How does the work done in moving the coil or magnet get transformed into electrical energy in the form of current in the coil? I read somewhere that:(but don't remember where)

In electromagnetic induction, the mechanical energy spent(ie,the work done) in moving the coil or the magnet,to change the magnetic flux is transformed into the electrical energy in the form of current in the coil.

The first question that arises in my mind is that is the above Statement correct? If yes, how is the work done in moving the coil or magnet is transformed into the electrical energy in form of current?
I am not able to understand how this can happen. I looked on the internet but couldn't find any useful insights. Could someone plz explain how this happens.
 A: You are right in being sceptic about the work done by the magnetic field. Lorentz force law
$$\vec F = q\vec v\times \vec B$$
basically says that the force being exerted on a moving charge by a magnetic field is always perpendicular to the velocity of the charge. Now you know that acceleration power is given by
$$P=\vec F\cdot\vec v$$
which says that only the component of the force that is parallel to the velocity will perform any work. But because Lorentz force is perpendicular to velocity, and so has no parallel component, the work done by the Lorentz force itself is zero. A homogeneous magnetic field will even get the charge to move on a circle (or a helix) with constant velocity.
However, the magic is in mentioning the word "induction". Because, if the magnetic field is not constant, the law of induction says that it will always be accompanied by a rotational electric field:
$$\vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t}$$
You might probably only know the law of induction yet in the form where it is integrated over a closed line (e.g. a conductive circuit):
$$\oint_{\partial A} \vec E\cdot d\vec s = -\frac{\partial}{\partial t}\int_A \vec B\cdot d\vec A=-\frac{\partial \Phi}{\partial t}$$
with the magnetic flux that goes through the area $A$ of the circuit
$$\Phi:=\int_A \vec B\cdot d\vec A$$
But this form is really just a reformulation of the differential equation above, since the area $A$ can be considered arbitrary.
For the induced electric field $E$ Coulomb's law (which you might only know from the force acting between point charges, but which can also be formulated for an arbitrary electric field, as follows) determines the force acting on the charge:
$$\vec F=q\vec E$$
and as you can see this is not anymore constrained to being perpendicular to velocity. Hence, the induced electric field is actually able to do work on the charge. Since the electric field due to induction is rotational, it will drag the charge through a closed circuit. Note, that the induced electric field does not even require that the charge is initially moving at a velocity at all. Even if it starts from $v=0$ it will be accelerated by the induced electric field, whereas the magnetic field only acts on the charge if it is already moving.
If you wonder whether this is violating conservation of energy (continually driving a charge around a circle), it does not, because the now rotating charge is itself generating a magnetic field, which weakens the original magnetic field (and with it, the induced electric field): this is called Lenz' rule. Speaking a little simplistically: the magnetic field energy is first (partially) transformed into an electric field due to induction, and the electric field energy is then transformed into kinetic energy of the charge due to Coulomb's law.
Btw., the total force acting on the charge in the presence of the electric and the magnetic field is given by
$$\vec F=q\vec E+q\vec v\times \vec B$$
wherein the electric field again performs work and the magnetic field does not.
A: Yes that statement is completely correct if you want to generate electricity then you will have to do the external work and that work will get converted into the electrical energy let's look at this diagram and understand:-

Here, loop and a movable rod is in the role and both are conductive in nature, so as we can see that the magnetic field is directed insided the plain of the paper or in the direction " - $\hat{k}$" and let's say that if we slide the rod towards the "$\hat{i}$" direction then according to the equation:-
$\vec{F} = q (\vec{V} × \vec{B})$
In this situation the direction of velocity of Positive Charges inside the moveable rod is in the  $\hat{i}$ direction and it is obvious because the moveable rod is itself moving in the same direction so it will let all the positive charges present inside to move in the same direction.
$\vec{F} = qVB \hat{j}$
So as we can see that the positive charges will experience the force in the $\hat{j}$ direction and similarly for all the charges present inside the conductor, so if there is some net movement of positive charges in some fixed direction then there will also be the current as well. So by doing some work in moving the rod will result in the generation of electricity or in other words the kinetic energy of rod will be converted into the kinetic energy of the positive charges. And we can also say that is obeys the Faraday's law hence Magnetic Flux is changing as the area of loop is increasing with time so it will induce a current in the whole loop.
