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Suppose we have a single open circuit wire, moving with speed $v$ inside an infinite and uniform magnetic field $B$. Wire has a length of $L$ and it moves perpendicular to magnetic field. In this case Lorentz force is used to calculate the EMF created in wire and it is: $$EMF=vBL$$

We know that this formula is derived using Lorentz force and also Lorentz force can be deduced from Maxwell-Faraday equation (explained here), so can we obtain the same above formula using Maxwell-Faraday equation?

$$∮_cE.dl=−\frac{d}{dt}∬_sB.ds$$

If yes, how ? Also here says that Faraday's law of induction can not be applied to a single wire:

The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux enclosed by the circuit.

This version of Faraday's law strictly holds only when the closed circuit is a loop of infinitely thin wire, and is invalid in other circumstances.

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  • $\begingroup$ What is Maxwell-Faraday equation? How would you derive Lorentz force from it? $\endgroup$
    – Ján Lalinský
    Jan 12, 2017 at 19:05
  • $\begingroup$ @JánLalinský Added one link and Maxwell-Faraday equation. $\endgroup$
    – user141851
    Jan 12, 2017 at 19:18

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yes, with can show it with Maxwell-faraday equation: $$∮_cE.dl= -\frac{d}{dt} ∬_sB.ds$$

We can suppose a contour with rectangular shape that our wire is one side of it. the wire side is moving and other side is constant so the area of contour is growing with V speed and if we write Maxwell-Faraday equation for this contour we have: enter image description here $$ emf=- \frac{d}{dt} ∬_sB.ds= -\frac{d}{dt} ∬_sB.dx.dy= -BL \frac{d}{dt} ∫dx=-BL \frac{d}{dt} (x)=-VBL $$

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  • $\begingroup$ If the width of the rectangular is fixed, $\frac{d}{dt}∫dx$ becomes zero. Therefore one side must be fixed and the other side (the wire) must be moving to have a time changing width. $\endgroup$
    – user141851
    Jan 13, 2017 at 8:37
  • $\begingroup$ yes your right my suppose was wrong so if we consider your correction then can show that the result is the same. $\endgroup$
    – afs_mp
    Jan 13, 2017 at 9:08

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