Resistive Characteristics Of a Coil in Changing Magnetic Field

wayneh

Joined Sep 9, 2010
17,153
I think the reality of your wave pulse is going to depend on the exact geometry of the magnet and the coil, and neither is simple. The magnet in particular will not be uniform, and will not be a uniform position from the coil. So you'd have to break it all up and do a finite element analysis. Not easy to set up, even with plenty of processing power to do the work.

In other words, I would not expect a precise sine wave. There's no reason to expect one. For a rotating system, maybe.
 

BR-549

Joined Sep 22, 2013
4,938
You bet......wave form will be asymmetrical, unless constant velocity of magnet thru coil or periodic movement.
 
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Thread Starter

positive8

Joined Sep 27, 2016
61
"Aha, is it? Picture a coil and a resistor place side by side"

Draw a picture.
Visualize a square, please. The coil and the resistor respectively occupy the vertical walls, the upper and lower parallels are
simple wire.
That is the complete setup. To me, when I make resistance measurements on the unpowered circuit, it seems parallel.

To reduce the question to an even more fundamental level, is a single-shot waveform formed within a coil in a previously inactive simple circuit, as described ,affected by the external resistance of the solitary resistor outside the coil?
Without component values it is hard to evaluate or double check your observations. For the coil, how about the diameter and the number of turns. For the magnet, some indication of field strength, velocity, and frequency of motion.
I think the reality of your wave pulse is going to depend on the exact geometry of the magnet and the coil, and neither is simple. The magnet in particular will not be uniform, and will not be a uniform position from the coil. So you'd have to break it all up and do a finite element analysis. Not easy to set up, even with plenty of processing power to do the work.

In other words, I would not expect a precise sine wave. There's no reason to expect one. For a rotating system, maybe.

Typically, using a cylindrical neodymium magnet of 2 cm diameter travelling with sub millimeter clearance through a 100-500 turn coil(2 cm) , usually in to 10-20 m/s range reasonably constantly velocity. Using a magnetic field calculator tool, and/or working backwards through the equations knowing all other variables, I get a field value of about .55T. The magnetic field at the "rim" of the cylinder top "pinches" to a higher .7T or so, and it is this focused area of the field which passes with 0.5mm of the closest wires in the coil (then again, this is the case with every cylindrical magnet).
The waveform, upon close visual examination is quite symmetrical, though not expected to be perfect, obviously, due to variations in field strength and even the extremely minor deceleration of the magnet in passing.
I can check all this, as was suggested, by comparing a single-shot waveform from a signal generator versus the coil shot across the same circuit. Which I will...but was just curious if a single-shot waveform generated under these circumstances across a quiescent circuit would exhibit behavior other than nominally projected by the boilerplate resistance/reactance phenomena.
 

Thread Starter

positive8

Joined Sep 27, 2016
61
I give up. The rest of you and the TS are on your own in the absence of what has been requested. Buona fortuna.
Thanks! You were still useful. Just so you know, I went to a physics forum and got a precise answer within 5 minutes. This was in fact almost exclusively a physics question, little to do with everyday electrical engineering. My bad for not going to the right pros.............
 

BR-549

Joined Sep 22, 2013
4,938
Faraday's induction equation tries to describe the process and results of induction, but fails to show the cause. Putting a B field thru a loop does not cause a voltage. It causes current. The pressure of this current causes the voltage. This is why the coil has to be conductive and closed. No voltage on a glass ring, because no current.

A magnetic field induces a perpendicular force on charge. It makes the charge move sideways. If a B field is moved thru a close conductor, all the electrons will move in the same direction. This one, same direction causes electron pressure....voltage. The resistance will determine how much current for how much pressure.

You can not measure the resistance of the circuit, by measuring the resistance across the circuit. You must measure the resistance thru the circuit. i.e. break or cut circuit and measure with ohm meter. It's a series circuit and will have series resistance.
 

wayneh

Joined Sep 9, 2010
17,153
To me, when I make resistance measurements on the unpowered circuit, it seems parallel.
No, the resistance is in series with the EMF. There is only one path for current to flow, and no parallel path. I hope this didn't come from your physics "pros".
 

Thread Starter

positive8

Joined Sep 27, 2016
61
No, the resistance is in series with the EMF. There is only one path for current to flow, and no parallel path. I hope this didn't come from your physics "pros".
If you go back and read the previous comments on my question, you'll see that even on this forum "pros" disagree as to whether it's parallel or in series. And there's nuanced reasons for this that may have escaped you.

Actually, despite my repeated posing of the question from various forms, no-one on this forum was able to grasp the question as it was intended to be posed. This is no doubt entirely due to my inarticulation.

However, posed as a physics question to a physicist, it is accessible. The main thrust was, when stored energy is released from an inductor, does it function more as a "black box" separately from the remainder of the circuit when determining the instantaneous value of the peak discharge voltage or it is influenced by the total Z values of the remainder of the circuit.
 

wayneh

Joined Sep 9, 2010
17,153
The circuit is in series. The current is – and must be – identical in all points of the circuit. That's not required for a parallel circuit, which offers alternatives paths for current.

The waveform you observe across the resistor will depend on the resistance. More resistance means more voltage and less current. The EMF is the theoretical maximum voltage with no losses. Some of that EMF will be lost in the coil's resistance and the rest in the resistor. The lower the resistance of the resistor, the higher the current and the larger the voltage loss in both components.

As current flows in the coil, a magnetic field appears that opposes the one imposed by the magnet. The magnet will be decelerated by passing through the coil, in proportion to the current. So a shorted coil will exhibit the greatest braking on the magnet.
 

Thread Starter

positive8

Joined Sep 27, 2016
61
The circuit is in series. The current is – and must be – identical in all points of the circuit. That's not required for a parallel circuit, which offers alternatives paths for current.

The waveform you observe across the resistor will depend on the resistance. More resistance means more voltage and less current. The EMF is the theoretical maximum voltage with no losses. Some of that EMF will be lost in the coil's resistance and the rest in the resistor. The lower the resistance of the resistor, the higher the current and the larger the voltage loss in both components.

As current flows in the coil, a magnetic field appears that opposes the one imposed by the magnet. The magnet will be decelerated by passing through the coil, in proportion to the current. So a shorted coil will exhibit the greatest braking on the magnet.
But the inductor, in terms of its electrical behavior, will "perceive" its internal resistance as essentially infinite if it generates a single waveform as a high-speed magnet passes through it, no? Therefore, it acts as an open gap relative to whatever circuit it is attached to, and consequently becomes a power source, like a battery rather than a circuit component, during this brief period (I'm thinking). The large voltage that can be measured across the coil , as it exits, is then "perceived" by the remainder of the circuit (through which it then travels) as an external input, rather than characterized by the elements of the circuit as a whole. This is what I mean by the "black box" phenomenon.
While it is true the magnet would theoretically be slowed as you described, its mass and velocity (momentum) in this case is large enough for this to be negligible.

Is this flawed thinking? When I present this as a theoretical profile, everybody goes, yeah, that's technically true. But I don't know i that translates into hard-core engineering reality, though..........

Thanks for your response, btw.............
 

BR-549

Joined Sep 22, 2013
4,938
The magnetic field of the passing magnet............is not the magnetic field of the coil.

Why do you think an inductor is ever perceived as an infinite resistance?
 

Thread Starter

positive8

Joined Sep 27, 2016
61
The magnetic field of the passing magnet............is not the magnetic field of the coil.

Why do you think an inductor is ever perceived as an infinite resistance?
Induction can take place without the presence of a magnet, so that's immaterial. Extremely high resistances - effectively infinite as viewed by the remainder of the circuit - can be encountered. As an offhand example, a 1H inductor at 5MHz presents 31+ Mohms impedance. Mathematically, there is a period, however short, where the impedance is infinite as the inductor sheds stored energy, no? Practically speaking, many conventional circuits perceive inductors essentially as open circuits to high frequencies, thus there use as filters. Most tellingly-and this penetrates to the heart of my question - the amplitude of the voltage waveform does not change as the impedance changes over time as it discharges, leading me to the "black box" point of view. Nothing saying I'm right, just feeling my way., but this seems a legitimate question.
 

BR-549

Joined Sep 22, 2013
4,938
I re-read all your posts on this thread. Your circuit and induction characterization is foreign to me.

I shall retire.
 

Thread Starter

positive8

Joined Sep 27, 2016
61
I re-read all your posts on this thread. Your circuit and induction characterization is foreign to me.

I shall retire.
Seeee? That's what I said. It's a physics problem, not an electrical engineering one. Substantially different approaches. Just remember, you were the one who re-activated the thread, not me ;-)
 

Thread Starter

positive8

Joined Sep 27, 2016
61
I re-read all your posts on this thread. Your circuit and induction characterization is foreign to me.

I shall retire.
To put the very fundamental difference between electrical engineering and physics into sharp relief, as an example, it may interest you to know that capacitive coupling between parallel wires has a very prosaic explanation in terms of practical circuitry, but, in fact, within physics the effect is known to be caused by relativistic considerations, even though electron velocity is nominal at most! I refer you to "Concepts of Modern Physics" by A. Beiser for a short but illuminating treatment on this effect.
 

wayneh

Joined Sep 9, 2010
17,153
Until you can specify an experiment to test your hypothesis, what you're doing is neither science nor engineering. I can't pretend to understand your hypothesis but so far I have not seen any observations to support or refute it.
 

BR-549

Joined Sep 22, 2013
4,938
Good luck on your studies. I would refer you to Ampere and Weber. A current loop is almost a perfect description of a charge.

Study Parson's Magneton.
 
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