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A conducting rod AB of mass M and length L is hinged at its end A. It can rotate freely in the vertical plane (in the plane of the Figure). A long straight wire is vertical and carrying a current I. The wire passes very close to A. The rod is released from its vertical position of unstable equilibrium. The emf between the ends of the rod when it has rotated through an angle $θ$ is $\frac{μ_{0} I}{2 π \sin θ} \sqrt{n g L (1 - \cos θ)}$ . Find ‘n’

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answer is 3.

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Detailed Solution

$ω = \sqrt{\frac{3 g (1 - \cos θ)}{L}} ε = \frac{μ_{0} I}{2 π \sin θ} \sqrt{\frac{3 g (1 - \cos θ)}{L}}$

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