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Asked in GSEB Board March 2018 · Mutual inductance and induced EMF
Given: M = 5 mH = 5×10⁻³ H, I = I₀ sin ω t with I₀ = 10 A and ω = 100π rad s⁻¹.
Idea: the emf in the second coil is ε₂ = -M(dI)/(dt).
(dI)/(dt) = I₀ω cos ω t, so ε₂ = -MI₀ω cos ω t.
Its maximum value, when |cos ω t| = 1, is MI₀ω.
εₘₐₓ = 5×10⁻³×10×100π = 5π V (about 15.7 V).
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