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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 OAperpendicularly, so that the rod rotates clockwise at a uniform angular velocity ω. Find the force.
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Solution
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
\[de = Blv\]
\[de = B \times dr \times \omega r\]
\[ \Rightarrow e = \int_0^a \left( B\omega r \right)dr\]
\[ = \frac{1}{2}B\omega a^2\]
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
\[F = ilB\]
\[ = \frac{B a^2 \omega}{2R} \times a \times B\]
\[ = \frac{B^2 a^2 \omega}{2R}\]
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