#turns

Energy Stored in Inductor and Magnetic Energy

A coil of 130 turns and area 0.04 m² is in a field that increases from 0 to 0.05 T in 0.2 s. What is the induced emf?

**Energy in inductor** cannot change instantaneously because that would require infinite power, current through inductor continuous, voltage may jump, principle used in chokes, inductive kick, back emf, explaining why inductor opposes change in current. Δ Φ = B A = 0.05 × 0.04 = 0.002 Wb . ε = N (Δ Φ/Δ t) = 130 × (0.002/0.2) = 130 × 0.01 = 1.3 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 = ½ L I², result 1.3 V

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

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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.

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

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Self-Induction and Self-Inductance

A coil of 260 turns rotates at 75 rad/s in a 0.05 T field. If the area is 0.018 m², what is the maximum emf?

**Energy stored in inductor** U =½ L I², L inductance, I current, energy in magnetic field, density u = B²/(2μ₀), B=μ₀ n I inside solenoid, U = (B²/2μ₀)×volume, illustrating equivalence of circuit and field energy. ε₀ = N B A ω = 260 × 0.05 × 0.018 × 75 = 17.55 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 17.55 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Self-Induction and Self-Inductance

A coil of 320 turns rotates at 65 rad/s in a 0.08 T field. If the area is 0.015 m², what is the maximum emf?

**Self-inductance of solenoid** L = μ₀ N² A / l, N total turns, A cross-section, l length, for N=650 turns per meter means n=650 m⁻¹, if length 1 m N=650, A=0.014, L=4π×10⁻⁷×650²×0.014/1=0.00743 H, self-induced emf magnitude L |dI/dt|, dI/dt=12 A/s, e=0.089 V, opposes change. ε₀ = N B A ω = 320 × 0.08 × 0.015 × 65 = 24.96 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 24.96 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Self-Induction and Self-Inductance

A coil of 75 turns and area 0.05 m² is in a 0.12 T field that drops to zero in 0.25 s. What is the induced emf?

**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. Δ Φ = B A = 0.12 × 0.05 = 0.006 Wb . ε = N (Δ Φ/Δ t) = 75 × (0.006/0.25) = 75 × 0.024 = 1.8

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

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Magnetic Flux and Faraday's Laws of Induction

A coil of 90 turns experiences a magnetic flux change from 0 to 0.025 Wb in 0.05 s. What is the induced emf?

**Magnetic flux** Φ = B·A = B A cosθ, B magnetic field (T), A area (m²), θ angle between B and normal to area, unit Wb = T·m², Faraday's law induced emf e = -N dΦ/dt, N turns, negative sign Lenz's law indicating opposition, magnitude |e| = N |ΔΦ/Δt|, for 100 turns ΔΦ=0.03 Wb Δt=0.06 s e=100×0.03/0.06=50 V. ε = N (Δ Φ/Δ t) . Δ Φ = 0.025 Wb , Δ t = 0.05 s , N = 90 . ε = 90 × (0.025/0.05) = 90 × 0.5 = 45 V . Using Φ = B A cosθ, e = -N dΦ/dt =

Ref: NCERT > Physics Book > Electromagnetic Induction > Magnetic Flux and Faraday's Laws of Induction

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Rotational EMF and AC Generator

A coil of 160 turns and area 0.025 m² is rotated at 20 Hz in a 0.09 T field. What is the maximum emf?

**AC generator** principle same as rotating coil, N=200 turns A=0.04 m² B=0.1 T f=50 Hz ω=2π×50=314 rad/s, e₀= N B A ω =200×0.1×0.04×314=251.2 V, emf e= e₀ sin ωt, frequency equals rotation frequency, maximum when plane parallel to field. ω = 2π v = 2π × 20 = 40π rad/s . ε₀ = N B A ω = 160 × 0.09 × 0.025 × 40π = 45.24 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

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

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Magnetic Flux and Faraday's Laws of Induction

A coil of 110 turns experiences a magnetic flux change from 0 to 0.05 Wb in 0.1 s. What is the induced emf?

**Faraday's first law** emf induced when flux linking coil changes, second law magnitude proportional to rate of change, e = -dΦ/dt, for N turns e = -N dΦ/dt, flux Φ = B A cosθ, change can be due to B change, A change, or θ change, all produce emf. ε = N (Δ Φ/Δ t) . Δ Φ = 0.05 Wb , Δ t = 0.1 s , N = 110 . ε = 110 × (0.05/0.1) = 110 × 0.5 = 55 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N

Ref: NCERT > Physics Book > Electromagnetic Induction > Magnetic Flux and Faraday's Laws of Induction

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Mutual Induction and Mutual Inductance

A solenoid of 650 turns and length 1.2 m induces an emf of 2.6 V in a nearby coil when its current changes from 2 A to 6

**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 = 6 - 2 = 4 A , Δ t = 0.4 s . M = (ε/(Δ I/Δ t)) = (2.6/(4/0.4)) = (2.6/10) = 0.26 H . Using Φ = B

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

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Torque on Current Loop, Magnetic Moment and Galvanometer

A circular coil of 45 turns and radius \( 6 \, \text{cm} \) carries a current of \( 1.2 \, \text{A} \). What is the magn

**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 45 × 1.2/2 × 0.06) = (21.6 π × 10⁻⁶/0.12) = 1.8 π × 10⁻⁴ ≈ 5.65 × 10⁻⁴ T . Using F = q v B sinθ, F = I l

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A square loop of side \( 0.1 \, \text{m} \) with 25 turns carries \( 2 \, \text{A} \) in a magnetic field of \( 0.8 \, \

**Magnetic field due to long straight wire** at distance r is B = μ₀ I/(2π r), μ₀ = 4π×10⁻⁷ T·m/A, direction circular around wire given by right-hand grip rule. For I = 18 A, r = 0.15 m, B = 2×10⁻⁷×18/0.15 = 2.4×10⁻⁵ T, illustrating 1/r dependence. Torque tau = N I A B sin θ , where A = 0.1 × 0.1 = 0.01 m² . tau = 25 × 2 × 0.01 × 0.8 × sin 30° = 0.5 × 0.8 × 0.5 = 0.2 N m . Using F = q v B sinθ, F = I l B sinθ, B =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A square loop of side \( 0.18 \, \text{m} \) with 30 turns carries \( 2 \, \text{A} \) in a magnetic field of \( 0.4 \,

**Field at centre of circular loop** with N turns is B = μ₀ N I/(2R), R radius (m), direction along axis via right-hand rule, magnitude proportional to N I/R. For R = 0.09 m, N = 45, I = 1.2 A, B = 4π×10⁻⁷×45×1.2/(2×0.09) = 3.77×10⁻⁴ T, showing N enhancement. Torque tau = N I A B sin θ , where A = 0.18 × 0.18 = 0.0324 m² . tau = 30 × 2 × 0.0324 × 0.4 × sin 60° = 0.7776 × 0.866 = 0.6734 ≈ 0.67 N m . Using F = q v B sinθ, F = I l B

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop

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