Self-Induction and Self-Inductance

Self-Induction and Self-Inductance

A conducting disc rotates in a uniform magnetic field parallel to its axis. The induced emf between the center and rim a

**Self-induction** emf induced in coil due to change in its own current, e = -L dI/dt, L self-inductance (H), L = μ₀ N² A / l for solenoid, N turns, A area (m²), l length (m), μ₀=4π×10⁻⁷ H/m. For solenoid 650 turns/m means n=650, A=0.014 m², L = μ₀ n² A l? Actually per unit length? For length l, N=n l, L= μ₀ n² A l, if l=1 m, L=4π×10⁻⁷×650²×0.014=7.43×10⁻³ H, dI/dt=(3-6)/0.25=-12 A/s, e= -L×(-12)=0.089 V. Rotation causes radial charge separation via the magnetic force ( F = q v × B ), inducing an emf from the center to the rim. Using Φ =

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Self-Induction and Self-Inductance

A coil of 260 turns rotates at 75 rad/s in a 0.05 T field. If the area is 0.018 m², what is the maximum emf?

**Energy stored in inductor** U =½ L I², L inductance, I current, energy in magnetic field, density u = B²/(2μ₀), B=μ₀ n I inside solenoid, U = (B²/2μ₀)×volume, illustrating equivalence of circuit and field energy. ε₀ = N B A ω = 260 × 0.05 × 0.018 × 75 = 17.55 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 17.55 V follows, reflecting Faraday's law and Lenz's opposition.

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Self-Induction and Self-Inductance

A coil of 320 turns rotates at 65 rad/s in a 0.08 T field. If the area is 0.015 m², what is the maximum emf?

**Self-inductance of solenoid** L = μ₀ N² A / l, N total turns, A cross-section, l length, for N=650 turns per meter means n=650 m⁻¹, if length 1 m N=650, A=0.014, L=4π×10⁻⁷×650²×0.014/1=0.00743 H, self-induced emf magnitude L |dI/dt|, dI/dt=12 A/s, e=0.089 V, opposes change. ε₀ = N B A ω = 320 × 0.08 × 0.015 × 65 = 24.96 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 24.96 V follows, reflecting Faraday's law and Lenz's opposition.

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Self-Induction and Self-Inductance

A rod rotates at 10 rad/s in a 0.5 T field. If the length from the axis to the tip is 0.6 m, what is the emf induced?

**Energy stored in inductor** U =½ L I², L inductance, I current, energy in magnetic field, density u = B²/(2μ₀), B=μ₀ n I inside solenoid, U = (B²/2μ₀)×volume, illustrating equivalence of circuit and field energy. ε = (1/2) B ω R² . ε = (1/2) × 0.5 × 10 × (0.6)² = 0.9 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 0.9 V follows, reflecting Faraday's law and Lenz's opposition.

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Self-Induction and Self-Inductance

A conducting loop is stationary in a uniform magnetic field that increases in strength. The induced emf is generated by

**Self-inductance of solenoid** L = μ₀ N² A / l, N total turns, A cross-section, l length, for N=650 turns per meter means n=650 m⁻¹, if length 1 m N=650, A=0.014, L=4π×10⁻⁷×650²×0.014/1=0.00743 H, self-induced emf magnitude L |dI/dt|, dI/dt=12 A/s, e=0.089 V, opposes change. A changing magnetic field induces an electric field (per Maxwell’s equations), which drives the emf in the stationary loop. 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 Induced electric field follows,

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Self-Induction and Self-Inductance

A conducting rod moves perpendicular to a uniform magnetic field with constant velocity. What is true about the induced

**Self-induction** emf induced in coil due to change in its own current, e = -L dI/dt, L self-inductance (H), L = μ₀ N² A / l for solenoid, N turns, A area (m²), l length (m), μ₀=4π×10⁻⁷ H/m. For solenoid 650 turns/m means n=650, A=0.014 m², L = μ₀ n² A l? Actually per unit length? For length l, N=n l, L= μ₀ n² A l, if l=1 m, L=4π×10⁻⁷×650²×0.014=7.43×10⁻³ H, dI/dt=(3-6)/0.25=-12 A/s, e= -L×(-12)=0.089 V. The induced emf ( ε = B l v ) remains constant because the velocity, magnetic field, and rod length are constant, leading to a steady flux change rate.

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Self-Induction and Self-Inductance

A coil of self-inductance 1.8 H has its current increased from 3 A to 7 A in 0.5 s. What is the magnitude of the induced

**Energy stored in inductor** U =½ L I², L inductance, I current, energy in magnetic field, density u = B²/(2μ₀), B=μ₀ n I inside solenoid, U = (B²/2μ₀)×volume, illustrating equivalence of circuit and field energy. ε = L (Δ I/Δ t) . Δ I = 7 - 3 = 4 A , Δ t = 0.5 s . ε = 1.8 × (4/0.5) = 1.8 × 8 = 14.4 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 =

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Self-Induction and Self-Inductance

A conducting loop is compressed in a uniform magnetic field. The induced current flows to maintain what quantity against

**Self-inductance of solenoid** L = μ₀ N² A / l, N total turns, A cross-section, l length, for N=650 turns per meter means n=650 m⁻¹, if length 1 m N=650, A=0.014, L=4π×10⁻⁷×650²×0.014/1=0.00743 H, self-induced emf magnitude L |dI/dt|, dI/dt=12 A/s, e=0.089 V, opposes change. The current opposes the decrease in flux by maintaining the original flux direction, resisting the reduction in area per Lenz’s law. 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 Magnetic flux

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Self-Induction and Self-Inductance

A coil of 75 turns and area 0.05 m² is in a 0.12 T field that drops to zero in 0.25 s. What is the induced emf?

**Self-induction** emf induced in coil due to change in its own current, e = -L dI/dt, L self-inductance (H), L = μ₀ N² A / l for solenoid, N turns, A area (m²), l length (m), μ₀=4π×10⁻⁷ H/m. For solenoid 650 turns/m means n=650, A=0.014 m², L = μ₀ n² A l? Actually per unit length? For length l, N=n l, L= μ₀ n² A l, if l=1 m, L=4π×10⁻⁷×650²×0.014=7.43×10⁻³ H, dI/dt=(3-6)/0.25=-12 A/s, e= -L×(-12)=0.089 V. Δ Φ = B A = 0.12 × 0.05 = 0.006 Wb . ε = N (Δ Φ/Δ t) = 75 × (0.006/0.25) = 75 × 0.024 = 1.8

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Self-Induction and Self-Inductance

A square loop of side 16 cm rotates at 18 rad/s in a 0.2 T field. What is the maximum emf induced?

**Self-inductance of solenoid** L = μ₀ N² A / l, N total turns, A cross-section, l length, for N=650 turns per meter means n=650 m⁻¹, if length 1 m N=650, A=0.014, L=4π×10⁻⁷×650²×0.014/1=0.00743 H, self-induced emf magnitude L |dI/dt|, dI/dt=12 A/s, e=0.089 V, opposes change. A = (0.16)² = 0.0256 m² . ε₀ = N B A ω = 1 × 0.2 × 0.0256 × 18 = 0.09216 V ≈ 0.092 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

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Self-Induction and Self-Inductance

A coil of self-inductance 1.2 H has its current increased from 2 A to 6 A in 0.4 s. What is the magnitude of the induced

**Energy stored in inductor** U =½ L I², L inductance, I current, energy in magnetic field, density u = B²/(2μ₀), B=μ₀ n I inside solenoid, U = (B²/2μ₀)×volume, illustrating equivalence of circuit and field energy. ε = L (Δ I/Δ t) . Δ I = 6 - 2 = 4 A , Δ t = 0.4 s . ε = 1.2 × (4/0.4) = 1.2 × 10 = 12 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 =

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Self-Induction and Self-Inductance

A wheel with 6 spokes of 0.6 m each rotates at 45 rpm in a 0.5 T field. What is the induced emf?

**Self-inductance of solenoid** L = μ₀ N² A / l, N total turns, A cross-section, l length, for N=650 turns per meter means n=650 m⁻¹, if length 1 m N=650, A=0.014, L=4π×10⁻⁷×650²×0.014/1=0.00743 H, self-induced emf magnitude L |dI/dt|, dI/dt=12 A/s, e=0.089 V, opposes change. ω = 2π × (45/60) = 1.5π rad/s . ε = (1/2) B ω R² = (1/2) × 0.5 × 1.5π × (0.6)² = 0.8478 V ≈ 0.85 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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