Direction of current flow In one of my homework questions, I was given the following setup. I was asked to determine if the induced current flows from X to Y, or from Y to X.

From Lenz's Law I know that the current in the tube flows in the direction opposite to R to oppose the change in flux due to the rotation of the tube in the direction R. However, I am completely lost as to how to determine the direction of current flow across XY. It seems that the charges on the tube along XY are always parallel to the flux field of the magnet. Because of this, no emf should be induced across XY by Faraday's Law.
 A: Faraday's law fails here. Let's go back to basics. We use the Lorentz force.
And what is happening is as the rod rotates, the charges in it rotate too. However, the rod is neutral so there is no net current flowing. Now field of the bar magnet is towards left in the wire, the lorentz force applies on the protons and electrons inside the wire, causes the electrons only to move in the wire circularly as the force on the electron is towards centre of the winding(radially inwards) and the force on the protons is radially outwards which gets balanced by constraint forces of the wire.
Thus only, electrons flow.
This causes, a net current to flow and thus we see the effect as an EMF. After a certain instant, there is an accumulation of negative charges at one end after which no more accumulation will take place.
Now regarding the direction of the current flow, it could have flown both ways by this logic. To find direction, now use the numerical values which are give as only one direction will give the matching value of the current in the question as 
 the $\vec{v}$ for the electrons, you need to take it from the frame of the magnet .
A: The Lorentz force appears to supply the solution.
It acts in the direction q${\bf v} \times {\bf B}$. Recall though that conventional current acts in the opposite direction to electron flow and that electrons are negatively charged!
And Faraday's law doesn't fail, just hard to apply. If you construct a long line integral loop along the axis of the cylinder and back through the conductor there is a rate of change of flux linkage due to the cylinder rotation, no E-field along the axis and therefore an E-field in the conductor.
