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Magnetic field inside a Hollow asymmetrical cylinder

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data

    A cylinder of radius a has a hole drilled through it of radius b, the hole is parallel to the axis of the cylinder but is displaced from it
    by a distance d in the x-direction where d < a - b. The asymmetric hollow cylinder has a uniform current density J = Jzez
    Show that the magnetic field inside the hole is uniform and show that
    B = (μ0/2)Jzdey

    2. Relevant equations

    Amperes law?

    3. The attempt at a solution

    I am at a complete loss, I dont know how to use ampere's law for something that is asymmetrical. Any ideas on how I should approach it?
     
  2. jcsd
  3. Mar 8, 2012 #2
    Use Ampere's law twice. Get an expression for the magnetic field (in cartesian coordinates) for an intact cylinder and one for the displaced cylinder. Using superposition add both vector fields.
     
  4. Mar 9, 2012 #3

    turin

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    How is this going to result in B parallel or antiparallel to J? I am stumped.
     
  5. Mar 10, 2012 #4
    I think information given is only part of the question, that the current densities are antiparallel aligned along the z-axis and of the same magnitude.
     
  6. Mar 10, 2012 #5

    marcusl

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    B will not be parallel to J, it is perpendicular. (Remember the right hand rule?) In your post, you note that J is along z and you are to show that B is uniform and along y.
     
    Last edited: Mar 10, 2012
  7. Mar 10, 2012 #6
    I'm not saying B is parallel or antiparallel to J, but the original post doesn't give all the facts, rather a current density of [itex]\textbf{J}=J\textbf{e}_z[/itex] along the z-axis of a full cylinder plus the smaller clyinder displaced by d on the x axis of [itex]\textbf{J}=-J\textbf{e}_z[/itex] in the opposite direction. Adding the field of the full cylinder and the offset one should give the expression for B.
     
  8. Mar 10, 2012 #7

    marcusl

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    bobred, my comment was directed to turin, who did say that B is aligned with J in post #3.
     
  9. Mar 10, 2012 #8
    Appolpgies marcusl.
     
  10. Mar 11, 2012 #9

    turin

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    Sorry for the confusion, y'all. I was squinting when I first read the original post, and all I saw were z's. I tried to delete my post when I realized the ey in the expression for B, but apparently I'm not allowed to delete it.

    (Oh ya, and thanks to marcusl for replying to my question.)
     
  11. Mar 30, 2012 #10
    I was wondering if you could help me with the magnetic field on the plane of the two axes. would it be the same as the hole?
     
  12. Mar 31, 2012 #11

    rude man

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    The only way I can think of doing this is via Biot-Savart. A bit complex but doable that way.

    Maybe solving del x B = 0 could be done but I don't know how.

    I don't see how Ampere could be of any use. Any loop you put in the cavity will have a zero circulation integral of B dot dl.
     
  13. Mar 31, 2012 #12

    rude man

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    EDIT: my apologies also, bobred! I didn't get your idea at first. Now I think it's got great potential, so to speak :biggrin: .

    Put the large cylinder along z with cross-sectional center at x=y=0. Assume it's solid. Then the field is known everywhere inside the solid cylinder using Ampere, including where the hollow part sits.

    Then, remove the solid cylinder and replace it with a solid cylinder with the shape and location of the hole, same current density except the current goes along
    -z instead of +z. Again, the field is readily established using Ampere.

    Then superpose the two B fields.

    Elegant - thanks bobred!
     
    Last edited: Mar 31, 2012
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