Can the emf induced in a conductor passing through the magnetic field be explained via magnetic flux? Let's say I have a conducting rod passing through a constant magnetic field. I get positive charges in the rod moving with velocity will experience a force (eg. upwards) and move up to the top of the rod and hence the potential difference between the top and bottom of the rod will induce an emf.
The thing is, Faraday's law also states that the induced emf is proportional to the rate of change in magnetic flux linkage. Here's what I don't get. Is flux linkage a property inherent in closed loops or can be used for rods as well? I know that rods technically do not have a 'number of turns' but they do have an area.
If so, then there will be magnetic flux through the rod, but as it moves through the field, the magnetic flux will remain the same, so there will be no induced emf in the rod.
Would it be correct to say that there are 2 sources of induced emf in the rod: one from the separation of positive and negative charges and another one from induced emf via Faraday's law (which is doing nothing).
Does that mean that when a rod first enters a magnetic field, the magnetic flux through the rod will increase and there will be an emf induced? Would it mean there will be 2 contributing emfs in the rod, one from the separation of charges and one from the increase in flux linkage by Faraday's Law?
I can't find answers for this online. For a bit more about my knowledge level on this subject, I do not know anything about Maxwell's equations and what they might imply.
 A: The emf in a moving wire arises from the magnetic Lorentz forces acting on the free electrons, as they are carried along by the wire...
$$\vec F = (-e)\ \vec w \times \vec B$$
Here, $\vec w$ is the velocity of the free electron. This is the vector sum of the wire's velocity, $\vec v,$ and the drift velocity, $\vec {v_{dr}},$ of the electron through the wire. So
$$\vec F = (-e)\ \vec v \times \vec B\ +\ (-e)\ \vec {v_{dr}} \times \vec B$$
The second term on the right acts at right angles to the wire and implies that we have to do work as we push the wire. The first term on the right gives rise to the emf. If the wire has length $\ell$, and is being moved in a direction at right angles to itself, with $\vec B$ at right angles to plane containing the wire and $\vec B$, then the emf is $$\mathscr E =\frac{\text{work done on charge}}{\text{charge}}=\frac{-evB \ell}{-e}=vB \ell$$
Now let's deal with the flux approach. In my opinion, for a moving conductor this approach is not as fundamental as the Lorentz force approach just given.
For the flux approach we need to include the moving conductor in a circuit. But it's important that the rest of the circuit isn't moving with the conductor, or emf's in the conductor and the rest of the circuit will cancel out. This is why elementary textbooks have the conductor sliding on fixed rails. The circuit might be completed by a fixed voltmeter connected between the rails. As the conductor slides, the area of the circuit changes, so the rate of change of flux through the circuit is
$$\frac{d \Phi}{dt}=\frac{\ell v dt\ B}{dt}=\ell v\ B$$
So you see that the flux approach leads to the same emf as the Lorentz force approach. Why use the flux approach? (a) Because it automatically sums the emfs in any parts of the circuit that are moving, (b) It works both for a moving conductor (provided it is thought of as part of a circuit) and for changing flux density in a fixed circuit (the province of the Faraday-Maxwell law): rather neat.
