#coil calculation

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

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Energy Stored in Inductor and Magnetic Energy

A coil of 120 turns and area 0.04 m² is in a 0.09 T field that drops to zero in 0.3 s. What is the induced emf?

**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. Δ Φ = B A = 0.09 × 0.04 = 0.0036 Wb . ε = N (Δ Φ/Δ t) = 120 × (0.0036/0.3) = 120 × 0.012 = 1.44 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B

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

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Magnetic Flux and Faraday's Laws of Induction

A coil of 100 turns experiences a magnetic flux change from 0 to 0.03 Wb in 0.06 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 = 0.03 Wb , Δ t = 0.06 s , N = 100 . ε = 100 × (0.03/0.06) = 100 × 0.5 = 50 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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Magnetic Force on Moving Charge - Lorentz Force and Motion

A rectangular loop of area \( 0.03 \, \text{m}^2 \) with 20 turns carries \( 4 \, \text{A} \) in a field of \( 0.9 \, \t

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. tau = N I A B sin θ , θ = 90° to plane means sin 0° = 1 with normal. tau = 20 × 4 × 0.03 × 0.9 × 1 = 2.16 N m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A circular coil of radius \( 0.06 \, \text{m} \) with 25 turns carries \( 3 \, \text{A} \). What is the magnetic field a

**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. B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 25 × 3/2 × 0.06) = (30 π × 10⁻⁶/0.12) = 2.5 π × 10⁻⁴ ≈ 7.85 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B =

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

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