watts and meters.

Staff Emeritus
Gold Member
Yes, your equation for flux density is correct. It is actually very intuitive. The equation is saying that all of the energy emitted by the light source (each second) is spread out evenly over a sphere of radius 1 m. Therefore, the amount of power arriving per unit area at you (the observer) 1 metre away is just the total power divided by the surface area of this sphere. This is the flux density. The higher the flux of a source, the more light energy arriving in a given area, and the brighter the source will appear. If you go farther away, the sphere becomes larger, meaning that the same amount of energy is spread out over a much larger area, and the amount that you get is less. This is why isotropic sources appear fainter, their brightness diminishing with the inverse square of the distance.

You don't know the units???

You divided something in watts by something in metres^2. What do you think the resulting units are?

As for finding the magnitudes of the E and B fields: you almost certainly have something in your book or notes that gives a relation between the power of an EM wave, and the amplitude of its electric and magnetic fields. It may be helpful to look up the Poynting vector on Wikipedia. Particularly the section on plane waves. Although, this isn't a plane wave...

heycoa
The flux density at r = 1m (which you have correctly calculated) is the time averaged Poynting vector at r = 1m. So how is the amplitudes of E and B related to <S> (the time averaged Poynting vector). Also recall that the amplitudes of E and B themselves are related to each other, so you can represent <S> as being either a function of E^2 or B^2.

Also the units for <S> are just W/m^2 if you used watts and meters.

The equation I am looking at in the book would be the Poynting vector = c*ε0/2 * E02.

So I can solve for the E0, which would give me the amplitude of the electric field and magnetic field because they are equal. Am I correct?

Staff Emeritus
Gold Member
The equation I am looking at in the book would be the Poynting vector = c*ε0/2 * E02.

So I can solve for the E0, which would give me the amplitude of the electric field and magnetic field because they are equal. Am I correct?

Yes you can calculate E0 that way. But NO: E0 and B0 are not equal. They differ by a constant factor.

heycoa
Yes you can calculate E0 that way. But NO: E0 and B0 are not equal. They differ by a constant factor.

Yeah, youre right. Lol I had just gotten to the point of boxing my answer and I was like, "no, thats not right".

Its E0=c*B0

heycoa
Well thank you both so much! I very much appreciate your responses and your time!!!