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Magnetized infinite cylinder.

  1. Jun 19, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    Suppose I have an infinite cylinder with radius R, axis along the z axis and constant magnetization M[itex]\hat{z}[/itex]. I wish to find the magnetic field everywhere. (This is not a HW question per se, yet thought I might get some comments on my attempt at solving it nevertheless.)

    2. Relevant equations



    3. The attempt at a solution
    As M is constant and [itex]\vec{j}[/itex] = cM[itex]\hat{θ}[/itex], i.e. tangential current, may I consider this to be an infinite solenoid, whose magnetic field is zero everywhere, except for within the solenoid where it is equal to 4πnI/c (using c.g.s), where n is the density of wraps, which is equal to 4πM (in the positive z direction)?
     
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  3. Jun 19, 2013 #2

    TSny

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    That all looks correct to me.
     
  4. Jun 19, 2013 #3
    Okay, and if this were an infinite slab between z=0 and z=d (thickness d), with uniform polarization P[itex]\hat{z}[/itex]? How could I find its electric field?

    I do know that for a non-polarized slab it would be:

    2πρ(2z-d)[itex]\hat{z}[/itex] for 0≤z≤d; 2πρd[itex]\hat{z}[/itex] for z>d; -2πρd[itex]\hat{z}[/itex] for z<d

    And that here:

    ρ_bound = -[itex]\nabla[/itex][itex]\cdot[/itex]P = 0?
     
    Last edited: Jun 19, 2013
  5. Jun 19, 2013 #4
    I believe I made a mistake, didn't I? That relation between P and rho was for rho_bound, wasn't it? What about rho_free?
     
  6. Jun 19, 2013 #5
    I do know that:
    [itex]\nabla[/itex][itex]\cdot[/itex]E = 4πρfree, right?
     
    Last edited: Jun 19, 2013
  7. Jun 19, 2013 #6
    Nabla and E should be vectors of course. Should/could I use the relation in #5 to find E, or rather discontinuity?
     
  8. Jun 19, 2013 #7

    TSny

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    I don't know where you got these expressions. Why would there be any electric field if the slab is not polarized?

    Yes, for a uniform polarization, ρbound will be zero at all interior points of the slab. But there will be bound surface charges on each surface. The electric field can be found from these surface charges.
     
  9. Jun 19, 2013 #8

    TSny

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    [itex]\nabla[/itex][itex]\cdot[/itex]E = 4πρtotal = 4π(ρfree + ρbound)
     
  10. Jun 19, 2013 #9
    But wouldn't rho_bound (volumetric!) be zero here?
     
  11. Jun 19, 2013 #10
    So should -P be equal to the jump in electric field? Or is it P? How may I determine the direction of the normal vector?
     
  12. Jun 19, 2013 #11

    TSny

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    Yes, that's correct for points interior and exterior to the slab for this problem. ( I thought you were writing a general expression rather than an expression specific to this problem.) Now, what is ρfree for this problem?
     
    Last edited: Jun 19, 2013
  13. Jun 19, 2013 #12
    Other than expressing it via the terms in #8, I am not sure. Here's an attempt - may it also be zero? Are there free charges in this setting?
     
  14. Jun 19, 2013 #13

    TSny

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    You are free to define your own convention for the direction of the normal vector. If you think about the polarization as coming from a bunch of little electric dipoles uniformly spread in the slab, you should be able to see which one of the two surfaces of the slab has a positive surface charge density and which one has a negative charge density.
     
  15. Jun 19, 2013 #14

    TSny

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    From the way in which you stated the set-up, I assumed that there is no free charge anywhere. Of course, you could add free charge if you want!
     
  16. Jun 19, 2013 #15
    So that implies that E is constant, right? Now, is the following correct: 4*pi*sigma (=discontinuity in electric field) = P = -E? Within the slab, that is. And zero outside it?
     
    Last edited: Jun 19, 2013
  17. Jun 19, 2013 #16

    TSny

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    Yes, E will be constant inside the slab.

    I think you are missing a factor of 4*pi in front of P. I'm not sure what your sign convention means when you write -E. If you mean that the direction of E inside the slab will be opposite to the direction of P, then that's correct.
     
  18. Jun 19, 2013 #17
    Within the slab, E should be equal to 4*pi*Pi in the negative z direction, whereas it would be zero anywhere outside the slab. Right?
    I would like to challenge my understanding even further by replacing the polarization with a uniform magnetization, M in the (positive) x direction. Then, for starter, j should be equal to -Mc in the y direction, right?
     
  19. Jun 19, 2013 #18

    TSny

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    Right.

    Yes, jy = -Mc on the upper surface of the slab.
     
  20. Jun 19, 2013 #19
    Could I then simply substitute that value of j in the formula for the magnetic field of a slab, viz.:
    2*pi*j*d/c in the y direction (for z>d); -2*pi*j*d/c in the y direction (for z<d); 4*pi*j*z/c in the y direction (for z between 0 and d)?
     
  21. Jun 19, 2013 #20

    TSny

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    No, I don't think that's right. When I wrote jy = -Mc, I meant for jy to represent the surface current density on the upper surface of the slab. There will also be a surface current density on the lower surface. You should be able to see that there will not be any volume current density within the slab.
     
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