Electromagnetic Induction

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

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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 =

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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,

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AC Generator, Back EMF and Time Duration of EMF

In an AC generator, what happens to the induced emf when the rotational speed of the coil doubles?

**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. The emf is proportional to the angular speed ( ε = N B A ω ), so doubling the rotational speed doubles the 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 Doubles follows, reflecting Faraday's law and Lenz's opposition.

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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

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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 ω sinωt,

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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

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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

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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

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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 =

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AC Generator, Back EMF and Time Duration of EMF

A rectangular loop of 0.1 m × 0.2 m moves out of a 0.4 T field at 0.5 m/s along its shorter side. What is the emf?

**Circular loop deformed into straight wire** in field B=0.12 T radius 16 cm area πr²=0.0804 m² flux 0.00965 Wb drops to zero in 0.6 s e=0.0161 V, illustrating flux change due to area change induces emf, even without B change, area deformation changes Φ = B A cosθ. ε = B l v , l = 0.2 m . ε = 0.4 × 0.2 × 0.5 = 0.04 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

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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

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