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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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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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- #3

cnh1995

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This discussion might help.

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So: is there a non zero electric field inside the cable?

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marcusl

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"SC" means "superconductor"? "AC" means "alternate current"?A SC has AC impedance so it develops a voltage across the circuit (SC coil, for example).

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?The electric field inside the bulk material is still zero beyond a thin penetration depth.

This you have written is the self induced voltage, not that the applied one with a battery, is it correct?BTW, your formula for voltage is incorrect. It should be [tex]V=-L\frac{dI}{dt}[/tex]

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marcusl

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Yes."SC" means "superconductor"? "AC" means "alternate current"?

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.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?

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.This you have written is the self induced voltage, not that the applied one with a battery, is it correct?

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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?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.

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?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.

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marcusl

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True for AC (not true for DC, of course)The applied voltage at the SC ends is different than zero,

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.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?

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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?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.

Thank you.

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marcusl

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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

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

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- #15

marcusl

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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: [tex]\vec{E}=\Lambda \frac{\partial\vec J}{\partial t}[/tex]

where [itex]\Lambda[/itex] 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.

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Looking up in the net I found for the first London equation:

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: [tex]\vec{E}=\Lambda \frac{\partial\vec J}{\partial t}[/tex]

where [itex]\Lambda[/itex] 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.

∂j/∂t = n

where 'm' is the electron's mass, 'e' its charge in absolute modulus and n

Thanks.

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marcusl

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

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

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marcusl

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That's correct--it's nonsense.

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Daz

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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.

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Thanks to both for the answers.

Regards,

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Regards,

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