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#magnetic field

328 public questions tagged with this topic.

Lenz's Law, Eddy Currents and Applications

A coil of 180 turns and area 0.02 m² is rotated at 45 Hz in a 0.08 T field. What is the maximum 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 v = 2π × 45 = 90π rad/s . ε₀ = N B A ω = 180 × 0.08 × 0.02 × 90π = 81.43 V ≈ 81.4 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 > Lenz's Law, Eddy Currents and Applications

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Lenz's Law, Eddy Currents and Applications

A circular coil of radius 10 cm and 220 turns rotates at 60 rad/s in a 0.02 T field. What is the maximum emf induced?

**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. A = π r² = 3.14 × (0.1)² = 0.0314 m² . ε₀ = N B A ω = 220 × 0.02 × 0.0314 × 60 = 8.2992 V ≈ 8.3 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

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 rod rotates at 15 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.4 m, what is the emf induced?

**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. ε = (1/2) B ω R² . ε = (1/2) × 0.3 × 15 × (0.4)² = 0.36 V . Using Φ = B A cosθ, e = -N dΦ/dt =

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

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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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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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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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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 130 turns and area 0.03 m² is rotated at 30 Hz in a 0.08 T field. 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. ω = 2π v = 2π × 30 = 60π rad/s . ε₀ = N B A ω = 130 × 0.08 × 0.03 × 60π = 58.62 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, M = e/(dI/dt) and U = ½ L I², result 58.62 V follows,

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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Energy Stored in Inductor and Magnetic Energy

A wheel with 7 spokes of 0.45 m each rotates at 36 rpm in a 0.6 T field. 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. ω = 2π × (36/60) = 1.2π rad/s . ε = (1/2) B ω R² = (1/2) × 0.6 × 1.2π × (0.45)² = 0.383 V ≈ 0.38 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 0.38 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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