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Slab of material in magnetic field. Determine magnetic field

  1. Dec 4, 2014 #1
    1. The problem statement, all variables and given/known data

    A very large slab of material of thickness [itex]d[/itex] lies perpendicularly to a uniform magnetic field [itex]\vec H_0 = \vec a_zH_0[/itex], where [itex]\vec a_z[/itex] is the unit vector in the z-direction. Determine the magnetic field intensity (ignoring edge effect) in the slab:

    a) if the slab material has a magnetic permeability μ

    b) if the slab is a permanent magnet having magnetization vector [itex]\vec M_i = \vec a_zM_i[/itex]


    2. The attempt at a solution

    I really don't know how start with this. I just want some hints, would be very appreciated!

    I tried looking at the boundary conditions,

    [itex]μ_1H_{1n}=μ_2H_{2n}[/itex],

    and

    [itex]\vec a_{n2}\times(\vec H_1 - \vec H_2)=\vec J_s[/itex],

    but I couldn't get anywhere.. :( Help!
     
    Last edited: Dec 4, 2014
  2. jcsd
  3. Dec 4, 2014 #2

    rude man

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    a) There's a very simple equation relating B to H. What is it?
    b) There's a still simple equation relating B, H and M. What is it?
     
  4. Dec 4, 2014 #3
    [itex]\vec H=\vec B/μ \rightarrow \vec B_0/μ=μ_0\vec a_zH_0/μ[/itex]

    and

    [itex]\vec H =\vec B/μ_0- \vec M \rightarrow \vec H=μ_0\vec a_zH_0/μ_0-\vec M_i=\vec a_z(H_0-M_i)[/itex]

    Thank you!
     
    Last edited: Dec 4, 2014
  5. Dec 5, 2014 #4

    rude man

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    So we have B = μ0(H + M)
    where B, H and M are all vectors, not necessarily pointing in the same direction.

    In part (a) you have magnetic material but not a permanent magnet. So what is the relative direction between B and H? In other words, do they add or subtract?

    In part (b) you have to analyze the permanent magnet in terms of magnetizing or Amperian currents, which changes some of the vectors' directions with respect to each other, so they may not all just add. You mentioned one boundary condition which boils down to B1n = B2n. But there is also a boundary condition on H. What does H do at the boundary? If you draw an integration loop which crosses the surface, you know that ∫H ds around this path = 0 since there is no actual current piercing this loop. It's a bit difficult but you can figure out what the direction of H has to do to satisfy this integral (or you can look it up! :smile:).

    BTW H, B and M are all collinear but do not necessarily point in the same direction.
     
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