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A thin semicircular conducting ring (a half ring) of radius R falls, keeping its plane vertical, through a horizontal uniform magnetic field B. At the instant the diameter MQ is horizontal (configuration MNQ) and the ring moves with speed V, find the potential difference developed across the ends of the ring.

1. zero
2. B*V*pi*R² / 2 and M is at a higher potential
3. pi*R*B*V and Q is at a higher potential
4. 2*R*B*V and Q is at a higher potential

Correct answer: 2*R*B*V and Q is at a higher potential

Solution

Motional EMF between two endpoints equals B*V times the component of the end-to-end vector perpendicular to V. For a uniform field, only the straight-line separation of the endpoints matters, not the curved path. The two ends of the semicircular ring are separated by the diameter 2R, so the EMF = B*V*(2R) = 2*R*B*V. The polarity makes Q the higher-potential end.

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