Question on Poynting Vector calculation in wire.

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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
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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
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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
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Yes, that's right
 
  • #5
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Thanks.
 
  • #6
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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
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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
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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
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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
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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
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Right because I is flowing in -z direction.
Thanks.
 
  • #12
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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
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Right because I is flowing in -z direction.
Thanks.
No, current flow in +z direction same direction as E.
 
  • #14
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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
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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
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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
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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
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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
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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
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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
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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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