MH's ideal coil and voltage question

[partzman]

Where is MMF covered in the conversation in regards to EMF?

https://en.wikipedia.org/wiki/Magnetomotive_force

https://en.wikipedia.org/wiki/Electromotive_force

Since the conversion is from EMF into MMF within the inductor,, don't you think that it would make sense to view things in the correct sequence?

With ZERO resistance in the ideal inductor there is no mechanism to produce a counter EMF, only the conversion into MMF.

Now think about what happens when the EMF is no longer supplied,, the MMF will continue to keep the current flowing unchanged unless and until current can no longer flow unimpeded.

Open the circuit and you introduce an infinite resistance which provides for an infinite EMF, close the circuit with a zero resistance and there is no change, close the circuit with a resistance and the EMF will then manifest as it is converting the MMF in a relationship with the resistance.

This is all covered by the older formulas.

Convenience can lead to misunderstandings.

Good points and I agree.

pm

[picowatt]

That is correct in the fact that when I have asked for a model or math derivation, none was given. However with the statement that Emf=Cemf when di = .8A/s, the parallel model with L as shown has to be assumed for any di to flow thru L. Please diagram schematically how this would be accomplished otherwise.

I have no idea where you are getting the CEMF in parallel idea from.  The schematic of the model I and others have been discussing is exactly as you have drawn it in your right hand drawing.

My model is correctly noted and defined and "C)" is correct as compared to yours.  Please explain how your position of Emf=Cemf when di=.8A/s fits in my series connected model.

That is exactly what I have been trying to do.

I didn't say that. What I did say in B) is that di=(Emf-Cemf)*dt/L implying that Cemf=0. I don't have a problem with that as I use Cemf for feedback correction only and follow Faraday's law as I stated in B).

In your version of "C)", you clearly state that when di=.8A/s, the CEMF=0.  That is incorrect.  The CEMF can only equal zero when di=0.  When di=.8A/s the CEMF=4volts.

Your formula in "B)", states that when EMF=CEMF, di=0, which is correct and explains why the rate of change cannot exceed or be less than .8A/s.

With the correction to "C)" so that it reads "C)  When di=.8A/s, CEMF=4volts", everything else in your notations are exactly as I have been stating and di is maintained at .8A/s.

I will say here that I have somewhat conceded with this model that Emf/Cemf feedback is required to achieve the .8A/s di in this example which I am not absolutely convinced is the case. Even my own model needs the math fudge to make this happen so something is still wrong here but that is another subject

You are right because in the Cemf in "A" is missing the minus sign.  I don't see a problem with Cemf=0 due to di=Emf*dt/L using conventional current flow. Are you saying that Cemf is inherent in Faraday's original derivation of this formula?

I'm sorry PW but I can't seem to justify changing C) to what you state because it makes no electrical sense in the series connected configuration. However, I am certainly open to see how this can be logically achieved.

pm

I think you are under the impression that just because EMF=CEMF all current flow ceases.  What ceases is rate of change.  Being an ideal inductor, the Vsource depicted as the generator of CEMF has zero resistance.

In the step wise discussion of the feedback mechanism, when CEMF=EMF, di equals zero and any current flowing prior to EMF=CEMF continues to flow, it just does not increase or decrease.

However, CEMF can only equal EMF when di=.8A/s.  It is this equality that regulates di to .8A/s.

Again, other than your statement in "C)", I agree with everything else stated in your notations.  With that one correction to "C)", the feedback mechanism that regulates di works just as described.


PW

[minnie]

  What is the speed at which the inductor is responding?
  If virtual photons are involved, they would be considered as
  mass-less I presume and would "propagate??" at C.
  Just tryin' to get an idea of what's goin' on.
   Thank you men,
                  John.

[verpies]

What is the speed at which the inductor is responding?

That is an interesting question - especially in the near field case.

I don't know of an experiment that measures the near field propagation of magnetic field, but I do know of an experiment measuring the electric propagation speed.
http://www.pandualism.com/c/coulomb_experiment.html

[tinman]

That is an interesting question - especially in the near field case.

I don't know of an experiment that measures the near field propagation of magnetic field, but I do know of an experiment measuring the electric propagation speed.
http://www.pandualism.com/c/coulomb_experiment.html

When looking very closely(very narrow time divisions on the scope),i see current rising along with the voltage at T=0,and then the current drop's back down to 0,and then rises again from 0 to follow our exponential curve. So this is either the voltage is instantaneous across the inductor,and these little ceramic resistors have some sort of lag time,or current flows instantly due to winding capacitance,and then drops back to 0 when that capacitor is full,and then current starts to flow ?.

Not to sure on that one,but something is happening before current flow starts due to induction.

I will whip up a quick video to show you guys what i am seeing,and maybe one of you know what is going on?.


Brad