#coil

Lenz's Law, Eddy Currents and Applications

A coil of 60 turns and area 0.04 m² is in a 0.1 T field that drops to zero in 0.2 s. What is the induced emf?

**Solenoid second coil** experiences emf only when current in solenoid changes because flux linkage changes only then, steady current gives constant Φ, dΦ/dt=0, no emf, when current changes, dΦ/dt ≠0, emf induced, illustrating Faraday's law requirement of changing flux. Δ Φ = B A = 0.1 × 0.04 = 0.004 Wb . ε = N (Δ Φ/Δ t) = 60 × (0.004/0.2) = 60 × 0.02 = 1.2 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

Ref: NCERT > Physics Book > Electromagnetic Induction > Lenz's Law, Eddy Currents and Applications

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Energy Stored in Inductor and Magnetic Energy

A coil of 130 turns and area 0.04 m² is in a field that increases from 0 to 0.05 T in 0.2 s. What is the induced emf?

**Energy in inductor** cannot change instantaneously because that would require infinite power, current through inductor continuous, voltage may jump, principle used in chokes, inductive kick, back emf, explaining why inductor opposes change in current. Δ Φ = B A = 0.05 × 0.04 = 0.002 Wb . ε = N (Δ Φ/Δ t) = 130 × (0.002/0.2) = 130 × 0.01 = 1.3 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 1.3 V

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

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Energy Stored in Inductor and Magnetic Energy

A coil of 70 turns experiences a magnetic flux change from 0 to 0.02 Wb in 0.04 s. What is the induced emf?

**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. ε = N (Δ Φ/Δ t) . Δ Φ = 0.02 Wb , Δ t = 0.04 s , N = 70 . ε = 70 × (0.02/0.04) = 70 × 0.5 = 35 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 35 V

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

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Energy Stored in Inductor and Magnetic Energy

A coil with \( L = 0.3 \, \text{H} \) has its current increased from 0 to 6 A in 0.6 s. What is the energy stored?

**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. W = (1/2) L I² = (1/2) × 0.3 × (6)² = 0.15 × 36 = 5.4 J . 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 > Energy Stored in Inductor and Magnetic Energy

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Mutual Induction and Mutual Inductance

A coil of 130 turns and area 0.08 m² is in a 0.12 T field that drops to zero in 0.4 s. What is the induced emf?

**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. Δ Φ = B A = 0.12 × 0.08 = 0.0096 Wb . ε = N (Δ Φ/Δ t) = 130 × (0.0096/0.4) = 130 × 0.024 = 3.12 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 = ½ L I²,

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

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Energy Stored in Inductor and Magnetic Energy

A coil with \( L = 0.6 \, \text{H} \) has its current increased from 0 to 5 A in 0.5 s. What is the energy stored?

**Energy in inductor** cannot change instantaneously because that would require infinite power, current through inductor continuous, voltage may jump, principle used in chokes, inductive kick, back emf, explaining why inductor opposes change in current. W = (1/2) L I² = (1/2) × 0.6 × (5)² = 0.3 × 25 = 7.5 J . 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 7.5 J follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

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Mutual Induction and Mutual Inductance

A solenoid of 350 turns and length 0.7 m induces an emf of 1.2 V in a nearby coil when its current changes from 1 A to 4

**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 = 4 - 1 = 3 A , Δ t = 0.3 s . M = (ε/(Δ I/Δ t)) = (1.2/(3/0.3)) = (1.2/10) = 0.12 H . Using Φ = B

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

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Energy Stored in Inductor and Magnetic Energy

A coil with \( L = 0.1 \, \text{H} \) has its current increased from 0 to 4 A in 0.2 s. What is the energy stored?

**Energy in inductor** cannot change instantaneously because that would require infinite power, current through inductor continuous, voltage may jump, principle used in chokes, inductive kick, back emf, explaining why inductor opposes change in current. Energy: W = (1/2) L I² . W = (1/2) × 0.1 × (4)² = 0.05 × 16 = 0.8 J . 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 0.8 J follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

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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

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

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Mutual Induction and Mutual Inductance

A coil of 40 turns experiences a magnetic flux change from 0 to 0.01 Wb in 0.05 s. What is the induced emf?

**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 = N (Δ Φ/Δ t) . Δ Φ = 0.01 Wb , Δ t = 0.05 s , N = 40 . ε = 40 × (0.01/0.05) = 40 × 0.2 = 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

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

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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 =

Ref: NCERT > Physics Book > Electromagnetic Induction > Self-Induction and Self-Inductance

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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

Ref: NCERT > Physics Book > Electromagnetic Induction > Self-Induction and Self-Inductance

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