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#magnetic flux

40 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 metal ring is placed in a uniform magnetic field perpendicular to its plane. If the field strength decreases, the indu

**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. According to Lenz’s law, the induced current opposes the decrease in magnetic flux by generating a magnetic field in the same direction as the original field. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt,

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

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Lenz's Law, Eddy Currents and Applications

In a system of two coils, the mutual inductance depends on which property of the setup?

**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. Mutual inductance depends on the geometry (e.g., coil sizes, separation, orientation) and the medium’s permeability, affecting how much flux from one coil links with the other. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A

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 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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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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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 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 conducting loop is compressed in a uniform magnetic field. The induced current flows to maintain what quantity against

**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. The current opposes the decrease in flux by maintaining the original flux direction, resisting the reduction in area per Lenz’s law. 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 Magnetic flux

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 is placed near a current-carrying wire. If the current in the wire is suddenly switched off, what happens to the

**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. When the current is switched off, the magnetic field due to the wire collapses, reducing the magnetic flux through the coil from some value to zero. 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 Increases

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 conducting loop is rotated in a uniform magnetic field. The induced emf depends on the rate of change of what physical

**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. Faraday’s law states that the induced emf is proportional to the rate of change of magnetic flux through the loop, which varies as the loop rotates. 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 Magnetic

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 bar magnet is moved towards a coil of 50 turns at a constant speed. The magnetic flux through the coil increases from

**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. Given: N = 50 , Δ Φ_B = 0.02 - 0 = 0.02 Wb , Δ t = 0.1 s . Using Faraday's law: ε = -N (Δ Φ_B/Δ t) . ε = 50 × (0.02/0.1) = 50 × 0.2 = 10 V . Magnitude of emf = 10 V. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt =

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

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