# EMF in Faraday's Law

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?

That is just how we define the induced emf: It is just the integral of the electric field over the closed loop.

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.

It also makes sense dimensionally: The unit of the electric field (in SI) is kg⋅m⋅s−3⋅A−1, and the unit of the volt is kg·m2·s−3·A−1, which is just the former multiplied by distance.

• Ok, I think I got it, but is this way of thinking about it good? : I would say that the changing magnetic flux creates an electric field that obeys Faraday’s Law. Is this right? Commented Nov 14, 2019 at 17:54
• Yes that's correct. Commented Nov 14, 2019 at 20:59
• 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. Commented Nov 17, 2019 at 22:20
• A bit of clarification is necessary. A changing magnetic field is not required, as you're saying, however from the frame of reference of the loop, you can indeed say that a changing magnetic field induces an electric field in the loop. Commented Nov 17, 2019 at 23:38