Re: [Faraday's law ?]

["David Meeker" <dcm3c@xxxxxxx>]

--- 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