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Author Topic: Dose it take power to create a magnetic field  (Read 1819 times)

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Decided to start this thread in relation to Luc's claim.

First up,please know that i am not trying to prove anyone wrong here,but more so wanting to know if i am wrong.

So as i do not mix up Luc's word's,i will copy and paste his exact statements here.

Claim-->:Here is a video demo I made in December to demonstrate that the cost of power to create a magnetic field using an electromagnet is exactly the cost of wire resistance.
This is not coincidence but an exact science.Please present your argument if you feel this is incorrect.


So,dose it take energy/power to create a magnetic field ?,or is Luc correct when he says the magnetic field is created for free,and is backed by exact science?.

Now,i know that it takes no energy to maintain a magnetic field,and the I/R losses add up to be equal to the power being delivered to the coil once peak current flow through the coil has been reached,but i am interested only in the creation of the magnetic field-as Luc has stated.

I would be interested in hearing what others have to say.


Brad


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Yes, it takes no energy to maintain magnetic field (or current) in an ideal coil.
It takes energy to create a magnetic field in an ideal coil, but in such coil all of that energy can be recovered back.
It takes energy to create a magnetic field in a resistive coil, but in such coil not all of that energy can be recovered back.

Power is lower if the creation rate is slower, even if the final energy is the same.

This graph of energies and powers always applies to RL crcuits.
1) For charging coils in an RL circuit, the time on the horizontal axis runs forward,
2) For discharging coils in an RL circuit - the time on the horizontal axis runs backwards.
3) For zero resistance circuits - the time stands still ...and everything becomes constant.

The speed of the time is determined by the time constant Tau=L/R.

P.S.
The magnitude of the magnetic flux (Φ) is not shown on the graph but it can be calculated at any time by using the simple relationship Φ=i*L



   

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It takes energy to create a magnetic field in a resistive coil, but in such coil not all of that energy can be recovered back.

Power is lower if the creation rate is slower, even if the final energy is the same.

Thanks for your answer verpies.
I removed the part about an ideal coil,as we will never be working with ideal coils.


Anymore takers?.


Brad


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...
I removed the part about an ideal coil,as we will never be working with ideal coils.
...

Not sure...  we are not far from superconductors at room temperature.


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It's not as complicated as it may seem...
I think Luc's statement requires elaboration to make it more clear. Also, I would question that Luc is saying the field is created for free.
   
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It takes energy to create a magnetic field in an ideal coil, but in such coil all of that energy can be recovered back.

Verpies reply is very close to what I'm saying, just in a different way.

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?

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

No more debates are needed, it's now confirmed a magnetic field is free and it's only wire resistance that keep us away from this reality, plus the limitations (maybe purposeful?) of our technology at this time that prevents us from using this great phenomenon.
Who knows what else we'll discover once average experimenter like us have True super conductive wire to play with.

Regards
Luc
   
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Luc:  "it's now confirmed a magnetic field is free and it's only wire resistance that keep us away from this reality,"

    Sorry, Luc, I have a hard time believing this.  The B field stores energy, therefore one would say that energy should be required to produce it, beyond the wire resistance (loss). 
   So how is it
"it's now confirmed a magnetic field is free and it's only wire resistance that keep us away from this reality," 
How confirmed?
   
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Verpies confirmed it.

It takes energy to create a magnetic field in an ideal coil, but in such coil all of that energy can be recovered back.

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.
If we take the energy back the same way we put the energy in then the magnetic field time is double.

Regards
Luc

   

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Luc:  "it's now confirmed a magnetic field is free and it's only wire resistance that keep us away from this reality,"

    Sorry, Luc, I have a hard time believing this.  The B field stores energy, therefore one would say that energy should be required to produce it, beyond the wire resistance (loss). 
   So how is it
"it's now confirmed a magnetic field is free and it's only wire resistance that keep us away from this reality," 
How confirmed?

Ah,that is the comment i was waiting for-->the B field stores energy.

So,if it takes no energy to create the field,but the field will deliver energy when it collapses around the coil that created it,then we now have the !!creation!! of energy  :o

it's now confirmed a magnetic field is free

Where is this confirmation ?.


Brad.


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, plus the limitations (maybe purposeful?) of our technology at this time that prevents us from using this great phenomenon.
Who knows what else we'll discover once average experimenter like us have True super conductive wire to play with.

Regards
Luc

Quote
If the coil has no resistance then what ever power you put into it you can fully recover

If the coil has no resistance,then it has no inductance.
How is P/in calculated when R=0 ?.

Quote
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

You cannot create a magnetic field while the inductor is in a super conductive state.

Quote
No more debates are needed, it's now confirmed a magnetic field is free and it's only wire resistance that keep us away from this reality

I would like to see this confirmation Luc,as my tests i carried out last night clearly show that it takes energy to create the magnetic field.

In saying that,there may be a way of doing what you are trying to achieve--i will give it some more thought  O0


Brad


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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.

Brad

I guess you did not watch this video I posted on the first page of my topic.
In this video they demonstrate what happens when you have current circulating in a Real super conductor.
They demonstrate the Voltage falls to 0.000
So if the voltage is 0.000 how can there be any power dissipated across the coil? ... please explain where the power comes from.

https://www.youtube.com/watch?v=PifnlZigp6c

Cheers mate
Luc
   

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You cannot create a magnetic field while the inductor is in a super conductive state.
I would like to see this confirmation Luc,as my tests i carried out last night clearly show that it takes energy to create the magnetic field.

Brad

I think he may mean 'how much NET energy does it take to create a magnetic field?'
There is indeed a defined amount of energy required to establish the field, but the field collapse returns nearly all of this back into the system.

A perfect inductor and perfect capacitor would exchange energy forever without losses.
A nearly perfect inductor and capacitor will exchange energy with an extremely high Q factor.


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I guess you did not watch this video I posted on the first page of my topic.
In this video they demonstrate what happens when you have current circulating in a Real super conductor.
They demonstrate the Voltage falls to 0.000


https://www.youtube.com/watch?v=PifnlZigp6c

Cheers mate
Luc

Yes,i watched the video Luc,but i dont think you paid attention to what they did in that video.
What they showed was a complete waste of time and effort.

They clearly state at 5:40 in the video-->to charge the device,we use a 20 amp 5v power supply. Once connected to the circuit,we then submerge the device in liquid nitrogen.

So,it is exactly as i said it is.
First the coil has to be energised,and then you submerge it in liquid nitrogen.

Quote
So if the voltage is 0.000 how can there be any power dissipated across the coil? ... please explain where the power comes from.

But the voltage was not 0 when they charged the coil,as the coil was charged before being submerged in liquid nitrogen,and becoming super conductive. So,the power to charge the coil would have been the maximum power that there power supply could have delivered to the coil before it was made super conductive.

The video shows exactly what i said--you cannot charge the inductor once it becomes super conductive,as it will reject any magnetic field that tries to penetrate it.

Even if the voltage was 0 across the coil,it will never be 0 within the transformer of the power supply.
So even if the super conductor dose not dissipate any power,the power supply being used to power up the super conductor will,and it will dissipate the maximum amount of power it is capable of,because you just dropped a dead short across it.

The other video you posted,where the guy has a coil under some flat bar,which is sitting on the scales,clearly shows a massive increase in current draw when he cools the coil with liquid nitrogen to bring the resistance down. Before he cools the coil,there is no arcing of the wire when he touches the coil on the battery. When he cools the coil with liquid nitrogen to reduce the coils resistance,and then connects it to the battery,you can see massive arcing of the wire,which clearly means a massive increase of power input to the coil,as we could assume that the batteries voltage would remain fairly stable between the two tests.

Brad


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I think he may mean 'how much NET energy does it take to create a magnetic field?'
There is indeed a defined amount of energy required to establish the field, but the field collapse returns nearly all of this back into the system.

A perfect inductor and perfect capacitor would exchange energy forever without losses.
A nearly perfect inductor and capacitor will exchange energy with an extremely high Q factor.

And once you want to draw energy out of this perfect system,what happens ?

I fail to see any OU in this as Luc is claiming.

Brad


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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.
Please explain.

   
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Yes,i watched the video Luc,but i dont think you paid attention to what they did in that video.
What they showed was a complete waste of time and effort.

They clearly state at 5:40 in the video-->to charge the device,we use a 20 amp 5v power supply. Once connected to the circuit,we then submerge the device in liquid nitrogen.

That does not mean they applied 5v @ 20 amps. They are just confirming the size of power supply used. Also, they would of had to connect the circuit to the coil before placing it in the Liquid Nitrogen. So that statement is not saying the energized the coil before placing it in the Nitrogen. That would be pointless to do such a thing,

So,it is exactly as i said it is.
First the coil has to be energised,and then you submerge it in liquid nitrogen.

That is your saying. Nowhere did they say they energized the coil before placing it in the Nitrogen.

The other video you posted,where the guy has a coil under some flat bar,which is sitting on the scales,clearly shows a massive increase in current draw when he cools the coil with liquid nitrogen to bring the resistance down. Before he cools the coil,there is no arcing of the wire when he touches the coil on the battery. When he cools the coil with liquid nitrogen to reduce the coils resistance,and then connects it to the battery,you can see massive arcing of the wire,which clearly means a massive increase of power input to the coil,as we could assume that the batteries voltage would remain fairly stable between the two tests.

Brad

I did write that his video is not suited for any power evaluation. However, did you not consider that the large spark can be the now large inductive discharge?

Cheers
Luc
   

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Please explain.

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


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 author=gotoluc link=topic=3723.msg72573#msg72573 date=1550279495]




Quote
I did write that his video is not suited for any power evaluation. However, did you not consider that the large spark can be the now large inductive discharge?

Of course the discharge will be greater,as the input power increase a lot as well.

Quote
That does not mean they applied 5v @ 20 amps. They are just confirming the size of power supply used. Also, they would of had to connect the circuit to the coil before placing it in the Liquid Nitrogen. So that statement is not saying the energized the coil before placing it in the Nitrogen. That would be pointless to do such a thing,

That is your saying. Nowhere did they say they energized the coil before placing it in the Nitrogen.

Lets say your right.
We can already get an LED to flash using a normal coil when we disconnect the current.

I do agree that a super conductor will not dissipate any power,but you continually fail to take into account the power that the supply will be dissipating as heat. So how can any OU be had from this system if power is being dissipated by circuit components ?.

Brad


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The link below starts 3:44
Hit pause then click on wheel (settings) and select speed at 0.5 to slow it down.
Hit play and you can see he charges the coil when in Nitrogen. You will see a blue light when he powers the coil, then immediately when the blue light goes out you will see the LED flash, discharging the coil. He repeats the twice and each time it does the same thing.

https://youtu.be/PifnlZigp6c?t=224

Regards
Luc
   
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Lets say your right.
We can already get an LED to flash using a normal coil when we disconnect the current.

Correct but at a loss because you now agree that a super conductor will not dissipate any power... and that's all I'm bringing to your attention.

I do agree that a super conductor will not dissipate any power,but you continually fail to take into account the power that the supply will be dissipating as heat. So how can any OU be had from this system if power is being dissipated by circuit components ?.

Brad

True, power supplies loaded down to the millivolt range may not be efficient and why I mentioned 2500 Farad capacitors could be used. You can charge those in the millivolt range and they will deliver gobs of current efficiently.
I don't think there is anyone here that has any working experience with super conducting coils, so we still have no real knowledge what is truly needed to charge them.

Regards
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The link below starts 3:44
Hit pause then click on wheel (settings) and select speed at 0.5 to slow it down.
Hit play and you can see he charges the coil when in Nitrogen. You will see a blue light when he powers the coil, then immediately when the blue light goes out you will see the LED flash, discharging the coil. He repeats the twice and each time it does the same thing.

https://youtu.be/PifnlZigp6c?t=224

Regards
Luc

Ok,so he charges up the coil,and then disconnects the coil,and the LED flashes.

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

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


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Correct but at a loss because you now agree that a super conductor will not dissipate any power... and that's all I'm bringing to your attention.

True, power supplies loaded down to the millivolt range may not be efficient and why I mentioned 2500 Farad capacitors could be used. You can charge those in the millivolt range and they will deliver gobs of current efficiently.
I don't think there is anyone here that has any working experience with super conducting coils, so we still have no real knowledge what is truly needed to charge them.

Regards
Luc

Yes,i have always agreed that a superconductor will dissipate no power.
But the thread is about you claim that the magnetic field comes for free,where i state that it dose not.

Think about what your claim means--
If it takes no energy to create the field,but you can get energy back from that collapsing field,then you have just done what physics has told us we cannot do--create energy.

With an ideal power supply,and an ideal coil,the best we could do is 100%.
If we go over this 100%,then we now start to create energy--the thing we are told we cannot do.


Brad


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   Sorry, Luc, I have a hard time believing this.  The B field stores energy, therefore one would say that energy should be required to produce it, beyond the wire resistance (loss). 
   So how is it
it's now confirmed a magnetic field is free and it's only wire resistance that keep us away from this reality

How confirmed?

Let's look at it in another way.
We have a one Dollar bill which represents energy and we have a bottle of Coke which represents a magnetic field.
You go to a vending machine and insert your one Dollar bill and it dispenses you a Coke plus returns 4  x .25 cent quarters.
So you walk away with a Coke and one Dollar in quarters.
Was the Coke free?

Regards
Luc
   

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How confirmed?


Let's look at it in another way.
We have a one Dollar bill which represents energy and we have a bottle of Coke which represents a magnetic field.
You go to a vending machine and insert your one Dollar bill and it dispenses you a Coke plus returns 4  x .25 cent quarters.
So you walk away with a Coke and one Dollar in quarters.
Was the Coke free?

Regards
Luc

No,as you dissipated energy walking to the vending machine,and inserting and collecting money. :D

But here is something i want you all to think about.

If a superconductor has !!no!! resistance,it also has no inductance.
If it had inductance,then it would produce a BEMF.
If it produced a BEMF,then it would have resistance,as BEMF is an effective resistance to current flow.

If there is no inductance,then there is no collapsing magnetic field from which to capture the inductive kickback energy from.

The conservation of energy stands  O0


Brad


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Ok,so he charges up the coil,and then disconnects the coil,and the LED flashes.

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

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

Well, hopefully a 0ohm super coil has inductance/impedance so that an input could be applied in pulses without killing off the input to a dead short, etc. ;D Like if the super coil were used as a gen coil, would there be a voltage across that coil when we turn the gen??? Could there be a voltage across the coil? 

 Had some discussion some years back on 0ohm coils and their inductance. Would a perfect coil even allow current to flow ever? Like the magnet floating above a super conductor, could there be that sort of effect in a super coil and no current could flow?  Dont know because Ive never played with one, but my theory was that it cant pass current because the reverse emf of the self inductance, which impedes the input may be always equal to the input. ???  And I theorized that there may have to be some resistance in order for the reverse emf to always be lower than the input to allow for current and the mag field to rise. :o   I visualize the possibility of a chilled coil such as Luc is going to give a go, may have less voltage than at room temp. Maybe not. Hopefully not.  But it should provide more current, for sure.

Lol, just a thought from what I just wrote. The floating magnet on the super conductor...  If its floating and we push down on the magnet and it springs back upward against the push downwards, what if a super coil could somehow produce more reverse emf than the input. :o  That would be something.  Just connect the coil to the battery and the battery charges.  Would be sweet if it were that simple. ;)

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