# Violation in superconductivity

You just said it was generated by the electron spins inside the superconductor.
thats almost totally incomprehensible. the currents wouldbe inside the superconductor too.

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
Gold Member
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.

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.

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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.

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
Gold Member
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
Staff Emeritus
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.

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 doesnt collapse then I wouldnt expect to see any voltage. where would the enegy come from?

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
Staff Emeritus
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
Gold Member
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.

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 isnt the cooper pairs that produce the field. it would be the material itself.

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
Staff Emeritus
I am assuming that it isnt 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.

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.

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 thats because the magnetic field is collapsing through the opening.

ZapperZ
Staff Emeritus
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
Gold Member
I think I misunderstood what you were saying. yes you will observe a voltage when you open the switch but thats 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.

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.

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 cant 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
Staff Emeritus
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.

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 dont care.

ZapperZ
Staff Emeritus