A metal disc of radius ' R ' rotates with an angular velocity ' ' about an axis perpendicular to its plane passing through its centre in a magnetic field of induction ' B ' acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is (magnitude only)

Step-by-step solution

The correct option is (D). We can imagine the disc to be a collection of thin rods connected in parallel between the center of the disc and the rim. So if we calculate the induced emf on a thin rod rotating about its axis then this should be equal to that of the disc. The tiny motional emf developed across the element dr can be written as: dE = Bvdr Taking the velocity v = r , dE = B rdr On integrating across the rod, . E = ₀^ R B rdr = B r ^2 2 ]₀^ R = B R ^2 2