Question

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 length. The rod OA is made to rotate with a uniform angular speed ω as shown in the figure. Find the current in the rod when ∠ AOC = 90°.

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
$$\begin{gathered} de=Blv \\ de=B\times dr\times \omega r \\ \Rightarrow e={\int }_0^a\left(B\omega r\right)dr \\ =\frac{1}{2}B\omega a^2 \end{gathered}$$

As ∠AOC = 90°, the minor and major segments of AC are in parallel with the rod.
The resistances of the segments are $\frac{R}{4}$ and $\frac{3R}{4}$.
The equivalent resistance is given by
$R\text{'}=\frac{{\displaystyle \frac{R}{4}}\times {\displaystyle \frac{3R}{4}}}{R}=\frac{3R}{16}$
The motional emf induced in the rod rotating in the clockwise direction is given by
$e=\frac{1}{2}B\omega a^2$
The current through the rod is given by
$$\begin{gathered} i=\frac{e}{R\text{'}} \\ =\frac{B{a}^2\omega }{2R\text{'}} \\ =\frac{B{a}^2\omega }{2\times 3R/16} \\ =\frac{B{a}^2\omega \times 16}{2\times 3R}=\frac{8B{a}^2\omega }{3R} \end{gathered}$$