Figure shows a conducting circular loop of radius a placed in a uniform, perpendicular magnetic field `B.A` thick metal rod `OA` is pivoted at the centre `O` . The other end of the rod toucher the loop are connected by a wire `OC` of resistance `R.A` force is applied at the middle point of the rod `OA` perpendicularly, so that the rod rotates clockwise at a uniform angular velocity `omega` . Find the force.
(b) Consider the situation shown in the figure of the previous problem. Suppose the wire connecting `O` and `C` has zero resistance but the circular loop has a resistance `R` uniformly distributed along its lenght. The rod `OA` is made to rotate with a uniform angular speed `omega` as shown in the figure. Find the current in the rod when `angleAOC=90^(@)` .

Explanation

The rod `OA` can be replaced by a battery of emf
`e=(1)/(2)Bomegaa^(2)`
Magnetic force on loop `F_(m)=(B^(2)omegaa^(3))/(2R)` , acts at mid- point of rod and in upward direction.
Since rod rotates at constant angular velocity i.e. net torque about `O` is zero.
`F_(m)(a)/(2)=F.(a)/(2)` implies `F=F_(m)=(B^(2)omegaa^(3))/(2R)`
Resistance `prop` length `R_(CA)=R_(1)=(R)/(4) , R_(CBA)=R_(2)=(3R)/(4)`
`R_(1)` and `R_(2)` in parallel `R_(eq)=(R_(1)R_(2))/(R_(1)R_(2))=((R)/(4).(3R)/(4))/(R)=(3R)/(16)`
`I=(e)/(R_(eq))=((1)/(2)Bomegaa^(2))/((3R)/(16))=(8Bomegaa^(2))/(3R)`