The electromotive force generated by the rotation of the conductor bar around the o point is e1= e1=12bwl2,

If the direction of electromotive force generated by rotating the OM rod clockwise is M→O, the direction of electromotive force generated by the on rod is O→N, because two electromotive forces are connected in series, E=Bwl2.

The rotation time T4 is constant,

When the OM bar turns into the third quadrant, the direction of electromotive force generated by the OM and on bars becomes O→M, N→O, and the magnitude remains unchanged.

So the direction of E=Bwl2 is reversed, and so on, so A is right and BCD is wrong.

So choose a.