Rotating bar magnet : current induced in circuit 
I don't think this problem makes sense. The answer given is (a). Aren't the field lines parallel to the loop, what does rotation affect ? create atomic currents?
 A: The trick in such a problem is visualization. There is an emf between the two points of connection. There are a lot of ways to attack this and this Wikipedia page is very illuminating. Personally, I like to imagine a moving rod or a line segment that cuts through the magnetic field, which is seen as stationary and constant.
Now, there is a horizontal as well as a vertical distance between the points of connection. The vertical distance is along the magnetic field lines, so it does not cut through them. However, the horizontal distance does cut between the lines.


Now, the horizontal emf is, as detailed here, just $\frac{d\phi}{dt}=B\frac{dA}{dt}=B\frac{d\theta}{dt2\pi}*\pi R^2=B\omega \frac{R^2}{2}$, which is constant. So there is a DC current.
I highly recommend reading that Wikipedia page.
A: The magnetic field from the bar magnetic could be represented by magnetic field lines. This lines outside the magnet are bended between the poles and by this they cross the wire in its horizontal part. Since the magnet is rotating the field lines one by one crossing the horizontal part of the electrical circuit.
Now remember the Lorentz force $$\vec F = q \vec v \times \vec B $$
If a moving charge (an electron) goes through an external magnetic field (non-parallel to this magnetic field) then the charge gets deflected (perpendicular to the plane made by the electrons movement direction and the external magnetic field).
The Lorentz force is one of the three possibilities between the constituents magnetic field, relatively movement between this field and a charge and the reflection of the charge. Beside the Lorentz force the other phenomena are called induction of a current and induction of a magnetic field.
Due to this source the vector product from perpendicular vectors can be rewritten to
 $$ q \vec v = \dfrac {(\vec B \times \vec F)}{\|\vec {B}\|^2} $$
This is an equation that helps to understand that a current flows in the circuit and the answer is a).
For more details see on Wikipedia about the Homopolar generator:

