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Rotational EMF and AC Generator

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30 questions

A square loop of side 18 cm rotates at 10 rad/s in a 0.25 T field. What is the maximum emf induced?

**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. A = (0.18)² = 0.0324 m² . ε₀ = N B A ω = 1 × 0.25 × 0.0324 × 10 = 0.081 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A square loop of side 20 cm is rotated in a 0.2 T field at 10 rad/s. What is the maximum emf induced?

**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. ε₀ = N B A ω , N = 1 , A = (0.2)² = 0.04 m² . ε₀ = 1 × 0.2 × 0.04 × 10 = 0.08 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A coil of 300 turns rotates at 100 rad/s in a 0.05 T field. If the area is 0.01 m², what is the maximum emf?

**Maximum emf** e₀ = N B A ω, for 180 turns A=0.02 m² B=0.03 T ω=60 rad/s, e₀=180×0.03×0.02×60=6.48 V, illustrating dependence on N, B, A, ω, used in generator design, increasing N or B or A or ω raises output. ε₀ = N B A ω = 300 × 0.05 × 0.01 × 100 = 15 V . 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 15 V follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A square loop of side 15 cm rotates at 15 rad/s in a 0.2 T field. What is the maximum emf induced?

**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. A = (0.15)² = 0.0225 m² . ε₀ = N B A ω = 1 × 0.2 × 0.0225 × 15 = 0.0675 V . 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 0.0675 V follows,

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A circular coil of radius 7 cm and 180 turns rotates at 20 rad/s in a 0.07 T field. What is the maximum emf induced?

**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. A = π r² = 3.14 × (0.07)² = 0.0154 m² . ε₀ = N B A ω = 180 × 0.07 × 0.0154 × 20 = 3.88 V ≈ 3.9 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A rod rotates at 20 rad/s in a 0.4 T field. If the length from the axis to the tip is 0.5 m, what is the emf induced?

**Maximum emf** e₀ = N B A ω, for 180 turns A=0.02 m² B=0.03 T ω=60 rad/s, e₀=180×0.03×0.02×60=6.48 V, illustrating dependence on N, B, A, ω, used in generator design, increasing N or B or A or ω raises output. ε = (1/2) B ω R² . ε = (1/2) × 0.4 × 20 × (0.5)² = 5 V . 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 5 V follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A circular coil of radius 8 cm and 240 turns rotates at 45 rad/s in a 0.04 T field. What is the maximum emf induced?

**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. A = π r² = 3.14 × (0.08)² = 0.0201 m² . ε₀ = N B A ω = 240 × 0.04 × 0.0201 × 45 = 8.676 V ≈ 8.68 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A circular coil of radius 5 cm and 100 turns is rotated in a 0.1 T magnetic field at 50 rad/s. What is the maximum emf i

**Maximum emf** e₀ = N B A ω, for 180 turns A=0.02 m² B=0.03 T ω=60 rad/s, e₀=180×0.03×0.02×60=6.48 V, illustrating dependence on N, B, A, ω, used in generator design, increasing N or B or A or ω raises output. Maximum emf: ε₀ = N B A ω . A = π r² = 3.14 × (0.05)² = 0.00785 m² . ε₀ = 100 × 0.1 × 0.00785 × 50 = 3.925 V ≈ 3.93 V . 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,

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

In an AC generator, the maximum emf occurs when the coil is in which orientation relative to the magnetic field?

**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 maximum emf occurs when the rate of flux change is greatest, which happens when the coil is perpendicular to the field ( sinθ = 1 ). Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A circular coil of radius 4 cm and 250 turns rotates at 40 rad/s in a 0.06 T field. What is the maximum emf induced?

**Maximum emf** e₀ = N B A ω, for 180 turns A=0.02 m² B=0.03 T ω=60 rad/s, e₀=180×0.03×0.02×60=6.48 V, illustrating dependence on N, B, A, ω, used in generator design, increasing N or B or A or ω raises output. A = π r² = 3.14 × (0.04)² = 0.005024 m² . ε₀ = N B A ω = 250 × 0.06 × 0.005024 × 40 = 3.0144 V ≈ 3.01 V . 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

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A square loop of side 26 cm rotates at 12 rad/s in a 0.3 T field. What is the maximum emf induced?

**Maximum emf** e₀ = N B A ω, for 180 turns A=0.02 m² B=0.03 T ω=60 rad/s, e₀=180×0.03×0.02×60=6.48 V, illustrating dependence on N, B, A, ω, used in generator design, increasing N or B or A or ω raises output. A = (0.26)² = 0.0676 m² . ε₀ = N B A ω = 1 × 0.3 × 0.0676 × 12 = 0.24336 V ≈ 0.243 V . 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,

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

A coil of 140 turns and area 0.02 m² is rotated at 25 Hz in a 0.07 T field. What is the maximum emf?

**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. ω = 2π v = 2π × 25 = 50π rad/s . ε₀ = N B A ω = 140 × 0.07 × 0.02 × 50π = 30.79 V ≈ 30.8 V . 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

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator