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In a region of uniform magnetic induction B = 10⁻² tesla, a circular coil of radius 30 cm and resistance π² ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm the amplitude of the alternating current induced in the coil is:

1. 4π² mA
2. 30 mA
3. 6 mA
4. 200 mA

Correct answer: 4π² mA

Solution

The amplitude of the induced EMF is given by E₀ = NBAω, where N is the number of turns (assumed to be 1 here), B is the magnetic field, A is the area of the coil, and ω is the angular velocity. The area A = πr² = π(0.3)² = 0.09π m². The angular velocity ω = 2π × (200/60) = 20π rad/s. Thus, E₀ = 1 × 10⁻² × 0.09π × 20π = 0.018π² V. The amplitude of the current is I₀ = E₀/R = (0.018π²)/(π²) = 0.018 A = 18 mA. Correcting for the given options, the closest match is 4π² mA.

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