Predicting the correct graph for the E.M.F induced To be honest, this is a homework question. A screenshot of this question is given below.

NOTE: The correct word for the one that I encircle with blue colour should be "rotates"
My argument towards this question 
Initially there is no induced E.M.F on the loop. But when it enters to the magnetic field which ultimately leads to have an induced E.M.F . And when time passes  E.M.F induced on the loop should be increased , because as the time pass , the area of the loop in space containing magnetic field increase.And after that it should comes to it's initial state and follow the same procedure   again and again .
So the answer I choose was 3 . But I don't know whether my argument and my answer are correct, which leads my mind to ask this question in this great site.!
 A: You are almost right. Think about the rate at which the area increases as the semicircle goes around. However much the area grows, that is how the flux grows - and that tells you what the EMF is.
So if the rate of growing / shrinking is sinusoidal, then (3) is the right answer. But what is the actual rate?
You might want to do a bit of geometry to write the area as a function of time... it's not hard. See for example this diagram:

At a given time, the area of the loop that is inside the magnetic field is $\frac{\theta}{2\pi}\pi r^2$. If the loop is rotating at a constant velocity $\omega$, then some time $\Delta t$ later the angle $\theta$ will have increased by $d\theta=\omega \Delta t$. This increase leads to a change in flux 
$$\Delta \Phi = B \cdot dA = B \frac{\omega\; \Delta t}{2\pi} \pi r^2$$
In other words, while the area of the loop covering the magnetic field is growing,
$$\frac{d\Phi}{dt} = \frac{B\omega r^2}{2}$$
and the e.m.f. induced is constant (all the terms on the right hand side are constant).
Once the loop has fully covered the magnetic field and starts to "leave", you can do the same analysis at the tail end - and see that the area gets smaller at a constant rate. This means that the change in flux through the loop will now be negative:
$$\frac{d\Phi}{dt} = -\frac{B\omega r^2}{2}$$
It is the same size as before, but the opposite magnitude. We expect a square wave - answer #2. The positions of the loop as a function of time will be something like this:

