Shows a conducting circular loop radius a placed in a uniform, perpendicular magnetic field B. A metal rod OA is pivoted at the centre O of the loop. The other A of the rod touches the loop. The rod OA and the loop are resistanceless but a resistor having a resistance R is connected between O and a fixed point C on the loop . the rod OA is made to rotate anticlockwise at a small but uniform angular speed `omega` by an external force. Find (a) the current in the resistance R and (b) the torque of the external force needed to keep the rod rotating with the constant angular velocity omega.

Explanation

The emf between the ends of the rotating rod is `epsilon = int d epsilon = int_0 ^a Bomegax dx =(1)/(2) B omega a^2`
The positive charges of the rod will be pushed towards O by the magnetic field
Thus, the rod may be replaced by a battery of `emf = (1)/(2) B omega a ^2` with the positive terminal towards O
The equivalent circuit diagram is shown in the circular loop joins A to C by a resistanceless path
(a) The current in the resistance R is `i = (epsilon)/(R) = (B omega a ^2)/(2 R)`
(b) the force on the rod due to the magnetic field is F=iaB
As the force is uniformly distributed over OA, it may be assumed to act at the middle point of OA
The torque is, therefore, `Gamma =(iaB) (a)/(2) = (B^2 omega a^4) /(4 R)` in clockwise direction
To keep the rod rotating at uniform angular velocity, an external torque `(B^2 omega a^4)/(4R)` in anticlockwise direction is needed.