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 0.131 m2. Surrounding the solenoid near the center of its length is a second coil of 50 turns. What is the magnitude of the induced emf in the second coil? If the resistance of the second coil is 0.00409 ohm, what is the induced current? I have no idea how to do this.

Expert Answers

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

Ԑi = N(Δφ/Δt)


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

Approved by eNotes Editorial Team

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