Question on Poynting Vector calculation in wire.

In summary, the conversation discusses finding the Poynting vector for a long conducting wire with DC current. The book considers the electric field (E) to be on the surface of the wire, but it can be calculated at any point where E is not zero. The Poynting Theorem is used to verify the energy balance in the wire, with the surface integral representing the electromagnetic energy flowing into the wire per unit time. The time derivative of this energy flow is equal to the change in total energy inside the wire, and in this case, results in the electric and magnetic energy terms vanishing. The correct equation for energy balance also includes the term for ohmic dissipation. The conversation also clarifies the direction of current flow
  • #1
yungman
5,718
241
This start out as an example in the book:

Find Poynting Vector of a long conducting wire with radius = b and conductivity [itex] \sigma [/itex] with DC current through the wire. Assume wire in z direction and use cylindrical coordinates.

Since DC current, current distribute evenly inside the wire.

[tex] \vec I \;=\; \vec J \pi b^2 \;=\; \sigma \vec E \pi b^2 \;\Rightarrow\; \vec E \;=\; \hat z \;\frac {I}{\sigma \pi b^2} [/tex]

[tex] \vec H \;=\; \hat {\phi} \;\frac {I}{2\pi b} \;\hbox { on the surface of the wire.}[/tex]

[tex] \hbox {Poynting vector = }\; \vec E X \vec H \;=\; \hat z \;X\; \hat{\phi} \;\frac {I^2}{2\sigma \pi^2 b^3}[/tex]

This is all nice and good. MY question is: The book consider E is on the surface of the conductor. From the calculation about, I see that the E is evenly inside the wire also since it is DC current. Why the book calculate as if the E is on the surface or in the air right above the surface of the wire?
 
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  • #2
You can calculate at any point in space where E is not zero but what is usually done in a case like the example is that you want to know flux of energy through an element of surface and in this case one usually calculate the flux of energy into the conductor and you do that integrating the Poynting vector over the conductor's surface
 
  • #3
Thanks for the response.

What you are saying is the book choose the surface of the wire to calculate the Poynting vector of the surface.

In this case where the I is DC, I can choose to find the Poynting vector along an arbitrary surface say at radius equal a < b ( a surface inside the wire ) and calculate the Poynting vector for surface of radius = a.

For surface of radius = a< b inside the wire, do I use

[tex]I_a = \frac { I \pi a^2}{\pi b^2} \;\hbox { and } \; \vec H = \frac {I_a}{2\pi a}[/tex]

To find the Poynting vector at surface radius = a?
 
  • #4
Yes, that's right
 
  • #5
Thanks.
 
  • #6
This is the related question on the same problem. This part is the verify Poynting Theorem on the surface of the wire assume the wire on z-axis:

[tex] -\int_{S'} \vec P \cdot d \vec {S'}\;=\; \int_{v'} \frac{\mu}{2} H^2 dV' \;+\; \int_{v'} \frac{\epsilon}{2} E^2 dV' + \int_{v'} \sigma E^2 dV' [/tex] (1)

Where the first two terms on the right are the electric and magnetic energy enter into the section of wire. The third term is the ohmic dissipation due to the EM.

For E and H on the surface of the wire:

[tex] \vec E = \hat z \frac {I}{\pi b^2 \sigma} \;\hbox { and }\; \vec H = \hat {\phi} \frac {I}{2\pi b} \;\Rightarrow\; \vec P = (-\hat r)\frac{I^2}{2\pi^2b^3\sigma} [/tex]

[tex] -\int_{S'} \vec P \cdot d \vec {S'}\;=\; -\int_0^L (-\hat r)\frac{I^2}{2\pi^2b^3\sigma} \cdot \hat r (2\pi b) dl' \;=\; \frac {I^2 L}{\pi b^2\sigma}\;=\;I^2R [/tex]

My understand is [itex] -\int_{S'} \vec P \cdot d \vec {S'} [/itex] is the total power entering into the section of wire length equal to L which include the electric and magnetic energy together with the power dissipated in the wire due to ohmic loss as shown in (1). But in the answer, only the term of the ohmic dissipation appeared!

Where is the magnetic and electric energy? Does that mean both energy goes in from one side of the wire and leaving on the other side of the wire and net energy does not change in the section of wire?
 
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  • #7
The surface integral is the electromagnetic energy flowing into the conductor per unit time and this is equal to d/dt(total energy inside the conductor). The time derivative make your electromagnetic energy terms vanish.
 
  • #8
facenian said:
The surface integral is the electromagnetic energy flowing into the conductor per unit time and this is equal to d/dt(total energy inside the conductor). The time derivative make your electromagnetic energy terms vanish.

I forgot to put in the time derivative on the EM energy term! Can you explain why the time derivative of inflow of EM energy per unit time is zero?
 
  • #9
I must be missing something here in a sign, but doesn't crossing E into H here result in the Poynting vector out from the wire?
 
  • #10
John94N said:
I must be missing something here in a sign, but doesn't crossing E into H here result in the Poynting vector out from the wire?

[tex] \hat z \;X\; \hat {\phi} \;=\; -\hat r \;\hbox { which is point inward.}[/tex]
 
  • #11
Right because I is flowing in -z direction.
Thanks.
 
  • #12
I think the energy from H and E is flowing into the wire is equal to the power dissipated as heat from I squared R flowing out from the wire, think of an incandescent bulb
 
  • #13
John94N said:
Right because I is flowing in -z direction.
Thanks.

No, current flow in +z direction same direction as E.
 
  • #14
If electric field vector is positive at the tip and negative at the base then positive current must flow from tip to base or in reverse direction to E. Correct? This must be so if EXH is in the minus r direction.
 
  • #15
John94N said:
base then positive current must flow from tip to base or in reverse direction to E. Correct?

No, current mus flow in the same direction as E. Remember tha current is defined as flow of positive charges.
 
  • #16
John94N said:
If electric field vector is positive at the tip and negative at the base then positive current must flow from tip to base or in reverse direction to E. Correct? This must be so if EXH is in the minus r direction.

You are thinking of electron flow, current is opposite direction.
 
  • #17
yungman said:
Can you explain why the time derivative of inflow of EM energy per unit time is zero?

the correct (1) equation for energy balance is
[tex] -\int_{S'} \vec{P} \cdot d \vec {S'}\;=\;\frac{d}{dt}\left(\int_{v'} \frac{\mu}{2} H^2 dV' \;+\; \int_{v'} \frac{\epsilon}{2} E^2 dV' \right)+ \int_{v'} \sigma E^2 dV' [/tex]
 
  • #18
facenian said:
the correct (1) equation for energy balance is
[tex] -\int_{S'} \vec{P} \cdot d \vec {S'}\;=\;\frac{d}{dt}\left(\int_{v'} \frac{\mu}{2} H^2 dV' \;+\; \int_{v'} \frac{\epsilon}{2} E^2 dV' \right)+ \int_{v'} \sigma E^2 dV' [/tex]

Holly! How can I miss this! So both terms don't have t component so the derivative respect to t is zero! Just that simple?!:redface::eek:

Thanks

Alan
 
  • #19
Sorry to get hung up on this but if current flow is defined as positive charges then current on a wire in +z has E in +z. By right hand rule H is clockwise around wire and then E X H, right hand rule again, must be out from wire, which must be wrong but I don't see why.
 
  • #20
John94N said:
Sorry to get hung up on this but if current flow is defined as positive charges then current on a wire in +z has E in +z. By right hand rule H is clockwise around wire and then E X H, right hand rule again, must be out from wire, which must be wrong but I don't see why.

No, if your thumb is direction of current and H is direction of your 4 fingers which is in counter clock wise direction. So if thumb is z, 4 fingers is in [itex]\phi[/itex]. Then propagation is in [itex] \hat z X \hat{\phi} \;= -\hat r[/itex].
 
  • #21
John94N said:
Right because I is flowing in -z direction.
Thanks.

I think the energy from H and E is flowing into the wire is equal to the power dissipated as heat from I squared R flowing out from the wire, think of an incandescent bulb

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1. How do you calculate the Poynting Vector in a wire?

The Poynting Vector in a wire can be calculated using the formula S = ExH, where S is the Poynting Vector, E is the electric field, and H is the magnetic field. This formula represents the energy flow per unit area in the direction of the electromagnetic wave.

2. Why is the Poynting Vector important in wire calculations?

The Poynting Vector is important in wire calculations because it represents the energy flow in the wire due to the electric and magnetic fields. It helps us understand how energy is transferred through the wire and can be used in various applications, such as designing efficient electrical circuits.

3. What factors affect the Poynting Vector in a wire?

The Poynting Vector in a wire is affected by the strength and direction of the electric and magnetic fields, as well as the properties of the wire itself, such as its conductivity and material composition.

4. How is the Poynting Vector related to the speed of an electromagnetic wave in a wire?

The Poynting Vector is directly related to the speed of an electromagnetic wave in a wire. The magnitude of the Poynting Vector is equal to the product of the electric and magnetic field strengths, which are also related to the speed of the wave. Therefore, the faster the wave, the larger the Poynting Vector will be.

5. Can the Poynting Vector be negative in a wire calculation?

Yes, the Poynting Vector can be negative in a wire calculation. A negative Poynting Vector indicates that the energy is flowing in the opposite direction of the electromagnetic wave, which can occur in cases of reflection or refraction. This does not violate the conservation of energy, as the total energy within the system remains constant.

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