Correct graph for induced E.M.F

The EMF that will be induced in the crossbar will be proportional to the magnetic flux that is "cut" by the crossbar as it moves around the wheel. So to be able to answer the question we need to look at the rate that the flux is being cut at each instance around the wheel.

Probably the best way to think about this is looking at the wheel from the side and visualising the crossbar as a point that is moving around a circle, like the image you have added to the question.

Now to be able to find the rate at which the flux is being cut as it goes around the wheel it is important to understand that the only way that the flux can be cut is by the point on the circle (the crossbar) moving vertically, and that any movement in the horizontal direction does not cause any emf to be induced, due to no flux being cut.

You will be able to find this information by looking at the gradient at each point on the circle. Ultimately what you are doing is considering how much the point moves over an infinitesimally small distance (ds), considering the horizontal and vertical components separately (think gradient = dy/dx, it might even be helpful to think about the Pythagoras theorem)

ok so to explain this I have drawn the above image illustrating the crossbar at three particular points, and I will now explain each one separately.

A

Right at the top of the wheel/circle the gradient will be perfectly horizontal, so if the point on the circle moves the infinitesimally small distance (ds), all of that movement will be horizontally. This means that no flux will be cut and hence no emf will be induced.

B

if you then move further round the circle, say 45 degrees as shown here. The vertical component is no longer zero. Looking again at the infinitesimally small distance ds we can see this happen. in this case the point has moved an equal distance vertically and horizontally (that is dx = dy). This means that some flux will be cut and hence emf will be induced.

C

If we then move round to where the gradient will be perfectly vertical (at 90 degrees from the start) we get the complete opposite case of A, where all of the motion is part of the vertical component. So at this point more emf will be induced in comparison to B, as the value dy is larger, in fact the largest value it can be.

Now we can put this all together we can understand how the emf is going to change as it moves around the circle. We know that the vertical component increases as the point moves from 0 to 90 degrees, where at 90 degrees it will reach it maximum.

Im going to leave the rest of answering the question to you, I believe that what I have provided here should be enough for you to be able to answer the question yourself.