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Asked in GUJCET 2020 · Peak and instantaneous EMF
Given: N = 1000, A = 0.10 m², B = 0.01 T, f = 0.5 rev s⁻¹.
Angular speed: ω = 2π f = 2π×0.5 = π rad s⁻¹.
Idea: a coil rotating in a uniform field generates ε = NABω sin ω t, so the peak emf is ε₀ = NABω.
ε₀ = 1000×0.10×0.01×π = π ≈ 3.14 V.
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