# Confusion about the direction of the vectors: motional EMF

I'm working through an example with motional EMF and I'm having trouble understanding the directions of vectors so that I can apply induction law.

The magnetic circuit seems complex because the circuit is used to analyze other situations but the air gap 3, the coil 3 and the single open loop coil are the ones relevant to this. We neglect dispersion and the magnetic reluctance of iron. Section S of the circuit is constant and it's a square of side a. The air gaps have thickness $\delta$. All of the coils have the same number of turns. The open coil has a current $i_0$ that is zero.

Now we have i2=1A and i3=i0=0 which originates a flux $\phi=-1mH$ and $B_3=-0.625 T$.

Now my question is about the next paragraph:

"The motional induction electric field only exists when the coil 0 passes the air gap 3 (I understand that, because only there B is not zero). On that conditions, the elecric field as an orthogonal direction to the figure plan, the same as the current i0 (I think I can also see that...). At the bottom side of the coil we will have $u_0=Bva$ (now that is what I don't understand, what is the direction of B? Is it the same as v? Why? I can't see the direction of the vectors!)."

Basically my question is about the direction of the vectors while applying the induction law. Can someone help me clarify it? I only need a small draw or some brief explanation.

4. The reason that the motion of the coil in the air gap generates a voltage is because as the coil moves through the air gap it cuts across the flux $\phi_3$, getting ever less flux as it moves. This is why the B field is made irrelevant -- all you need to know is that the coil is seeing all of $\phi_3$ when it's inside the core, and none when it's outside, and the change must generate some voltage.