Re: [femm] [Faraday's law ?]

[David Meeker <dmeeker@xxxxxxxxxxxxxxxxx>]

you se-ho wrote:

  To Dear Everybody:

How are you ? Thanks for Dr. Meeker and Dr. Gregory:

>ip = (M/L) isrc

If I understand correctly, these inductance (M and L) is a function
of frequency and logical conclusion to calculate M and L is to use
numerical method ? (My structure is on the order of micrometers. When
frequency is high, L is smaller than low freq. value and R is larger
than low freq.) When I calculated with just low frequency approximation,
the induced current at the passive loop was much smaller.
 

As frequency increases, you start running into eddy current effects which
tend to lower the apparent inductance and raise the apparent resistance. 
(At high frequencies, the type of model used by femm starts to break down,
and you have to move to one that doesn't neglect displacement currents,
i.e. a "full wave" solution.)

It is often possible to include the effects of eddy currents analytically, especially if your geometry is relatively simple.  The "bible" is Richard Stoll's Analysis of Eddy Currents, although this is a fairly hard-to-find book.  If your geometry is too complicated, you can resort to numerical methods.  Depending on what you are doing, small skin depth boundary condition methods can help make the computations a lot faster.

Dave.
 

 
Thank you very much for your kind consideration.
Sincerely,
SE-HO YOU

>
>--- In femm@xxxx, "you se-ho" <sxy28@xxxx> wrote:
> > Let's say there is one thin conductor whose current
> > is time-varying and there is conductor loop whose
> > conductivity is high (i.e good conductor), then
> > Faraday's law says there is induced emf and current
> > in the passive loop. Current is time-varying (let's
> > say as always sinusoid), to get the emf voltage, we
> > need to calculate d(flux)/dt. Therefore in the
> > _expression_ there is always frequency term. HERE IS
> > MY QUESTION. Does it mean induced (magnitude of)
> > current is linear function of source frequency ? I
> > think answer is NO because if it is true, the induced
> > current can be higher than source current.  If
> > somebody knows anything, I would like to here any
> > explanation.  Books, papers Ok, too.
>
>Consider a case where you've got a "source" loop where the current is
>controlled so that you have a constant current amplitude regardless of
>frequency, and a "passive" closed loop.  Assume that any skin effects
>in the "passive" loop are negligible for the frequency of interest.
>Then, you can write an electric circuit equation for the passive loop
>that looks like:
>
>L dip/dt - M disrc/dt + R ip = 0
>
>where ip is the passive loop current and isrc is the source current.
>L is the self-inductance of the passive loop, M is the mutual
>inductance between the source and passive loops, and R is the
>resistance of the passive loop.  The (M disrc/dt) part is the voltage
>that is induced in the loop--the induced _voltage_ is proportional to
>frequency because the derivative.  However, the impedance of the
>passive loop increases with frequency, balancing out the increase in
>voltage with respect to frequency and limiting the current that flows
>in the passive loop.
>
>At "high enough" frequencies, we could neglect the resistance all
>together, because this impedance becomes very small in comparison to
>the inductance at high frequencies.  Then, we just have:
>
>L dip/dt - M disrc/dt = 0
>
>which we could integrate to yield:
>
>L ip - M isrc = 0
>
>and re-arrange to get:
>
>ip = (M/L) isrc
>
>so that the current in the passive loop goes to a constant level as
>the frequency increases. The "physical" interpretation is that as you
>increase the frequency, the flux produced by the current in the
>passive coil becomes equal in magnitude and opposite in direction to
>the flux that links the loop from the "source" loop, so the net flux
>linking the passive loop goes to zero.
>
>Dave Meeker
>--
>http://members.aol.com/_ht_a/dcm3c
>

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