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A rod of length l rotates with a small uniform angular velocity ω about an axis through its centre, perpendicular to the rod. A uniform magnetic field B is directed parallel to the rotation axis. Find the potential difference between the centre of the rod and one of its ends.
- zero
- (1/8) ω B l²
- (1/2) ω B l²
- B ω l²
Correct answer: (1/8) ω B l²
Solution
The motional emf between the rotation centre and a point at distance L is (1/2) B ω L². The centre-to-end distance is L = l/2, so the potential difference = (1/2) B ω (l/2)² = (1/2) B ω l²/4 = (1/8) B ω l².
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