#circular loop

AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 15 cm is deformed into a straight wire in a 0.15 T field in 0.6 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 × (0.15)² = 0.0106 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.0106/0.6) = 0.01767 V ≈

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 12 cm is deformed into a straight wire in a 0.2 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 = 0.2 × π × (0.12)² = 0.00904 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00904/0.5) = 0.01808 V ≈

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 13 cm is deformed into a straight wire in a 0.18 T field in 0.6 s. What is the induced emf?

**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. Initial flux: Φ = B A = 0.18 × π × (0.13)² = 0.00956 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00956/0.6) = 0.01593 V ≈ 0.016 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)

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 11 cm is deformed into a straight wire in a 0.2 T field in 0.5 s. What is the induced emf?

**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. Initial flux: Φ = B A = 0.2 × π × (0.11)² = 0.0076 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.0076/0.5) = 0.0152 V ≈ 0.015 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)

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Magnetic Flux and Faraday's Laws of Induction

A circular loop is placed in a uniform magnetic field, and the field strength is increased steadily. What determines the

**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. According to Faraday’s law, the induced emf depends on the rate of change of magnetic flux, which increases with the rate at which the magnetic field strength changes. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v =

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

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 18 cm is deformed into a straight wire in a 0.14 T field in 0.7 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 = 0.14 × π × (0.18)² = 0.01425 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.01425/0.7) = 0.02036 V ≈

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AC Generator, Back EMF and Time Duration of EMF

A circular loop of radius 16 cm is deformed into a straight wire in a 0.12 T field in 0.6 s. What is the induced emf?

**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. Initial flux: Φ = B A = 0.12 × π × (0.16)² = 0.00965 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.00965/0.6) = 0.01608 V ≈ 0.016 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)

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Torque on Current Loop, Magnetic Moment and Galvanometer

A circular loop of radius \( 0.08 \, \text{m} \) with 60 turns carries a current of \( 0.75 \, \text{A} \). What is the

**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 60 × 0.75/2 × 0.08) = (18 π × 10⁻⁶/0.16) = 1.125 π × 10⁻⁴ ≈ 3.53 × 10⁻⁴ T .

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A circular loop of radius \( 0.2 \, \text{m} \) with 10 turns carries \( 1.5 \, \text{A} \). What is the magnetic field

**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. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 10 × 1.5/2 × 0.2) = (6 π × 10⁻⁶/0.4) = 1.5 π × 10⁻⁵ ≈ 4.71 × 10⁻⁵ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

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Torque on Current Loop, Magnetic Moment and Galvanometer

What is the direction of the magnetic field at the center of a circular current-carrying loop?

**Magnetic moment of loop** m = N I A (A·m²), potential energy U = -m·B = -N I A B cosθ, torque tends to align m with B. For square side 0.18 m, A = 0.0324 m², N=30, I=2 A, B=0.4 T, θ=60°, τ =30×2×0.0324×0.4×sin60° =0.7776×0.866=0.673 N·m, illustrating large torque for modest parameters. Using the right-hand rule, if the fingers curl along the direction of the current in the loop, the thumb points perpendicular to the plane of the loop, indicating the field direction at the center. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r),

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 circular loop of radius \( 0.09 \, \text{m} \) with 45 turns carries a current of \( 1.2 \, \text{A} \). What is the m

**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 45 × 1.2/2 × 0.09) = (21.6 π × 10⁻⁶/0.18) = 1.2 π × 10⁻⁴ ≈ 3.77 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ

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