Just this question is only for me because I have given any clarifications on the Faraday-Neumann-Lenz for my students of an high school.
We know that the magnetic flux, where $\:S$ is a regolar surface, (with all the hypotesis of regularity, $B\in\mathcal C^1$, etc.), is given by the formula:
$$\Phi_S(\mathbf{B})=\int_S \mathbf{B}\cdot d\mathbf{S} \tag 1$$
Now, the emf inducted if given by the formula:
$$\mathcal E_{\text{ind}}=-\frac{d\Phi_S(\mathbf{B})}{dt}=-\frac{d\displaystyle\int_S \mathbf{B}\cdot d\mathbf{S} }{dt} \tag 2$$
If $S$ is a planar surface and $\mathbf{B}$ must be an uniform field magnetic, the $(1)$ become:
$$\mathcal E_{\text{ind}}=-\frac{d(BS\cos \alpha)}{dt}$$ where the $\alpha$ is the angle between $\mathbf{B}$ and $\mathbf{S}$ and $\mathcal E_{\text{ind}}=\mathcal E_{\text{ind}}(t)$. If $B$ is a function that not depends from the time $t$, therefore (of course),
$$\mathcal E_{\text{ind}}=0$$
If $B=B(t)$ we have:
$$\mathcal E_{\text{ind}}=-S\cos \alpha\: \frac{d B(t)}{dt} \tag 3$$ and I can find easily the induced electromotive force and the induced current.
How could I find mathematically the induced electromotive force (emf) and the induced current starting from the formula $(2)$?
$$\mathcal E_{\text{ind}}=-\frac{d\Phi_S(\mathbf{B})}{dt}=-\frac{d\displaystyle\int_S \mathbf{B}\cdot d\mathbf{S} }{dt} \tag 2$$
The standard thing to do here is to take the derivative "under the integral sign" and get $$\mathcal E_{\text{ind}} = - \int_S \frac{\partial\mathbf B}{\partial t}\cdot d\mathbf S.$$ At this juncture you might use Faraday's law (Maxwell's equations) to convert this to a flux integral of $\text{curl}\,\mathbf E$.