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Author Topic: Dose it take power to create a magnetic field  (Read 1491 times)

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What would the time domain plot of the current look like from t=0 to t=1 when the slug hits the coil?
How about something like this ?
   

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It's not as complicated as it may seem...
As far as I know we won't know till someone physically tests it and that's more what this discussion should be about.
Proper and accurate testing of such may not be trivial.

You are certainly welcome to your opinion as to what the discussion here should be about, but I suggest that it not be limited as long as the discussion is on topic. Agreed?
   

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Yes, and shorting the coil is good for it. Shorting prevents the coil from losing energy.
It is the opposite behavior of a capacitor. Capacitors lose energy when shorted.  Don't confuse the response of coils and capacitors to shorting !!!

You can input some energy and charge an isolated coil with it, then short the coil to maintain the magnetic field and current in it ...and when you are done you can recover the input energy back. 
If it is a resistive coil, then only a part of the input energy can be recovered back.  The less the resistance and the shorter the time periods (with respect to Tau=L/R), the more energy can be recovered back.

BTW: Electrostatics can mechanically attract objects, too, but (unlike coils) it does not suffer the penalty of holding a capacitor charged for a long time - the resistance does not waste the energy continuously in electrostatics.

"Yes, and shorting the coil is good for it. Shorting prevents the coil from losing energy."

Ah.  If this is true, then that is the answer to my question....

If we build a field on the super coil with input and then simultaneously disconnect input and short the coil when the coils field gets to where we want it to be and it holds its currents and mag field steady, forever I suspect, then that holds the key to my question, 'Would a super coil even conduct or even produce a mag field?' 

If after we disconnect the input and short the coil, and the current is maintained, how is it maintained without a changing field to do so? That would imply that a normal coil with a magnet should produce continuous current, just not as much current, but seems like there should be something.  If the super coil all charged up and shorted and the currents in the coil are producing some perfect reflection(exact opposition) to the field that would normally collapse, then Id have to say that the coil will not take on input for the very same reason!!! ???  Perfect impedance!!  ;)  If so, then how did we charge the coil in the first place?

To me, 'if' the coil can build a field 'over time according to its inductive value', then it should collapse to drive the currents till the field is diminished just the same.   ;)

Does that make sense?

Mags
   

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Proper and accurate testing of such may not be trivial.

You are certainly welcome to your opinion as to what the discussion here should be about, but I suggest that it not be limited as long as the discussion is on topic. Agreed?

Indeed  O0

Poynt or verpies.

Can either or both of you explain as to why the value of the BEMF is slightly less than the applied EMF at T=0 ,(time of connection) ?.

I am also interested in what verpies says about there never being a current flow from T=0 in a super conductor,due to the infinite time constant.


Brad


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If after we disconnect the input and short the coil, and the current is maintained, how is it maintained without a changing field to do so?
Think of it like this:  The current and associated flux are changing, but infinitely slowly ...and infinitely slow change is equivalent to no change at all.  There can be no change if the RL circuit's time is frozen.

That would imply that a normal coil with a magnet should produce continuous current, just not as much current, but seems like there should be something.
Normal coils have the L/R time constant on the order of milliseconds, so current that flows in response to the externally changing flux of the permanent magnet (Lenz law) decays pretty quickly.  In superconducting coils it decays for years... or never.  That is the difference.

If the super coil all charged up and shorted and the currents in the coil are producing some perfect reflection(exact opposition) to the field that would normally collapse, then Id have to say that the coil will not take on input for the very same reason!!! ???
Yes

If so, then how did we charge the coil in the first place?

I already answered this:
You could, however, charge the coil with energy by opening its circuit, putting a magnet in it, shorting it, and pulling the magnet out ...or shorting it, heating it up until it loses superconductivity, putting a magnet in it, freezing it until it gains superconductivity and pulling the magnet out.
In both scenarios, the superconducting coil will be left with a perpetually circulating electric current that will maintain the same flux, that the magnet was providing before it was pulled out.  This is an example of a "pull out technique", that does work.


To me, 'if' the coil can build a field 'over time according to its inductive value', then it should collapse to drive the currents till the field is diminished just the same.   ;)
Yes, but the charging time does not have to be the same as the discharging time.  I wrote about this here already, too.
« Last Edit: 2019-02-17, 00:32:58 by verpies »
   

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Can either or both of you explain as to why the value of the BEMF is slightly less than the applied EMF at T=0 ,(time of connection) ?.
I don't know that it is less.  I see it as simultaneous reaction. My personal take on electromagnetism is quite unusual as I view it through the prism of motion cancellation*.
However, the conventional dogma is that the reaction proceeds at c according to retarded potentials.

I am also interested in what verpies says about there never being a current flow from T=0 in a super conductor,due to the infinite time constant.
I think I already did in so many words.
The electrically driven current tries to increase but with a perfect superconductor that process happens infinitely slowly.  With an imperfect superconductor, the increase might happen over e.g. 10 years.  How would you distinguish that from constantness in the first minute of the experiment?
You might also conceptualize this as you did before, namely as an equal but opposing BEMF that decays for 10 years ...or never.

* The way I see it, matter is a 3D motion which is responsible for gravity and the flow of time, electric current is a 1D motion and when it cancels part of this 3D motion, it leaves a 2D imbalance, which is the changing magnetic flux and related phenomena.  Conversely, a changing magnetic flux is a 2D motion and when it cancels part of the 3D motion, it leaves a 1D imbalance, which is the electric current. I don't know if this helps you to understand the universe my way...but you asked.  We don't have to talk about my point of view anymore.
   
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You are certainly welcome to your opinion as to what the discussion here should be about, but I suggest that it not be limited as long as the discussion is on topic. Agreed?

No problem and agreed

So I put together a similar test to what Brad did, trying to compare a 10 Ohm Resistive load to a 10 Ohm Inductive load and my result measures quite differently.
Please have a look and let me know what you think or if I'm missing something

https://youtu.be/AwPxagq5SQI

Regards
Luc
   

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No problem and agreed

So I put together a similar test to what Brad did, trying to compare a 10 Ohm Resistive load to a 10 Ohm Inductive load and my result measures quite differently.
Please have a look and let me know what you think or if I'm missing something

https://youtu.be/AwPxagq5SQI

Regards
Luc

That is because your test is completely different to mine.

My test show exactly how much power was used to !create! the magnetic field,while yours just shows power being burned off across resistors and the inductor. Where is your non inductive coil ?

What you are doing is subtracting the inductive discharge from your input power in the inductor test,while you only calculate input power in your resistive test  ???
How dose this tell you how much power is used to !create! the magnetic field?

So in your inductor test,you have calculated P/in with your scope,and then calculated the P/out from the inductive kickback with the scope (magnetic field now gone),and the scope has subtracted your inductive kickback P/out from the P/in and you still end up with 600mW P/in.

So as the inductive kickback energy(energy stored in the magnetic field) has already been subtracted from the P/in,where exactly is the extra 600mW going ?
You have 255mA RMS flowing through your .1 ohm CVR,so that is dissipating 6.5mW,so your still missing 593.5mW. Could it be that this missing 593.5mW is used to !create! the magnetic field?.


Brad
« Last Edit: 2019-02-17, 02:02:21 by TinMan »


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That is because your test is completely different to mine.

Yes I agree

My test show exactly how much power was used to !create! the magnetic field,while yours just shows power being burned off across resistors and the inductor. Where is your non inductive coil ?

I'm not convinced your test validated what you think or wrote. From what I read (after your test) Poynt and Verpies did not seem to agree it proved what you claimed and why I attempted a different way of testing.
I'm quite sure a wire wound resistor qualifies as a non Inductive resistive coil.  After all, it's only the resistance value that needs to compare and be the same.
Poynt and or Verpies can correct me if I'm wrong about this.

What you are doing is subtracting the inductive discharge from your input power in the inductor test,while you only calculate input power in your resistive test  ???
How dose this tell you how much power is used to !create! the magnetic field?

Yes, exactly what I want. By substantiating the Inductive recovery which actually had to go through the 10 Ohms of resistance for a second time, give me a rough estimate of a balance of power less the Inductive part. This balance could represents the bulk of the power dissipated in the coils resistance. Which would need to be calculated by someone with the qualification and inclination to do so.
If that's an incorrect approach, then Poynt or Verpies can recommend the correct test.

So in your inductor test,you have calculated P/in with your scope,and then calculated the P/out from the inductive kickback with the scope (magnetic field now gone),and the scope has subtracted your inductive kickback P/out from the P/in and you still end up with 600mW P/in.

Exactly

So as the inductive kickback energy(energy stored in the magnetic field) has already been subtracted from the P/in,where exactly is the extra 600mW going ?
You have 255mA RMS flowing through your .1 ohm CVR,so that is dissipating 6.5mW,so your still missing 593.5mW.
Could it be that this missing 593.5mW is used to create the magnetic field?.

Brad

How can the extra 600mW be all for for creating the magnetic field since Nothing has yet been deducted for the coils resistive losses?
Poynt or Verpies can possibly figure out how much of the left over power goes to resistance losses since Frequency, duty cycle, Voltage, average and rms Current data is all there below with scope shots.

Cheers
Luc

ADDED

Test Frequency 400Hz
Inductive duty on time 50%
Resistive duty on time 10%
Both 10 Ohms DC Resitance
Coil Inductance 130mH
Wire wound Resistor Inductance 8uH
« Last Edit: 2019-02-17, 03:40:24 by gotoluc »
   

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Woa
How come your post has gone like 50 foot wide Luc ?


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Yes, and the answer is unclear if that can be done without affecting the electromagnetic field.
I am certain that the magnetic flux, which penetrates an ideal shorted coil, will NOT change due to any external disturbances, such as motions of ferromagnetic objects in the vicinity of the coil. See this principle in action in this video.
It is also clear to me that these external disturbances to such coil will disturb its current.

What I am not certain about, is how the disturbed current of a pre-charged col, will affect the coil's subsequent discharge of energy.
Electric current alone, is NOT energy.
   
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Woa
How come your post has gone like 50 foot wide Luc ?

Don't know mate?... I don't see it like that in my browser!
Is it still like that for you or anyone else?

Cheers
Luc
   
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I am certain that the magnetic flux, which penetrates an ideal shorted coil, will NOT change due to any external disturbances, such as motions of ferromagnetic objects in the vicinity of the coil. See this principle in action in this video.
It is also clear to me that these external disturbances to such coil will disturb its current.

What I am not certain about, is how the disturbed current of a pre-charged col, will affect the coil's subsequent discharge of energy.
Electric current alone, is NOT energy.

Well, I'm glad we are on the same page and now can focus on the task of real physical test to get to the truth of the matter.

Thanks for sharing all your knowledge in this matter. Without your help I would of given up on this discussion.

Kind regards
Luc
   

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Don't know mate?... I don't see it like that in my browser!
Is it still like that for you or anyone else?

Cheers
Luc

No,it's good now.
It was the only post on the page like it,all the rest were normal.


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It's not as complicated as it may seem...
I am certain that the magnetic flux, which penetrates an ideal shorted coil, will NOT change due to any external disturbances, such as motions of ferromagnetic objects in the vicinity of the coil. See this principle in action in this video.
Perhaps the quantity of flux or flux density does not change (perhaps), but there is most certainly a disturbance in the coil's magnetic field via flux path and/or density change in a non-symmetrical fashion. This will (or should) generate a superimposed emf across the coil's finite resistance and burn off energy until the disturbance (changing flux path or density) ceases. After the event the coil's energy should therefore be reduced due to the heat loss in the resistance.

Quote
It is also clear to me that these external disturbances to such coil will disturb its current.
Agreed, and I envision it slightly different than you have drawn. In order to burn off some extra energy in the resistance, the current would have to increase during the event. After the event, the slowly declining current (appearing flat) will have stepped down a level.

Quote
What I am not certain about, is how the disturbed current of a pre-charged col, will affect the coil's subsequent discharge of energy.
Can you clarify this question?
   
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What I am not certain about, is how the disturbed current of a pre-charged col, will affect the coil's subsequent discharge of energy.
Electric current alone, is NOT energy.
My take on it.
Edit.  You are talking about a perfect shorted coil, therefore zero resistance.
If the disturbance came about because of movement of a nearby PM, then (assuming no further movement) the energy recovered when the coil is fully discharged from its new current value down to zero is one half the flux-linkage change from its previous holding value down to whatever value the PM is supplying multiplied by that new current value.
If the disturbance came from movement of a nearby magnetically soft material that sees only the field from the coil then it is a bit more complex, but can be found if the new value of inductance due to that nearby material is known.  Then it is simply 1/2Li2 where L is that new value and i is the current at the start of the discharge
Smudge
   

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Perhaps the quantity of flux or flux density does not change (perhaps), but there is most certainly a disturbance in the coil's magnetic field via flux path and/or density change in a non-symmetrical fashion.
I have some experience with Type2 superconducting coils and I can tell you that the magnetic flux density (B) definitely changes in response to the approach of a permanent magnet and can be picked up by a magnetometer probe in the bore or near it.
However, none of the surface integration exercises that I've done on it indicated that the magnetic flux (Φ), which penetrates the bore, changes in response to that external field disturbance.

This will (or should) generate a superimposed emf across the coil's finite resistance and burn off energy until the disturbance (changing flux path or density) ceases.
After the event the coil's energy should therefore be reduced due to the heat loss in the resistance.
It doesn't have the time to burn off the energy of the disturbance before the discharge stage, because the resistance is so small and the time constant is so long.
To date, the time constant of a SC magnet has been proven empirically to be at least 23 years by the Belgium’s Membach Geophysical Station and the theoretical estimates are at 100000 years.
...and superconductors chilled below the "vortex glass temperature" are claimed to truly have zero resistance.

In order to burn off some extra energy in the resistance, the current would have to increase during the event.
But the current, after the attraction of the soft ferromagnetic slug, decreases. 
This can be even demonstrated with resistive coils, that attract slugs (or their core halves) during time periods, which are much shorter than their initial time constant (Tau).

Agreed, and I envision it slightly different than you have drawn.
After the event, the slowly declining current (appearing flat) will have stepped down a level.
Could you draw that in MSPaint ?

Quote from: verpies
What I am not certain about, is how the disturbed current of a pre-charged coil, will affect the coil's subsequent discharge of energy.
Can you clarify this question?
It was not a question, I was merely writing what I was uncertain about.
The scenario, to which my statement referred to, begins with a SC solenoidal coil getting pre-charged with current, next: getting shorted, next: attracting a soft ferromagnetic slug into its bore from afar (which disturbs the coil's current but not flux), and finally discharging that disturbed current into a capacitor.
   

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If the disturbance came from the movement of a nearby magnetically soft material that sees only the field from the coil then it is a bit more complex, but can be found if the new value of inductance due to that nearby material is known.  Then it is simply 1/2Li2 where L is that new value and i is the current at the start of the discharge
Will this relationship be preserved, too ?: 
½Φ2/L1 == ½Φ2/L2
where L1 is the inductance before the disturbance and L2 is the inductance after the disturbance and Φ is the flux penetrating the shorted superconducting coil before and after the disturbance (before discharge, though) ?
   
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Will this relationship be preserved, too ?: 
½Φ2/L1 == ½Φ2/L2
where L1 is the inductance before the disturbance and L2 is the inductance after the disturbance and Φ is the flux penetrating the shorted superconducting coil before and after the disturbance (before discharge, though) ?
I can't see that could hold as L1 is not equal to L2.
   

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I can't see that could hold as L1 is not equal to L2.
Yes, L1 is smaller than L2
   

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It's not as complicated as it may seem...
I have some experience with Type2 superconducting coils and I can tell you that the magnetic flux density (B) definitely changes in response to the approach of a permanent magnet and can be picked up by a magnetometer probe in the bore or near it.
However, none of the surface integration exercises that I've done on it indicated that the magnetic flux (Φ), which penetrates the bore, changes in response to that external field disturbance.
Doesn't a dynamic change in flux density change the coil's inductance? Wouldn't a dynamic change in inductance change or create an emf?

Here is a thought experiment that could be performed on the bench. Using the attached setup, pipe a DC current through the coil, taking care to keep the core well below saturation (it most likely won't anyway with an air gap). Insert a CVR in the coil circuit for monitoring. Ideally, rig something up that can mechanically vibrate the bottom "keeper" piece of the core in a vertical fashion to vary the air gap.

1) Will a varying emf be detectable across the CVR?
2) If there IS an induced emf, does it extend both above and below the steady state CVR voltage?

Quote
It was not a question, I was merely writing what I was uncertain about.
The scenario, to which my statement referred to, begins with a SC solenoidal coil getting pre-charged with current, next: getting shorted, next: attracting a soft ferromagnetic slug into its bore from afar (which disturbs the coil's current but not flux), and finally discharging that disturbed current into a capacitor.
I'm still not clear on the scenario. "finally discharging that disturbed current into a capacitor" does not make sense to me. What is the disturbed current? After the attracted slug event is over, there is a new "steady state" condition of lower energy, and if de-energized at some point after, would do so according to the known electrical laws.
   
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Hi Poynt,

I actually have such a test device ready to go.
It's a double ferrite E core with top E fixed and the bottom falls with gravity once the cores are demagnetizes.
Does the test frequency matter?...  how much of a swing in Inductance do you want, 10%, 20% 30%?
With a 1mm air gap (adjustable) the inductance swing from low to high about 3 times the value.
So the coil is powered at low inductance then the bottom E core gets magnetically pulled up so Inductance rises to maximum then Inductive discharge is collected and E core falls back down to start the cycle over.

Let me know if it sounds suitable for the test you have in mind.

Regards
Luc
   

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It's not as complicated as it may seem...
I think what you're referring to is a little different than what I described Luc.

The coil is fed with a fixed, steady DC voltage. The keeper is then mechanically shifted closer to the core, then back, ideally by something that can do it at about 100Hz or so. The idea is to vary the gap mechanically (not allowing the magnetic field to draw it in per se). So in this case the keeper would need to be held firmly by the mechanical device.
   
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I think what you're referring to is a little different than what I described Luc.

The coil is fed with a fixed, steady DC voltage. The keeper is then mechanically shifted closer to the core, then back, ideally by something that can do it at about 100Hz or so. The idea is to vary the gap mechanically (not allowing the magnetic field to draw it in per se). So in this case the keeper would need to be held firmly by the mechanical device.

Yep, that's not what I've got.
I did build one of those a few years ago but took it apart after the I cores on the rotor came apart. Actually have a video of it when it blew up.

Regards
Luc
   

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Doesn't a dynamic change in flux density change the coil's inductance?
Not with an air coil. It might change the inductance of a cored coil if the flux density moves part of the core above the BH knee.

Wouldn't a dynamic change in inductance change or create an emf?
I would think so

Here is a thought experiment that could be performed on the bench.
It's a good experiment. The keeper can be glued to a woofer (a LF speaker).
I would expect an opposing emf as the keeper approaches and aiding emf when it departs.

P.S.
"Disturbed current" is the current flowing after the external flux disturbance.
   
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