Mutual Induction and Mutual Inductance

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.

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Mutual Induction and Mutual Inductance

A circular coil of radius 8 cm and 150 turns rotates at 25 rad/s in a 0.06 T field. What is the maximum emf induced?

**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. A = π r² = 3.14 × (0.08)² = 0.0201 m² . ε₀ = N B A ω = 150 × 0.06 × 0.0201 × 25 = 4.5225 V ≈ 4.52 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

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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

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Mutual Induction and Mutual Inductance

A coil is wound tightly around a core material. If the current through it changes rapidly, the induced emf opposing this

**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. This is self-induction, where a changing current in a coil induces an emf that opposes the change, proportional to the coil’s self-inductance. 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 Self-induction follows, reflecting Faraday's law and Lenz's opposition.

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Mutual Induction and Mutual Inductance

A coil of 130 turns and area 0.08 m² is in a 0.12 T field that drops to zero in 0.4 s. What is the induced 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. Δ Φ = B A = 0.12 × 0.08 = 0.0096 Wb . ε = N (Δ Φ/Δ t) = 130 × (0.0096/0.4) = 130 × 0.024 = 3.12 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,

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Mutual Induction and Mutual Inductance

A magnet is dropped through a vertical copper tube. The magnet’s fall is slower than expected due to what effect?

**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. The falling magnet induces currents in the copper tube, which create a magnetic field opposing the magnet’s motion, slowing its fall. This is an application of Lenz’s law. 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,

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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.

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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

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Mutual Induction and Mutual Inductance

A conducting rod is stationary in a varying magnetic field. The induced emf in the rod arises due to which component of

**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. For a stationary rod ( v = 0 ), the emf is induced by the electric field generated by the time-varying magnetic field, not the magnetic force ( q v × B ), which requires motion. 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

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Mutual Induction and Mutual Inductance

A coil of self-inductance 0.5 H has its current increased from 1 A to 4 A in 0.25 s. What is the magnitude of the induce

**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 = L (Δ I/Δ t) . Δ I = 4 - 1 = 3 A , Δ t = 0.25 s . ε = 0.5 × (3/0.25) = 0.5 × 12 = 6 V . Using Φ = B

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Mutual Induction and Mutual Inductance

A coil of 40 turns experiences a magnetic flux change from 0 to 0.01 Wb in 0.05 s. What is the induced 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. ε = N (Δ Φ/Δ t) . Δ Φ = 0.01 Wb , Δ t = 0.05 s , N = 40 . ε = 40 × (0.01/0.05) = 40 × 0.2 = 8 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

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Mutual Induction and Mutual Inductance

A wheel with 9 spokes of 0.5 m each rotates at 55 rpm in a 0.6 T field. What is the induced emf?

**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. ω = 2π × (55/60) = (11π/6) rad/s . ε = (1/2) B ω R² = (1/2) × 0.6 × (11π/6) × (0.5)² = 0.4328 V ≈ 0.43 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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