Current in a superconducting circuit

[Kiara]

This may seem a stupid question, but I can't find information anywhere else. What equation does one use to calculate the current present in a superconducting circuit, as I=V/R fails?

 

[sophiecentaur]

In practice, the source of the current will have some resistance. This will introduce a finite limiting value for your Current.

 

[Kiara]

makes sense. so tell me if I've got this right, or if I'm dreadfully wrong:
since the resistance of copper is 16.78 nΩ·m at room temperature, I can take an ordinary 9V battery to obtain a current of 536 million amps, run the copper wire to a superconducting circuit, and keep that current of 536 million amps? it sounds kinda ridiculous...

 

[Vanadium 50]

You are making at least three errors:

• You are using resistivity instead of resistance.
• You are ignoring the internal resistance of your 9V battery.
• You can only "keep that current" until your battery dies. With a typical 9V battery, if you could really get 500 MA out of it, this would be a microsecond or two.

 

[K^2]

V = IR still holds for superconductor. Voltage across superconductor is always zero. You cannot physically apply a voltage differential across it.

 

[willem2]

V = IR still holds for superconductor. Voltage across superconductor is always zero. You cannot physically apply a voltage differential across it.

You can still have a voltage diffferential because of Inductance

$$V = L \frac {dI}{dt}$$

 

[DrDu]

makes sense. so tell me if I've got this right, or if I'm dreadfully wrong:
since the resistance of copper is 16.78 nΩ·m at room temperature, I can take an ordinary 9V battery to obtain a current of 536 million amps, run the copper wire to a superconducting circuit, and keep that current of 536 million amps? it sounds kinda ridiculous...

In practice, there is an upper limit on the current in superconductors. When you surpass it, the superconductor will become a normal conductor.

 

[willem2]

V = IR still holds for superconductor. Voltage across superconductor is always zero. You cannot physically apply a voltage differential across it.

You can still have a voltage diffferential because of Inductance

$$V = L \frac {dI}{dt}$$

 

[K^2]

I'm not sure that defining inductance for a superconductor loop is that trivial. Meisner effect is going to prevent magnetic field in the superconductor. That means an additional surface current that counters the field. And I'm not sure what that's going to do to the total field from the coil, and subsequently, inductance.

 

[Phrak]

I'm not sure that defining inductance for a superconductor loop is that trivial. Meisner effect is going to prevent magnetic field in the superconductor. That means an additional surface current that counters the field. And I'm not sure what that's going to do to the total field from the coil, and subsequently, inductance.

No, the Meissner effect is the explusion of magnetic flux from the interior of the superconductor media. A single loop superconducting ring, for instance, has total flux passing through the loop proportional the the current. The applied voltage is, of course, zero for a closed ring.

 

[Born2bwire]

No, the Meissner effect is the explusion of magnetic flux from the interior of the superconductor media. A single loop superconducting ring, for instance, has total flux passing through the loop proportional the the current. The applied voltage is, of course, zero for a closed ring.

Yeah. As I recall, SQUIDs are interesting devices that rely on the ability for a superconducting loop to respond to an externally applied magnetic field. So if you want to read up on an application and behavior of this take a look into these devices.

 

[Vanadium 50]

Getting back to the original question, DrDu is right - there is an upper limit to the current (actually, the current density), and beyond that, the superconductor becomes normal again. However, if you are really talking about a battery, you'll have maybe an ohm of internal resistance. For 9V, that means that the battery can't possibly source more than 9A. That is, by far, the most important factor.

 

[ZapperZ]

I may be missing something here, but isn't this rather, er, trivial? Why can't one just measure the induced magnetic field in a closed loop around a superconducting wire segment? Isn't that how one measures the current using a loop ammeter?

Zz.

 

[K^2]

No, the Meissner effect is the explusion of magnetic flux from the interior of the superconductor media. A single loop superconducting ring, for instance, has total flux passing through the loop proportional the the current. The applied voltage is, of course, zero for a closed ring.

Right, but to expel magnetic flux from conductor itself, there is going to be an eddy current in the conductor. It isn't obvious to me why that eddy current isn't going to affect the magnetic field generated by the applied current.

 

[f95toli]

I'm not sure that defining inductance for a superconductor loop is that trivial. Meisner effect is going to prevent magnetic field in the superconductor. That means an additional surface current that counters the field. And I'm not sure what that's going to do to the total field from the coil, and subsequently, inductance.

It is more or less that trivial. The total inductance of a superconducting loop will in most cases (well below Tc, Jc etc) be very close to the geometric inductance. There is an additional contribution from the kinetic inductance but that will typically only be a percent or two of the total L (unless there is a Josephson junction in the loop as in the case of SQUIDs, but the Josephson inductance is not usually considered to be part of the inductance of the loop itself).
Also, the Meissner effect has nothing to do with it; the kinetic inductance is -in the two fluid model- related to the relative amount of quasiparticles in the bulk, which it is related to the temperature (the kinetic inductance goes like sqrt(1-(T/Tc)^2)))

When I design superconducting circuits (SQUIDs and microwave circuits) I nearly always just use normal simulation packages; in most cases my simulations agree very well with experiments even if I completely neglect the kinetic inductance.

 

[K^2]

No, I wouldn't expect kinetic inductance to matter a whole lot here either. That's not what I'm thinking about.

Consider just a straight superconducting wire with another wire in parallel to it. If I run current through the first wire, it will induce magnetic field at location of the second wire. Meisner effect will result in eddie currents through that other wire. These will make field inside second wire zero, but also distort magnetic field around that wire. Will that not affect inductance of the first wire? It seems like it should. It may be a minor contribution, but I don't see any obvious way to show that. Now I close these two wires at one end, and I have a loop now. Same logic should apply. Meisner effect in one half of the loop will affect current in the other half. So why does this not have any (significant) impact on the inductance?

 

[f95toli]

The current distribution in a superconducting wire is obviously different from that of an ordinary metallic wire; this is important at high frequencies where almost all the current will flow at the corners if you have a nearly-rectangular conductor (which is typically the case for a SQUID, since they are patterned from thin films). So yes, from that respect I guess one could say that there is an effect. But this is not that different from a very good normal conductor where the current will be concentrated at the surface due to the skin effect (which in turn effects the inductance even for a metallic inductor, note that the geometric inductance is frequency dependent even for a loop made from a normal conductor); so the end result is that the inductance is more or less just the usual geometric inductance.

When I do my calculations I usually assume that the conductance is about 100 times that of gold which seems to work quite welll; this is a phenomenological parameter that takes into account surface roughness, impurities etc which are important at the frequencies I am working at (4-8 GHz).

 

[K^2]

Hm. Fair enough.

 

[Phrak]

Right, but to expel magnetic flux from conductor itself, there is going to be an eddy current in the conductor. It isn't obvious to me why that eddy current isn't going to affect the magnetic field generated by the applied current.

This may be true, but for a given ring of superconductive material, the total flux remains proportional the current. The details depend upon the super conductor; type I or type II. Type I materials allow no penetration of flux, but in type II materials, lowered through the critical temperature, flux is pinned in the material. It's trapped. These regions are called flux vortexes. Where there is a vortex there is a circulation of current. The critical temperature has a softer knee than type I materials, and the eye of the vortex, as far as I know, is no longer superconductive, thus the admission of flux, so the answer as to whether or not a superconductor admits flux kinda depends upon how you ask it.

 

[cabraham]

This may seem a stupid question, but I can't find information anywhere else. What equation does one use to calculate the current present in a superconducting circuit, as I=V/R fails?

Fails? How so? Ohm's law simply states that V = I*R. In a superconductor, R = 0, so that I*R is zero, meaning that V = 0. No failure there. That is, of course, as long as I is finite. In superconductivity, there is a limit for current before the conductor everts to normal non-super conductivity. As long as that limit is not exceeded, R = 0, I*R = 0, & V = 0.

No conflict there at all. As far as "I = V/R" goes, since V is zero, & R is zero, you have a "0/0" expression, which is indeterminate. The equation must be placed in a determinate form. Then results can be obtained.

Claude

 

[Bob S]

The current distribution in a superconducting wire is obviously different from that of an ordinary metallic wire; this is important at high frequencies where almost all the current will flow at the corners if you have a nearly-rectangular conductor (which is typically the case for a SQUID, since they are patterned from thin films). So yes, from that respect I guess one could say that there is an effect. But this is not that different from a very good normal conductor where the current will be concentrated at the surface due to the skin effect (which in turn effects the inductance even for a metallic inductor, note that the geometric inductance is frequency dependent even for a loop made from a normal conductor); so the end result is that the inductance is more or less just the usual geometric inductance.

The inductance of a superconducting coil is similar to an identical ordinary coil at high frequencies (when skin effect is important), but there is no consequent ac loss equivalent to the eddy current losses at high frequencies in normal conductors, when the currents are concentrated at the surface. .

The attached thumbnail shows inductance measurements made on a 7-meter long superconducting magnet used in the Fermilab Tevatron. The frequency range covers 10 Hz to 5000 Hz. The two sets of points were made at room temperature (top points) and at 4.2 kelvin (bottom points). The ≈ 3 mH difference in inductance is due to the magnetic flux exclusion from the superconducting coil in the magnet.

AC loss measurements were made at the same time as the inductance measurements, and they were the same; no change from room temperature to superconducting. The loss of inductance in both measurements at frequencies above ≈ 50 Hz were due to eddy currents in the non-magnetic stainless steel cryostat.

Bob S

 

[Kiara]

whoa. lots of facts.

but referring back to my original problem... I'm trying to build a small superconducting electromagnet. If I want only 2600 A to flow through the magnet, then how do I 1) supply current and 2) monitor it? I was considering that if a 1.5 V battery had an amp-hour rating of 2.5 A/h, then it could theoretically discharge 9000 A in one second (actual will be less of course), which would be more than enough to supply the needed current. the excess current would then be dealt with with a circuit breaker or resistors... is this sound?

 

[Bob S]

but referring back to my original problem... I'm trying to build a small superconducting electromagnet. If I want only 2600 A to flow through the magnet, then how do I 1) supply current and 2) monitor it? I was considering that if a 1.5 V battery had an amp-hour rating of 2.5 A/h, then it could theoretically discharge 9000 A in one second (actual will be less of course), which would be more than enough to supply the needed current. the excess current would then be dealt with with a circuit breaker or resistors... is this sound?

Many superconducting magnets run in the persistent mode; the superconducting magnet leads are shorted together and the (persistent) current continues flowing in the magnet and slowly decays with time L/R constants of hours, days, or months. The circuit used to "pump up" the circulating current is called a flux pump. Here is one description:

http://www.google.com/url?sa=t&sour...x_z2DQ&usg=AFQjCNH9yzd3bOe8cN5Wp8fmea8E-Wi8rw

This technique is used to maintain the current in MRI magnets, for example. The magnetic energy stored in the magnet is ½LI², and if a 1.5 volt battery is used to pump up the current, the pumping rate is dI/dt = V/L. So if for example the magnet inductance is 15 mHy (milliHenrys) and the battery voltage is 1.5 volts, then dI/dt = 100 amps per second. So about 26 seconds would be required to reach 2600 amps.

[added] Here is another article on flux pumps

http://www.google.com/url?sa=t&sour...k4z3DQ&usg=AFQjCNFW45Oxhb0fu_d0AmTkCMk9wwsfNw

Bob S

 

[Vanadium 50]

whoa. lots of facts.

You should read them carefully.

I was considering that if a 1.5 V battery had an amp-hour rating of 2.5 A/h, then it could theoretically discharge 9000 A in one second (actual will be less of course), which would be more than enough to supply the needed current.

Before we go to superconductivity, let's consider regular electronics. A physical 9V battery looks like an ideal 9V battery with a 1 Ohm resistor in series. That's what we mean by internal resistance. What is the maximum current such a device can source?