Doubt on induced voltage
When a wire moves across the B field, a voltage is induced across it. This emf will be proportional to the field (constant), velocity (constant) and length of wire perpendicular to the velocity (different segments, different values).

I will label the segments from X to Y as A, B, C, D, E, F and G. 



Segments A, C, E, and G will give rise to an e.m.f. since the wire is moving at right angles to its length. The first three of these give rise to an emf of the same polarity - the last one (G) will reduce the emf a bit. 

So we're looking at either (3) or (4). But which is it? For this we need to know the polarity. Now the B field is out of the paper (that is the meaning of the little dot in a circle at the edge of the diagram - you are looking at the tip of an arrow; if the field was into the paper you would see a cross).

A positive charge moving from left to right in the page would feel a downward force due to the B field. This follows from the expression for the Lorentz force:

$$F = q(\vec{E} + \vec{v}\times\vec{B})$$

With \$\vec{v}\$ pointing to the right, and \$\vec{B}\$ pointing towards you, the force \$\vec{F}\$ must point down. If the positive charge is moved down, we are left with a negative potential at the top of the wire.

This means that (3) is the correct answer.

Just to clarify - the point (1) corresponds to the moment that segment C enters the field (t=0); at (2), segment A enters - increasing the e.m.f. A very short time later (3), E enters the field. We now have the largest amount of wire "all pointing the same way". Finally, at time (4) segment G enters the magnetic field; this will reduce the e.m.f. as it is pointing the other way (following the wire from X to Y, A, C and E point down; but G points up).