#mutual inductance

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

View Details

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

View Details

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

View Details

Mutual Induction and Mutual Inductance

A solenoid of 400 turns and length 0.5 m induces an emf of 0.8 V in a nearby coil when its current changes from 2 A to 4

**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 = M (Δ I/Δ t) . Δ I = 4 - 2 = 2 A , Δ t = 0.2 s . M = (ε/(Δ I/Δ t)) = (0.8/(2/0.2)) = (0.8/10) = 0.08 H . 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

View Details

Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.2 H has a current change of 5 A/s in the primary coil. What is the induced emf in th

**Mutual inductance calculation** M = e₂/(dI₁/dt), for 200 turns length 0.5 m nearby coil e=0.5 V dI=2 A dt=0.2 s dI/dt=10 A/s, M=0.5/10=0.05 H, depends on geometry, orientation, number of turns, area, separation, coupling coefficient k = M/√(L₁ L₂) ≤1. ε = M (dI/dt) = 0.2 × 5 = 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 1 V follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

View Details

Mutual Induction and Mutual Inductance

A solenoid of 350 turns and length 0.7 m induces an emf of 1.2 V in a nearby coil when its current changes from 1 A to 4

**Mutual inductance** M = N₂ Φ₂₁/I₁, flux linking coil 2 due to current in coil 1, emf in 2 e₂ = -M dI₁/dt, M = μ₀ N₁ N₂ A / l for coaxial solenoids, unit henry (H), same as self-inductance. For solenoid 500 turns length 1 m induces 1.5 V when current 0 to 3 A in 0.2 s, dI/dt=15 A/s, M = e/(dI/dt)=1.5/15=0.1 H. ε = M (Δ I/Δ t) . Δ I = 4 - 1 = 3 A , Δ t = 0.3 s . M = (ε/(Δ I/Δ t)) = (1.2/(3/0.3)) = (1.2/10) = 0.12 H . Using Φ = B

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

View Details

Mutual Induction and Mutual Inductance

A solenoid of 600 turns and length 0.8 m induces an emf of 2 V in a nearby coil when its current changes from 1 A to 4 A

**Mutual inductance calculation** M = e₂/(dI₁/dt), for 200 turns length 0.5 m nearby coil e=0.5 V dI=2 A dt=0.2 s dI/dt=10 A/s, M=0.5/10=0.05 H, depends on geometry, orientation, number of turns, area, separation, coupling coefficient k = M/√(L₁ L₂) ≤1. ε = M (Δ I/Δ t) . Δ I = 4 - 1 = 3 A , Δ t = 0.2 s . M = (ε/(Δ I/Δ t)) = (2/(3/0.2)) = (2/15) = 0.133 H ≈ 0.13 H . 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 > Mutual Induction and Mutual Inductance

View Details

Mutual Induction and Mutual Inductance

A solenoid of 250 turns and length 0.8 m induces an emf of 0.75 V in a nearby coil when its current changes from 0 to 2.

**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 = M (Δ I/Δ t) . Δ I = 2.5 - 0 = 2.5 A , Δ t = 0.25 s . M = (ε/(Δ I/Δ t)) = (0.75/(2.5/0.25)) = (0.75/10) = 0.075 H . 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

View Details

Mutual Induction and Mutual Inductance

A solenoid of 550 turns and length 1.1 m induces an emf of 1.65 V in a nearby coil when its current changes from 0 to 3

**Mutual inductance calculation** M = e₂/(dI₁/dt), for 200 turns length 0.5 m nearby coil e=0.5 V dI=2 A dt=0.2 s dI/dt=10 A/s, M=0.5/10=0.05 H, depends on geometry, orientation, number of turns, area, separation, coupling coefficient k = M/√(L₁ L₂) ≤1. ε = M (Δ I/Δ t) . Δ I = 3 - 0 = 3 A , Δ t = 0.3 s . M = (ε/(Δ I/Δ t)) = (1.65/(3/0.3)) = (1.65/10) = 0.165 H . 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

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

View Details

Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.3 H has a current change of 5 A/s in the primary coil. What is the induced emf in th

**Mutual inductance calculation** M = e₂/(dI₁/dt), for 200 turns length 0.5 m nearby coil e=0.5 V dI=2 A dt=0.2 s dI/dt=10 A/s, M=0.5/10=0.05 H, depends on geometry, orientation, number of turns, area, separation, coupling coefficient k = M/√(L₁ L₂) ≤1. ε = M (dI/dt) = 0.3 × 5 = 1.5 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 1.5 V follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

View Details

Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.22 H has a current change of 7 A/s in the primary coil. What is the induced emf in t

**Mutual inductance calculation** M = e₂/(dI₁/dt), for 200 turns length 0.5 m nearby coil e=0.5 V dI=2 A dt=0.2 s dI/dt=10 A/s, M=0.5/10=0.05 H, depends on geometry, orientation, number of turns, area, separation, coupling coefficient k = M/√(L₁ L₂) ≤1. ε = M (dI/dt) = 0.22 × 7 = 1.54 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 1.54 V follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

View Details

Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.25 H has a current change of 6 A/s in the primary coil. What is the induced emf in t

**Mutual inductance** M = N₂ Φ₂₁/I₁, flux linking coil 2 due to current in coil 1, emf in 2 e₂ = -M dI₁/dt, M = μ₀ N₁ N₂ A / l for coaxial solenoids, unit henry (H), same as self-inductance. For solenoid 500 turns length 1 m induces 1.5 V when current 0 to 3 A in 0.2 s, dI/dt=15 A/s, M = e/(dI/dt)=1.5/15=0.1 H. ε = M (dI/dt) = 0.25 × 6 = 1.5 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,

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

View Details