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#emf calculation

37 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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Lenz's Law, Eddy Currents and Applications

A solenoid of 450 turns and length 0.9 m induces an emf of 1.8 V in a nearby coil when its current changes from 2 A to 5

**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 (Δ I/Δ t) . Δ I = 5 - 2 = 3 A , Δ t = 0.3 s . M = (ε/(Δ I/Δ t)) = (1.8/(3/0.3)) = (1.8/10) = 0.18 H . 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 self-inductance 0.6 H has its current decreased from 4 A to 1 A in 0.15 s. What is the magnitude of the 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. ε = L (Δ I/Δ t) . Δ I = 1 - 4 = -3 A , Δ t = 0.15 s . ε = 0.6 × (-3/0.15) = 0.6 × (-20) = -12 V , magnitude = 12 V. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, 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 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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Energy Stored in Inductor and Magnetic Energy

A coil of 80 turns is placed in a magnetic field that increases from 0 to 0.03 Wb/m² in 0.2 s. If the coil area is 0.05

**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. Flux change: Δ Φ = B A = 0.03 × 0.05 = 0.0015 Wb . ε = N (Δ Φ/Δ t) = 80 × (0.0015/0.2) = 80 × 0.0075 = 0.6 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 L I², result

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 5 spokes of 0.4 m each rotates at 30 rpm in a 0.7 T field. 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. ω = 2π × (30/60) = π rad/s . ε = (1/2) B ω R² = (1/2) × 0.7 × π × (0.4)² = 0.1759 V ≈ 0.176 V . 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 > 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 0.03 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 coil of self-inductance 1.0 H has its current decreased from 5 A to 2 A in 0.15 s. What is the magnitude of the induce

**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 = L (Δ I/Δ t) . Δ I = 2 - 5 = -3 A , Δ t = 0.15 s . ε = 1.0 × (-3/0.15) = 1.0 × (-20) = -20 V , magnitude = 20 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 > Energy Stored in Inductor and Magnetic Energy

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Rotational EMF and AC Generator

A rod rotates at 20 rad/s in a 0.4 T field. If the length from the axis to the tip is 0.5 m, what is the emf induced?

**Maximum emf** e₀ = N B A ω, for 180 turns A=0.02 m² B=0.03 T ω=60 rad/s, e₀=180×0.03×0.02×60=6.48 V, illustrating dependence on N, B, A, ω, used in generator design, increasing N or B or A or ω raises output. ε = (1/2) B ω R² . ε = (1/2) × 0.4 × 20 × (0.5)² = 5 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 5 V follows, reflecting Faraday's law and Lenz's opposition.

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

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AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of sides 15 cm and 6 cm moves out of a 0.4 T field at 1.5 m/s perpendicular to the shorter side. What

**Back emf** in motor opposes applied voltage, e_b = N B A ω sin ωt, reduces net current, at start ω=0 e_b=0 current large, as speed increases e_b increases limiting current, power conversion mechanical, principle of motor and generator reciprocity. ε = B l v , l = 0.15 m . ε = 0.4 × 0.15 × 1.5 = 0.09 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 0.09 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Induced EMF Due to Change in Magnetic Field

A wheel with 4 spokes of 0.5 m each rotates at 48 rpm in a 0.8 T field. What is the induced emf?

**Field decreasing to zero** induces emf trying to maintain field, current direction such that its field adds to original. For 150 turns area 0.06 m² B 0.14 T to zero in 0.3 s, e=150×0.06×0.14/0.3=4.2 V, as earlier, showing linear dependence on N, A, ΔB/Δt. ω = 2π × (48/60) = 1.6π rad/s . ε = (1/2) B ω R² = (1/2) × 0.8 × 1.6π × (0.5)² = 0.5024 V ≈ 0.5 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)

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

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