Question

A 2.50 m length of wire is held in an east–west direction and moves horizontally to the north with a speed of 15.6 m/s. the vertical component of earth's magnetic field in this region is 40.0 μt directed downward. calculate the induced emf between the ends of the wire and determine which end is positive.

Answer 1

The electromotive force on the free electrons in the wire is given by
E=vBl
where v is the velocity, B is the field component perpendicular to the wire and l is the length of the wire.  So we get

E=15.6*40E-6*2.5m=1.56mV

Now the force that causes this to happen is given by F=q(v X B) where q is the charge on the electron which is negative.  the cross product of the velocity and magnetic field points in a direction that is westward.  Since the electrons are negative they are pushed to the east, leaving the west end of the wire with a positive charge .