#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 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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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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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

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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 =

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Lenz's Law, Eddy Currents and Applications

A magnet is moved towards a coil, inducing a current. If the magnet’s speed increases, what happens to the induced curre

**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. Faster motion increases the rate of flux change, resulting in a larger induced emf and, assuming constant resistance, a larger induced current. 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,

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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 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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Self-Induction and Self-Inductance

A conducting disc rotates in a uniform magnetic field parallel to its axis. The induced emf between the center and rim a

**Self-induction** emf induced in coil due to change in its own current, e = -L dI/dt, L self-inductance (H), L = μ₀ N² A / l for solenoid, N turns, A area (m²), l length (m), μ₀=4π×10⁻⁷ H/m. For solenoid 650 turns/m means n=650, A=0.014 m², L = μ₀ n² A l? Actually per unit length? For length l, N=n l, L= μ₀ n² A l, if l=1 m, L=4π×10⁻⁷×650²×0.014=7.43×10⁻³ H, dI/dt=(3-6)/0.25=-12 A/s, e= -L×(-12)=0.089 V. Rotation causes radial charge separation via the magnetic force ( F = q v × B ), inducing an emf from the center to the rim. Using Φ =

Ref: NCERT > Physics Book > Electromagnetic Induction > Self-Induction and Self-Inductance

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 14 cm is deformed into a straight wire in a 0.15 T field in 0.5 s. What is the induced emf?

**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. Initial flux: Φ = B A = 0.15 × π × (0.14)² = 0.00923 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00923/0.5) = 0.01846 V ≈

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