The mutual inductance between two coils is $0.15 H$ . If a current of $20 A$ in the primary coil is cut off in $0.05 s$ , find the emf induced in the secondary coil. If the secondary coil has $60$ turns, then find the flux-change in it.

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Solution

Emf induced is $E = L \frac{- Δ I}{Δ t} = 0.15 \times \frac{20}{0.05}$

$⟹ E = 60 V$

$E = \frac{Δ ϕ}{Δ t}$

$⟹ Δ ϕ = 60 \times 0.05 = 3 w e b e r$

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The mutual inductance between two coils is 0.15H . If a current of 20A in the primary coil is cut off in 0.05s , find the emf induced in the secondary coil. If the secondary coil has 60 turns, then find the flux-change in it.

Emf induced is E=L−ΔIΔt=0.15×200.05⟹E=60V E=ΔϕΔt⟹Δϕ=60×0.05=3weber

The mutual inductance between two coils is $0.15 H$ . If a current of $20 A$ in the primary coil is cut off in $0.05 s$ , find the emf induced in the secondary coil. If the secondary coil has $60$ turns, then find the flux-change in it.

Emf induced is $E = L \frac{- Δ I}{Δ t} = 0.15 \times \frac{20}{0.05}$

$⟹ E = 60 V$

$E = \frac{Δ ϕ}{Δ t}$

$⟹ Δ ϕ = 60 \times 0.05 = 3 w e b e r$