1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Poynting's Theorem.

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data

    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





    3. 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. jcsd
  3. May 15, 2009 #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.
     
  4. May 15, 2009 #3

    dx

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  5. May 15, 2009 #4
    Pfft, yeah what he said.

    sorry for the bogus post.
     
  6. May 15, 2009 #5

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  7. May 16, 2009 #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.
     
  8. May 16, 2009 #7

    dx

    User Avatar
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook