Poynting Vector Energy Delivered

In summary, the Poynting vector can be calculated by multiplying the square of the magnetic field (measured in teslas) by the area (1m^2) and dividing by two times the product of the vacuum permeability and the speed of light. When dealing with a material, the relative permittivity and relative permeability must be taken into account, and the equation should be modified by inserting a factor of n (the square root of the product of the relative permittivity and relative permeability) in the denominator.
  • #1
sanitykey
93
0

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:
Physics news on Phys.org
  • #2
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
Thanks so much! This was confusing me for so long but i get it now, you're a star ^_^
 

What is the Poynting Vector Energy Delivered?

The Poynting Vector Energy Delivered is a measure of the energy flow in an electromagnetic field. It represents the amount of energy that is delivered per unit area per unit time.

How is the Poynting Vector Energy Delivered calculated?

The Poynting Vector Energy Delivered is calculated by taking the cross product of the electric field and magnetic field vectors at a specific point in space.

What are the units of measurement for the Poynting Vector Energy Delivered?

The units of measurement for the Poynting Vector Energy Delivered are watts per square meter (W/m2).

What is the significance of the Poynting Vector Energy Delivered in electromagnetic fields?

The Poynting Vector Energy Delivered is significant because it represents the direction and rate of energy transfer in an electromagnetic field. It helps us understand how energy is transmitted and received in different systems.

How is the Poynting Vector Energy Delivered used in practical applications?

The Poynting Vector Energy Delivered is used in a variety of practical applications, such as designing antennas, understanding the energy transfer in power transmission systems, and calculating the radiation exposure from different sources.

Similar threads

Replies
8
Views
10K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
4
Replies
118
Views
12K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
8
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
903
  • Advanced Physics Homework Help
Replies
1
Views
2K
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
780
  • Advanced Physics Homework Help
Replies
1
Views
2K
Back
Top