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#Faraday's law

113 public questions tagged with this topic.

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 conducting loop is placed in a uniform magnetic field with its plane parallel to the field lines. Why is no emf induce

**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. When the plane is parallel to the field, the flux through the loop is zero ( Φ = B A cos 90° = 0 ), so changing the field strength does not alter the flux, resulting in 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

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

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AC Generator, Back EMF and Time Duration of EMF

A loop of 0.2 m × 0.1 m moves out of a 0.5 T field at 2 m/s along its longer side. How long does the emf last?

**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. Time = distance/velocity, distance = width along motion = 0.1 m. t = (0.1/2) = 0.05 s . Using Φ = B A cosθ, e = -N dΦ/dt = -N A

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 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 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 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 170 turns and area 0.07 m² is in a field that decreases from 0.1 T to 0 in 0.5 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 = 0.1 × 0.07 = 0.007 Wb . ε = N (Δ Φ/Δ t) = 170 × (0.007/0.5) = 170 × 0.014 = 2.38 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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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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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 10 cm is deformed into a straight wire in a 0.1 T field. If the flux change occurs in 0.2 s, w

**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 = 0.1 × π × (0.1)² = 0.00314 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00314/0.2) = 0.0157 V ≈

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

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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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