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Asked in GUJCET 2021 · Mutual inductance and induced EMF
Given: M=1.5 H, Δ I=20-0=20 A, Δ t=0.5 s.
Idea: the flux linked with the second coil is proportional to the current in the first, N₂Φ₂=MI₁, so Δ(N₂Φ₂)=M Δ I₁.
Δ(N₂Φ₂)=1.5×20.
So the change of flux linkage is 30 Wb. The 0.5 s is needed only for the emf, (30)/(0.5)=60 V.
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