A very long, straight solenoid with a cross-sectional area of 1.94 cm^2 is wound with 87.2 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)=(0.160A/s^2)t^2. A secondary winding of 5.0 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid.
A) What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?

Given Information:  

current = I = 3.2 A

Area of solenoid = A = 1.94 cm² = 0.94x10⁻³ m²

Turns per cm at primary side = n = 87.2 turns/cm = 8720 turns/m

current at time 0 = i(t) = (0.160 A/s²) t²

Number of turns secondary side = N = 5 turns

Required Information:  

Induced emf at secondary side = ξ = ?  

Induced emf at secondary side ξ = -7.605x10⁻⁶ V

Explanation:

From the Faraday's law the induced EMF ξ is given by  

ξ = -NΔΦ/Δt  

t can be found using the given i(t)

3.2 = 0.160 t²

t² = 3.2/0.160

t = √3.2/0.160

t = 4.472 seconds

change in flux ΔΦ is given by

ΔΦ = BA

ΔΦ  = μ₀nIA

ΔΦ  = 4πx10⁻⁷*8720*3.2*0.194x10⁻³

ΔΦ  = 6.802x10⁻⁶ Tm²

So the induced emf in the secondary winding is

ξ = -NΔΦ/Δt  

ξ = -5*6.802x10⁻⁶/4.472

ξ = -7.605x10⁻⁶ V

The negative sign indicates that the induced emf opposes the change that produced it in the first place.

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