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# Re: [Faraday's law ?]

- To: femm@yahoogroups.com
- Subject: Re: [Faraday's law ?]
- From: "you se-ho" <sxy28@hotmail.com>
- Date: Mon, 23 Jul 2001 18:03:02
- Bcc:

`
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.

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

``

>     --- In femm@y..., "you se-ho" <sxy28@h...> 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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