#maximum emf

Lenz's Law, Eddy Currents and Applications

A coil of 170 turns and area 0.015 m² is rotated at 35 Hz in a 0.1 T field. What is the maximum 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. ω = 2π v = 2π × 35 = 70π rad/s . ε₀ = N B A ω = 170 × 0.1 × 0.015 × 70π = 56.03 V ≈ 56 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 > Lenz's Law, Eddy Currents and Applications

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Energy Stored in Inductor and Magnetic Energy

A coil of 300 turns rotates at 70 rad/s in a 0.07 T field. If the area is 0.012 m², what is the maximum 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. ε₀ = N B A ω = 300 × 0.07 × 0.012 × 70 = 17.64 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 17.64 V 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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Energy Stored in Inductor and Magnetic Energy

A square loop of side 20 cm rotates at 20 rad/s in a 0.15 T field. What is the maximum emf induced?

**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. A = (0.2)² = 0.04 m² . ε₀ = N B A ω = 1 × 0.15 × 0.04 × 20 = 0.12 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, result 0.12 V 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 230 turns rotates at 85 rad/s in a 0.03 T field. If the area is 0.02 m², what is the maximum 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. ε₀ = N B A ω = 230 × 0.03 × 0.02 × 85 = 11.73 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 11.73 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Mutual Induction and Mutual Inductance

A circular coil of radius 8 cm and 150 turns rotates at 25 rad/s in a 0.06 T field. What is the maximum emf induced?

**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. A = π r² = 3.14 × (0.08)² = 0.0201 m² . ε₀ = N B A ω = 150 × 0.06 × 0.0201 × 25 = 4.5225 V ≈ 4.52 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 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.

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

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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.

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

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Self-Induction and Self-Inductance

A coil of 150 turns and area 0.01 m² is rotated at 40 Hz in a 0.06 T field. What is the maximum 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. ω = 2π v = 2π × 40 = 80π rad/s . ε₀ = N B A ω = 150 × 0.06 × 0.01 × 80π = 22.62 V

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

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Induced EMF Due to Change in Magnetic Field

A coil of 280 turns rotates at 80 rad/s in a 0.06 T field. If the area is 0.01 m², what is the maximum emf?

**Uniform field change** in coil produces emf proportional to area and turns, for circular coil radius 0.16 m area πr²=0.0804 m², B 0.12 T deformed to wire in 0.6 s, ΔΦ=0.12×0.0804=0.00965 Wb, e=0.00965/0.6=0.0161 V, illustrating area change also induces emf. ε₀ = N B A ω = 280 × 0.06 × 0.01 × 80 = 13.44 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 13.44 V follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Induced EMF Due to Change in Magnetic Field

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Induced EMF Due to Change in Magnetic Field

A circular coil of radius 9 cm and 200 turns rotates at 50 rad/s in a 0.03 T field. What is the maximum emf 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. A = π r² = 3.14 × (0.09)² = 0.0254 m² . ε₀ = N B A ω = 200 × 0.03 × 0.0254 × 50 = 7.62 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,

Ref: NCERT > Physics Book > Electromagnetic Induction > Induced EMF Due to Change in Magnetic Field

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Induced EMF Due to Change in Magnetic Field

A coil of 100 turns and area 0.02 m² is rotated at 50 Hz in a 0.05 T field. What is the maximum emf?

**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. ω = 2π v = 2π × 50 = 100π rad/s . ε₀ = N B A ω = 100 × 0.05 × 0.02 × 100π = 31.4 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,

Ref: NCERT > Physics Book > Electromagnetic Induction > Induced EMF Due to Change in Magnetic Field

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Rotational EMF and AC Generator

A circular coil of radius 8 cm and 240 turns rotates at 45 rad/s in a 0.04 T field. What is the maximum emf induced?

**Rotational emf** when coil area A rotates with angular speed ω in uniform field B, flux Φ = B A cos ωt, emf e = -N dΦ/dt = N B A ω sin ωt, maximum e₀ = N B A ω, frequency = ω/2π. For square side 22 cm area 0.0484 m² N=1 ω=14 rad/s B=0.15 T, e₀=1×0.15×0.0484×14=0.1016 V, sinusoidal. A = π r² = 3.14 × (0.08)² = 0.0201 m² . ε₀ = N B A ω = 240 × 0.04 × 0.0201 × 45 = 8.676 V ≈ 8.68 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

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