sweet springs said:
Thanks. I am happy to know that Feynman is not alone.
I think flux rule contains more than Maxwell's ##\nabla \times E=-\frac{\partial B}{\partial t}##. Is that right?
Faraday's law is equivalent to Maxqwll's ##\nabla \times E=-\frac{\partial B}{\partial t}##. Is that right ?
I am still confused.
The point is that most textbooks confuse the subject by not clearly stating that the naive "flux rule" holds true only if the surface and its boundary you integrate Faraday's Law (in SI units ;-)),
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B},$$
over is at rest. Then and ONLY Then you have
$$\int_{\partial f} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{\mathrm{d}}{\mathrm{d} t} \int_{f} \mathrm{d}^2 \vec{f} \cdot \vec{B}=-\frac{\mathrm{d}\Phi}{\mathrm{d} t}.$$
If you have the general case of a moving surface and boundary, you get an additional term when bringin the partial derivative wrt. ##t## out of the integral, which you can lump to the left-hand side, leading to the complete and correct electromotive force:
$$\int_{\partial f} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})=-\frac{\mathrm{d}}{\mathrm{d} t} \Phi.$$
In any case it's more save to stay with the fundamental laws, which are the local Maxwell equations, including the relativistic (!) constitutive equations (in the most simple approximation you may use the linear-response approximation a la Minkowski). Then and only then everything is consistent and frame independent, as it must be.
Nevertheless particularly this chapter of the Feynman Lectures vol. 2 is a gem of textbook literature and should be carefully studied by any serious physics student!