Question

Derive an expression for instantaneous induced emf in an A.C. generator.

Answer 1

In the above figure 

N and S are poles of a magnet, θ is the angle between the directions of magnetic field B and area vector A. 

When the coil is rotated in the magnetic field, the flux linked with the coil varies. At any’ instant of time ‘t’, A cosθ is the component of area vector along the direction B. 

The magnetic flux linked with the coil at any instant of time ‘t’ is given by 

Φ_B = B x component of area vector along the field direction. 

For 1 turn Φ_B = BA cos θ 

For n turns Φ_B = nAB cos θ 

Φ_B = nABcosωt….(1) [∵ θ = ωt] 

Where ‘ω’ is the angular velocity of the coil at time t. From the Faraday’s second law,

e = \(\frac{\mathrm{d} \phi}{\mathrm{d} \mathrm{t}}\)

e = -\(\frac{\mathrm{d}}{\mathrm{dt}}\) [nAB cos cot]

[From (l) Φ = nABcosωt]

e = (-n AB) [-sin ωt) × ω 

e = nABωsinωt

e = e₀ sin(ωt) 

where e₀ is the peak value of emf = nAB ω