What is the difference between EMF induced in Faraday's law and the potential difference due to electric field? What is the difference between emf induced in Faraday's  law and the potential difference due to electric field?
If we thake a conducting circular loop in a changing magnetic field then we have an emf induced in it.
But in electric circuits the potential difference between the two points of a conducting wire is zero.
Also explain how this emf causes the electric current to flow.
 A: Yes this issue often causes confusion. The basic thing you need to think about here is the electric field. In the first instance, remove any conducting wire and just suppose there is a region of space where there is a magnetic field, uniform spatially (i.e. uniform direction and size), but changing with time. Let's say it is increasing. In this scenario there is also an electric field in that region of space. The electric field in this situation runs in circular loops around the magnetic field lines.
Ok so far so good: we have a changing magnetic field, and, in that same region of space, also an electric field running in circular loops.
Now suppose you put a conducting wire in a loop in that same region, following the direction of the electric field lines, but for the moment do not close the circuit. That is, you have a loop of wire but with a gap in it so that it does not close. What will happen?
The electrons in the wire will be pushed by the electric field, and they will move, such that they start to pile up on one side of the gap. This imbalance in the charge distribution in the wire causes a counter-balancing electric field. The electrons keep moving until this counter-balancing electric field (caused by the electrons) is equal and opposite to the one caused by the changing magnetic field. Thus when the system settles down the net electric field inside the conducting wire is zero (I am assuming the rate of change of the $B$ field here is constant).
At this point there is a build-up of negative electric charge on one side of the gap in the wire loop, and a corresponding positive charge on the other side of the gap. Also there is potential difference across that gap: it is equal to the e.m.f. which you can calculate using Faraday's law. So if you were to now connect a resistor or a light bulb or something like that across the gap, then a current would flow.
When a resistor is connected across the gap, there is a potential difference across the resistor and an electric field inside the resistor. There is no electric field inside the conducting wire (if we assume zero resistance of the wire), and the electric field just outside it is also affected by the distribution of charge in the wire.
On electric potential difference
In the above I did not mention the concept of potential difference until near the end. This is because in electromagnetism it is best to regard the fields and the charges as the main idea, and then concepts such as potential difference come in as useful tools to help in calculating and in getting insight.
Electric potential comes into its own, as a concept, in static conditions, because then we can find a function $V(x,y,z)$ such that the electric field can be written as
$$
{\bf E} = - {\bf \nabla} V.
$$
(This is a standard shorthand notation for a gradient. In terms of components it means:
$$
E_x = - \frac{\partial V}{\partial x},\\
E_y = - \frac{\partial V}{\partial y},\\
E_z = - \frac{\partial V}{\partial z}.)
$$
In non-static cases, as for example in the presence of a changing magnetic field, things are not so simple, because now the electric field is such that it can point around a loop, and this means there is no function $V$, having a single value at each location, such that $\bf E$ is its gradient. However we can still investigate quantities such as
$$
-\int_{P_1}^{P_2} {\bf E} \cdot d{\bf l}
$$
where  $P_1$ and $P_2$ are two points in space (or possibly the same point) and the integral is taken along a path running between those points. In a static problem, this integral would give the potential difference between $P_2$ and $P_1$. In a non-static problem, we may choose to give another name to the result of this integral. It is often called 'electromotive force' or 'emf' (this is arguably not a very clear name but is adopted for historical reasons). But this is just a name. You could call it a potential difference if you like, as long as you realise that you are really talking about the integral of electric field along a given path. What you know is that if a charge were to move along that path, then the net energy given to the charge by the electric field is minus the amount given by this integral. This can be deduced because the force on such a charge would be
$$
{\bf f} = q ( {\bf E} + {\bf v} \times {\bf B} )
$$
and therefore the work done when the charge moves through a displacement $d{\bf r}$ is
$$
{\bf f} \cdot d{\bf r} = q {\bf E} \cdot d{\bf r}.
$$
Here the magnetic field term does not contribute because for a charge moving along a path described by $\bf r$ we have that $\bf v$ and $d{\bf r}$ are parallel so the magnetic force is perpendicular to $d{\bf r}$.
The main point of this final section of my answer is to say that 'potential difference' and 'emf' are different words for essentially the same thing, namely the integral of electric field along a path. The reason to have two terms is that the first one (potential difference) draws attention to a useful property of static fields, and the second one draws attention to the fact that the situation under consideration is not static so we have to proceed a little more carefully in our reasoning. In particular, for a non-static case we should not assume that there is any function $V(x,y,z)$ (having a single value at each point $(x,y,z)$) which gives the electric field as its gradient.
A: One possible source of confusion is that $\mathbf E = -\nabla V$
if there is no vector potential present, no changing magnetic field. Otherwise the relation is: $$\mathbf E = -\nabla V - \frac{\partial \mathbf A}{\partial t}$$
Where $\mathbf A$ is the vector potential.
From Maxwell induction equation:
$$\frac{\partial \mathbf B}{\partial t} = -\nabla \times \mathbf E$$
In the case, let's suppose that the magnetic field is in z-direction, and the circuit in the xy plane. The equation applies in the z-direction only:
$$\frac{\partial B_z}{\partial t} = -(\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y})$$
If we mistakenly take $E_x = -\frac{\partial V}{\partial x}$ and $E_y = -\frac{\partial V}{\partial y}$, (without the vector potential) we would get:
$$\frac{\partial B_z}{\partial t} = -(\frac{\partial^2 V}{\partial x \partial y} - \frac{\partial^2 V}{\partial y \partial x}) = 0 \implies B_z = cte$$
But this is not true according to the hypothesis! And that is why we are puzzled by how is it possible for $\oint E.dl \neq 0$. Really,  there is no difference in the potential for the same point for example (the beggining and end of the loop integral).
If we put the correct terms for the E-fields and express $B_z$ also in terms of the vector potential, all terms cancel and the equation is fulfilled.
The conclusion is: the $emf$ can not be explained by the notion of a (scalar) $V$ potential. For example: a linear growing B-field induces a constant E-field in the wire loop. If there is only a resistor in the loop, the points just before and after it must have the same potential because they are joined by a virtually zero resistance wire (the rest of the loop). But there is a current in the resistor anyway, due to the electric field.
A: $\oint \vec{E}\cdot \vec{dl} = 0$
Is when there is an absence of a changing magnetic field. Which is what assume you mean when you say "in circuits"
This means that on a CLOSED loop, the emf is zero across this loop.
In static electric circuits, the potential difference is NOT zero about any 2 points, this is just false  If this were the case, no current would be produced.
It is only zero on CLOSED curves, where the start and end are at the same point. There is a potential difference from the positive to the negative terminal  in a circuit ,because they are not at the same point, the terminals are separated by a distance. Emf is circuits is measured with respect to a non closed curve. The closed curve is zero, because inside the battery the field points in the opposite direction cancelling out the field outside
In both cases, where the Emf is produced by the field in the battery or via faradays law, It is the electric field that is the cause of the emf. The differences in these electric fields is that one has curl and one does not.
An emf is just a fancy way of saying there exists a non zero electric field on the chosen path. Where there is an E field, there is a force on charges causing them to accelerate and produce a current.
Search up drude model of conductivity to understand why it is a CONSTANT current.
Potential difference across a zero resistance wire
A: A time varying magnetic field induces a rotational (non-conservative) electric field, which we will call $\vec{E}_{induced}$. If a conductor is present within that time-varying magnetic field, electrons in the conductor will rearrange, and that re-arrangement will cause a second irrotational (conservative) electric field to be present. We will call this the reaction field $\vec{E}_{reaction}$.
The total electric field is the sum of the induced field and the reaction field.
$$\vec{E}_{total}=\vec{E}_{induced}+\vec{E}_{reaction}$$

What is the difference between emf induced in Faraday's law and the potential difference due to electric field?

The EMF induced along a curve C which begins as point A and ends at point B (possibly the same as A) is
$$ \mathscr E_{induced} = \int_C \vec{E}_{induced} \cdot d\vec{\ell}$$
or in the case where curve C is a closed curve
$$ \mathscr E_{induced} = \oint_C \vec{E}_{induced} \cdot d\vec{\ell}$$
In this case, the induced EMF is equal to the rate of change of flux enclosed by the loop C.
$$\mathscr E_{induced} = -\frac{d\Phi}{dt}$$
If we integrate the total electric field along the curve C, we will get the voltage drop through that path C.
$$ V_C = \int_C \vec{E}_{total} \cdot d\vec{\ell}$$
This voltage drop $V_C$ is what will be used if you want to apply Ohm's law to a wire in the shape of the curve C.
$$V_C=IR$$
where $I$ is the current through the wire, and $R$ is the resistance through the wire.
$V_C$ also represents the work per charge associated with moving a test charge from A to B along curve C (assuming the electric field does not vary over the time the charge is moved from A to B).
Note that because we are working in a regime where there is a time varying magnetic field, the total electric field has a rotational component, and hence is not conservative. This means that the voltage drop along one path will generally be different from the voltage drop along a different path. That is, if the path begins at A, and ends at B, the voltage drop between A and B is not path-independent. Therefore, the notion of a potential difference between A and B doesn't really apply in this case. (At least not a notion of potential difference that can be used in Ohm's law calculations). We can however, speak of the voltage drop from A to B along a specific path C.
