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A metallic rod of length l is attached to a string of length 2l and rotated with angular speed omega on a horizontal table, the far end of the string being fixed. A uniform vertical magnetic field B is present. The emf induced between the two ends of the rod is:

1. 2*B*omega*l²/2
2. 3*B*omega*l²/2
3. 4*B*omega*l²/2
4. 5*B*omega*l²/2

Correct answer: 5*B*omega*l²/2

Solution

The rotating rod spans radii 2l to 3l. Integrating B*omega*r dr (or using the difference of squares formula) gives emf = (1/2)*B*omega*((3l)² - (2l)²) = 5*B*omega*l²/2.

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