MH's ideal coil and voltage question

[MileHigh]

Much to your chagrin Mr. Wannabee Ignorance Enforcer, indeed some people now understand what true resonance really is, and they understand what it is not.  Resonance is NOT a timing delay device that results in better synchronous timing for a machine at one specific operational frequency.  Stomp your feet or bang your head against the wall as much as you want, the truth will not change.  I know that the spread of knowledge upsets you, so just go back to your hate and your polishing.

[3Kelvin]

OFF-Topic+++OT++OT

Hello Together,
i try to understand the ideal coil , ideal source question.

My math skills are very rusty and due to i need a refresh to understand the scientific discussion in this thread.

My question:
What is the meaning of "dA" ?
So far, i be able to remember it is a change of the area like a2*b2 - a1*b1 = Delta A [Rectangle]

What relationship dA is mentioned due to a magnetic field?
What kind of relation is dA in the context of magnetic fields?

Sry for the double question,
my English is not so good.
Try to become better

Thx for your answers.

Love + Peace
3K

[partzman]

I don't wish to add fuel to any fire here but I do have a logical question.  I will refer to the quoted portion below from Wikipedia's Counter-Electromotive Force.

Quote: "The counter-electromotive force (abbreviated counter EMF, or CEMF),[1] also known as the back electromotive force, is the voltage, or electromotive force, that pushes against the current which induces it. CEMF is the voltage drop in an alternating current (AC) circuit caused by magnetic induction (see Faraday's law of induction, electromagnetic induction, Lenz's Law). For example, the voltage drop across an inductor is due to the induced magnetic field inside the coil.[1][2] The voltage's polarity is at every moment the reverse of the input voltage.[1][3]"

Now I ask this, in a simple circuit with a supply having an inductor connected across it, how can you have CEMF in the inductor that is in opposite polarity to the applied supply voltage as stated above when the supply goes from 0 to some amount at T0 and remains for a period of time? Where is this CEMF hiding? How do I measure it? Or does it really exist in this example?

Curious minds wish to know!

pm

[poynt99]

Going back to my simplified example...

Given:
L=5H
V=+4V

What happens:
At t<0, XL=0
At t=0, XL=infinity, and the current rises from 0A linearly at 0.8A/s.

At t=1, XL=5 Ohms
At t=2, XL=2.5 Ohms
At t=4, XL=1.25 Ohms
i.e. with every doubling of "t", "XL" goes to 1/2 of previous value. XL never reaches 0 Ohms.

At t>0 when I_L begins rising, an induced cemf of -4V is produced, based on Vcemf=L x di/dt. The cemf remains steady at -4V as long as di/dt stays at 0.8A/s. This of course occurs simultaneous with the application of 4V and the rise of I_L.

The question is, if emf=cemf, would it not makes sense that this result is created through an equalizing process? The amps/s rate ultimately being determined by the applied voltage and the inductance values.

The equalization I refer to comes about via the simultaneous process of an applied emf that wants to drive a current, vs. a reactionary process that wants to lower that current. To me this is very much like a negative feedback mechanism commonly used in linear amplifiers. Your amplifier may have an inherent gain of 1000, but through the application of negative feedback, the gain is reduced to some desired level, such as 100. In the case of our self-inductance, the negative feedback mechanism is the self-induced current and B field, which happens to oppose the B field resulting from the applied voltage. It is not quite an exact analogy, but conceptually similar. It all happens in real time, simultaneously, and only the end result is observable.

Perhaps it comes down to an applied emf, vs. an induced E field. We know the applied emf is 4V, but can we break the induced cemf (equivalent to the E field) down any further? Well, for a multi-turn inductor we can divide the cemf by the number of turns to obtain the induced cemf per turn. To me this would represent the actual value of the E field circulating around the inductor. So if we have a total cemf of 4V, and a 1000-turn coil, the actual induced E field would be 4/1000 = 4mV. This does not sound like much, but with R=0, 4mV could drive a significant current in a single loop, and each loop would carry the same current. I'm sure the induced current can be derived from the E field (or B field) and rate of change.

Some things to perhaps think about anyway. Sorry there are no definitive answers here. Still thinking about this.

[ramset]

Miles
quote
"They know what resonance really is..."
---------------------------------------------------

Oh goody we can put that in with Electricity , magnetism and gravity  then....

I profoundly disagree Miles
the day we truly "Know what resonance is "and how to make it work in all materials for our benefit.

will be a very big day.

as Smokey 2 was teaching you earlier in this thread.
oh yeah
he's another one of your CLICKS

well done... well done Miles ..
I can feel the breeze of you patting yourself on the back from here...