Re: [femm] [A few questions]

[dcm3c@xxxxxxx]

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