If a piece of metal of length $l$ is moving with a speed $v$ in a region where there is a uniform magnetic field $B$ perpendicular to it, there will be a potential difference across its terminals equal to $lvB$ which is known as motional EMF. This can be shown and understood in terms of magnetic and electric forces on the free charges in the metal.

How can one calculate such EMF from Faraday's law, $\displaystyle\mathcal{E} = \left|\frac{d\Phi_B}{dt}\right|$?

(where $\Phi_B$ is the magnetic flux $\int \bf{B}\cdot d\bf{a}$)

(If $B$ is not changing, then the change in the magnetic flux must be due to change in an area, but the area of what? What are the boundaries of this area?)

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    $\begingroup$ Minor point: the potential difference (integral of electric field) is not the same thing as emf; in a piece of metal, electromotive force is actually opposite to electric field. They have the same absolute values if the electric current does not flow, but they have opposite signs (the corresponding intensities cancel each other). $\endgroup$ – Ján Lalinský Jan 16 '14 at 3:27
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    $\begingroup$ You're describing a "homopolar generator". There is a discussion of how Faraday's law does (not) apply to homopolar generators at: en.wikipedia.org/wiki/… $\endgroup$ – Steve Byrnes Mar 17 '14 at 14:42
  • $\begingroup$ Related.physics.stackexchange.com/questions/146628/faradays-paradox $\endgroup$ – Paul Dec 24 '14 at 11:09

The general Faraday law of emf formulated with magnetic flux is meant mostly for closed circuits made of thin wire, which can be assigned area without problem.

For other situations, magnetic flux may not have sense. Moving piece of metal is still subject to magnetic electromotive intensity, but it has to be calculated for any point of the metal as $$ \mathbf E^* = \mathbf v \times \mathbf B $$ where $\mathbf v$ is the velocity of the metal element.

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  • $\begingroup$ You should define E*... $\endgroup$ – Steve Byrnes Mar 17 '14 at 14:42
  • $\begingroup$ $\mathbf E^*$ in general is electromotive intensity, i.e. force per unit charge that acts on the mobile charge and excites electric current. In this case, it is due to motion through magnetic field and is approximately given by the above expression. $\endgroup$ – Ján Lalinský Mar 17 '14 at 20:17

I think your question answers itself, indeed the area of what? Faraday's law is for a closed loop of wire, thus Faraday's law is inappropriate and we should look for an alternative, as you have done by considering the Lorentz force. If the metal were a closed loop of circumference $l$ then Faraday's law would be valid. The forces on the electrons from the top and the bottom would both act in the same direction but in opposite orientations around the loop and thus no EMF around. Also the force on the protons would be counter to that on the electrons and thus no overall force on the loop, thus we conclude that Faraday's Law saying that there would be no net EMF is correct.

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