Past Year Prelim Paper - EMI

Just to clarify some doubts or misconceptions, please refer to the 2007 R_JC Prelim Paper (P1)

32) PQ is placed on the same vertical plane as the wire and is moved at constant speed v away from the wire.
How will the magnitude of the induced e.m.f. in PQ vary and which end will be at a higher potential?

(1) The Magnetic Field Strength B of a wire obviously decreases with distance away from the wire. Since ε = Blv, if B decreases, the magnitude of the e.m.f. decreases.

(2) If you use Fleming's RHR to find the direction of "induced current" (there is no current since there is no closed loop, hence the parenthesis) you would think that the direction of current is from Q to P and hence Q is at a higher potential.

However it is important to note that there is no current. but there is separation of charge.

The rod moves to the right. so imagine the rod as one electron. According to Fleming's LHR, the magnetic force F_m acting on this particle will be downwards. Hence if it was possible to move, the particle would move in a circle downwards.

Now consider the fact that the movement of the electron is limited by the physical dimensions of the rod. So after a certain time, the electrons would more or less accumulate at Q and hence the ε direction would point from P to Q.

Another way to look at it would be that if you were to connect this rod to an external circuit, electrons would flow from Q, through the outer circuit into P. Hence current will still flow from P to Q but via the outer circuit.

Like some of you have mentioned, the rod acts like a "battery". only that the driving "force" this time is not due to an electric field, but due to the magnetic field/force.