FEMM Manual - Finite Element Method Magnetics

where the first bracketed term in (2.10) is the contribution from the turns of coil 2 that are
pointed out of the page and the second term is the contribution from the turns of coil 2 that
are pointed into the page. To evaluate mutual inductance with FEMM, one substitutes values
into (2.10). First, run the model with only “coil 1” turned on. Then, integrate A over the
volume in which the second coil lies (although the second coil is not turned on). For planar
problems, you will typically have to make two separate integrations–one over the region
where the turns in “coil 2” are pointed out of the page (i.e. that part of the coil in which
a positive current results in current flowing in the out-of-the-page direction), and one over
the region in which the turns in “coil 2” are pointed into the page. Add these two results
together to get the total A1dV2 integral. Lastly, multiply the integral result times n2/(i1a2) to
get mutual inductance.
• Magnetic field energy This selection calculates the energy stored in the magnetic field
in the specified region. This integral can be used as an alternate method of getting inductance
for problems that are linear (at least not heavily saturated). Denoting E as the energy stored
in the magnetic field, inductance can be obtained by solving the formula:
E = Li2
2
(2.11)
In the case of nonlinear materials, the energy is computed via:
Z �Z B
W = H(B ′ )dB ′
�
dV (2.12)
to take proper account of the energy under nonlinear conditions
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• Magnetic field coenergy For linear problems, coenergy is numerically the same as energy.
For nonlinear problems, coenergy is defined as:
Z �Z H
Wc = B(H ′ )dH ′
�
dV (2.13)
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Coenergy can be used in an alternative method of force and torque computation. To compute
force via coenergy, currents are held constant, and the position of the object upon which the
force is desired is perturbed slightly. The force can then be estimated by:
F = Wc(p+δ) −Wc(p)
δ
(2.14)
where p denotes the initial position, p+δ denotes the perturbed position, and δ is the magnitude
of the perturbation. The component of force determined in this way acts along the
direction of the perturbation–one has to perform two such operations to get both horizontal
and vertical components of the force.
• Hyst. and/or Laminated eddy current losses. This selection is typically used to
obtain the core losses produced in laminated iron sections in harmonic problems.
• Resistive losses This selection integrates the i 2 R losses due to currents flowing in the
“z” direction (or θ direction, if you are evaluating an axisymmetric problem).
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