Let us be a bit more specific:

In a single open wire there will not be any current, since there is no closed loop (which is a fairy tale for children: If you switch on the H-field very fast, there will be a very transient electric field inducing some compensating current in some parts of the wire; this is a high frequency problem and details will depend strongly on geometry).

If we consider a closed wire, first we have to determine a surface integral of ANY (whichever you like) simply connected surface, that has the wire as its boundary, integrating BVec Scalarproduct SurfacenormalVec * surfaceElement.

This gives the magnetic flux passing through the surface.

If this flux changes in time, either because your B-field (or H-field) changes itself, or due to mechanical movement of the field against the loop, you will introduce an electric field along the wire.

The integral of this E-field along the wire is the electric voltage.

The current you are looking for depends on the total resistance of the wire, simply by Ohms law (fairy tale for children: if there is a homogeneous current density across the wire section, which its not true even for relaxed frequencies and wires not really thick).

So: Constant field, infinitely extended, constant velocity => No current?

At a first glance yes, at the second no:

We missed the second term: vVec x BVec in the Lorentz-force driving the charges in the loop. Consider a e.g. circular loop moving through a homogeneous H-Field, i.e. B-Field, perpendicular to the circle's normal. This produces a Lorentz-force, which is in some part of the circle tangential, somewhere perpendicular, somewhere reverse tangential, somewhere perpendicular again. Ooops, that cancels out.

But not if the B-field only partially fills the circles cross section. Now you get a net force, a net E-field (force divided by elementary charge, and a net current). You absorb electric power, you heat up your wire, and you need some force to push the circle through the B-field. Roughly thats the reason, why eddy current brakes work as they do.

Hope that helps. Best regards. Hans-Walter

Yes, it does. A time varying magnetic field is always accompanied by a spatially-varying, non-conservative electric field and vice-versa (Faraday's law). Therefore, how a magnetic field will interact with an electric wire or a pair of parallel electric wires will produce an induced electromotive force and current.

But what exactly is a "moving magnetic field"? If you mean simply a time-varying field then the answer is encoded in Maxwell equations, especially in their differential form.

But if your idea is related to magnetic monopoles, then - sorry - no such object has been detected until today. The somewhat artificial objects called so and observed in spin-ice materials are not the true monopoles.

Thank you for all the answers.

to Marek : of course, I know that a time varying magnetic field generates an electric fiels. What I mean is really a moving field. Imagine a permanent magnet moving parrallely to an elongated wire.

If one has a magnetic field with no electric field then a subsequent boost or change of frame will result in an electric field. If there are stationary charges present that are free to move these electric fields will then cause current.

Thanks. How can I estimate the current, depending on the speed, the magnetic field's amplitude, the distance to the wires ?

Well, the net E field in a good conductor is very nearly zero. The E field comes from relativity or in other words the standard (v x B) term (you need to check units etc) the ohms law gives the reverse E field the rest you'll have to work out based on your problem.

Now I understand! So let's go to your figure. In my opinion the arrow symbolizing electric current is incorrect. Think about the not moving cross section of the wire, perpendicular to velocity vector. While magnet is moving, the flux through this area will change, first increasing then (after the magnet passes) decreasing. This will create an electric field in cross section plane, forming closed lines. Consequently, the current will flow in this plane too. This means the estimate should consider, in the first place, the magnetic flux and its changes. The speed of its change will depend on the distance between the magnet and a wire, and on magnet's strength. The flux nevertheless will not be uniform and so will be the current density. To complicate things even more, we should be aware of the additional field created by the induce current. In short: the best estimate will be obtained from experimental data rather than from calculations. The idea is already realized in stationary (training) bikes - as magnetic brake. It's effectiveness is easily controlled by the distance between permanent magnets and a rotating, conductive wheel.

I suggest looking at things in the rest frame of the magnet. Two equal and opposite currents; one from the fixed atoms and one from in the opposite direction from the valence electrons. The E field experienced by the electrons due to -q(v x B) in the magnet frame will generate a current flow component along (v x B) in proportion to the conductivity of the wire. The current component of the fixed atoms remains unchanged because the atoms are fixed. By dealing with this in the magnet frame one has time independent B fields and a much simpler physical picture.

Thank you for all your answers. My idea was the following ; imagine we put wires inside a road (longitudinal or transverse wires? to be decided) and stick permanent magnets under cars moving along this road. Then if the movement of the cars generates a current in the wires, we should be able to store this current and use it elsewhere (for de-icing or any other needs related or not to the road). I guess there will be a force applied to the cars that will tend to slow them down, but it may be negligible...

Is this correct?

Fabrice,

comparing your idea to magnetic breaks in training bike one can see that noticeable effects appear only when the distance between magnet(s) and a flying wheel is small enough, i.e. certainly less than 1 cm.  I don't think such a distance would be practical in vehicles on the road.  But you can also consider this situation from energetic point of view.  What power (per square meter) is needed to keep the road ice free at temperature, say, -5C?  Assume 100% efficiency and some realistic car density on the road and you will see what a fraction of their engines' power will be required to melt the ice.

If you consider the automobiles to be conductive elements, then their motion through a magnetic field would create an EMF proportional to their conductivity and their velocity, and back EMF as well. The question then arises as to how their discontinuity would affect the resulting EMF (that is, they will not act as a continuous conductive element). Hence, one might suggets that they might generate a discontinuous EMF with interims proportional to the spacing of the vehicles.