#rectangular loop

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 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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AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of sides 40 cm and 20 cm moves out of a 0.9 T field at 1 m/s perpendicular to the longer side. What i

**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 = 0.9 × 0.2 × 1 = 0.18 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 loop of 0.45 m × 0.2 m moves out of a 0.6 T field at 1.5 m/s along its longer side. How long does the emf last?

**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. Time = distance/velocity, distance = width along motion = 0.2 m. t = (0.2/1.5) = 0.1333 s ≈ 0.13 s . 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

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AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of sides 30 cm and 12 cm moves out of a 0.6 T field at 1 m/s perpendicular to the shorter side. What

**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 = B l v , l = 0.3 m . ε = 0.6 × 0.3 × 1 = 0.18 V . Using Φ = B A cosθ, e = -N

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AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of sides 15 cm and 6 cm moves out of a 0.4 T field at 1.5 m/s perpendicular to the shorter side. What

**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. ε = B l v , l = 0.15 m . ε = 0.4 × 0.15 × 1.5 = 0.09 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 0.09 V 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 0.3 m × 0.45 m moves out of a 0.5 T field at 0.9 m/s along its shorter side. What is the 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. ε = B l v , l = 0.45 m . ε = 0.5 × 0.45 × 0.9 = 0.2026 V ≈ 0.203 V . Using Φ = B A cosθ,

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Motional EMF - Rod and Rectangular Loop

A loop of 0.32 m × 0.16 m moves out of a 0.35 T field at 2 m/s along its longer side. How long does the emf last?

**Rectangular loop moving out of field** emf e = B l v, l side perpendicular to motion cutting field lines, e constant while partially in field, zero when fully out, duration t = (side parallel to motion)/v. For 0.38×0.55 m loop B=0.65 T v=0.7 m/s along shorter side 0.38 m, l=0.55 m (side cutting), e=0.65×0.55×0.7=0.25 V, lasts t=0.38/0.7=0.54 s. Time = distance/velocity, distance = width along motion = 0.16 m. t = (0.16/2) = 0.08 s . 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,

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Motional EMF - Rod and Rectangular Loop

A loop of 0.35 m × 0.1 m moves out of a 0.25 T field at 2.5 m/s along its longer side. How long does the emf last?

**Motional emf** for rod length l moving with velocity v perpendicular to uniform field B, e = B l v (V), B in T, l in m, v in m/s, direction given by right-hand rule, positive end where positive charges accumulate due to q v×B force. For l=0.9 m, v=1.2 m/s, B=0.5 T, e=0.5×0.9×1.2=0.54 V. Time = distance/velocity, distance = width along motion = 0.1 m. t = (0.1/2.5) = 0.04 s . 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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Motional EMF - Rod and Rectangular Loop

A rectangular loop of 0.36 m × 0.5 m moves out of a 0.55 T field at 0.8 m/s along its shorter side. What is the emf?

**Rectangular loop moving out of field** emf e = B l v, l side perpendicular to motion cutting field lines, e constant while partially in field, zero when fully out, duration t = (side parallel to motion)/v. For 0.38×0.55 m loop B=0.65 T v=0.7 m/s along shorter side 0.38 m, l=0.55 m (side cutting), e=0.65×0.55×0.7=0.25 V, lasts t=0.38/0.7=0.54 s. ε = B l v , l = 0.5 m . ε = 0.55 × 0.5 × 0.8 = 0.22 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt,

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Motional EMF - Rod and Rectangular Loop

A rectangular loop of 0.24 m × 0.38 m moves out of a 0.3 T field at 0.6 m/s along its shorter side. What is the emf?

**Rectangular loop moving out of field** emf e = B l v, l side perpendicular to motion cutting field lines, e constant while partially in field, zero when fully out, duration t = (side parallel to motion)/v. For 0.38×0.55 m loop B=0.65 T v=0.7 m/s along shorter side 0.38 m, l=0.55 m (side cutting), e=0.65×0.55×0.7=0.25 V, lasts t=0.38/0.7=0.54 s. ε = B l v , l = 0.38 m . ε = 0.3 × 0.38 × 0.6 = 0.0684 V ≈ 0.068 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B

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Motional EMF - Rod and Rectangular Loop

A rectangular loop of sides 28 cm and 14 cm moves out of a 0.7 T field at 1.2 m/s perpendicular to the longer side. What

**Motional emf** for rod length l moving with velocity v perpendicular to uniform field B, e = B l v (V), B in T, l in m, v in m/s, direction given by right-hand rule, positive end where positive charges accumulate due to q v×B force. For l=0.9 m, v=1.2 m/s, B=0.5 T, e=0.5×0.9×1.2=0.54 V. ε = B l v , l = 0.14 m . ε = 0.7 × 0.14 × 1.2 = 0.1176 V ≈ 0.118 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L

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