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A conducting rod of length ℓ is attached to a string of length 2ℓ and is rotated on a horizontal table about the fixed end of the string with angular speed ω. If a uniform magnetic field B acts vertically through the region, what is the induced emf between the two ends of the rod?
- Bωℓ²
- 3Bωℓ²/2
- 2Bωℓ²
- 5Bωℓ²/2
Correct answer: 5Bωℓ²/2
Solution
The rod spans radius 2L to 3L. EMF = B*omega*integral(r dr) from 2L to 3L = B*omega*(9L^2-4L^2)/2 = 5*B*omega*L^2/2.
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