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A circular coil of resistance 10 ohm, radius 10 cm and 100 turns is spun at 100 revolutions per second about a diameter that is perpendicular to a uniform magnetic field of 10 mT. Taking pi² = 10, the peak (amplitude) value of the induced current in the coil is nearly:

1. 2 A
2. 200 A
3. 0.002 A
4. none of these

Correct answer: 2 A

Solution

Peak EMF e0 = N B A omega = N B (pi r²)(2 pi f) = 100 * 0.01 * pi*(0.1)² * 2 pi * 100. Compute: 100*0.01 = 1; pi*0.01 = pi*0.01; times 2*pi*100 = 200 pi. So e0 = 1 * (0.01 pi) * (200 pi) = 2 pi² = 2*10 = 20 V. Peak current = 20/10 = 2 A.

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