#current change

Energy Stored in Inductor and Magnetic Energy

A solenoid of 700 turns per meter and area 0.015 m² has a current drop from 8 A to 5 A in 0.3 s. What is the self-induce

**Magnetic energy density** u = B²/(2μ₀), for B=0.5 T, u=0.25/(2×4π×10⁻⁷)=0.25/(2.513×10⁻⁶)=99471 J/m³, large, but volume small, total energy moderate. Inductor stores energy in field, released when current interrupted causing spark. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (700)² × 0.015 × 1 = 0.00923 H . ε = L (Δ I/Δ t) = 0.00923 × (5 - 8/0.3) = 0.00923 × (-10) = 0.0923 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

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Energy Stored in Inductor and Magnetic Energy

A solenoid of 550 turns per meter and area 0.016 m² has a current drop from 7 A to 4 A in 0.3 s. What is the self-induce

**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (550)² × 0.016 × 1 = 0.00608 H . ε = L (Δ I/Δ t) = 0.00608 × (4 - 7/0.3) = 0.00608 × (-10) = 0.0608 V ≈ 0.061 V . Using

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Mutual Induction and Mutual Inductance

A solenoid of 400 turns and length 0.5 m induces an emf of 0.8 V in a nearby coil when its current changes from 2 A to 4

**Solenoid carries steady current** second coil experiences emf only when current in solenoid changes because dΦ/dt ≠0 only when I changes, steady current gives constant flux, no induction, illustrating Faraday's law requires changing flux, not static field. ε = M (Δ I/Δ t) . Δ I = 4 - 2 = 2 A , Δ t = 0.2 s . M = (ε/(Δ I/Δ t)) = (0.8/(2/0.2)) = (0.8/10) = 0.08 H . 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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Mutual Induction and Mutual Inductance

A coil of self-inductance 0.5 H has its current increased from 1 A to 4 A in 0.25 s. What is the magnitude of the induce

**Mutual inductance** M = N₂ Φ₂₁/I₁, flux linking coil 2 due to current in coil 1, emf in 2 e₂ = -M dI₁/dt, M = μ₀ N₁ N₂ A / l for coaxial solenoids, unit henry (H), same as self-inductance. For solenoid 500 turns length 1 m induces 1.5 V when current 0 to 3 A in 0.2 s, dI/dt=15 A/s, M = e/(dI/dt)=1.5/15=0.1 H. ε = L (Δ I/Δ t) . Δ I = 4 - 1 = 3 A , Δ t = 0.25 s . ε = 0.5 × (3/0.25) = 0.5 × 12 = 6 V . Using Φ = B

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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 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 coil of self-inductance 0.8 H has its current decreased from 6 A to 3 A in 0.2 s. What is the magnitude of the 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. ε = L (Δ I/Δ t) . Δ I = 3 - 6 = -3 A , Δ t = 0.2 s . ε = 0.8 × (-3/0.2) = 0.8 × (-15) = -12 V , magnitude = 12 V. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B

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Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.3 H has a current change of 5 A/s in the primary coil. What is the induced emf in th

**Mutual inductance calculation** M = e₂/(dI₁/dt), for 200 turns length 0.5 m nearby coil e=0.5 V dI=2 A dt=0.2 s dI/dt=10 A/s, M=0.5/10=0.05 H, depends on geometry, orientation, number of turns, area, separation, coupling coefficient k = M/√(L₁ L₂) ≤1. ε = M (dI/dt) = 0.3 × 5 = 1.5 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 1.5 V follows, reflecting Faraday's law and Lenz's opposition.

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Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.3 H has a current change of 6 A/s in the primary coil. What is the induced emf in th

**Mutual inductance** M = N₂ Φ₂₁/I₁, flux linking coil 2 due to current in coil 1, emf in 2 e₂ = -M dI₁/dt, M = μ₀ N₁ N₂ A / l for coaxial solenoids, unit henry (H), same as self-inductance. For solenoid 500 turns length 1 m induces 1.5 V when current 0 to 3 A in 0.2 s, dI/dt=15 A/s, M = e/(dI/dt)=1.5/15=0.1 H. ε = M (dI/dt) = 0.3 × 6 = 1.8 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,

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Induced EMF Due to Change in Magnetic Field

A coil of self-inductance 1.5 H has its current decreased from 7 A to 4 A in 0.3 s. What is the magnitude of the induced

**Induced emf due to B change** e = -N A dB/dt, N turns, A area (m²), dB/dt rate of change of field (T/s). For 110 turns area 0.035 m² B 0.09 T to 0 in 0.5 s, dB/dt=0.18 T/s, e=110×0.035×0.18=0.693 V, direction opposes decrease via Lenz's law. ε = L (Δ I/Δ t) . Δ I = 4 - 7 = -3 A , Δ t = 0.3 s . ε = 1.5 × (-3/0.3) = 1.5 × (-10) = -15 V , magnitude = 15 V. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v

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Self-Induction and Self-Inductance

A coil of self-inductance 0.9 H has its current increased from 1 A to 4 A in 0.25 s. What is the magnitude of the induce

**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. ε = L (Δ I/Δ t) . Δ I = 4 - 1 = 3 A , Δ t = 0.25 s . ε = 0.9 × (3/0.25) = 0.9 × 12 = 10.8 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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Mutual Induction and Mutual Inductance

A solenoid of 650 turns and length 1.2 m induces an emf of 2.6 V in a nearby coil when its current changes from 2 A to 6

**Mutual inductance** M = N₂ Φ₂₁/I₁, flux linking coil 2 due to current in coil 1, emf in 2 e₂ = -M dI₁/dt, M = μ₀ N₁ N₂ A / l for coaxial solenoids, unit henry (H), same as self-inductance. For solenoid 500 turns length 1 m induces 1.5 V when current 0 to 3 A in 0.2 s, dI/dt=15 A/s, M = e/(dI/dt)=1.5/15=0.1 H. ε = M (Δ I/Δ t) . Δ I = 6 - 2 = 4 A , Δ t = 0.4 s . M = (ε/(Δ I/Δ t)) = (2.6/(4/0.4)) = (2.6/10) = 0.26 H . Using Φ = B

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