1. Limited time only! Sign up for a free 30min personal 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!

Describe the magnetic field anywhere around plate

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data
    On a thin infinite metal plate, which is ##d## thick and has got electric conductivity ##\sigma ##, two electrodes are applied. The distance between the electrodes is ##r## and the potential is ##U##.

    Describe the magnetic field anywhere in space. Ignore the contribution of the lines that bring the current to electrodes.


    2. Relevant equations



    3. The attempt at a solution

    I don't really know what would be the smartest way to start this problem.

    Basically, if I understand the problem correctly, I can imagine the two electrodes as two electric charges and on that plane I should get currents like this http://www.phy.ntnu.edu.tw/ntnujava/snapshotejs/twopointcharges_2_20090117112750.gif . Now for a given point in 3D space I would have to somehow integrate over the entire metal plate. That's if I am not completely wrong.

    Hmmm.. Why do you suggest? How should I start?
     
  2. jcsd
  3. Jul 9, 2014 #2

    ZetaOfThree

    User Avatar
    Gold Member

    Well, modeling the electrodes as point charges seems a little bit complicated... plus the potential difference between two point charges would be infinite. As a general rule of thumb when solving physics problems, try to make things as simple as possible. Did they suggest a way to model the electrodes at all?
     
  4. Jul 9, 2014 #3

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The only thing relevant for the magnetic field is the currents flowing in the plate as a result of the applied potential. Some questions to get you started: What is the electric potential in the plate going to be? How does this relate to the current density? How does the current density relate to the magnetic field? And yes, somewhere there will be an integral over the metal plate.
     
  5. Jul 10, 2014 #4
    Nono, they haven't. That was my interpretation of the problem which I though would simplify it, but according to you it does right the opposite.

    I agree.

    What do you mean by "What is the electric potential going to be?" ? ?
     
  6. Jul 10, 2014 #5

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    From the problem, it is stated that the electrodes have a potential difference U. Now, the potential is only relevant up to a constant so we may use 0 and U (or +-U/2 for that matter, other choices seem annoying). It is this potential difference that drives the currents. If you prefer, you could work in terms of the electric field instead.
     
  7. Jul 10, 2014 #6
    Aha, good point. Let's take +- U/2, just because it looks nice.

    now to answer your other two questions;
    -Hmmm, current density ##j## is now ##j=\sigma E##, where ##E## is electric field. And I am tempted to say that ##E=U/r## but this doesn't seem right. I wouldn't expect the current density to be constant on the plane, therefore I think it should be ##E=U/d##, where ##d## is (hmmm; this may be hard to explain with my knowledge of english) the length of the curve along which the current flows.
    Or is it not?

    -Relation between the current density and magnetic field is described with Amper's law.
     
  8. Jul 10, 2014 #7
    Is that the exact question?

    Are the electrodes "infinitesimally small" or do they have some radius? Seems to me that the problem might not be mathematically well defined unless the electrodes have some nonzero size.

    :frown:
     
  9. Jul 11, 2014 #8
    Yes, that is the exact question. But we were told that the problem has general parameters and if we wish to illustrate the solution or maybe even come up with a numerical result, we are allowed to determine meaningful parameters.

    Basically, that is the only reason why I originally suggested modeling electrodes as point charges. Which turned out to be nonsense.

    So my guess, and answer to you would be that we could probably limit ourselves on some shape of electrodes.
     
  10. Jul 11, 2014 #9
    Ah I see. That's probably good :smile:

    I don't think it's complete nonsense, but if you're going to make the analogy work, you'll have to make it more exact.

    What's ##\nabla \cdot {\bf J}##?
     
  11. Jul 11, 2014 #10
    ##\nabla \cdot {\bf J}=-\frac{\partial \rho}{\partial t}##
     
  12. Jul 11, 2014 #11
    Which is what, given that the charge doesn't build up anywhere on the plate?

    (except at the electrodes, maybe)
     
  13. Jul 11, 2014 #12
    Well it has to be some kind of electric current. ##I=\frac{de}{dt}=\frac U R## where ##R=\frac{1}{\sigma}\frac d S## .
     
  14. Jul 11, 2014 #13
    Well, it's what the equation says - the rate of change of the charge density is minus the divergence of the current density :smile:

    But what is the rate of change of charge density on the plate? Do we have blobs of charge in the middle of the plate getting bigger and bigger over time, or do we have a static situation?
     
    Last edited: Jul 11, 2014
  15. Jul 11, 2014 #14
    :D

    Static situation, of course.
     
  16. Jul 11, 2014 #15
    Great! :smile:

    That means that [itex]\frac{\partial \rho}{\partial t} = 0[/itex]. Which in turn means that [itex]\nabla \cdot {\bf J} = 0[/itex].

    So what's [itex]\nabla \cdot {\bf E}[/itex] ?

    (remembering that [itex]{\bf J} = \sigma {\bf E}[/itex])
     
  17. Jul 11, 2014 #16
    I would say that ##\nabla \cdot E=\frac{\rho }{\varepsilon _0}## but since ##\nabla \cdot {\bf J} = 0##, I would say it also has to be ##0##.
     
  18. Jul 11, 2014 #17
    I agree.

    So last question - what's [itex]\nabla^2 \phi[/itex], where [itex]\phi[/itex] is the potential?

    (remembering that [itex]{\bf E} = -\nabla \phi[/itex])
     
  19. Jul 12, 2014 #18
    It's ##0## , but i don't really get it what are you trying to say...
     
  20. Jul 12, 2014 #19

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    So ##\nabla^2 \phi = 0## in the plate. This means that the potential is a harmonic function and that in order to find the current you must solve Laplace equation. In order to do that you need some boundary conditions ...
     
  21. Jul 12, 2014 #20

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    By the way, this shows that if you have a conducting material, there are no internal residual charge densities in a stationary state.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted