Answered

an emf of 19.0 mv is induced in a 527-turn coil when the current is changing at the rate of 10.0 a/s. what is the magnetic flux through each turn of the coil at an instant when the current is 4.60 a? (enter the magnitude.)



Answer :

The magnitude of the magnetic flux through each coil when the current is 4.00A is 15.58 T.m²

N = 527 turn

Δi/Δt  = 10.0 A/s

ϵ = 19.0 mV = 19 x 10⁻³ V

I = 4.60 A

From ϵ = L(Δi/Δt), we have:

ϵ = L(Δi/Δt)

19 x 10⁻³ V = L (10.0 A/s)

L = 19 x 10⁻³V/10.0 A/s

L = 19 x 10⁻² H

From L = NΦ/I, we have:

Φ = Li/N

Φ = 19 x 10⁻² H (4.60 A)/527

Φ = 15.58 μT.m²

So, the magnitude of the magnetic flux is 15.58 T.m²

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