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Maxwell Stress Tensor -> Force between magnets and perfect iron

  1. Apr 9, 2014 #1
    (this is not a hw)

    Assume you have a magnet of dimensions x_m, h_m, remanent flux density Br, and coercive field density Hc. The magnet is placed in a magnetic "C" structure (perfect iron) such that it is connected on one side but there is an airgap on the other side.

    xx..... xx

    Stack length is 1 m for simplicity.

    I know how to calculate gap flux density as a function of airgap length. I am struggling, however, with using the Maxwell Stress tensor to calculate the force between the magnet and the structure through the airgap.

    This is what I tried:

    Bm = Br / (1+Br*g(h_m*u0*Hc))

    \nabla\cdot\sigma_{xyz}= \frac{1}{\mu_{0}} \begin{pmatrix}
    \frac{\partial 0.5B_{x}^{2}}{\partial x} & \frac{\partial B_{x}B_{y}}{\partial x} & \frac{\partial B_{x}B_{z}}{\partial x}\\

    \frac{\partial B_{y}B_{x}}{\partial y} & \frac{\partial 0.5B_{y}^{2}}{\partial y} & \frac{\partial B_{y}B_{z}}{\partial y}\\

    \frac{\partial B_{z}B_{x}}{\partial z} & \frac{\partial B_{z}B_{y}}{\partial z}& \frac{\partial 0.5B_{z}^{2}}{\partial z} \\


    care only about one direction, which simplifies the equation to 1/2u0 * dBm^2/dx.

    Now integrate over volume:

    F = \int_{V} \mathbf{f} \mathrm{d} V = \int_{V} (\nabla\cdot\sigma)\mathrm{d} V = \oint_{S} (\mathbf{f}\cdot\mathbf{n})\mathrm{d} A

    But the numbers consistently come out wrong - with respect to a FEA simulation. Do I use the MST incorrectly?

    Thank you.
  2. jcsd
  3. Apr 9, 2014 #2

    Meir Achuz

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    The derivation of the MST requires a linear material (constant mu). Therefor, the MST cannot be applied to ferromagnets.
  4. Apr 9, 2014 #3
    Thank you. I forgot to mention that the iron has ideal (linear) magnetic properties such that:

    B_{iron} = \mu_{r} \cdot \mu_{0} \cdot H_{iron}

    Also, the permanent magnet has a linear loading curve:

    B_{mag} = B_{R} \cdot (1-H_{mag}/H_{c})
  5. Apr 10, 2014 #4
    Found the problem: I used the divergence theorem incorrectly.

    Volume integral of field divergence is equal to the closed surface integral of the field itself, not its divergence.

    F = \int_{V} \mathbf{f} \mathrm{d} V = \int_{V} (\nabla\cdot\sigma)\mathrm{d} V = \oint_{S} (\sigma\cdot\mathbf{n})\mathrm{d} A = \oint_{S} \sigma \cdot\mathrm{d} \vec{A}
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