Faraday‘s law of induction integral form

Question

If we want to derive the integral form of Faraday‘s law of induction from $$\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t} $$ we get with Stokes: $$\int_{\partial A}(\vec{E}+\vec{v}\times\vec{B})\:\mathrm{d}\vec{r}=-\frac{\mathrm{d}}{\mathrm{d} t}\int_A \vec{B}\:\mathrm{d} \vec{A}$$ But in many textbooks we find: $$U_{ind}= -\frac{\mathrm{d}}{\mathrm{d} t}\int_A \vec{B}\:\mathrm{d} \vec{A} $$ where $$U_{ind}:=\int \vec{E} \:\mathrm{d}\vec{r} $$ So my question is now if they referring always to a static surface or is the general expression for $U_{ind}$: $$U_{ind}= \int_{\partial A}(\vec{E}+\vec{v}\times\vec{B})\:\mathrm{d}\vec{r} $$ Which would make more sense according to the lorentz force.

Answer 1

Transformer EMF is defined as the EMF generated due to an induced electric field about some closed curve

Which is equal to,

$\int \vec{E} \cdot \vec{dl}=-\iint \frac{\partial \vec{B}}{\partial t} \cdot \vec{da}$

This is only the same as,

$\int \vec{E} \cdot \vec{dl}=-\frac{d}{dt}\iint \vec{B} \cdot \vec{da}$

Provided the surface enclosing your chosen curve does NOT change in time

In the below formula, if $\vec{da}$ IS changing in time, this formula gives a non zero emf, whereas the former, DOES NOT.

This discrepancy between these two formulas for when the surface IS changing in time is due to "Motional EMF"

Which is what you've talked about, which is an emf due to the magnetic lorentz force. This fact can be derived from faradays law, and the maxwell Faraday equation.

I think the confusion stems from what you define $U_{ind}$, they meant TRANSFORMER emf, but you mean the total induced emf. OR, this book meant total emf but were describing a situation where the closed curve is not changing with time.