Confusion about emf definition

Question

The emf is supposed to be the energy gained by charges passing through a closed loop, just as in the case where we have a circuit where the potential drop equals the potential gain thanks to a battery. (Not that I understand the mechanism behind batteries: I tried asking here, but got no satistfying answer. If anyone can clarify this it would be great as well)

However, in electromagnetic induction, this work is provided by the magnetic field, with the formula $d\epsilon=\textbf{VxB}\cdot\textbf{dl}$, with $V$ being the speed of the particle and $dl$, the vector of a small line of current. Supposedly, if we take the case of a rectangular coil aligned with the xy axes that moves with speed $\textbf{vi}$ and a field with direction $\textbf{k}$, there's only work in the sides aligned with the y-axis, because $\textbf{VXB}$ is always $\textbf{ixk=j}$, and work is $\textbf{F} \cdot \textbf{dl}$. Yet this doesn't make sense... we see that the particles in the wire will experience a force upwards, and so the vector product will always be changing. What am I missing?

To make matters worse, if we only consider a vertical rod in the plane and move it in the x axis, the magnetic field makes particles move in the y direction, creating an electric field. This field creates an electric potential that would make the rod behave just as a battery. So, in this case, the emf for a closed circuit would be the electrostatic field... But isn't this impossible, since the work it would do along a line is zero? I'm rather confused, because the field should indeed be conservative, as it's created by charges.

I'm mixing up everything I have learned before... What's a sensible explanation for emf, especially in the induction case?

Answer 1

However, in electromagnetic induction, this work is provided by the magnetic field, with the formula $d\epsilon=\textbf{VxB}\cdot\textbf{dl}$, with $V$ being the speed of the particle and $dl$, the vector of a small line of current.

No, that's all wrong.

The formula you give is for motional EMF, not induced EMF. Motional EMF is due to motion of conductor in magnetic field, and $\mathbf v$ in this formula is not velocity of any particle, but it is the velocity of the conductor element described by $d\mathbf l$.

When current due to motional EMF increases, the work is being done by the conductor body on the current (and work of same magnitude and opposite sign is done on the conductor by the current, and the conductor decelerates).

the particles in the wire will experience a force upwards, and so the vector product will always be changing.

It may be changing, in particular it may increase in magnitude as current increases. Direction of the motional force per unit charge won't be changing much unless $\mathbf v$ is changing.

This field creates an electric potential that would make the rod behave just as a battery. So, in this case, the emf for a closed circuit would be the electrostatic field...

No, the electric potential gradient will act against the motional EMF in the rod. If equilibrium is reached (very likely unless the rod accelerates too much), then both forces cancel each other out and we have a moving rod with zero current and polarized ends.