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I'm trying to wrap my head around Faraday and Lenz's law.

Mathematically it implies that

$$\oint_{\partial s} (\vec E +\vec v \times \vec B) = \frac{d \phi}{dt}$$ where $$ \phi = \int \int \vec B \cdot\vec {dA}.$$ But aren't magnetic fields supposed to have zero flux? So isn't $\phi$ supposed to be zero?

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  • $\begingroup$ @Drjh , OMG, it is really alarming that your comment and two answers contain the same mistake. The term with the magnetic field must be there, if the right-hand side of the equation contains the time derivative of the flux. $\endgroup$ Commented Feb 3, 2021 at 15:39

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First, Faraday's Law is $$ \mathcal{E} = - \frac{d}{dt} \Phi$$ (with a minus sign).

Second, the EMF is defined as $$ \mathcal{E} = \int_{\mathcal{C}} \mathbf{f}\,d\mathbf{s}$$ where $\mathbf{f}$ is the force per unit of charge. What you have written is correct in the case where the only force is the electromagnetic force. The term with $\mathbf{v}\times\mathbf{B}$ will be perpendicular to the path for wires (or if your path follows a linear current), since it's perpendicular to velocity.

Third, as they've already stated in comments, magnetic flux is $0$ only for closed surfaces. In general, flux is not $0$ for any surface. Think of the magnetic field of a solenoid and chose the surface to be just inside it: all the field lines will pass through the surface in the same direction. By the way, in closed surfaces, you won't be able to apply Faraday's Law since the surfaces have to be bounded, as you have written in your integral.

Forth, you didn't ask explicitly but Lenz's Law is contained on the minus sign of Faraday's Law. It just says that the induced EFM will induce a magnetic field which goes against the change in flux.

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  • $\begingroup$ The term ${\bf v}\times{\bf B}$ cannot be dropped. There is no reason the velocity is always aligned with ${\mathrm d}{\bf l}$. The magnetic term is part of the EMF. $\endgroup$ Commented Feb 3, 2021 at 15:34
  • $\begingroup$ Your are right. I was assuming a current in a wire. I'll correct it, thanks. $\endgroup$
    – Gilgamesh
    Commented Feb 3, 2021 at 19:35

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