It's good to keep in mind that the fundamental physical laws underlying classical electromagnetics are the local Maxwell equations, i.e., the Maxwell equations in differential form. Much of the problems students have with the Law of Induction is that it's often represented in incomplete form. The fundamental Faraday law reads
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$
Using Stokes's integral Law for an arbitrary surface ##S## with boundary ##\partial S##, gives
$$\int_{\partial S} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_S \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
Now comes the tricky business. To get the usual Faraday Law in integral form, you want to take out the time derivative from the surface integral on the right-hand side. Often people don't discuss this carefully enough. If you have a moving surface, there's an additional term. Taking this properly into account, you'll get the one and only correct integral form of Faraday's Law of Induction
$$\int_{\partial S} \mathrm{d} \vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_S \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
Here ##\vec{v}=\vec{v}(t,\vec{x})## is the velocity field of the boundary of the surface. This tells you that you have to define the electromotive force in Faraday's law including the complete Lorentz force per unit charge and not only the electric piece. Of course, if the surface under consideration is not moving, then ##\vec{v}=0##, but only then it's correct to forget the magnetic part of the Lorentz force!