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Poynting Vector Energy Delivered

  • Thread starter sanitykey
  • Start date
  • #1
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Homework Statement



If B0 = 5*10^-7 [T], relative permittivity = 2 and relative permeability = 1.15 how much energy is delivered to an area of 1m^2 by the wave in one second?

Where B0 is a magnetic field measured in teslas.

Homework Equations



< P > = ((B0^2)*c*(1)) / (2*(vacuum permeability))

(1 is the area given to me)

The Attempt at a Solution



My problem here is that I don't know how to use the relative permittivity and relative permeability. I've been googling for a few hours to try and find a relation between them and the poynting vector but i can't find anything at all.

I know c = 1/root(vacuum permittivity*vacuum permeability) but that doesn't really help me.

I'm pretty sure that's the equation I have to use but i'm thinking i need to *modify* it. My best guess is that maybe i should use c/n instead so the equation would be:

< P > = ((B0^2)*c*(1)) / (2*(vacuum permeability)*n)

n = root(relative permittivity * relative permeability)

Can anyone confirm if this is right and if not how would you use the relative permittivity and relative permeability?

Thanks
Richy

edit: c = 1/root(vacuum permittivity*vacuum permeability) (forgot the 1 over originally whoops)
 
Last edited:

Answers and Replies

  • #2
diazona
Homework Helper
2,175
6
Close, but you missed one thing: when "converting" EM equations from vacuum to a material, you just convert the permittivity and permeability from their vacuum values to their in-material values everywhere they occur in the equation. So if you start with
[tex]\langle P\rangle = \frac{B_0^2 c A}{2 \mu_0} = \frac{B_0^2 A}{2 \mu_0\sqrt{\mu_0\epsilon_0}}[/tex]
then in the material, your equation becomes
[tex]\langle P\rangle = \frac{B_0^2 A}{2 \mu\sqrt{\mu\epsilon}} = \frac{B_0^2 c A}{2 n \mu}[/tex]
where [tex]\mu = 1.15\mu_0[/tex] and [tex]\epsilon = 2\epsilon_0[/tex].

So basically I think you're right about inserting the factor of n, but you also need to add a factor to the permeability that appears in the denominator.
 
  • #3
93
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Thanks so much! This was confusing me for so long but i get it now, you're a star ^_^
 

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