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# Re: [femm] [A few questions]

- To: femm@yahoogroups.com
- Subject: Re: [femm] [A few questions]
- From: dcm3c@aol.com
- Date: Mon, 6 Aug 2001 09:01:45 EDT

In a message dated 8/5/01 8:28:22 PM Eastern Daylight Time, sxy28@hotmail.com
writes:

> To Dear All:
> How are you? I have a few questions.
> (1) In the manual, denominator in equation (44) should be r^(n+1). It is
> obvious mistake, but it did not change from previous manual.

Whoops--I'll correct this equation.

> The thing I wanted to ask is
> How I can use the mixed boundary condition. If I check the axi1.fem file,
> the c0 coefficient is 6962160. Is this from (1/(mu_0*4.5 inch * 0.0254) ? I
> would like to make sure that I know correctly. If the problem is the planar
> in rectangular coordinate, Can I still use same equation (from the center
> of
> area of interest to most outside boundary) ?

Yes, you have interpreted it correctly. Eq. (44) is just a typo. The
result, eq. (46), is correct. The boundary condition works regardless of
whether you are using polar or rectangular coordinates--you always get the
same answer regardless of the coordinate system. However, you do have to
have a circular outer boundary for this to work right.

> The thing that I really want to ask is that in the manual it said when
> n=1, the objects in the solution region look like a dipole when view from a
> long distance. I would like to more explanation for this.

The idea is that a dipole has a field distribution of the same form that is
assumed in the derivation of the asymptotic boundary condition.

> (2) In the manual, it says resistive losses from i^2*R. I think it should be
> 0.5*I^2*R. Nevertheless, I think FEMM gives correct value.

The instantaneous power is R*i^2. For magnetostatic calculations, the loss
reported is R*i^2. For AC problems, i varies sinusoidally, so the average
loss over a cycle ends up being (1/2)*R*i^2, so that's what's reported.

> (3) In the layout, if I choose entire domain with " define area over which
> integration performed", and calculate A dot J integration. Should it be
> same
> as magnetic field energy ?

Well, the A dot J integration usually yields 2*(stored energy), although
there are circumstances where this isn't true. For example, if you are doing
a magnetostatic problem with nonlinear materials, this won't be the case.
Also, if you are using asymptotic boundary conditions, some of the energy is
stored "outside the domain", so that the A dot J will yield a (slightly)
higher energy than integrating "stored energy" over the entire domain.
Dave.

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