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#induced emf

120 public questions tagged with this topic.

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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Lenz's Law, Eddy Currents and Applications

A solenoid of 800 turns per meter and area 0.013 m² has a current change from 9 A to 6 A in 0.2 s. What is the self-indu

**Eddy currents** are circulating currents induced in bulk conductor by changing flux, oppose motion, cause damping, heating, energy loss, minimized by laminating core into thin sheets insulated, increasing resistance, reducing eddy current magnitude, used in induction heating and braking. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (800)² × 0.013 × 1 = 0.01045 H . ε = L (Δ I/Δ t) = 0.01045 × (6 - 9/0.2) = 0.01045 × (-15) = 0.15675 V ≈ 0.157 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt =

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

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Lenz's Law, Eddy Currents and Applications

A coil of 80 turns and area 0.05 m² is in a 0.1 T field that drops to zero in 0.25 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 × 0.05 = 0.005 Wb . ε = N (Δ Φ/Δ t) = 80 × (0.005/0.25) = 80 × 0.02 = 1.6 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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Lenz's Law, Eddy Currents and Applications

A solenoid with mutual inductance 0.25 H has a current change of 4 A/s in the primary coil. What is the induced emf in t

**Lenz's law** induced current direction opposes change in flux causing it, e = -N dΦ/dt negative sign, conservation of energy. Magnet moved towards coil south pole first, approaching south pole increasing flux into coil with south polarity, coil face nearest magnet becomes south pole to repel, opposing approach, so face becomes south pole, repelling magnet. ε = M (dI/dt) = 0.25 × 4 = 1 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

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

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Lenz's Law, Eddy Currents and Applications

A wheel with 8 spokes of 0.55 m each rotates at 50 rpm in a 0.4 T field. What is the induced emf?

**Eddy currents** are circulating currents induced in bulk conductor by changing flux, oppose motion, cause damping, heating, energy loss, minimized by laminating core into thin sheets insulated, increasing resistance, reducing eddy current magnitude, used in induction heating and braking. ω = 2π × (50/60) = (5π/3) rad/s . ε = (1/2) B ω R² = (1/2) × 0.4 × (5π/3) × (0.55)² = 0.3166 V ≈ 0.317 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

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 120 turns and area 0.04 m² is in a 0.09 T field that drops to zero in 0.3 s. What is the induced emf?

**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. Δ Φ = B A = 0.09 × 0.04 = 0.0036 Wb . ε = N (Δ Φ/Δ t) = 120 × (0.0036/0.3) = 120 × 0.012 = 1.44 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B

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 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

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 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 conducting loop is placed in a magnetic field that remains constant in magnitude and direction. No emf is induced beca

**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. Emf is induced only when magnetic flux changes. A constant field with a stationary loop results in no flux change, hence no emf. 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 No change in magnetic flux 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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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 15 cm is deformed into a straight wire in a 0.15 T field in 0.6 s. What is the induced 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. Initial flux: Φ = B A = 0.15 × π × (0.15)² = 0.0106 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.0106/0.6) = 0.01767 V ≈

Ref: NCERT > Physics Book > Electromagnetic Induction > AC Generator, Back EMF and Time Duration of EMF

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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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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 12 cm is deformed into a straight wire in a 0.2 T field in 0.5 s. What is the induced 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. Initial flux: Φ = B A = 0.2 × π × (0.12)² = 0.00904 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00904/0.5) = 0.01808 V ≈

Ref: NCERT > Physics Book > Electromagnetic Induction > AC Generator, Back EMF and Time Duration of EMF

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