A thin wire AC shaped as a semi-circle of diameter d rotates with a constant angular frequency ω in a uniform magnetic field of induction B→. The vector ω→ is parallel to B→and the rotation axis XY passes through the end A of the wire and is perpendicular to the diameter AC (see figure). B→ is directed left to right in the plane of the paper. The value of the line integral I=∫E→.dr→ taken along the wire from the point A to the point C will be ωBd2n. The value of n is

It is clear that AC traces out a circle as the given wire rotates about axis XY. The emf developed by induction between the ends A and C of the given semicircular wire isthe same as if a straight wire AC were rotating perpendicular to field B.Hence flux linked ϕB=B12AC2θwhere θ is the angle traced in time t.By Faraday's law, the induced emf e=∮E→.dl=e=-dϕBdtBut we also have, ∮E.dl=e=-dϕBdt∮E.dr=-ddt12d2θB=-12d2Bdθd=-B2ωd2But we desire only the ∫CAE.dr which will be 12 or ∮E.drHence ∫CAE.dr=-B4ωd2