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Electric Charges And Fields

Question
CBSEENPH12039076

A metallic rod of ‘L’ length is rotated with angular frequency of ‘ straight omega ’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.

Solution

Emf is induced is given by, epsilon space equals fraction numerator d phi subscript B over denominator d t end fraction equals straight d over dt left parenthesis BA right parenthesis straight space equals straight space straight B straight space dA over dt
where, dA over dt comma spaceis the rate of change of area of the loop formed by the sector OPQ.

At any instant of time t, let  be the angle between the rod and the radius of the circle at P.
Area of the sector OPQ = straight R squared cross times fraction numerator straight theta over denominator 2 straight pi end fraction equals 1 half straight R squared straight theta ;R is the radius of the circle.
Therefore, emf induced is, Error converting from MathML to accessible text. 
                                     straight epsilon equals straight space 1 half BR squared dθ over dt equals BωR squared over 2