Poynting's Theorem: Joule Heating in Cylindrical Conductor

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In summary, the power dissipated to heat in length l of the conductor is P = IV. The attempt at a solution was to evaluate the integral term but was not correct.f
  • #1
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Homework Statement



A long cylindrical conductor of radius 'R' and electrical conductivity σ is immersed in a uniform electric field 'E' directed along its axis, imposed by a distant battery. By evaluating an appropiate term in the integral form of the poynting theorm, show that the power dissipated to heat (the joule heating) in length 'l' of the conductor is:

P = IV





The Attempt at a Solution



I think the integral term which is asking to evaluate is:

∫jf . E dV where jf is the conduction current density and dv is the whole volume of the cylinder.
 
  • #2
That won't be enough. The theorem relates (in this particular case) the time rate of change of the energy density to to the integral you have given. More generally, you'd also have to integrate over the poynting vector itself. In this case, however, the poynting vector is zero because of the absence of a magnetic field.

The time rate of change in the energy density is proportional to the power.
 
  • #3
The Poynting vector is not zero. There's a magnetic field due to the current.

The electromagnetic energy flowing through a surface S is

[tex] \int_{S} S \cdot dA [/tex]

where the S in the integrand is the Poynting vector. Evaluate this over the surface of the wire (segment of length l) to find the energy flowing into it. This must equal the energy dissipated as heat because the current is steady.
 
  • #4
Pfft, yeah what he said.

sorry for the bogus post.
 
  • #5
The electromagnetic energy flowing through a surface S is

[tex] \int_{S} S \cdot dA [/tex]

where the S in the integrand is the Poynting vector. Evaluate this over the surface of the wire (segment of length l) to find the energy flowing into it. This must equal the energy dissipated as heat because the current is steady.

In the interest of accuracy,

[tex] \int_{S} S \cdot dA [/tex]

is the electromagnetic energy flowing through a surface S per unit time.

And, it must equal the power (not energy) dissipated as heat.
 
  • #6
I'm still confused on how to solve this, I tried evaluating the integral:

∫ S. dA and i got S = 4(pi)R² . E²√(permittivity/permeability)


Do I need to equate this to the joule heating term? and is this term simply jf . 4/3(pi)R³ ?


Thanks.
 
  • #7
I'm still confused on how to solve this, I tried evaluating the integral:

∫ S. dA and i got S = 4(pi)R² . E²√(permittivity/permeability)

That's not correct. If you do it correctly, you should find the rate of flow of energy into the segment of length L to be LE(πr²)j = LEI. Show your work so I can see where you went wrong.
 

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