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Two Coaxial Linear Magnetic Materials

  1. Oct 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider a two-layered cylindrical wire with inner-layer permeability μ1 and outer-layer permeability μ2. A line current I runs through the center in the z direction. Calculate the bound currents and the magnetic field produced by the bound currents.

    2. Relevant equations

    [1] ∫ B⋅dl = Iμ0
    [2] ∫ H⋅dl = Ifree
    [3] B = μ0 (H+M)
    [4] B = μH
    [5] Jb = ∇x M
    [6] Kb = M x n

    3. The attempt at a solution

    Using equation 2 and symmetry, I come up with
    H = I/(2πs)

    Using equation 4, I found the inner and outer material B fields. These point in the φ direction.
    B=Iμ1/(2πs) in the inner material
    B=Iμ2/(2πs) in the outer material.

    Plugging B and H into equation 3, I found the inner and outer material M fields. These point in the φ direction.
    M= I(μ10 - 1) / (2πs) in the inner material
    M= I(μ20 - 1) / (2πs) in the outer material

    Plugging M into equation 5, I calculate that Jb = 0 in both the inner and outer material.

    I expected some bound volume current, so this result is strange to me. If there is no bound volume current and I draw an amperian loop within the inner material, the enclosed total current must be equal to I. If that's the case, I should be able to use equation 1 to find that
    B = Iμ0/(2πs)
    but I already calculated a different inner B field above.

    How can I reconcile the different B values in this inner material?
     
  2. jcsd
  3. Oct 27, 2016 #2

    Charles Link

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    For a little bit, it seemed you had found a case where ## \oint M \cdot dl ## is non-zero and where ## \nabla \times M=0 ##. The ## I_{free} ## is uniformly distributed through the ## \mu_1 ## material. Thereby, the current enclosed in your ## B \cdot dl ## loop and ## H \cdot dl ## loop is proportional to ## r^2 ## and thereby you will not have a vanishing curl and you will get a finite ## J_b ##. ## \\ ## If you did work the problem as the case where you have a very thin inner inner conductor (of radius "a" )down the center carrying the entire current, I think you would find there is a magnetic current at the boundary where the ## \mu_1 ## material begins that supplies the necessary difference between ## \mu_o \oint H \cdot dl ## and ## \oint B \cdot dl ##.
     
    Last edited: Oct 27, 2016
  4. Oct 28, 2016 #3

    Charles Link

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    By magnetic current, I mean "magnetic surface current". In general it is magnetic surface currents (and magnetic currents ## J_b ## ) that cause the magnetic field ## B ## in materials, along with the ## H ## from free currents.
     
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