Violation in superconductivity

In summary, electrons flow forever in a superconducting wire because of zero resistance. The kinetic energy is effectively the potential applied to the superconductor, which allows for no loss in energy. This can be compared to adding water into a circular pipe with no resistance, creating a perpetual flow. However, in a perfect superconductor, there are other losses besides resistance that prevent the current from flowing forever. The current can be maintained for years due to the persistent current in superconducting magnets, which can be measured by the induced magnetic field. The Meissner effect is not applicable in this situation as there is no external magnetic field. Extracting energy from the system will cause the current to dissipate, making it not perpetual motion
  • #1
spidey
213
0
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?
 
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  • #2
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.
 
  • #3
why do you think that they travel forever?
 
  • #4
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.
 
  • #5
granpa said:
why do you think that they travel forever?

because of zero resistance..
 
  • #6
I got it..Thanks blindnz & cubeleg..
 
  • #7
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.
 
  • #8
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.
 
  • #9
granpa said:
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.
 
  • #10
how do they know that the current is still flowing?
 
  • #11
granpa said:
how do they know that the current is still flowing?

By measuring the induced magnetic field!

Zz.
 
  • #12
the magnetic field wouldn't be expected to collapse (even if the current dies) due to the meissner effect.
 
  • #13
granpa said:
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.
 
  • #14
granpa said:
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)
 
  • #15
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.
 
  • #16
granpa said:
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.
 
  • #17
ZapperZ said:
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?
 
  • #18
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.
 
  • #19
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?
 
  • #20
ZapperZ said:
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.

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

You just said it was generated by the electron spins inside the superconductor.
 
  • #21
granpa said:
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.
 
  • #22
Vanadium 50 said:
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.
 
  • #23
granpa said:
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.
 
  • #24
granpa said:
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.
 
  • #25
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.
 
  • #26
Vanadium 50 said:
You just said it was generated by the electron spins inside the superconductor.

thats almost totally incomprehensible. the currents wouldbe inside the superconductor too.
 
  • #27
f95toli said:
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?
 
  • #28
granpa said:
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.
 
  • #29
ZapperZ said:
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.
 
Last edited:
  • #30
f95toli said:
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.
 
  • #31
Vanadium 50 said:
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.
 
  • #32
granpa said:
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)
 
  • #33
granpa said:
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.
 
  • #34
f95toli said:
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?
 
  • #35
ZapperZ said:
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.
 

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