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Asked in NEET 1988 · Peak and instantaneous EMF
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.
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