Say the turbine is 10 feet in diameter (that's the radius of the dashed circle). Also say that the coil has 100 turns and has a square cross-section with a length of 10 feet and a height of 6 feet. Say that the magnetic rotor has the same height but is only 2 feet wide, it has a magnetic field strength is 0.1T, and it is rotating at 60Hz (note this is not the angular frequency). A typical turbine supplies 10kW of power. Use Faraday's law to find the induced emf in the coil and the amount of induced current.

a. ε = 21,014sin(120πt) V

b. 0.476cosec(120πt)  

Explanation:

a. Induced emf

We know the induced emf, ε = -dΦ/dt where Φ = magnetic flux through coil = NABcosθ where N = number of turns of coil =, 100, A = area of coil = 10 ft × 6 ft = 60 ft² = 60 × 1 ft² = 60 × (0.3048)² m² = 5.574 m², B = magnetic field strength = 0.1 T and θ = angle between B and normal to A = ωt.

So, Φ = NABcosθ = 100 × 5.574 m² × 0.1 T cosθ  = 55.74cosθ Tm²

So, ε = -dΦ/dt = ε = -d(55.74cosθ Tm²)/dt = -d(55.74cosθ Tm²)/dθ × dθ/dt = -55.74 ×(-sinθ) Tm²)/dθ × ω (ω = dθ/dt = angular frequency of shaft = 2πf where f = frequency of rotor = 60 Hz )

ε = 55.74sinθ Tm²) × 2πf

ε = 55.74sinθ Tm²) × 2π(60 Hz)

ε = 6689πsinθ V

ε = 21,014sinθ V

ε = 21,014sinωt V

ε = 21,014sin(2πft) V

ε = 21,014sin(2π(60 Hz)t) V

ε = 21,014sin(120πt) V

b. Current in coil

Since power P = Iε where I = current and ε = induced emf = 21,014sinθ V.

Since power, P = 10 kW = 10000 W

I = P/ε

= 10000 W/21,014sinθ V

= 0.476/sinθ

= 0.476cosecθ

= 0.476cosecωt

= 0.476cosec(120πt)  

The maximum current is obtained when θ = 90°

I =  10000 W/21,014sin90 V

I = 10000 W/21,014 V

I = 0.476 A

I = 476 mA

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