Re: [femm] Electric circuits vs. material current density

[David Meeker <dmeeker@xxxxxxxxxxxxxxxxx>]

Fernando Seco wrote:

> Dear David & FEMM users,
>
> I'm trying to simulate the magnetic fields in an iron tube
> created by a solenoid wrapped around it, which carries a total
> current of 40 A @ 1 kHz. FEMM provides two ways to simulate this:
>
> a) Put the copper coil in a circuit and specify a total
> current of 40 A; edit the copper material properties and set
> J = 0 MA/m^2. This is done in file DYNFE05.FEM.
>
> b) Do not make an electric circuit, but put a current
> density of J = 8 MA/m^2 in the copper (times 5 mm^2 cross section
> should give the total 40 A current). This is in file DYNFE06.FEM.
>
> (I'm attaching both files if you want to try it)
>
> When I actually perform the simulations, the result is
> that the magnetic field is twice larger in the first case (I was
> expecting roughly the same values). If you use a block integral to
> compute the total current in the solenoid you get I~=40 A in the
> first case, but only I=~14+j*14 A (|I|=20 A) in the second. That
> seems to explain the different computed magnetic fields.
>
> Could it be that in the first case you give the peak
> current and in the second the RMS current? But that would give
> only a difference of sqrt(2), not 2, I think.
>

In all cases, the results are peak values.

>
> My question is: what is the physical meaning of both
> types of simulation, and which is the "correct" one to actually
> describe a real solenoid?

In both of your simulations, what you have really modeled is a 1-turn solid copper
tube carrying current, rather than a coil with a lot of turns. There are some
results that you probably didn't intend to have, due to currents that are induced in
the coil region of your problem. You can get the result that you probably intended
by setting the conductivity in the coil region to zero. If the coil conductivity is
set to zero, the current will be evenly distributed over the coil's cross-section and
won't change as a function of frequency.

Anyhow, the first problem (the one with J set in the copper region), the J that you
are setting is the "source current density"-- that is, the current density that would
be there in the absence of induced currents. Setting the source current is, in some
sense, like setting the applied voltage. Now, when you run the coil at 1 kHz, there
is non-negligible inductive impedance so that extra voltage is required to get the
same net current. But since setting J implies a constant voltage, the total current
is less at 1000 Hz than at 0 Hz.

In the second problem (the one with the J=0 and a "circuit" property specifying the
current), the coil region still looks like a solid bar because of the nonzero
conductivity. However, the program is now solving for an applied voltage drop that
forces there to be a prescribed total amount of current in the coil region. It's
like closing a feedback loop that selects the applied voltage to realize a commanded
current. In fact, you can choose View | Circuit Properties off of the femmview main
menu when you are looking at the results of your second problem to display the
voltage that was required to make the 40 Amps.

Dave Meeker


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