I am really confused in the definition of EMF $$\varepsilon$$ in Faraday's Law. The example I saw was a loop of wire moving with velocity $$\vec{v}$$ inside a uniform magnetic field $$\vec{B}$$, which does not complete the whole area of the loop, hence leading to a change in magnetic flux through the wire $$\Phi_B$$. The EMF is defined as the work per unit charge. In this case $$\vec{F_B} = q \vec{v} \times \vec{B}$$. Then the EMF is given by: $$\oint_C \frac{\vec{F_B}}{q} \cdot d\vec{s} =\oint_C (\vec{v} \times \vec{B}) \cdot d\vec{s}$$ Now Faraday's Law says: $$\varepsilon = -\frac{d\Phi_B}{dt}$$ Until here, I understood everything, but here is the part I don't get. This equation is rewritten in terms of the electric and magnetic fields using the following relationships: $$\Phi_B = \iint_S \vec{B} \cdot d\vec{A}$$ $$\varepsilon = \oint_C \vec{E}\cdot d\vec{s}$$ The part I don't understand is the definition of $$\varepsilon$$, the EMF. Why is it the closed integral of the electric field? As far as I know, the electric field is the electric force per unit charge, but in this case, the force is magnetic. How is this definition true?

To see that this makes sense, consider the definition of emf: It is the work done on an electric charge over some path. Imagine a unit charge inside an electric field $$\vec E$$. To move this charge an infinitesimal distance $$d\vec s$$, we need energy $$\vec E . d\vec s$$. To get the energy over the whole loop, we simply integrate that term.
• Your answer is not fully correct. In the frame where the wire is moving in the presence of a static field $\bf B$ there is no electric field. Neither the changing flux induces an electric field in that frame. The electric field appears in the rest frame of the wire. – GiorgioP Nov 17 '19 at 22:20