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# [Faraday's law ?]

- To: femm@yahoogroups.com
- Subject: [Faraday's law ?]
- From: "you se-ho" <sxy28@hotmail.com>
- Date: Sat, 21 Jul 2001 21:57:35
- Bcc:

`
To Dear Everybody:`
`How are you? it seems like that I always ask something not contributing anying. Feeling guilty, I would like to ask
onemore and promise as time goes, I will contribute my knowledge
to this very useful group.`
`While I try to model crosstalk thing (and waiting Cornelius tutorial,I requested tutorial to him in person thinking to send e-mail to group is not good), I would like to understand more basics.`
`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.`

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

``

> `SE-HO: I think what you call cross-talk is what I call mutual (inductance)
> so it will be covered in the summary. However, the general process
> is that you have to calculate the coupling coefficient (can be done
> with FEMM), then calculate the induced current in the second coil
> by hand. It can not be done automatically in a 2D model. For a final
> picture of the fields emanating from the 2 coils you have to put
> a current source on either coil, using the values you calculated
> by hand.`

>     RUKMAL: sri I did not come back to your last post, too busy but havent
>     forgotten

>     a nice weekend to everyone, regards,
>     Cornelius

> ``

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