Maxwell Stress Tensor -> Force between magnets and perfect iron

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 5K views
SunnyBoyNY
Messages
63
Reaction score
0
(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.

xxxxxxxx
xx... xx
xx...gg
xx...mm
xx...mm
xxxxxxxx

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:

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

[itex] \nabla\cdot\sigma_{xyz}= \frac{1}{\mu_{0}} \begin{pmatrix}<br /> \frac{\partial 0.5B_{x}^{2}}{\partial x} & \frac{\partial B_{x}B_{y}}{\partial x} & \frac{\partial B_{x}B_{z}}{\partial x}\\ <br /> <br /> \frac{\partial B_{y}B_{x}}{\partial y} & \frac{\partial 0.5B_{y}^{2}}{\partial y} & \frac{\partial B_{y}B_{z}}{\partial y}\\<br /> <br /> \frac{\partial B_{z}B_{x}}{\partial z} & \frac{\partial B_{z}B_{y}}{\partial z}& \frac{\partial 0.5B_{z}^{2}}{\partial z} \\ <br /> \end{pmatrix}<br /> [/itex]

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

Now integrate over volume:

[itex] 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 [/itex]

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

Thank you.
 
Physics news on Phys.org
Meir Achuz said:
The derivation of the MST requires a linear material (constant mu). Therefor, the MST cannot be applied to ferromagnets.

Thank you. I forgot to mention that the iron has ideal (linear) magnetic properties such that:

[itex] B_{iron} = \mu_{r} \cdot \mu_{0} \cdot H_{iron}[/itex]

Also, the permanent magnet has a linear loading curve:

[itex] B_{mag} = B_{R} \cdot (1-H_{mag}/H_{c})[/itex]
 
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.

[itex] 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}[/itex]