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A semicircular wire of radius R is spun at a constant angular velocity omega about an axis that passes through one of its ends and is perpendicular to the plane of the semicircle. The whole arrangement is in a uniform magnetic field of magnitude B perpendicular to the plane. Find the EMF induced between the two ends of the wire.

1. B*omega*R²/2
2. 2*B*omega*R²
3. it is variable
4. none of these

Correct answer: 2*B*omega*R²

Solution

For a rigid conductor rotating in a uniform field, the EMF between its ends equals that of the straight chord connecting them, because the induced EMF depends only on endpoint coordinates. The chord joining the two ends of a semicircle is the diameter, length 2R. A straight rod of length L rotating about one end gives EMF = (1/2)*B*omega*L². With L = 2R: EMF = (1/2)*B*omega*(2R)² = (1/2)*B*omega*4R² = 2*B*omega*R².

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