Question

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 touches the loop at A. The centre O and a fixed point C on 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 ω. Find the force.

Answer 1

Calculation of the emf induced in the rotating rod:
It is given that the angular velocity of the disc is ω and the magnetic field perpendicular to the disc is having magnitude B.

Let us take an element of the rod of thickness dr at a distance r from the centre.
Now,
Linear speed of the element at r from the centre, v = ωr
$$\begin{gathered} de=Blv \\ de=B\times dr\times \omega r \\ \Rightarrow e={\int }_0^a\left(B\omega r\right)dr \\ =\frac{1}{2}B\omega a^2 \end{gathered}$$
Because it is connected to resistance R, the current in the circuit containing the rod, wire and circular loop is given by
$i=\frac{B{a}^2\omega }{2R}$
The direction of the current is from point A to point O in the rod.
The magnitude of the force that is applied on the rod is given by
$$\begin{gathered} F=ilB \\ =\frac{B{a}^2\omega }{2R}\times a\times B \\ =\frac{B^2{a}^2\omega }{2R} \end{gathered}$$