Solve EM Radiation Homework: Compute E & B with Monochromatic Point Source

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
qoqosz
Messages
19
Reaction score
0

Homework Statement


We are given monochromatic point source of EM radiation which power is P=100W. The task is to compute E(r) and B(r). We can assume that r is large enough to treat wave as a plane wave.


Homework Equations





The Attempt at a Solution



First of all - what for do we assume that for big r it is plane wave?
My solution to this task is:

In a sphere of radius r and thickness dr there is an amount of energy W: [tex]W = P dt = \frac{1}{\epsilon \mu} EB 4 \pi r^2 dr[/tex]
Then [tex]P = \frac{1}{\epsilon \mu} EB 4 \pi r^2 c \iff EB = \ldots[/tex] and so on... I can easily calculate values of E and B but still - what for is the mentioned assumption?
 
Physics news on Phys.org
First, why is there a factor of [itex]1/\epsilon[/itex] in your expression? Do the units make sense?

If the fields aren't those of a plane wave, then the Poynting vector [itex]\textbf{S}=\frac{1}{\mu_0}\textbf{E}\times\textbf{B}[/itex] does not necessarily point in the radial direction and have magnitude [itex]EB[/itex]. The fields themselves could also depend on the polar and azimuthal angles,

[tex]\implies P(r)=\oint\textbf{S}\cdot d\textbf{a}=\int_0^{\pi}\int_0^{2\pi}\textbf{S}\cdot\hat{\textbf{r}}r^2\sin\theta d\theta d\phi\neq\frac{4\pi r^2}{\mu_0}EB[/tex]

in general.
 
Ok, thanks. I used [tex]\frac{1}{\mu \epsilon} EB[/tex] as an energy density not an energy flux.
 
qoqosz said:
Ok, thanks. I used [tex]\frac{1}{\mu \epsilon} EB[/tex] as an energy density not an energy flux.

But that doesn't even have units of energy density...
 
You're right - I made stupid mistake :( Should be: [tex]\frac{1}{\mu c} EB[/tex]