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.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

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