Induction in closed loops

I'm trying to understand the law of current induction in closed loops.

I just saw this example, of a non-moving circuit in a constant magnetic field $$B$$. On it, a sliding (conducting metal bar) which moves with a velocity $$v$$.

In this case, I saw that the EMF generated is $$BLv$$, since this is the flux change in either the right or the left loops.

So first of all, I'm wondering why we can deduce the EMF from each loop separately? Don't they affect one another? What if the flux change wasn't equal in both of the loops? I've drawn such a case here:

There's a non moving black half-closed loop, above it the moving bar and above it another non-moving half-closed red loop.

So in this case, the flux change in the black loop is different than the flux change in the red loop. What would then be the EMF?

Summing up, I'm trying to understand why in the first example only one loop was considered when calculating the EMF, and how would that logic apply in a case such as the second case, where there is no symmetry in the flux of the two loops.

Thanks!

Each loop (even a loop consisting of multiple smaller loops) has an EMF along it given by Faraday's law of induction, also called the flux rule: $$\mathcal{E}=-\frac{d}{dt}\iint\limits_S\vec{B}·d\vec{A}\ .$$ It is okay for two adjacent loops to have different EMFs.
• Yes, the EMF can be different for each loop. Recall what EMF is: it is the integral of force per unit charge along a wire. In this case the moving bar is the only thing that contributes to EMF. Note that the red loop contains a shorter section of the bar, hence the magnitude of the EMF along the red loop is smaller. In effect a "smaller battery" is connected to the red loop. Kirchoff's voltage law still applies in the sense that $\oint{\vec{E}·d\vec{\ell}}=0$. Much like in a battery, the charges are driven by a non-electric force, which is the magnetic Lorentz force in the bar in this case.