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Moving a magnet close to a conductor induces a current. If it consists of a superconducting material with resistance $R=0$, then my textbook says:

Then the induced current will continue to flow even after the induced emf has disappeared.

This makes sense physically - there is no resistance to stop charge flow. But then the book draws this conclusion:

Thanks to this persistent current, it turns out that the flux through the loop is exactly the same as it was before the magnet started to move, so the flux through a loop of zero resistance never changes.

If the flux $\Phi$ never changes in a superconductor, from Faraday's law this means - from what I have learned - that no electromotive force $\mathcal{E}$ is induced:

$$\mathcal{E}=-\frac{\mathrm{d} \Phi}{\mathrm{d}t}=0 \:\:\:\:\text{ when }\frac{\mathrm{d} \Phi}{\mathrm{d}t}=0$$

My conclusion is therefor: There would never be induced any current at all. Current can never be induced in a superconductor loop. Is this the case or am I misunderstanding my book?

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  • $\begingroup$ How that is Ampere's law?thats i think Faraday's law? $\endgroup$ – Paul Jan 14 '15 at 17:01
  • $\begingroup$ @Paul It is Faraday's law, not Ampere's law. I have edited the question to address that. $\endgroup$ – Dave Coffman Jan 14 '15 at 17:06
  • $\begingroup$ Out of curiosity, what textbook are you using? $\endgroup$ – Dave Coffman Jan 14 '15 at 17:09
  • $\begingroup$ It is Faradays law, yes. Sorry for mistyping. @DaveCoffman my textbook is University Physics by Young and Freedman, 13th edition. A great book, but this part was unclear. $\endgroup$ – Steeven Jan 14 '15 at 18:15
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My conclusion is therefor: There would never be induced any current at all. Current can never be induced in a superconductor loop. Is this the case or am I misunderstanding my book?

One cannot conclude this for the reasons you have given; it does not follow from the fact that the total flux does not change that the current cannot change.

For a non-zero, finite resistance, there must be an emf to sustain a circulating current. However, for the case of zero resistance, there can be a current without emf (zero resistance) and further, a changing current.

Since the net flux through the surface bounded by the superconducting loop is just the magnetic flux due to the magnet plus the magnetic flux due to the current through the superconductor,

$$\Phi = \Phi_m + \Phi_i$$

imposing the condition

$$\frac{d\Phi}{dt} = 0$$

implies that

$$\frac{d\Phi_m}{dt} = -\frac{d\Phi_i}{dt} $$

Thus, if the magnet is moved, causing $\frac{d\Phi_m}{dt} \ne 0$ then, it must be the case that $\frac{d\Phi_i}{dt} \ne 0$, i.e., that the current circulating through the superconductor is changing.

Since the net flux is not changing, there is no emf around the loop enclosing the surface. Nonetheless, since the resistance is zero, the current is independent of the emf and so one cannot conclude that the current is zero or unchanging.

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  • $\begingroup$ Since the net flux is not changing, there is no emf around the loop enclosing the surface.- but Faraday's Law that is $$\varepsilon = -\frac{d\Phi}{dt}\;.$$ only talks of the flux due to the moving magnet & not due to the induced current, isn't it? $\endgroup$ – user36790 Dec 13 '15 at 14:44
  • $\begingroup$ @user36790, the right hand side is the (negative) time rate of change of magnetic flux period, i.e., the actual (net, total) magnetic flux. Consider, for example, coupled inductors. $\endgroup$ – Alfred Centauri Dec 13 '15 at 15:27
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Electromotive force $\mathcal{E}$ can be interpreted as the unit thermodynamic work $\mathrm{d}W$ that as to be performed by the energy source to move a unit charge $\mathrm{d}q$ through one loop move, namely : $$ \mathcal{E}=\frac{\mathrm{d}W}{\mathrm{d}q} $$ In usual conductors, the Ohm's law $\;\textbf{j}=\sigma\textbf{E}\;$ ensures that such work will be dissipated. This implies that a non null emf $\mathcal{E}$ as to be applied constantly to maintain the charge flow through the conductor.

However, in a superconductor, the electric current is no more described by the usual Ohm's law since there is no more dissipation. Indeed, London equations gives you that : $$ \textbf{j}=-\frac{nq}{mc}\textbf{A} $$ where $\textbf{A}$ is the vector potential, $n$ the carriers density, $m$ the mass of charge carriers and $c$ the speed of light. In a supraconductor, current is only here to screen the magnetic field inside the material, not to dissipate energy. Then, you don't need anymore to maintain $\mathcal{E}\neq 0$ in order to move charges in the material (i.e. to have current).

To sum up :

  • In a conductor : $\mathcal{E}=0\;\rightarrow\;\textbf{j}=0$

  • In a superconductor : $\mathcal{E}=0\;\nrightarrow\;\textbf{j}=0$

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Feynman:

In a ''perfect conductor'' there is no resistance whatever to the current. So if currents are generated in it,they can keep going forever .In fact,the slightest emf would generate an arbitrarily large current -which really means that there can be no emf at all.Any attempt to make a magnetic flux go through such sheet generates currents that creat opposite B fields -all with infintesimal emf's,so with no flux entering. If we have a sheet of a perfect conductor and put an electromagnet next to it,when we turn on the current in the magnet ,currents called eddy currents appear in the sheet,so that no magnetic flux enters.

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