# Electric field inside perfect conductor with inductance

A constant voltage V is applied at the two ends of a cable which is a perfect conductor (e.g. a superconductor). Since the cable, anyway, has a non zero inductance L, the current inside it is not infinite but it is proportional to time t: i(t) = V/L * t.
In this situation, how much it is the electric field E inside the cable? Is it zero anyway? If it is, how can I reconcile it with the fact there is a finite voltage V at its ends?

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Maybe my was a naive question but just a bit help?

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Sorry, but I didn't get much help from that discussion because there no one mentioned the impedance of the superconducting cable. We have: V = Z*I where the voltage V across the superconducting cable, its impedance Z and the current I inside of it are complex numbers. Since Z is non zero (at least because of its inductance: Z = iωL) V is non zero too.
So: is there a non zero electric field inside the cable?

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marcusl
Gold Member
A SC has AC impedance so it develops a voltage across the circuit (SC coil, for example). The electric field inside the bulk material is still zero beyond a thin penetration depth. BTW, your formula for voltage is incorrect. It should be $$V=-L\frac{dI}{dt}$$

A SC has AC impedance so it develops a voltage across the circuit (SC coil, for example).
"SC" means "superconductor"? "AC" means "alternate current"?
The electric field inside the bulk material is still zero beyond a thin penetration depth.
But I have to grasp this: it's zero because it's the sum of the applied voltage (with a battery, e.g.) and the self-induced voltage with opposite sign?
BTW, your formula for voltage is incorrect. It should be $$V=-L\frac{dI}{dt}$$
This you have written is the self induced voltage, not that the applied one with a battery, is it correct?

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marcusl
Gold Member
"SC" means "superconductor"? "AC" means "alternate current"?
Yes.
But I have to grasp this: it's zero because it's the sum of the applied voltage (with a battery, e.g.) and the self-induced voltage with opposite sign?
No. A changing current through a SC coil produces a voltage given by the equation I wrote, due to the coil's inductance. In this regard, SC and copper coils behave similarly. Inside of the SC material (wire) itself, however, the field is zero for the same reason that it is zero inside of any perfect conductor: charges arrange themselves to exactly cancel any externally applied fields.
This you have written is the self induced voltage, not that the applied one with a battery, is it correct?
Yes, but they are the same in this case since losses in a SC are (essentially) zero. Integrating this equation allows you to find the time-varying current in terms of a time-varying voltage, so that your equation requires an integral to be correct.

A changing current through a SC coil produces a voltage given by the equation I wrote, due to the coil's inductance. In this regard, SC and copper coils behave similarly. Inside of the SC material (wire) itself, however, the field is zero for the same reason that it is zero inside of any perfect conductor: charges arrange themselves to exactly cancel any externally applied fields.
The applied voltage at the SC ends is different than zero, but the field E inside of it (and so the line integral of E along a path taken inside the SC) is zero. This because E is not conservative in this case, is it correct?
Yes, but they are the same in this case since losses in a SC are (essentially) zero. Integrating this equation allows you to find the time-varying current in terms of a time-varying voltage, so that your equation requires an integral to be correct.
If I apply a voltage V to the SC ends with a device which keeps V constant and we can neglect SC losses, where can we consider applied, along the circuit, the self induced voltage?

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marcusl
Gold Member
The applied voltage at the SC ends is different than zero,
True for AC (not true for DC, of course)
but the field E inside of it (and so the line integral of E along a path taken inside the SC) is zero. This because E is not conservative in this case, is it correct?If I apply a voltage V to the SC ends with a device which keeps V constant and we can neglect SC losses, where can we consider applied, along the circuit, the self induced voltage?
The field is non-zero on the surface of the conductor, for an alternating current, so the self-induced voltage is the integral of the surface fields.

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True for AC (not true for DC, of course) The field is non-zero on the surface of the conductor, for an alternating current, so the self-induced voltage is the integral of the surface fields.
The case I mean to discuss is that of a constant voltage V applied to a cable with zero resistance and zero capacitive impedance during the transitory. If the voltage generator ha zero internal resistance (let' neglect other kinds of impedance too), the current inside the cable is not constant: it increases linearly with time; if the voltage generator has a non zero internal resistance Ri (neglecting other forms of impedance) the current increases (at the beginning, almost linearly with time) till it reaches a constant value V/Ri, but what happens, however, at the first instants of times? Here the current it's certainly not continuous, but I don't even define it as "alternate". What happens here?
Thank you.

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marcusl
Gold Member
In this case you have a simple L-R circuit, where all of the resistance R is internal to the source. This is commonly solved with a differential equation (time domain) or Laplace transform (frequency domain). See Wikipedia "RL circuit," for instance.

In this case you have a simple L-R circuit, where all of the resistance R is internal to the source. This is commonly solved with a differential equation (time domain) or Laplace transform (frequency domain). See Wikipedia "RL circuit," for instance.
I have already solved it in the first post. My question is still the same: How much it is the electric field E inside the SC cable in this situation? If it's zero, why, since the applied voltage at its ends is non zero?

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marcusl
Gold Member
My answer is the same: E inside the bulk SC material is zero. Fields are excluded from the bulk material, even at DC, by the Meissner effect, which is unique to superconductors.The E field, and hence voltage, is supported on the surface within a small penetration depth, often called the London penetration depth.

The London penetration depth shouldn't refer to magnetic field only?

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marcusl
Gold Member
No, the electric and magnetic fields are related through the vector potential, which is the real driver of behavior in the SC.

We can examine London's first equation to see what happens if one tries to establish a voltage, hence apply an E field, along an SC: $$\vec{E}=\Lambda \frac{\partial\vec J}{\partial t}$$
where $\Lambda$ is a scale factor. Since a constant voltage quickly ramps the current to infinity, we conclude that one cannot impose a voltage across the bulk part of an SC wire.

No, the electric and magnetic fields are related through the vector potential, which is the real driver of behavior in the SC.

We can examine London's first equation to see what happens if one tries to establish a voltage, hence apply an E field, along an SC: $$\vec{E}=\Lambda \frac{\partial\vec J}{\partial t}$$
where $\Lambda$ is a scale factor. Since a constant voltage quickly ramps the current to infinity, we conclude that one cannot impose a voltage across the bulk part of an SC wire.
Looking up in the net I found for the first London equation:
∂j/∂t = nse2E/m
where 'm' is the electron's mass, 'e' its charge in absolute modulus and ns the number of superconducting carriers per unit volume. To make a little computation I need an approximate value of ns (an order of magnitude would be enough). Do you have any idea?
Thanks.

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marcusl
Gold Member
You have written 1/Λ. The value of Λ is approximately 2e-21, in SI units.

You have written 1/Λ. The value of Λ is approximately 2e-21, in SI units.
So, with a potential difference of 1V in a 1m of SC cable and so an electric field E = 1V/m, we would have, in theory, approximately 500,000 A of current in 10^(-15) s, independently if the cable is wrapped to form a solenoid, if the electric field were present inside of the cable?

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marcusl
Gold Member
That's correct--it's nonsense.

Daz
Gold Member
lightarrow: try googling “SC magnet controller” or “SC magnet charging time”. You’ll see plenty of examples where superconducting solenoids have a measurable, macroscopic DC voltage across their (superconducting) terminals for extended periods of time (hours or even days.)

Of course, no one doubts that the field inside the superconductor is zero. It would be nice to see a detailed analysis, but my guess is it’s all down to the back EMF induced by the time-varying magnetic field. I suppose that in this regard there is no practical difference between the situation you describe and the ideal inductor we study in basic electricity.

Thanks to both for the answers.
Regards,

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