An AC generator consists of a coil of 800 turns and a cross-sectional area of 2.5 m 2 . The coil is placed in a uniform
Emf e=NBAω sinωt, peak e0=NBAω, calculate per given B,A,N.
Ref: NCERT Physics Class 12 Chapter 3 Current Electricity
#AC generator
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The working of A.C Generator is based on the principle of
AC generator based on electromagnetic induction, rotating coil in magnetic field.
Ref: NCERT Physics Class 12 Chapter 4 Moving Charges and Magnetism
In an AC generator, what happens to the induced emf when the rotational speed of the coil doubles?
**Back emf** in motor opposes applied voltage, e_b = N B A ω sin ωt, reduces net current, at start ω=0 e_b=0 current large, as speed increases e_b increases limiting current, power conversion mechanical, principle of motor and generator reciprocity. The emf is proportional to the angular speed ( ε = N B A ω ), so doubling the rotational speed doubles the emf. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U = ½ L I², result Doubles follows, reflecting Faraday's law and Lenz's opposition.
Ref: NCERT > Physics Book > Electromagnetic Induction > AC Generator, Back EMF and Time Duration of EMF
In an AC generator, the frequency of the output voltage depends on which factor?
**AC generator** principle same as rotating coil, N=200 turns A=0.04 m² B=0.1 T f=50 Hz ω=2π×50=314 rad/s, e₀= N B A ω =200×0.1×0.04×314=251.2 V, emf e= e₀ sin ωt, frequency equals rotation frequency, maximum when plane parallel to field. The frequency equals the rotational speed of the coil in revolutions per second, as each full rotation produces one cycle of emf ( f = (ω/2π) ). Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U = ½ L I², result Rotational
Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator
In an AC generator, increasing the number of turns in the coil affects what aspect of the output?
**Magnetic flux** Φ = B·A = B A cosθ, B magnetic field (T), A area (m²), θ angle between B and normal to area, unit Wb = T·m², Faraday's law induced emf e = -N dΦ/dt, N turns, negative sign Lenz's law indicating opposition, magnitude |e| = N |ΔΦ/Δt|, for 100 turns ΔΦ=0.03 Wb Δt=0.06 s e=100×0.03/0.06=50 V. The emf ( ε = N B A ω sinω t ) is directly proportional to the number of turns, so more turns increase the amplitude of the induced emf. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l
Ref: NCERT > Physics Book > Electromagnetic Induction > Magnetic Flux and Faraday's Laws of Induction
In an AC generator, the emf varies sinusoidally with time. What is the primary reason for this variation?
**Rotational emf** when coil area A rotates with angular speed ω in uniform field B, flux Φ = B A cos ωt, emf e = -N dΦ/dt = N B A ω sin ωt, maximum e₀ = N B A ω, frequency = ω/2π. For square side 22 cm area 0.0484 m² N=1 ω=14 rad/s B=0.15 T, e₀=1×0.15×0.0484×14=0.1016 V, sinusoidal. The emf varies sinusoidally because the coil rotates in a uniform magnetic field, causing the angle between the magnetic field and the coil’s area vector to change continuously, following a sine function. Using Φ = B A cosθ, e = -N dΦ/dt = -N A
Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator
