AC Generator, Back EMF and Time Duration of EMF

AC Generator, Back EMF and Time Duration of EMF

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.

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AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of 0.1 m × 0.2 m moves out of a 0.4 T field at 0.5 m/s along its shorter side. What is the emf?

**Circular loop deformed into straight wire** in field B=0.12 T radius 16 cm area πr²=0.0804 m² flux 0.00965 Wb drops to zero in 0.6 s e=0.0161 V, illustrating flux change due to area change induces emf, even without B change, area deformation changes Φ = B A cosθ. ε = B l v , l = 0.2 m . ε = 0.4 × 0.2 × 0.5 = 0.04 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

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AC Generator, Back EMF and Time Duration of EMF

A coil with high self-inductance is suddenly disconnected from a battery. The resulting high voltage spike is due to wha

**Circular loop deformed into straight wire** in field B=0.12 T radius 16 cm area πr²=0.0804 m² flux 0.00965 Wb drops to zero in 0.6 s e=0.0161 V, illustrating flux change due to area change induces emf, even without B change, area deformation changes Φ = B A cosθ. The rapid current drop induces a large back emf ( ε = -L (dI/dt) ) due to self-inductance, causing a voltage spike. 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

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AC Generator, Back EMF and Time Duration of EMF

A rod rotates at 15 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.4 m, what is the emf induced?

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. ε = (1/2) B ω R² . ε = (1/2) × 0.3 × 15 × (0.4)² = 0.36 V . Using Φ = B A cosθ, e = -N dΦ/dt =

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AC Generator, Back EMF and Time Duration of EMF

A loop of 0.2 m × 0.1 m moves out of a 0.5 T field at 2 m/s along its longer side. How long does the emf last?

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. Time = distance/velocity, distance = width along motion = 0.1 m. t = (0.1/2) = 0.05 s . Using Φ = B A cosθ, e = -N dΦ/dt = -N A

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 15 cm is deformed into a straight wire in a 0.15 T field in 0.6 s. What is the induced emf?

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. Initial flux: Φ = B A = 0.15 × π × (0.15)² = 0.0106 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.0106/0.6) = 0.01767 V ≈

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AC Generator, Back EMF and Time Duration of EMF

In an AC generator, slip rings are used instead of a split ring commutator. What is the primary reason for this design c

**Circular loop deformed into straight wire** in field B=0.12 T radius 16 cm area πr²=0.0804 m² flux 0.00965 Wb drops to zero in 0.6 s e=0.0161 V, illustrating flux change due to area change induces emf, even without B change, area deformation changes Φ = B A cosθ. Slip rings allow continuous contact with the coil, preserving the alternating nature of the emf, unlike a split ring commutator which rectifies it to DC. 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

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 12 cm is deformed into a straight wire in a 0.2 T field in 0.5 s. What is the induced emf?

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. Initial flux: Φ = B A = 0.2 × π × (0.12)² = 0.00904 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00904/0.5) = 0.01808 V ≈

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 10 cm is deformed into a straight wire in a 0.1 T field. If the flux change occurs in 0.2 s, w

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. Initial flux: Φ = B A = 0.1 × π × (0.1)² = 0.00314 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00314/0.2) = 0.0157 V ≈

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 14 cm is deformed into a straight wire in a 0.15 T field in 0.5 s. What is the induced emf?

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. Initial flux: Φ = B A = 0.15 × π × (0.14)² = 0.00923 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00923/0.5) = 0.01846 V ≈

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 13 cm is deformed into a straight wire in a 0.18 T field in 0.6 s. What is the induced emf?

**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. Initial flux: Φ = B A = 0.18 × π × (0.13)² = 0.00956 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00956/0.6) = 0.01593 V ≈ 0.016 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)

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AC Generator, Back EMF and Time Duration of EMF

A loop of 0.3 m × 0.12 m moves out of a 0.4 T field at 2 m/s along its longer side. How long does the emf last?

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. Time = distance/velocity, distance = width along motion = 0.12 m. t = (0.12/2) = 0.06 s . Using Φ = B A cosθ, e = -N dΦ/dt = -N A

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