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Author Topic: Dose it take power to create a magnetic field  (Read 1489 times)
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Ok, so he charges up the coil ,and then disconnects the coil,and the LED flashes.

You forgot to add he charges up the coil "in the Liquid Nitrogen" because only minutes ago you said it's not what they did or even possible to charge a zero resistance coil. But that's okay mate, I'm patient, we're here to learn right.

I have to ask-->what is so great about that?
We can do exactly the same with a standard coil.

Did you get the part the coil will hold it's magnetic field indefinitely or until used. Do standard coils do that?

Now,lets say the coils resistance was 0 ohm's-->how would you calculate the power used to energise that coil?,so as you can get the little LED to flash

That question is one i am asking you Luc--how would you calculate the energy used to power up that coil,when the resistance value is 0

Brad

That my friend is the golden buzzer question, how do you measure 0.000 volts x 20 Amps max???
As far as I know, you can't and I don't know how others who have not done the experiment themselves are able to confirm the coil actually uses power. 
One thing we both know is you definitely wouldn't want to use the power supply they used to charge a zero resistance coil since the PS itself will dissipate way more power then they can get in their feeble Inductance value.

I'm taking the time to share my thoughts (instead of just keeping it to myself) to make us think as a group on other possibilities never yet explored.
I can see it's doing just that ;)

Cheers
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It's not as complicated as it may seem...
If the coil has no resistance,then it has no inductance.
How is P/in calculated when R=0 ?.
You cannot create a magnetic field while the inductor is in a super conductive state.


I thought it was self explanatory.

1-If the coil has no resistance,then it has no inductance.
Inductance in a coil creates the BEMF-dose it not ?
BEMF is part of the effective resistance value,where the wire resistance is the other part of the effective resistance of an inductor. So,if there is 0 resistance,there is neither any wire resistance,nor is there any BEMF resistance.If there is no BEMF resistance,then there is no inductance.

2-How is P/in calculated when R=0 ?.
This is a straight forward question,so i fail to see the problem here.

3-You cannot create a magnetic field while the inductor is in a super conductive state
Is that statement not correct?,as a true superconductor will reject any magnetic field that tries to penetrate it.


Brad
My goodness.

Don't confuse resistance and reactance. Yes, you are confusing them!

The statement that if a coil has no resistance, it has no inductance is incorrect. All coils have inductance.

The "effective resistance" reference you make is in terms of the inductor's reactance, and as long as current is changing in the coil, it then by default exhibits inductance, and inductive reactance.

The Pin (or more properly in this case, Energy) of a DC supply energizing a coil, whether it be ideal or non-ideal is measured via the same method we always use; with a scope (for Energy, integrating power over time).

Superconducting electromagnets do indeed create magnetic fields.
   

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Quote
Don't confuse resistance and reactance. Yes, you are confusing them!

Not at all. I never mentioned reactance.My statement was clear--inductance causes BEMF,and BEMF is an effective circuit resistance.Reactance has no resistive properties--it is the non resistive component of  impedance,and impedance only comes with inductance,and inductance creates that impedance-->BEFM,and BEMF is an effective resistance(impedance) to current flow.

Quote
The statement that if a coil has no resistance, it has no inductance is incorrect. All coils have inductance.

Once again,very wrong.
I have just finished testing a coil that has no inductance,against an identical coil,but which has inductance
The current trace through this coil is vertical,unlike the coil with inductance which is on a linear incline.
One coil produces a BEMF,and the other dose not.
One coil produces a magnetic field,the other dose not.
One coil has inductance,the other dose not.
One coil dissipates more power than the other--can you guess which one dissipates more power?.

Quote
The "effective resistance" reference you make is in terms of the inductor's reactance, and as long as current is changing in the coil, it then by default exhibits inductance, and inductive reactance.

And if the current dose not change over time,but go's straight to it's maximum value at T=0,then there is no reactance,no inductance,no magnetic field,and no BEMF.
It is inductance that creates BEMF-the coil self induces. The impedance is a result of the self induced BEMF.
This impeades(resists) current flow.

If the superconductive inductor has inductance,then it will also have BEMF.
If it has BEMF,then it's effective resistance value is not 0,as the BEMF will resist the flow of current.

Quote
The Pin (or more properly in this case, Energy) of a DC supply energizing a coil, whether it be ideal or non-ideal is measured via the same method we always use; with a scope (for Energy, integrating power over time).

So,when a superconducting coil,that has 0 resistance is dropped across a battery,and because of the 0 resistance no voltage exists across the inductor,we then calculate power by multiplying the current by 0 volt's?  ???

Quote
Superconducting electromagnets do indeed create magnetic fields.

But not without cost,which is what this thread is about.


Quote
My goodness.

Indeed.

To impede-->HINDER, IMPEDE, OBSTRUCT, BLOCK mean to interfere with the activity or progress of. HINDER stresses causing harmful or annoying delay or interference with progress.   IMPEDE implies making forward progress difficult by clogging, hampering, or fettering

To resist--> counteract or repel ,mean to set against something,to hinder progress,to fight against,to limit,


Brad
« Last Edit: 2019-02-16, 05:14:39 by TinMan »


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Id say that if the super coil has inductance, then we should be able to see a voltage across the coil from the point of connecting the input and as long as the current and field rises, as long as the input can handle the ever increasing currents.

Verpies said  "Yes, it takes no energy to maintain magnetic field (or current) in an ideal coil."

How is that?  Would the coil need to be shorted after input is taken away?

Mags


   

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If the coil has no resistance then what ever power you put into it you can fully recover = no cost of power = Free magnetic field because in the end you got a magnetic field PLUS you recovered all the power. Does that not qualify for a free Magnetic field?
It would if recovering the energy back would not decrease the magnetic field back to zero.

Now that you created a free magnetic field, what will you do with it?... what ever work you get out of it, would that part not qualify as OU? (this will stir the pot) ;D
Note, that you can get mechanical work out of the magnetic field ONLY when the coil is charged. After the coil is discharged (e.g. by recovering the input energy back) the magnetic field decreases back to zero and can no longer perform any work.

So whatever work you want to do, you have to do after the coil is charged and before it is discharged back to zero...
« Last Edit: 2019-02-16, 07:08:11 by verpies »
   

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Verpies said:  "Yes, it takes no energy to maintain magnetic field (or current) in an ideal coil."

How is that?  Would the coil need to be shorted after input is taken away?
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.
   

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Id say that if the super coil has inductance, then we should be able to see a voltage across the coil from the point of connecting the input and as long as the current and field rises, as long as the input can handle the ever increasing currents.

Verpies said  "Yes, it takes no energy to maintain magnetic field (or current) in an ideal coil."

How is that?  Would the coil need to be shorted after input is taken away?

Mags

That is correct Mags.
And if we can see a voltage across the coil,then that means that there is a resistance through the winding's.
If there is a resistance through the windings,then it is not superconductive.

The only way there can be no resistance,is if there is no inductance-->no BEMF (or CEMF) if you wish,to provide that resistance.
I suspect !hair splitting! will arrive soon  C.C

But common sense seems to be thrown out here when it suits the needs of a few  C.C

After some conversations with a few who were there-right in the mix of the TPU,any claims from others will now require !on the bench! proof before i believe what they have to say.
Getting there information via book's,or from what they were taught to be correct,will no longer cut the cake.
If it's good enough for me to do at the request of other(as i have always done),then it is good enough for others to do the same in return-->do for others as you ask them to do for you.

From now on,for me to pay any attention anyway, you will be required to present your argument by way of a presented video of your !on the bench! test results.

I have put this one to bed in my next video--to create a magnetic field dose take energy--there is no free magnetic field created with an inductor.
The same applies to superconductors--energy is required to create the magnetic field.-,and only that amount of energy can be collected back from the collapse of that magnetic field.
There is no OU to be had.



Brad


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Dose it take power/energy to create a magnetic field?

Inductive V non-inductive coil

https://www.youtube.com/watch?v=vcUfMQv7aAI

Enjoy


Brad


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-there is no free magnetic field created with an inductor.
There is if the inductor has a permeable core, but it exists within the inter-atomic space of the core material.  Classical theory says that within the material B and H are in opposite directions, but that is nonsense when you look at that inter-atomic space.  Unfortunately after over 20 years looking for OU I have not yet found a practical way of getting access to that free energy.  It could be possible if the material goes through a phase change (not an electrical phase angle change but a material phase change) such as going above the Curie temperature.  If the relative permeability drops to unity then that free magnetic energy stored inside the material becomes available, but cycling material around its Cure temperature is not very practical.
Smudge.
   

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There is if the inductor has a permeable core, but it exists within the inter-atomic space of the core material.  Classical theory says that within the material B and H are in opposite directions, but that is nonsense when you look at that inter-atomic space.  Unfortunately after over 20 years looking for OU I have not yet found a practical way of getting access to that free energy.  It could be possible if the material goes through a phase change (not an electrical phase angle change but a material phase change) such as going above the Curie temperature.  If the relative permeability drops to unity then that free magnetic energy stored inside the material becomes available, but cycling material around its Cure temperature is not very practical.
Smudge.

And in order to cycle the core material around it's Cure temperature,you would end up using more energy than that which is contain'd within that core materials magnetic field.

So once again,the conservation of energy stands.


Brad


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So you put energy in, then you get it all back and end up with a magnetic field. Sounds like a free magnetic field to me?
However, I don't think it would be free if the magnetic field instantly ended as you're taking the energy back. Of course that doesn't happen.
Yes, the magnetic field generated by the coil cannot end instantly any more than it can be built-up instantly.
It takes time to discharge even an ideal coil. The discharge of such coil into an ideal capacitor is 100% efficient and takes the time tD=½π*SQRT(L*C) or ¼ of the typical underdamped LC oscillation period.
The smaller the discharge capacitor - the faster the coil discharges but the higher the capacitor's final voltage because ½L*i2 = ½C*V2, and this transforms to V = i*SQRT(L / C). - the latter does not account for diode and resistive losses.
« Last Edit: 2019-02-16, 15:20:27 by verpies »
   

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The statement that if a coil has no resistance, it has no inductance is incorrect. All coils have inductance.
I agree.
Changing the resistance does not change the inductance ...but it does change the time constant Tau=L/R.
If the resistance is decreased to zero then this time constant becomes infinitely large, LR circuit's time stops ...and everything becomes constant.

The "effective resistance" reference you make is in terms of the inductor's reactance, and as long as current is changing in the coil, it then by default exhibits inductance, and inductive reactance.
I agree. The BEMF that opposes the change of current flow is also known as "inductive reactance".  In an ideal coil the BEMF opposes any change to current flow forever, because of the infinite time constant.

The Pin (or more properly in this case, Energy) of a DC supply energizing a coil, whether it be ideal or non-ideal is measured via the same method we always use; with a scope (for Energy, integrating power over time).
Yes, the famous equation E=½Li2 comes from this integration.  It is worth noting, that voltage does not appear in this equation ...only current and inductance.
However, the electric energy converted by the coil into magnetic flux can also be measured with a ruler and magnetometer (Hall, GMR, AMR, TMR, etc...) and calculated as E=½Φ2/L, without inserting any Current Sensing Resistors into the coil's circuit.

Superconducting electromagnets do indeed create magnetic fields.
Yes, but it is kind of hard to push any electric current into them when the Tau=∞, because the current would have to rise infinitely slow ...and "infinitely slow" is an euphemism for "never".
That is why hot pumping or flux trapping is used to charge them.
« Last Edit: 2019-02-16, 19:19:32 by verpies »
   

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My statement was clear--inductance causes BEMF, and BEMF is an effective circuit resistance.
It would be more accurate to say that BEMF causes an "effective resistance" to changes of electric current.
However, this "effective resistance" IS the "inductive reactance".

Q:So what is the difference between resistance and reactance?
A: The distinction is very simple: resistance opposes the instantaneous current (i) and reactance opposes the change of current (di/dt).

...impedance only comes with inductance,
Not only - impedance also comes with capacitance.

Formally, impedance = resistance + reactance.  (Z=R+X). The first component is commonly called the real component, the second component is called the imaginary component, and their sum is called the complex quantity (...or complex impedance).
Also, the quality factor of a coil is equal to Q=X/R.

...and inductance creates that impedance-->BEMF
Actually, pure inductance creates only the imaginary component of the impedance. Pure inductance does not create the real component of impedance.

...and BEMF is an effective resistance(impedance) to current flow.
Not precisely - BEMF causes an opposition to changes of electric current flow.  That is very different than "an opposition to the current flow".

The statement that if a coil has no resistance, it has no inductance is incorrect. All coils have inductance.
Once again,very wrong.
I have just finished testing a coil that has no inductance,against an identical coil,but which has inductance
The current trace through this coil is vertical,unlike the coil with inductance which is on a linear incline.
But that only proves, that it takes electric energy to create magnetic energy.
It doesn't prove that zero resistance causes zero inductance, which is the implication that Poynt objects to.

One coil produces a BEMF,and the other dose not.
That only proves that one coil has inductance and reactance, while the other does not.

One coil produces a magnetic field,the other dose not.
That only proves that one coil has inductance while the other does not.  Inductance is best conceptualized as the ratio of magnetic flux to electric current, L=Φ/i

One coil dissipates more power than the other--can you guess which one dissipates more power?.
Of course, the one that has less inductance, because inductance is responsible for conversion of electric current into the magnetic flux...and the more electric current gets converted to magnetic flux, the less electric current is converted to heat by the resistance.

And if the current dose not change over time,but go's straight to it's maximum value at T=0,then there is no reactance,no inductance,no magnetic field,and no BEMF.
Yes

It is the inductance that creates BEMF-the coil self induces. The impedance is a result of the self induced BEMF.
This is correct but not precise enough.  Since impedance consists of two components: resistance and reactance (Z=R+X) and pure inductance is capable of presenting reactance only, then it would be better to write that:
"The reactance is a result of the self-induced BEMF"

If the superconductive inductor has inductance,then it will also have BEMF.
Yes, but the BEMF in a superconducting coil is also a victim to the infinite time constant Tau=L/R and as a consequence it perpetually opposes any change of current.  Because of this infinity, the normal proportionality of BEMF to the rate of change of current (di/dt) degenerates to the proportionality to the absolute value of current (i). 
Such is the fate of the very concept of "change" when the RL circuit's time stops due to Tau=∞.
No time flow = No change = No BEMF decay.

If it has BEMF,then its effective resistance value is not 0, as the BEMF will resist the change to the flow of current.
Yes, but the proper name for the "effective resistance" IS "inductive reactance".
Also, when Tau<∞, BEMF does not oppose the absolute value of the current, it opposes the change of this current.  I allowed myself to make this addition to your sentence in green color in your quote above.

So,when a superconducting coil,that has 0 resistance is dropped across a battery,and because of the 0 resistance no voltage exists across the inductor, we then calculate power energy by multiplying the current by 0 volt's?  ???
We can calculate the energy stored in the coil by measuring its magnetic field with a magnetometer (Hall, GMR, AMR, TMR, etc...)* and calculating the energy as E=½Φ2/L, without disturbing the coil's circuit.
Alternately, we can calculate the energy stored in the coil only by considering only its inductance and current E=½Li2, but since a superconductive coil has an infinitely long time constant Tau, the current from the battery will take infinite time to rise if such coil is charged by such electric means.
You could, however, charge the coil with energy by opening its curcuit, 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 impede-->HINDER, IMPEDE, OBSTRUCT, BLOCK mean to interfere with the activity or progress of. HINDER stresses causing harmful or annoying delay or interference with progress.   IMPEDE implies making forward progress difficult by clogging, hampering, or fettering
To resist--> counteract or repel ,mean to set against something,to hinder progress,to fight against,to limit,
As you can see both of these words have similar meaning in non-technical jargon.
However, in electrical engineering impedance is the sum of resistance and reactance (Z=R+X).
And the most significant difference between resistance and reactance is that the former is proportional to current (i) and the latter is proportional to the rate of change of current (di/dt).

* Since magnetometers measure the Magnetic Flux Density (B) and not the total Flux (Φ), their measurements need to be integrated over an area (in sq.meters) in order to yield Flux.
« Last Edit: 2019-02-17, 10:45:25 by verpies »
   

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 author=verpies link=topic=3723.msg72618#msg72618 date=1550328037]



Well i knew the !splitting of hairs! was coming-it always dose.

Quote
It would be more accurate to say that BEMF causes an "effective resistance" to changes of electric current.
However, this "effective resistance" IS the "inductive reactance".

That is what i said.
If there is no inductance,there is no reactance,and there is no BEMF.
Quote: reactance-the non-resistive component of impedance in an AC circuit, arising from the effect of inductance or capacitance or both and causing the current to be out of phase with the electromotive force causing it.
So BEMF is induced by the inductance,and reactance is also a result of inductance.
So i was correct in what i said.

Quote
Not precisely - BEMF causes an opposition to changes of electric current flow.  That is very different than "an opposition to the current flow".

As i said,splitting hairs.

Quote
But that only proves, that it takes electric energy to create magnetic energy.

Please read thread title.

Quote
It doesn't prove that zero resistance causes zero inductance,

Once again-if a coil has inductance,it will produce a BEMF.
BEMF is an effective resistance--or quote: The resistance of an ideal inductor is zero. The reactance of an ideal inductor, and therefore its impedance, is positive for all frequency and inductance values. The effective impedance (absolute value) of an inductor is dependent of the frequency and for ideal inductors always increases with frequency.
So if you wish to call it effective impedance,then so be it.
If a coil has an effective resistance,it dose not have 0 resistance.

Quote
That only proves that one coil has inductance and reactance, while the other does not.
That only proves that one coil has inductance while the other does not.  Inductance is best conceptualized as the ratio of magnetic flux to electric current, L=Φ/i

I totally agree,but to quote poynt :All coils have inductance.
As my video showed,not all coils have inductance.
It also shows that the coil that has no inductance,has no BEMF,which is why the current trace is vertical--there was no opposition to the flow of current,as there was no BEMF.
BEMF/CEMF Quote:-->Counter-electromotive force (abbreviated counter EMF or simply CEMF), also known as back electromotive force (or back EMF), is the electromotive force or "voltage" that opposes the change in current which induced it.

Quote
BEMF does not resist the absolute value of the current,

That is correct.
BEMF creates a voltage across the coil that is of the same polarity across the coil to that of the source,where the value of that voltage is slightly less than that of the source at T=0.
This reduces the potential difference between the source and the coil(BEMF),which means less current will flow.
As time passes,the BEMF value decreases,and the potential difference increases,and so then dose the current.

Quote
Yes
This is correct but not precise enough.

Good enough for me and most others here.
As i said,splitting hairs was bound to come.

"The reactance is a result of the self-induced BEMF"

Quote
Yes,

Thank you  O0

Quote
but the BEMF in a superconducting coil is also a victim to the infinite time constant Tau=L/R.  Because of this infinity the normal proportionality of BEMF to the rate of change of current (di/dt) degenerates to the proportionality to current (i). 
Such is the fate of the very concept of "change" when the RL circuit's time stops due to Tau=∞.

So are you saying that the current will remain at 0 at T=0 ,due to the time constant now being infinite?.


Brad


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Quote
.
That is why hot pumping or flux trapping is used to charge them.

Indeed  O0


Quote
Yes, but it is kind of hard to push any electric current into them when the Tau=∞, because the current would have to rise infinitely slow ...and "infinitely slow" is an euphemism for "never"

So from T=0,nothing happens?

Brad


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Once again-if a coil has inductance,it will produce a BEMF.
Yup

BEMF is an effective resistance--or
quote: The resistance of an ideal inductor is zero. The reactance of an ideal inductor, and therefore its impedance, is positive for all frequency and inductance values. The effective impedance (absolute value) of an inductor is dependent of the frequency and for ideal inductors always increases with frequency.
I never wrote that.

So if you wish to call it effective impedance,then so be it.
I do not wish it.
I wish everybody called the "effective resistance" of a pure inductor an "inductive reactance" ...and the "effective resistance" of a resistive inductor a "complex impedance".

If a coil has an effective resistance,it dose not have 0 resistance.
If my wish was fulfilled then the "effective resistance" in this sentence would be changed to "inductive reactance" and "resistance" to "complex impedance", effectively making it:
"If a coil has an inductive reactance, then it does not have 0 impedance"  ...because Impedance = Resistance + Reactance (Z=R+X).

I would even go further and improve it like this:
"If a coil has an inductive reactance (or resistance), then it does not have 0 impedance"  ...because Impedance = Resistance + Reactance (Z=R+X).

The devil is in the details...

BEMF creates a voltage across the coil that is of the same polarity across the coil to that of the source,where the value of that voltage is slightly less than that of the source at T=0.
This reduces the potential difference between the source and the coil(BEMF),which means less current will flow.
As time passes,the BEMF value decreases,and the potential difference increases,and so then dose the current.
Yes and in a superconducting coil the BEMF decreases infinitely slowly, so the potential difference increases infinitely slowly, too (=never), so it stays at zero forever...and current never flows.

So are you saying that the current will remain at 0 at T=0 ,due to the time constant now being infinite?.
Not only at t=0 but at t>0, too.
   
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to create a magnetic field dose take energy--there is no free magnetic field created with an inductor.

I agree. The Law of Conservation of Energy applies to standard Inductors. Best I ever got was 98% efficient.

The same applies to superconductors--energy is required to create the magnetic field.-,and only that amount of energy can be collected back from the collapse of that magnetic field.
There is no OU to be had.

Brad

That's a confusing statement when you insist it cost to produce a magnetic field "energy is required to create the magnetic field and only that amount of energy can be collected back"

Sounds like you just wrote:  energy in, minus, energy out = 0 energy... so where is the cost of energy exactly?

Also, why are you not considering that once the magnetic field is created you can do anything you wish with it for any length of time, plus it oppose any change while magnetized.
Could we not have a piece of Iron attract to it = work... and at TDC discharge to collect back all our input energy?
Why could that not be a possibility of OU?... I'm sure there are many other possibilities.

I don't understand, you insist it cost to produce a magnetic field wet have never worked with a super conducting or even Cryogenic coils to see the reality.

Who are you and what have you done to the experimenter we know a TinMan Power?

Cheers
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It's not as complicated as it may seem...
Sounds like you just wrote:  energy in, minus, energy out = 0 energy
In a way, yes. The "collection" of the stored energy (via the magnetic field), can either be returned to the source, or transferred to a load.

Quote
... so where is the cost of energy exactly?
If the energy is returned to the source, then the net energy used (or "cost") is zero. If the stored energy is transferred to a load to perform work, then the "cost" of the energy is that which came from the source (assuming no losses).

Quote
Also, why are you not considering that once the magnetic field is created you can do anything you wish with it for any length of time, plus it oppose any change while magnetized.
Could we not have a piece of Iron attract to it = work... and at TDC discharge to collect back all our input energy?
Why could that not be a possibility of OU?... I'm sure there are many other possibilities.
It sounds like an assumption is being made that there is no loss of energy if the magnetic field performs some work, and that one would still be able to recover all the original applied energy after such work is performed.

Is that a safe assumption?
   

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It sounds like an assumption is being made that there is no loss of energy if the magnetic field performs some work, and that one would still be able to recover all the original applied energy after such work is performed.

Is that a safe assumption?
I do not know.
I am leaning towards Smudge on this. Especially regarding work on soft permeable materials, their internal magnetization, etc...

If a superconducting coil, with current already circulating in it (because it is precharged and shorted) attracts a slug of a soft ferromagnetic from afar, then what happens ?
   

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It's not as complicated as it may seem...
I do not know.
I am leaning towards Smudge on this. Especially regarding work on soft permeable materials, their internal magnetization, etc...
Is some work not required to polarize (or partially polarize) a magnetic domain?

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If a superconducting coil with current already circulating in it (because it is shorted) attracts a slug of a soft ferrite from afar, then what happens ?
First, I think it is important to note that "superconductor" does not imply 0Ω. There will always be some finite resistance, no matter how small. Just like absolute zero can never be achieved.

The energy expended in attracting the slug will be lost and manifest as a commensurate reduction of the circulating current and associated magnetic field.
   

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Is some work not required to polarize (or partially polarize) a magnetic domain?
I think so, but is this work recoverable when the domain depolarizes (relaxes) ?

The energy expended in attracting the slug will be lost and manifest as a commensurate reduction of the circulating current and associated magnetic field.
I agree that the circulating current (i1) will decrease to (i2) after the slug is attracted but the total flux penetrating the coil (Φ) will remain the same. Yes, some of the flux will come from the coil (@ i2) and some of it will come from the slug, but the net result will be the same.

But what will happen upon discharge of the coil?
The current i2 will be lower, but the inductance will increase from L1 → L2.
Will ½L1i12 == ½L2i22 ?
Will ½Φ2/L1 == ½Φ2/L2 ?
« Last Edit: 2019-02-16, 20:56:35 by verpies »
   

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It's not as complicated as it may seem...
I agree that the circulating current (i1) will decrease to (i2) after the slug is attracted but the total flux penetrating the coil (Φ) will remain the same. Yes, some of the flux will come from the coil (@ i2) and some of it will come from the slug, but the net result will be the same.

But what will happen upon discharge of the coil?
The current will be lower, but the inductance will increase from L1 → L2.
Will ½L1i12 == ½L2i22 ?
Will ½Φ2/L1 == ½Φ2/L2 ?

Even though L may increase, I don't think there will be a commensurate decrease in i or maintenance of flux.

iow:

W1>W2
   

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Even though L may increase, I don't think there will be a commensurate decrease in i or flux.
I think we can wonder about i but not that the flux penetrating a shorted superconducting coil will stay constant in response to a disturbance by an external flux source.  ...like in this video ?
Will the proportion L=Φ/i be broken ?

iow: W1>W2
Any other reason than CoE ?
« Last Edit: 2019-02-16, 20:54:29 by verpies »
   

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It's not as complicated as it may seem...
I think we can wonder about i but not that the flux penetrating a shorted superconducting coil will stay constant.  ...like in this video ?
Any other reason than CoE ?
What would the time domain plot of the current look like from t=0 to t=1 when the slug hits the coil?
   
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It sounds like an assumption is being made that there is no loss of energy if the magnetic field performs some work, and that one would still be able to recover all the original applied energy after such work is performed.

Is that a safe assumption?

Yes, and the answer is unclear if that can be done without affecting the electromagnetic field.
Verpies has said the BEMF work the other way around in a SC coil and fights against changes to its current?
My suggestion is the simplest approach of using the magnetic field. I'm sure there are other ways we can think of that may be less invasive.
As far as I know we won't know till someone physically tests it and that's more what this discussion should be about.

Regards
Luc
   
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