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In a region of 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

Asked in NEET 1988 · Peak and instantaneous EMF

Answer: (3) 6 mA

Step-by-step solution

The peak emf of a coil rotating in a uniform field is ε₀=NBAω, with N=1.

A=π r²=π(0.3)²=0.09π m².

ω=(200×2π)/(60)=(20π)/3 rad s⁻¹.

ε₀=10⁻²×0.09π×(20π)/3=0.06π²×10⁻¹=6π²×10⁻³ V.

I₀=(ε₀)/R=(6π²×10⁻³)/(π²)=6×10⁻³ A=6 mA.

The resistance was chosen as π² precisely so that it cancels.

Why the other options are wrong

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