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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?

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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.

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  • $\begingroup$ 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? $\endgroup$ – Nicolás Maíllo Gómez Nov 14 '19 at 17:54
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    $\begingroup$ Yes that's correct. $\endgroup$ – Al Nejati Nov 14 '19 at 20:59
  • $\begingroup$ 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. $\endgroup$ – GiorgioP Nov 17 '19 at 22:20
  • $\begingroup$ 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. $\endgroup$ – Al Nejati Nov 17 '19 at 23:38

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