Flux of the Poynting vector

In summary, to calculate the flux of Poynting's vector through a surface, you will need to use the formula dE/dt=∫S⋅dΩ and follow the steps outlined above.
  • #1
enerieire
8
0
Moved from technical forums, missing template
I have some problems in calculating the flux of Poynting's vector through a surface.
I have that:

S=c/4π⋅H^2⋅n (S: Poynting's vector, n: versor of the direction of the propagation of the em wave)

H=1/cR A'
but: A=(1/cR) d'; A'=(1/cR) d'' ==> H= (1/cR)^2 d''
==> S= (c/4π)⋅ 1/(cR)^4 d''^2 nI have to find that the result of the flux of S (I don't know through which surface) is:

dE/dt=∫S⋅dΩ=...=-2/(3⋅c^3) |d''|^2 (d is magnetic dipole momentum)

How is it calculated?
Thanks to everyone who will try to help me.
 
Physics news on Phys.org
  • #2


Hello,

To calculate the flux of Poynting's vector through a surface, you will need to use the formula dE/dt = ∫S⋅dΩ, where dE/dt is the time derivative of the electric field, S is Poynting's vector, and dΩ is the surface element.

First, you will need to determine the direction of the electric field, which is given by the versor n. This will give you the direction in which the electromagnetic wave is propagating.

Next, you will need to calculate the magnitude of Poynting's vector using the formula S=c/4π⋅H^2⋅n, where c is the speed of light, H is the magnetic field, and n is the direction of propagation.

To calculate the magnetic field H, you will need to use the formula H=1/cR A', where R is the distance from the source of the electromagnetic wave, and A' is the magnetic vector potential.

The magnetic vector potential A' can be calculated using the formula A'=(1/cR) d', where d' is the magnetic dipole moment. Substituting this into the formula for H, we get H=(1/cR)^2 d''.

Now, substituting this value of H into the formula for Poynting's vector, we get S= (c/4π)⋅ 1/(cR)^4 d''^2 n.

Finally, to find the flux of Poynting's vector through a surface, we will integrate S over the surface element dΩ. This will give us the total energy flux through the surface.

The final result for the flux of Poynting's vector will be dE/dt=∫S⋅dΩ=...=-2/(3⋅c^3) |d''|^2. This is the result you were given in the forum post.

I hope this explanation helps you understand how to calculate the flux of Poynting's vector through a surface. If you have any further questions, please let me know.
 

1. What is the Flux of the Poynting vector?

The Flux of the Poynting vector refers to the flow of electromagnetic energy through a given surface. It is a measure of the rate at which energy is transferred through a specific area in a specific direction.

2. How is the Flux of the Poynting vector calculated?

The Flux of the Poynting vector is calculated by taking the dot product of the electric field and magnetic field vectors at a given point, and then multiplying it by the surface area and the cosine of the angle between the surface and the direction of the energy flow.

3. What is the significance of the Flux of the Poynting vector?

The Flux of the Poynting vector is significant because it helps us understand the flow of electromagnetic energy, which is crucial in many fields, including optics, telecommunications, and electrical engineering. It also helps us calculate the intensity and power of electromagnetic radiation.

4. How does the Flux of the Poynting vector relate to the laws of conservation of energy?

The Flux of the Poynting vector is directly related to the laws of conservation of energy. According to these laws, energy cannot be created or destroyed; it can only be transferred from one form to another. The Flux of the Poynting vector represents the transfer of energy in the form of electromagnetic radiation, and its magnitude remains constant throughout this transfer, thus obeying the laws of conservation of energy.

5. Can the Flux of the Poynting vector be negative?

Yes, the Flux of the Poynting vector can be negative. This occurs when the angle between the surface and the direction of the energy flow is greater than 90 degrees, resulting in a negative cosine value. A negative Flux value indicates that the energy is flowing out of the surface instead of into it.

Similar threads

  • Advanced Physics Homework Help
4
Replies
118
Views
12K
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
2
Replies
50
Views
4K
Replies
1
Views
563
  • Introductory Physics Homework Help
Replies
7
Views
754
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
902
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
Back
Top