I'll explain the problem. [tex]\cdots[/tex] [tex]\cdots[/tex] [tex]\cdots[/tex] [tex]\cdots[/tex] [tex]\cdots[/tex] [tex]\cdots[/tex] [tex]\ldots[/tex]=====|====== [tex]\cdots[/tex] [tex]\cdots[/tex] || LOOP | LOOP || --> V [tex]\cdots[/tex] [tex]\cdots[/tex] ||=====|=====|| [tex]\cdots[/tex] [tex]\cdots[/tex] <-x-> ..| [tex]\cdots[/tex] [tex]\cdots[/tex] [Dotted area indicates field] In the usual discussion of motional emf, all authors go through the usual loop moving through a magnetic field which is perpendicular to the plane in which the loop is moving. Usually, this loop have resistance, say R and starts out with a bit of the loop already outside the field. The emf can be shown to be Bhv wherein B is the strength of the magnetic field, h is the length of the rod and v is the velocity with which the loop is being pulled along. Therefore the current can be calculated as BLV/R. Now he introduces a new twist in the question. Assume the rod has no resistance, i.e. that it is a perfect conductor. I have to show that the rod will now execute simple harmonic motion. This being said, I got the answer. Qualitatively, we can say that the current will not shoot to infinity as the self-inductance will not allow that to happen. The back EMF associated with the loop will save the day. In this case the flux can be taken to be Bhx where x is the length of the loop still inside the field. Refer to my "diagram". The flux for this loop is also given by LI where I is the current flowing through the loop and L is its self-inductance. Equating the two, we have current I = Bhx/L The force on this loop is given by (IhB). So, mx'' = -Kx, where K is B2h2/mL. It is quite obvious that the rod will execute simple harmonic motion. Hence proved. Now what confuses me is that the above equation holds for any loop with or without resistance. He says something here in the problem that in general cases the self-inductance thing is negligible. I don't get that. Shouldn't all loops execute SHM?