Skip to content

Lenz's Law, Eddy Currents and Applications

Latest questions in this category.

16 questions

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

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 =

Ref: NCERT > Physics Book > Electromagnetic Induction > 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

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

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,

Ref: NCERT > Physics Book > Electromagnetic Induction > 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

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

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

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

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

Ref: NCERT > Physics Book > Electromagnetic Induction > 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

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