The current in an air-core solenoid is reduced from 3.99 A to zero over 5.9s. The solenoid has 2000 turns per meter and a cross-sectional area of...

In this case you should apply the law of electromagnetic induction of Faraday.

Ԑi = N(Δφ/Δt)

Where:

Ԑi, is the EMF induced in the coil.

N, is the number of turns of the coil where the EMF is induced.

Δφ = BS, is the variation that occurs in the magnetic flux. In this case, we will consider that the cross section S is equal in both coils.

Δt, is the time during which the flux...

In this case you should apply the law of electromagnetic induction of Faraday.

Ԑi = N(Δφ/Δt)

Where:

Ԑi, is the EMF induced in the coil.

N, is the number of turns of the coil where the EMF is induced.

Δφ = BS, is the variation that occurs in the magnetic flux. In this case, we will consider that the cross section S is equal in both coils.

Δt, is the time during which the flux is changing.

Let's call #1 to the largest coil and #2 the smaller coil. Then we see that the variation of the magnetic flux in the coil #1, induces an EMF in the coil #2.

So, we can write the Faraday law, for coil #2, in the following way:

Ԑi = N2(Δφ2/Δt) = N2[Δ(B1S2)/Δt)]

The magnetic field of the coil #1 is:

B1 = μ0nI, where μ0 is the magnetic permeability of vacuum, n is the number of turns per unit length of the coil and I is the current.

Substituting in the equation of the EMF and considering that only varies the current, we have:

Ԑi = N2[Δ(μ0nIS2)/Δt)] = (N2μ0nS2)(ΔI/Δt)

Ԑi = (50)(4π*10^-7)(2*10^3)(0.31)(3.99/5.9)

Ԑi = 1.11*10^-2 V

The current is calculated by applying OHM's law:

I = Ԑi/R = (1.11*10^-2)/(4.09*10^-3)

I = 2.7 A