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Suppose we have a circular loop of wire, and we put a long perfect solenoid inside it which is connected to an AC voltage source so that the magnetic field inside it starts to vary by time, does this varying magnetic field induce an emf in our loop according to Faraday's law of induction? (suppose that the radius of the solenoid is much smaller than that of the circular loop)

Does this show some kind of non-locality like the Aharonov-Bohm effect? Or am I just misunderstanding the concept?

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You have some misunderstanding. Locality just means that a disturbance can only affect it nearest local space. Put it simply, it means that is the disturbance propagating at a finite speed outward.

In your case, there is EMF induced in the loop since the Faraday's law of induction states that the change of magnetic field creates a curl of electric field surrounding it. This disturbance, however, propagate at the speed of light so it is a completely local effect and it has nothing to do with non-locality. I am not quite sure about the AB effect, but the phase acquiring should not be related to non-locality since the potential used is usually static.

Edit: To clarify what I said, considering the integral form of the Faradays law: $$\int_{\partial\Sigma}\vec{E}\cdot d\ell=-\int_{\Sigma}\frac{\partial\vec{B}}{\partial t}\cdot dA$$ which means that the changing magnetic field in the center solenoid (even it ideal and infinitely long) create a non-zero intergrated electric field along any loops outside the solenoid. Though it only creates an electric field outside, but if you put an actual wire loop, the potential difference will drive the current through this wire. The energy is conserved because the current flow through this wire will create an exact opposite B-field of solenoid.

Using Faraday's law along, you would not come up with the finite propagating speed. You need all Maxwells equation to derive it.

A final word of the non-locality is that many classical theory are non-local. The Newtons law of gravitation is surely one of them. Another example, the diffusion is also non-local since the probability of a particle go to anywhere is never zero for any infinitesimal time $dt\to 0$.

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yeah that's right, but the magnetic flux is confined in a region of space because of the perfect solenoid, so any electric field generated would be confined in that region too. –  shakibv Jan 14 '13 at 9:23
    
No, even the magnetic field is confined, the electric field is not. Take a look of the integral form of Faraday's law, if you take a loop of wire outside, the induced EMF still equal to the change of magnetic flux inside the loop. It can be easily understood by the analogue of changing current in infinitely long wire with electric current, the current is confined, but the magnetic field is still currounding it. –  hwlau Jan 14 '13 at 9:32
    
@hwlau: To apply the integral form of Faraday's law, you have to prove it works in case of stepwise field dependence. If outside the solenoid the magnetic field is zero, its time derivative is zero too, so the electric field can only be "potential". Otherwise we have to admit a non zero propagating magnetic field too. –  Vladimir Kalitvianski Jan 14 '13 at 11:02
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But the magnetic field is zero where the conductor is. –  Larry Harson Jan 14 '13 at 11:42
    
@VladimirKalitvianski I have no idea what you are talking about. Electric field is more fundamental then charge, and it can certainly propagate with the speed of light. –  hwlau Jan 16 '13 at 19:14
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The magnetic field $\mathbf{B}$ outside the solenoid only can probably be different from zero due to radiation (the other Maxwell equations including boundary conditions on the solenoid surface). Otherwise the electric field is zero too.

The analogy with a long wire does not work because the main term responsible for the magnetic field is the current $\mathbf{j}$, not the time derivative of $\mathbf{E}$:$$\nabla \times \mathbf{B} = \frac{\partial \mathbf{E}}{\partial t} + \mathbf{j}$$

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I'm convinced that this phenomena may not be a non-local one (non-local in quantum mechanics context) but how can one justifies this? maybe by writing magnetic field as $\mathbf{B} = \nabla \times \mathbf{A}$ and then argue that the vector potential $\mathbf{A}$ is not just a purely mathematical construct, and giving it some physical credits (the magnetic field $\mathbf{B}$ is zero in the place of wire loop but the vector potential $\mathbf{A}$ is not) ? –  shakibv Jan 15 '13 at 6:43
    
@shakibv I did not understand your question. In QM we have a wave theory where every single point in space participates in creating the total wave; that's why all space points are important. Even when we use a boundary condition for our wave - it is a simplified solution of the total wave equation implying certain interactions. In this respect QM is very different from Classical Mechanics with local forces. –  Vladimir Kalitvianski Jan 16 '13 at 12:45
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