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Consider a rectangular circuit with one side being a rod of length $L$, which is free to move, and a constant magnetic field perpendicular to it. The rod moves at a constant velocity $v$, expanding the area of the circuit (this system). Let there be some resistor $R$ in the circuit.

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One can use Faraday's law to calculate the current $I$ from the E.M.F induced by the changing magnetic flux. But the same result follows from Lorentz force considerations, ignoring Faraday:

  • In a stationary configuration (assuming constant current), the total force on the charge carriers in the rod, in the direction along its axis, must vanish. This gives $vB=E$.

  • We deduce that the electric field is constant along the rod, and so there is a voltage $\mathcal E=EL=BLv$. We can calculate the current $I$ and get the same result as we would have gotten using Faraday's law.

  1. Does this mean that there is some way to deduce Faraday's law from the other Maxwell equations plus the Lorentz force? If so, how can we show this generally?

  2. Or maybe this can be done only when there are no time-varying magnetic fields?

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I've done the math, and will answer my own question.

It is important to distinguish between Faraday's law of induction, $\mathcal E=- \frac{d\phi_B}{dt}$ and the Maxwell–Faraday equation, $\nabla \times E=-\frac{\partial B}{\partial t}$. The latter is not a consequence of the Lorentz force, while the former is, in some cases.

The change in magnetic flux can be due to a time-varying magnetic field, or due to a time-varying area (as in the example cited in the question).

In the latter case (e.g the moving rod), the Maxwell–Faraday equation is irrelevant, as $\frac{\partial B}{\partial t}=0$. We can, however, use Lorentz-force considerations to deduce Faraday's law of induction for these cases. However, we cannot derive Faraday's law of induction as a general result using only Lorentz-force considerations. If we consider a static loop of wire, in the background of a time-varying magnetic field, Lorentz-force considerations do not imply a current. The Maxwell–Faraday equation is a fundamentally new result.

For more information on the mathematical derivation, see here.

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    $\begingroup$ Nice answer. This is discussed at great length in Griffiths Chapter 7 (perhaps this is what you're reading). Indeed, as Griffiths points out, that Faraday's Law should have two ostensibly different "reasons" was one of the motivations for Einstein to develop SR (see Griffiths Chapter 12). $\endgroup$
    – EE18
    Feb 7, 2023 at 17:07
  • $\begingroup$ @EE18 Thanks! I'll read it :) $\endgroup$
    – Rd Basha
    Feb 7, 2023 at 17:08

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