A horizontal straight wire 10 m long extending from east to west is falling with a speed of `5.0 "m s"^(-1)`, at right angles to the horizontal component of the earth's magnetic field, `0.30 xx 10^(-4) "Wb m"^(-2)`.
(a) What is the instantaneous value of the emf induced in the wire ?
(b) What is the direction of the emf ?
(c) Which end of the wire is at the higher electrical potential ?

Explanation

(a)`epsilon=Bvl`
`=(0.3xx10^(-4))(50)(10)`
`=1.5xx10^(-3)`V
(b)Here during the fall of wire, free electrons acquire vertically downward velocity which is perpendicular to `vecB_h`. Hence magnetic force `vecF_m= e(vecB_h xx vecv)` is exerted on the free electrons in a direction from east to west. Hence eastern end of wire becomes positive by losing electrons whereas western end of wire becomes negative by gaining electrons. Direction of this emf is away from positive terminal i.e. in a direction from west to east. (through the wire)
(c ) As explained above, eastern end of wire is at higher electric potential.