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Ampere's law for a closed ring bar magnet

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data

    A long bar magnet is bent into the form of a closed ring. If the intensity of magnetisation is M, and ignoring any end effects due to the join, find the magnetic field H and the induction B:

    (a) Inside the material of the magnet
    (b) just outside


    2. Relevant equations

    Ampere's law: [tex]\oint\vec{H}.d\vec{l}=\mu_{0}I[/tex]
    [tex]\vec{B}=\mu_{0}(\vec{H}+\vec{M})[/tex]

    3. The attempt at a solution

    I've done problems like this far too many times but after a while I always go rusty. So there's no current flowing, so I=0. And we integrate around a closed loop (would this be in the plane of the magnet or perpendicular to it?) of radius r. Usually for this sort of thing we end up with something like H.(2.pi.r) = 0, so H would be zero.

    I don't know how to modify it for just outside the magnet.

    As I say, I've done problems like this before but I think I've probably just always gone through the motions rather than really understanding what I'm doing. Any help would be greatly appreciated!
     
  2. jcsd
  3. May 20, 2010 #2

    _k_

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    This form of Ampere's law don't have [tex]\mu_0[/tex] on the right side.
    And the way you do it for inside is correct (you integrate along the loop inside that goes around the whole length of the magnet, that is in the plane of the loop created from magnet. Imagine how the standard bar magnet filed lines look like, now bend them to create a circle. You want to integrate along that circle.)
    For the outside you do exactly the same thing. Only M = 0.

    Or you can look at it this way: if the magnetization is M then magnetization current on the surface of the magnet is [tex] M \times n [/tex], where n is normal of the surface. And solve it as standard toroid with current on its surface.
     
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