Violation in superconductivity

[spidey]

How electrons flow forever in a superconducting wire?
From where electrons getting kinetic energy?
Is this perpetual motion? if so, can we extract work and build a perpetual motion machine?

 

[blindnz]

The electrons flow forever because there is zero resistance, the kinetic energy is effectively the potential you apply to the superconductor. From then on there is no resistance so there is no loss.

Think of it like water in a circular pipe (cliche) if you add water into a pipe that had no resistance. The energy for the movement is the water you added in and since there is nothing disrupting this flow all the energy is conserved so the movement continues forever.

 

[granpa]

why do you think that they travel forever?

 

[cubeleg]

In a superconductor there is no losses of electric energy, so once you provide into a coil a energy it is trapped and accumulated in current. Of course there are losses, due to several factors, so the current won't flow forever. But in a perfect superconductor the current will stay forever.

About the question of extracting work of this, I would say that it is not possible. For example an electric engine based in superconducting coils will be more efficient than a copper based one, due to the absent of resistive losses. But there are other losses besides the resistive, for examples losses in the ferromagnetic. Also if you extract some energy from the coil (by rotating another coil) it will loss energy.
Hope this help.

 

[spidey]

why do you think that they travel forever?

because of zero resistance..

 

[spidey]

I got it..Thanks blindnz & cubeleg..

 

[Enthalpy]

If superconductors have any resistance, it is damned low. Measurements have been made to a huge accuracy (with superconducting hyperfrequency cavities) and no resistance has been detected. Coils are used routinely and nobody observes a resistance. So to any experimentally meaningful accuracy, zero resistance.

 

[granpa]

I don't know of any proof that current flows forever or even for a short time after the emf is removed. the persistence of the magnetic field around the coil is probably due to the meissner effect.

 

[marcusl]

I don't know of any proof that current flows forever or even for a short time after the emf is removed. the persistence of the magnetic field around the coil is probably due to the meissner effect.

The current in superconducting magnets (such as are used for MRI machines) is routinely maintained for years after the initial drive is removed.

 

[granpa]

how do they know that the current is still flowing?

 

[ZapperZ]

how do they know that the current is still flowing?

By measuring the induced magnetic field!

Zz.

 

[granpa]

the magnetic field wouldn't be expected to collapse (even if the current dies) due to the meissner effect.

 

[ZapperZ]

the magnetic field wouldn't be expected to collapse (even if the current dies) due to the meissner effect.

What Meissner effect? No one is subjecting the superconductor to an external field. So no Meissner effect.

The "induced" magnetic field is due to the persistent current in the superconductor. Haven't you use a clip-on ammeter before? Same concept.

Zz.

 

[Vanadium 50]

the magnetic field wouldn't be expected to collapse (even if the current dies) due to the meissner effect.

Like Zapperz said, there is no external magnetic field, so there is no Meissner effect.

There are at least two other problems with this. One is that to generate a magnetic field there has to be a current somewhere. The only two possibilities are in individual atoms or in the bulk flow of electrons - i.e. what we call a current. If you decide it's not in the current, it must be in the atoms - but the problem with that is that the exact same atoms are in the exact same configuration when the superconductor is slightly above Tc as slightly below.

The other is that the Meissner effect essentially states that superconductors are perfect diamagnets. This idea of a magnetic field unsupported by currents is not a property of diamagnets. So it doesn't have these properties to begin with. (And, like ZapperZ said, the conditions aren't right for it to apply anyway)

 

[granpa]

the magnetic field is obviousely created by a current. but the current dies and the miessner effect prevents it from collapsing. the field is sustained by the spin of the electrons.

unless you believe in perpetual motion.

 

[ZapperZ]

the magnetic field is obviousely created by a current. but the current dies and the miessner effect prevents it from collapsing. the field is sustained by the spin of the electrons.

unless you believe in perpetual motion.

This is incorrect.

The persistent current does continue because there is nothing to dissipate the supercurrent, by definition. It is not "perpetual motion" because as soon as you try to use it to do work (i.e. power a motor), you are using it up and it will start to dissipate.

Again, there's no Meissner effect. These is the supercurrent. The magnetic field generate by the supercurrent is external to the superconductor, not inside the superconductor.

Zz.

 

[granpa]

The magnetic field generate by the supercurrent is external to the superconductor, not inside the superconductor.

Zz.

yes. so? what does that have to do with anything?

 

[Vanadium 50]

This is no more perpetual motion than "a body in motion remains in motion" is. As ZapperZ points out, if you try and extract energy from the system, the current dissipates.

By having the electron spins generate the field in your theory, you are essentially saying a superconductor is a ferromagnet. One problem with this is that magnetic order and superconductivity are at odds with each other: for years it was thought they were mutually exclusive, and even now our exceptions are rare and exotic: UGe2 under pressure, for example. (It is also worth noting it is neither a good ferromagnet nor a good superconductor)

The other problem is that superconductors are perfect diamagnets - that's what your Meissner effect is saying. Ferromagnets have their induced magnetism pointing in the other direction.

 

[granpa]

suppose I put a wire inside a tube of regular conductor and then bend the whole thing into a ring. then I create a regular current in the outer conductor and after it is established I lower the temp of the inner wire to the point that it bocomes supercondicting. then I allow the current in the outer wire to stop. what happens to the magnetic field?

 

[Vanadium 50]

Again, there's no Meissner effect. These is the supercurrent. The magnetic field generate by the supercurrent is external to the superconductor, not inside the superconductor.

yes. so? what does that have to do with anything?

You just said it was generated by the electron spins inside the superconductor.

 

[ZapperZ]

yes. so? what does that have to do with anything?

It means that there's no Meissner effect! You've been invoking that, and we keep telling you why it isn't there! It has everything to do with what you are claiming.

Zz.

 

[granpa]

This is no more perpetual motion than "a body in motion remains in motion" is. As ZapperZ points out, if you try and extract energy from the system, the current dissipates.

By having the electron spins generate the field in your theory, you are essentially saying a superconductor is a ferromagnet. One problem with this is that magnetic order and superconductivity are at odds with each other: for years it was thought they were mutually exclusive, and even now our exceptions are rare and exotic: UGe2 under pressure, for example. (It is also worth noting it is neither a good ferromagnet nor a good superconductor)

The other problem is that superconductors are perfect diamagnets - that's what your Meissner effect is saying. Ferromagnets have their induced magnetism pointing in the other direction.

it is not saying that superconductors are ferromagnets. ferromagnetism is not the same as diamagnetism. and how on Earth is diamagnetism a problem.

ok. maybe the meissner effect is something slightly different. what I meant to say is that the superconductor is a perfect diamagnet so the field can't collapse.

 

[Vanadium 50]

suppose I put a wire inside a tube of regular conductor and then bend the whole thing into a ring. then I create a regular current in the outer conductor and after it is established I lower the temp of the inner wire to the point that it bocomes supercondicting. then I allow the current in the outer wire to stop. what happens to the magnetic field?

If you haven't exceeded Jc or Hc for the superconductor, you don't have to "allow the current in the outer wire to stop". It's already stopped, and the current has moved into the superconductor. (If I understand what you are describing correctly, you have essentially two paths in parallel, so the current will take the path with 0 resistance.)

You had a magnetic field before the inner core went superconducting and you have a magnetic field after.

 

[ZapperZ]

it is not saying that superconductors are ferromagnets. ferromagnetism is not the same as diamagnetism. and how on Earth is diamagnetism a problem.

ok. maybe the meissner effect is something slightly different. what I meant to say is that the superconductor is a perfect diamagnet so the field can't collapse.

This is getting sillier by the minute.

The fact that we see persistent current even after the potential is turned off should settle all question regarding the OP.

Zz.

 

[f95toli]

The Meissner effect is -by definition- when a superconductor exp ells an EXTERNAL field. It has nothing to do with the field generated BY a superconducting coil.
A superconducting electromagnet is essentially just an ordinary coil with the strength of the field being proportional to the amount of current you drive into it with the heat-switch open (all the usual formulas for the B field apply); once the switch is closed the current just keeps flowing.
If you hook up an multimeter to the coil in parallel to the part of the magnet wire that is driven normal when you open the heat switch you will see a current flowing through the circuit, but since part of the circuit is now resistive this current will decay with a time constant L/R.

Moreover, the fact that there is a current flowing in the coil is the reason why there is so much energy stored in superconducting magnets (LI^2/2, L being the inductance). If part of the wire is driven normal and there is nowhere to dump the energy the coil will start to heat up VERY quickly (this in known as a quench). Since most magnets are cooled by liquid helium this lead to the production of a LOT of gas and can quite literally result in an explosion.

 

[granpa]

You just said it was generated by the electron spins inside the superconductor.

thats almost totally incomprehensible. the currents wouldbe inside the superconductor too.

 

[granpa]

The Meissner effect is -by definition- when a superconductor exp ells an EXTERNAL field. It has nothing to do with the field generated BY a superconducting coil.
A superconducting electromagnet is essentially just an ordinary coil with the strength of the field being proportional to the amount of current you drive into it with the heat-switch open (all the usual formulas for the B field apply); once the switch is closed the current just keeps flowing.
If you hook up an multimeter to the coil in parallel to the part of the magnet wire that is driven normal when you open the heat switch you will see a current flowing through the circuit, but since part of the circuit is now resistive this current will decay with a time constant L/R.

Moreover, the fact that there is a current flowing in the coil is the reason why there is so much energy stored in superconducting magnets (LI^2/2, L being the inductance). If part of the wire is driven normal and there is nowhere to dump the energy the coil will start to heat up VERY quickly (this in known as a quench). Since most magnets are cooled by liquid helium this lead to the production of a LOT of gas and can quite literally result in an explosion.

the energy is stored in the magnetic field. where else would it be?

 

[f95toli]

thats almost totally incomprehensible. the currents wouldbe inside the superconductor too.

So?
Of course superconductors can carry currents; that is sort of the whole point.
The fact that there is no FIELD inside the superconductor (well, in type I superconductors) does not mean that a superconductor can't carry current.

 

[granpa]

This is getting sillier by the minute.

The fact that we see persistent current even after the potential is turned off should settle all question regarding the OP.

Zz.

and just how do you 'see' this current? I know of no direct evidence that such a current continues long after the potential is turned off.

 

[granpa]

So?
Of course superconductors can carry currents; that is sort of the whole point.
The fact that there is no FIELD inside the superconductor (well, in type I superconductors) does not mean that a superconductor can't carry current.

the field extends a short distance into the skin of the superconductor. the current also flows mainly along the skin.

 

[granpa]

If you haven't exceeded Jc or Hc for the superconductor, you don't have to "allow the current in the outer wire to stop". It's already stopped, and the current has moved into the superconductor. (If I understand what you are describing correctly, you have essentially two paths in parallel, so the current will take the path with 0 resistance.)

You had a magnetic field before the inner core went superconducting and you have a magnetic field after.

no. the inner wire and the outer tube are 2 different circuits.

 

[f95toli]

and just how do you 'see' this current? I know of no evidence that such a current continues long after the potential is turned off.

As I wrote above: Hook up a multimeter in parallel to the switch, open switch, observe current decay.

I do this on a pretty regular basis since I work with superconducting magnets and have built a few of the myself (low field ones, with a simple heat switch). The test above is what I use to test the quality of the joint where I crimp the two ends of the wire that makes up the coil together. My joints are pretty bad compared to what you find in commercial magnets (they are welded) but even in my coils the current does not change much (a few percent) over a period of a few hours, the only reason why it drops at all is because of the joint (which is a "weak" superconductor, some of the supercurrent is probably carried by Josepshon current and is not completely dissipationless)

 

[ZapperZ]

and just how do you 'see' this current? I know of no evidence that such a current continues long after the potential is turned off.

Holy cow! When I asked you about the clamp on ammeter, you have no clue what I was asking?

I-can-dectect-the-current-by-measuring-the-magnetic-field-produced-by-the-supercurrent!

Which part of that do you not understand?

Furthermore, I can show you tunneling between two identical superconductors where at zero potential bias, there is a sharp peak in the conductance due to Josephson tunneling, i.e. the tunneling of the supercurrent across the junction even when there's no net bias across the junction. This is evidence that the persistent supercurrent can occur without any potential bias.

http://arxiv.org/PS_cache/cond-mat/pdf/9809/9809398v1.pdf

This is something that I have measured myself, not just something I read.

Zz.

 

[granpa]

As I wrote above: Hook up a multimeter in parallel to the switch, open switch, observe current decay.

I do this on a pretty regular basis since I work with superconducting magnets and have built a few of the myself (low field ones, with a simple heat switch). The test above is what I use to test the quality of the joint where I crimp the two ends of the wire that makes up the coil together. My joints are pretty bad compared to what you find in commercial magnets (they are welded) but even in my coils the current does not change much (a few percent) over a period of a few hours, the only reason why it drops at all is because of the joint (which is a "weak" superconductor, some of the supercurrent is probably carried by Josepshon current and is not completely dissipationless)

a voltage meter? if the magnetic field doesn't collapse then I wouldn't expect to see any voltage. where would the energy come from?

 

[granpa]

I-can-dectect-the-current-by-measuring-the-magnetic-field-produced-by-the-supercurrent!

Zz.

the magnetic field can be supported without a current by means of the electron spin. what part of that do you not understand.

 

[ZapperZ]

the magnetic field can be supported without a current by means of the electron spin. what part of that do you not understand.

And you have to show me the origin of the "electron spin" that can produce THAT high of a magnetic field. Calculate the total number of electrons in a supercurrent, and tell me you can get THAT high of a magnetic field produced in those magnets that we use at the LHC or RHIC.

That's absurd!

Not to mention, the Cooper Pairs in most superconductors have ZERO net magnetic moment since they form singlet pairing state!

Zz.

 

[f95toli]

Also, you would need to explain why the field generated by these "spins" is proportional the current injected into the solenoid (BEFORE the switch is closed, meaning the solenoid is just an electromagnet with zero dc-resistance, part of a circuit with an external power supply), if these were two different effects (ordinary electromagnet-electron spin) you would expect the field to change when the switch is closed and the solenoid is put into persistent current mode (and it doesn't).

This is incidentally a test first performed by Kammerlingh-Onnes almost 100 years ago.

 

[granpa]

And you have to show me the origin of the "electron spin" that can produce THAT high of a magnetic field. Calculate the total number of electrons in a supercurrent, and tell me you can get THAT high of a magnetic field produced in those magnets that we use at the LHC or RHIC.

That's absurd!

Not to mention, the Cooper Pairs in most superconductors have ZERO net magnetic moment since they form singlet pairing state!

Zz.

I am assuming that it isn't the cooper pairs that produce the field. it would be the material itself.

 

[granpa]

Also, you would need to explain why the field generated by these "spins" is proportional the current injected into the solenoid (BEFORE the switch is closed, meaning the solenoid is just an electromagnet with zero dc-resistance, part of a circuit with an external power supply), if these were two different effects (ordinary electromagnet-electron spin) you would expect the field to change when the switch is closed and the solenoid is put into persistent current mode (and it doesn't).

This is incidentally a test first performed by Kammerlingh-Onnes almost 100 years ago.

the spin is proportional to the field that tries to collapse.

 

[ZapperZ]

I am assuming that it isn't the cooper pairs that produce the field. it would be the material itself.

Then tell me why this occurs ONLY IN THE SUPERCONDUCTING STATE, and not in the normal state, i.e. you can't get any persistent current (and the resulting induced magnetic field) in the normal state!

I still want you to calculate the total magnetic moment of ALL the electrons in the material and tell me you can get anything above 1 Tesla.

Zz.

 

[granpa]

Then tell me why this occurs ONLY IN THE SUPERCONDUCTING STATE, and not in the normal state, i.e. you can't get any persistent current (and the resulting induced magnetic field) in the normal state!
.

because superconductivity=superdiamagnetism. its the diamagnetism that drives the electrons to form cooper pairs.

 

[granpa]

As I wrote above: Hook up a multimeter in parallel to the switch, open switch, observe current decay.

I do this on a pretty regular basis since I work with superconducting magnets and have built a few of the myself (low field ones, with a simple heat switch). The test above is what I use to test the quality of the joint where I crimp the two ends of the wire that makes up the coil together. My joints are pretty bad compared to what you find in commercial magnets (they are welded) but even in my coils the current does not change much (a few percent) over a period of a few hours, the only reason why it drops at all is because of the joint (which is a "weak" superconductor, some of the supercurrent is probably carried by Josepshon current and is not completely dissipationless)

I think I misunderstood what you were saying. yes you will observe a voltage when you open the switch but that's because the magnetic field is collapsing through the opening.

 

[ZapperZ]

because superconductivity=superdiamagnetism. its the diamagnetism that drives the electrons to form cooper pairs.

Whaaat?!

You just told me that this has nothing to do with the Cooper pairs. Now you are using THEM to justify the presence of the magnetic field!

Please make valid citations to back up ALL of your claims here, because I think you know what's coming next.

Zz.

 

[f95toli]

I think I misunderstood what you were saying. yes you will observe a voltage when you open the switch but that's because the magnetic field is collapsing through the opening.

Why would I measure the voltage? It is much better to use the multimeter as an ammeter and measure the current directly instead.
And, as I wrote above, what I (and everyone else) observe is that the current coming out from the solenoid is initially (it obviously decays, but pretty slowly if the initial current is a couple of amps and a good multimeter is used) just a few percent lower than the current I injected into the solenoid a few hours earlier.

Also, you DO realize that you are essentially proposing a new theory for conventional superconductivity? The BCS theory is perhaps the best verified theory there is in condensed matter physics, so you will find it very difficult to persuade people that it is wrong.
Both me and ZapperZ and have spent many years working with superconductors so it is not like we don't know anything anything about the topic.

 

[granpa]

Whaaat?!

You just told me that this has nothing to do with the Cooper pairs. Now you are using THEM to justify the presence of the magnetic field!

Please make valid citations to back up ALL of your claims here, because I think you know what's coming next.

Zz.

you're playing it a little fast arent you? I said just the opposite.

 

[granpa]

Why would I measure the voltage? It is much better to use the multimeter as an ammeter and measure the current directly instead.
And, as I wrote above, what I (and everyone else) observe is that the current coming out from the solenoid is initially (it obviously decays, but pretty slowly if the initial current is a couple of amps and a good multimeter is used) just a few percent lower than the current I injected into the solenoid a few hours earlier.

Also, you DO realize that you are essentially proposing a new theory for conventional superconductivity? The BCS theory is perhaps the best verified theory there is in condensed matter physics, so you will find it very difficult to persuade people that it is wrong.
Both me and ZapperZ and have spent many years working with superconductors so it is not like we don't know anything anything about the topic.

you can't put an ammeter is parrallel with a closed superconducting ring and measure anything. ammeters have internal resistance. no current would flow through it.I very much doubt that there is any conflict with BCS theory.

 

[ZapperZ]

you're playing it a little fast arent you? I said just the opposite.

Who can tell anymore since none of what you said made any sense, nor backed by any physics that we know. For instance, do you also deny you said this:

its the diamagnetism that drives the electrons to form cooper pairs.

I would like you, before you do anything else, show me valid citation to back this claim. This will be the last request I will ask. Show me how this doesn't conflict with BCS theory.

Zz.

 

[granpa]

Who can tell anymore since none of what you said made any sense, nor backed by any physics that we know. For instance, do you also deny you said this:
I would like you, before you do anything else, show me valid citation to back this claim. This will be the last request I will ask. Show me how this doesn't conflict with BCS theory.

Zz.

what is diamagnetism? its the tendency for the electrons in the material to orient themselves in such a way that they cancel out any field present at that point. its exactly what you would expect 2 electrons to do. they orient themselves so that they cancel out each others field.

it is upon you to show that there is any conflict.

finally: I am not trying to prove anything to anybody. you can believe anything you want. I really don't care.

 

[ZapperZ]

what is diamagnetism? its the tendency for the electrons in the material to orient themselves in such a way that they cancel out any field present at that point. its exactly what you would expect 2 electrons to do. they orient themselves so that they cancel out each others field.

This is nonsense. If that is true, then ALL metal can become superconductors. Yet, this is false and that only certain metals (and ceramics) can become superconductors. This clearly implies that the drive into superconductivity is governed by something BEYOND just the electron spin. You are also ignorant of the fact that the Ruthenates have spin TRIPLET state where the electrons have spin alligned PARALLEL to each other. This has falsified your faulty understanding of superconductivity.

I think I've read enough nonsense here. If the OP still has a question, he/she can PM me and we'll reopen this thread. Till then, this is done.

Zz.