In electrostatics, the electric field in a wire loop is conservative (why?) so an electron which makes a full turn around wire loses as much energy as it gains. In other words the emf of the circuit is zero, if I understand correctly
Electrostatic field is conservative because if we move a charge from point A to B, the work done is irrespective of the path taken.
The easiest way to prove this is by taking a point $q$ charge, taking it away to some point and then bringing it back to initial point (without increasing its kinetic energy).
For electrostatic field:
$W_{ext}=-q\Delta V=-q(\oint_{closed-loop}\vec E \cdot {d\vec{l})}=-q(-\frac{\mathrm{d}\phi }{\mathrm{d} t})=0 \\ \Rightarrow W=0$
This implies that irrespective of the path taken, work done is always $0$ in an electrostatic field when charge is brought back to initial position since $\frac{d \phi}{dt} $ is always $0$.
For induced electric field:
$W_{ext}=-q\Delta V=-q(\oint_{closed-loop}\vec E \cdot {d\vec{l}})=-q(-\frac{\mathrm{d}\phi }{\mathrm{d} t}) \neq 0 \\ \Rightarrow W \neq 0$
Here $\frac{d\phi}{dt}$ depend on the path taken.
Hence electrostatic field is conservative, but induced electric field is not.
In otherwords the emf of the circuit is zero, if I understand correctly.
Since there is no battery, there is no EMF. In the above case, we are making the charge move around. So there is no need of EMF.
If you were asking about a circuit with a battery, then the EMF is provided by the battery.
However, apparently in a circuit with an induced emf, an electron which makes one full loop can gain energy. So as electrons continue to travel around the circuit, they gain more and more energy. In otherwords, the induced emf is due to an induced electric field in the wire which is nonconservative. Does this make any sense?
The work done in changing the magnetic field is transferred to the electrons in the conductor, which gives them energy to move around. This is a consequence of Law of Conservation of Energy.
Now, main question:
... However, doesn't this contradict our definition (*) of induced emf?**
No. Because it is not rigorous to use Faraday's Law here in the first place.
$\oint_{c}\vec E \cdot {d\vec{l}}=-\frac{d}{dt}\int_{s} \vec B \cdot {d\vec{l}} $
is the common version of Faradays's Law. It is rigorously correct only if $\vec E$ represents the electric field in the rest frame of each segment $d\vec l$ of the path of integration. This is definitely not true is the case of motional EMF.
It is necessary to note that only a time-varying magnetic field induces a circulating non-conservative electric field in the rest frame of the laboratory. A changing flux does not induce an electric field. Hence in case of motional EMF, no electric field is induced in the laboratory frame(where rod is in motion). Hence technically $\oint_{c}\vec E \cdot {d\vec{l}} = 0$ here.
A potential difference is maintained across the rod by the EMF and Lorentz force maintains that EMF. Using $-\frac{d}{dt}\int_{s} \vec B \cdot {d\vec{l}} $ here is just a convenient way of calculating the EMF.
Source: A Student's Guide to Maxwell's Equations -- Daniel Fleisch
This is where it's a good time to "converse with the math". Let's look at the equation:
$$\mathcal{E} = \oint_\gamma \mathbf{E} \cdot d\mathbf{l}$$
which in this case equals $\mathcal{E}_\mathrm{ind}$. This is based on the work formula:
$$W = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$
Thus, what your question is, is essentially, asking "how can you have negative work". Looking at the above equation and recalling how a dot product behaves, there's only one way: $\mathbf{F}$ and the element of displacement $d\mathbf{r}$ must be aimed at cross purposes with each other. In the case of the integral for $\mathcal{E}_\mathrm{ind}$, the same goes only with $\mathbf{E}$ and $d\mathbf{l}$.
Hence, the answer to your question is: when $\mathbf{E}$ points opposite to $d\mathbf{l}$.
But what does that mean? Well, the key here is that we have to think a little more closely about the work formula. I believe what you are imagining it means is "the work done by the force as the force pulls the particle along with it". It actually is more general - in a work integral, the particle can be moved in any direction, including against the force. Of course, to make that motion happen in real life, you need to supply a source of contrary force, but that doesn't change the maths. This is why, say, in a more elementary example, you can talk of "negative work" done by the gravitational force when you lift an object off the floor.
(Why is it defined that way? Well, for one, because we often can't solve for the "real" trajectory the particle follows! If we stipulated that as a precondition, it would make work an extremely non-trivial concept!)
Where your mistake lies in, then, is in missing that. The displacement around the wire $d\mathbf{l}$ that we use to describe emf is not the displacement that necessarily occurs in reality to a real positive charge (after all, in many applications we aren't "really" dealing with positive charges anyways!).
Rather, it is a hypothetical one where we imagine that we grab a charge and move it around all the way through the circuit in a specific, fixed direction, and ask what the work - whether positive or negative - done by the electric force for that movement is. If we ask about the work done in the actual movements of charges, we will get a different answer, and yes, this one will always be positive, at least provided we don't get into looking at the situation in too-fine detail.
From an intuitive point of view, if you want to figure when the force is doing "positive" work and when it's doing "negative" work, imagine that you can feel the electric force tugging on the positive charge in your hand as you move it through the circuit. When you feel it helping you, i.e. the tug is with the motion of your hand, at that moment (i.e. that small increment $d\mathbf{l}$) the electric force is doing positive work, and you are doing negative work (to retard the charge, if you were to try and not naturally speed up your hand as you'd likely tend to, of course). When you feel it is fighting you, i.e. the tug is against the motion of your hand, the electric force is doing negative work, and you are doing positive work (to help it against the contrary pull). Total negative work, and hence negative emf, will be if, in moving the charge, you had to fight more often than flow.
Best Answer
First of all, An EMF doesn't have to be an electric field, It only is required that Work is being done on an object travelling in a certain path .
When there is a changing magnetic field, there is an induced electric field,
When there is a changing surface, because particles in the wire are in the presence of a magnetic field, they will experience a force q$ \vec{V} × \vec{B} $ This force is the cause of the EMF when the magnetic field is unchanging
Faradays law states the EMF about a CLOSED path, meaning your starting point A and final point B are the same point.
Potential difference as a concept isn't defined for faradays law as the field is non conservative, meaning $ \int \vec{E} \cdot \vec{dl}$ is PATH DEPENDANT meaning the work done by the field on an object changes depending the path I take.
meaning I physically cannot define potential difference between 2 points as I need to know the path inbetween, easy to visualise
Imagine I go clockwise about my path, starting at A and finishing at A, it gives me some value.
Now if I go anticlockwise, I would get a different result as my field is in the opposite direction
Mathematically conservative field are in the form $\nabla V$ Which means the curl is zero, which isn't the case in faradays law.
$\int \vec{E} \cdot \vec{dl}$ js defined however.. Just for the specific path you take, which is a closed path