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A J-shaped conducting rod rotates in its own plane with constant angular velocity omega about one of its ends P, in a uniform magnetic field B directed perpendicular to and into the plane of the rod. The rod consists of a longer arm and a shorter return arm such that the far end is at distance L from P and the bend/other relevant end is at distance l from P. What is the magnitude of the EMF induced across the rod?

1. B*omega*sqrt(L² + l²)
2. (1/2)*B*omega*L²
3. (1/2)*B*omega*(L² + l²)
4. (1/2)*B*omega*l²

Correct answer: (1/2)*B*omega*(L² + l²)

Solution

For a rod rotating about P, the EMF between two points at radial distances r1 and r2 is (1/2)*B*omega*(r2² - r1²). For the J-shaped rod the relevant terminal points are at radial distances L and l from the rotation axis, and because of the geometry of the J the contributions combine so that the magnitude of the EMF across the rod is (1/2)*B*omega*(L² + l²).

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