# Can current create voltage?

1. Jun 16, 2012

### nemesiswes

First off do superconductors act just like normal inductors except for resistance, the reason is because I know they expel magnetic fields and I was wondering if that has an effect on the inductance of the coil? Like for the same size and turns as a normal coil, will the superconductor have less inductance but still produce the same magnetic flux as the normal coil?

Basically I know that in a superconducting coil that if it is shorted to itself, it can have a huge current flow for a very long time because the resistance is very low to none. Now I also know when you connect that loop to say drive a load, the magnetic flux which the loop created will create a voltage to drive current across the load (returning energy). This is basically a very large inductor but superconducting.

Now what happens if the superconducting coil was a bifilar coil which nullifies inductance, well almost. When you connect the superconducting coil to a load, there is none to very little magnetic flux to induce a voltage in the coil to drive current across the load? So will the current create a voltage somehow to drive the current across the load.

2. Jun 16, 2012

### f95toli

It is more or less the same as for a regular conductor. In superconductors (and some other types of materials, e.g. ballistic semiconductors) you also have something called kinetic inductance, this will increase the inductance of a superconducting wire by say a percent or so; i.e. it is a small effect.

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Yes, if you suddenly connect a current-carrying superconducting loop of any type to a resistive load,the current will induce a voltage across that load.

3. Jun 16, 2012

### nemesiswes

So the current itself will create a voltage? I always thought that it was the magnetic flux built up around the coil that induces the voltage and so I figured the same for superconductors.

I am talking about a bifilar superconductor in opposing mode where the magnetic flux cancels out so the inductance is decreased. Thus if there is no magnetic flux, then shouldn't there be no voltage induced?

Plus how would the current create a voltage across the load if in superconductors, there is no voltage in a shorted superconducting loop. In other words, there can be no voltage or the current would infinite (since 0 resistance) and since power is current x voltage, the power would be zero, the real energy is stored in the magnetic flux isn't it?

So with no magnetic flux in a bifilar, how can a voltage be induced when the superconducting bifilar is connected to a load? Or in other words, how can energy be returned when there was no energy in the magnetic flux because there was no magnetic flux?

If like you said "It is more or less the same as for a regular conductor" then shouldn't it obey the same laws?

Ah , I just had a thought, lol, were you talking about the kinetic inductance caused by the electrons themselves, and that is why the current itself even with no magnetic flux will still induce a voltage across the load. It may be small but it is still a voltage that can drive atleast some current through the load.

Last edited: Jun 16, 2012
4. Jun 16, 2012

### f95toli

Energy will -of course- be conserved. However, there is no such thing as a loop completely without inductance, even a straight wire has some (small) inductance associated with it.

And yes, the kinetic inductance is due to the Cooper pairs in the superconductor, it has nothing to do with the geometry so that inductance contribution will always be there.

Also, note that even small inductances can give rise to a very large voltage if the change in current is sudden enough.

5. Jun 16, 2012

### nemesiswes

I know there is no such thing as no inductance and also that even small inductance's can give rise to huge voltages if the rate of change is fast enough. I was more interested in the energy stored and if no magnetic flux, then no energy can be stored in the flux.

I am curious though, how big can the kinetic inductance be for a superconductor carrying large currents?

6. Jun 16, 2012

### f95toli

Calculating the kinetic inductance (Lk)is far from trivial. Lk depends on the type of superconductor you are using, the temperature (Mattis-Bardeen theory) and the geometry. For "straight" thin-film structures (coplanar waveguides and similar structures) the kinetic inducance will typically be 1-2 percent of the geometric inductance, but the fraction will of course go down i you start winding coils since Lk will only depend on the length of wire you use, whereas the geometric inductance depends on how you wind the wire.

Btw, the energy stored will of course still be the usual LI^2/2 (with L=Lg+Lk); but that has nothing to do with how large the voltage can be. Compare it to static electricity, the voltage can be huge (kV) but the energies involved are tiny.

7. Jun 16, 2012

### nemesiswes

I'm not that great with math so even if I wanted to calculate it, I probably couldn't or it would take me awhile, lol.

Thanks for all the help though.

8. Jun 16, 2012

### Bob S

Real superconducting magnets can have both real and reactive impedances. AC resistances are due to eddy current losses, which are proportional to frequency squared. The inductance at higher frequencies drops because eddy currents shield portions of the magnetic volume. Due to the Meissner effect, dc magnetic fields are excluded from the interior of superconductors. However, as soon as the superconductor becomes normal, the magnet inductance increases due to penetration of the magnetic field into the superconductor. See Fig. 5 on page 13 of attached pdf. The inductance of a cold Tevatron main dipole magnet was continuously measured as it warmed up. Suddenly when the temperature reached about 10 kelvin, the innductance at all frequencies increased by about 4 milliHenrys.

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