How Much Energy Is Delivered by a Wave in 1 Second?

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SUMMARY

The discussion focuses on calculating the energy delivered by a wave in one second, given a magnetic field strength (B0) of 5*10^-7 T, relative permittivity of 2, and relative permeability of 1.15. The correct formula for energy delivery is < P > = (B0^2 * c * A) / (2 * μ), where μ is the modified permeability (μ = 1.15 * μ0) and ε is the modified permittivity (ε = 2 * ε0). The user confirmed that the Poynting vector equation must be adjusted to account for the material properties, leading to the conclusion that both permittivity and permeability need to be modified when transitioning from vacuum to material calculations.

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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:
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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
\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}}
then in the material, your equation becomes
\langle P\rangle = \frac{B_0^2 A}{2 \mu\sqrt{\mu\epsilon}} = \frac{B_0^2 c A}{2 n \mu}
where \mu = 1.15\mu_0 and \epsilon = 2\epsilon_0.

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.
 
Thanks so much! This was confusing me for so long but i get it now, you're a star ^_^
 

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