Brian Bi

In Faraday's law the derivative of the magnetic flux is a total or a partial derivative?

Faraday's law is a bit tricky. The differential form is the easiest to explain, but maybe not the easiest to understand:
$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$
That's a partial derivative.

Now, we can apply the Kelvin–Stokes theorem to rewrite it in an integral form. The Kelvin–Stokes theorem states:
$\oint_{\partial S(t)} \mathbf{A} \cdot d\mathbf{\ell} = \iint_{S(t)} \nabla \times \mathbf{A} \cdot d\mathbf{a}$
I think it's pretty clear that this doesn't alter the nature of the derivative:
$\oint_{\partial S(t)} \mathbf{E} \cdot d\mathbf{\ell} = -\iint_{S(t)} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{a}$

Here's the tricky part. The above equation is valid for any oriented surface $$S(t)$$ as a function of time. In the case that it doesn't move, we can cast it into a simple equivalent form using the Leibniz rule. Here's the form of the Leibniz rule for surface integration (from Wikipedia). Note that they call the surface $$\Sigma$$ instead of $$S$$:
$\frac{d}{dt} \iint_{\Sigma(t)} \mathbf{F}(\mathbf{r}, t) \cdot d\mathbf{A} = \iint_{\Sigma(t)} (\mathbf{F}_t(\mathbf{r}, t) - [\nabla \cdot \mathbf{F}(\mathbf{r}, t)]\mathbf{v})\cdot d\mathbf{A} - \oint_{\partial\Sigma(t)} [\mathbf{v} \times \mathbf{F}(\mathbf{r}, t)] \cdot d\mathbf{s}$
Now, if $$\Sigma$$ is moving, then this is complicated. But if it doesn't, then the terms involving $$\mathbf{v}$$ vanish, leaving:
$\frac{d}{dt} \iint_{\Sigma(t)} \mathbf{B} \cdot d\mathbf{a} = \iint_{\Sigma(t)} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{a}$
Therefore
$\oint_{\partial S} \mathbf{E} \cdot d\mathbf{\ell} = -\frac{d}{dt} \iint_S \mathbf{B} \cdot d\mathbf{a}$
This is an ordinary derivative you see on the right hand side. The integrand has both spatial and temporal dependence, but the integration destroys the former, leaving the surface integral on the right-hand side a function of $$t$$ alone.

Interestingly, we can do something with this even when the region is moving, that is, $$\mathbf{v} \neq \mathbf{0}$$. First, let's write $$\Phi(t) = \iint_{S(t)} \mathbf{B} \cdot d\mathbf{a}$$, the familiar quantity called flux. Furthermore, $$\nabla \cdot \mathbf{B} = 0$$, so one of the terms on the right-hand side goes away even though $$\mathbf{v} \neq \mathbf{0}$$. Thus:
$\frac{d\Phi}{dt} = \iint_{S(t)} \frac{\partial\mathbf{B}}{\partial t} \cdot d\mathbf{a} - \oint_{\partial S(t)} (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{\ell}$
Since
$\oint_{\partial S(t)} \mathbf{E} \cdot d\mathbf{\ell} = -\iint_{S(t)} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{a}$
as we found above, this allows us to write
$\frac{d\Phi}{dt} = -\oint_{\partial S(t)} \mathbf{E} \cdot d\mathbf{\ell} - \oint_{\partial S(t)} (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{\ell}$
We rewrite this slightly to obtain
$\oint_{\partial S(t)} (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot d\mathbf{\ell} = -\frac{d\Phi}{dt}$
Now the LHS is just the emf around the boundary of surface $$S$$. Thus we recover the familiar universal flux rule, as Griffiths calls it: $$\varepsilon = -\frac{d\Phi}{dt}$$. It's important to remember that this result was derived from considering both induced emf (which results from the Maxwell–Faraday law itself) and motional emf (which results from the Lorentz force on charges in the moving loop), rather than from the former alone. For this reason, some authors, including Griffiths, discourage referring to the universal flux rule simply as Faraday's law or the Maxwell–Faraday law.