Question

Consider the situation shown in the figure. Suppose the wire connecting O and C has zero resistance but the circular loop has a resistance Runiformly 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

\[de = Blv\]

\[de = B \times dr \times \omega r\]

\[ \Rightarrow e =  \int_0^a \left( B\omega r \right)dr\]

\[= \frac{1}{2}B\omega a^2\]

As ∠AOC = 90°, the minor and major segments of AC are in parallel with the rod.

The resistances of the segments are `R/4` and `(3R)/4.`

The equivalent resistance is given by

`R'=(R/4xx(3R/4))/R=(3R)/16`

The motional emf induced in the rod rotating in the clockwise direction is given by

`e=1/2Bomegaa^2`

The current through the rod is given by

\[i = \frac{e}{R'}\]

\[ = \frac{B a^2 \omega}{2R'}\]

\[ = \frac{B a^2 \omega}{2 \times 3R/16}\]

\[ = \frac{B a^2 \omega \times 16}{2 \times 3R} = \frac{8B a^2 \omega}{3R}\]