#circular coil

Induced EMF Due to Change in Magnetic Field

A circular coil of radius 9 cm and 200 turns rotates at 50 rad/s in a 0.03 T field. What is the maximum emf induced?

**Induced emf due to B change** e = -N A dB/dt, N turns, A area (m²), dB/dt rate of change of field (T/s). For 110 turns area 0.035 m² B 0.09 T to 0 in 0.5 s, dB/dt=0.18 T/s, e=110×0.035×0.18=0.693 V, direction opposes decrease via Lenz's law. A = π r² = 3.14 × (0.09)² = 0.0254 m² . ε₀ = N B A ω = 200 × 0.03 × 0.0254 × 50 = 7.62 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,

Ref: NCERT > Physics Book > Electromagnetic Induction > Induced EMF Due to Change in Magnetic Field

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Rotational EMF and AC Generator

A circular coil of radius 5 cm and 300 turns rotates at 35 rad/s in a 0.05 T field. What is the maximum emf induced?

**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. A = π r² = 3.14 × (0.05)² = 0.00785 m² . ε₀ = N B A ω = 300 × 0.05 × 0.00785 × 35 = 4.12375 V ≈ 4.12 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

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

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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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Solenoid, Toroid and Ampere's Law

A circular coil of 70 turns and radius \( 7 \, \text{cm} \) carries a current of \( 0.8 \, \text{A} \). What is the magn

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 70 × 0.8/2 × 0.07) = (22.4 π × 10⁻⁶/0.14) = 1.6 π × 10⁻⁴ ≈ 5.03 × 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 > Solenoid, Toroid and Ampere's Law

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Solenoid, Toroid and Ampere's Law

A circular loop of radius \( 0.13 \, \text{m} \) with 20 turns carries a current of \( 3.5 \, \text{A} \). What is the m

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 20 × 3.5/2 × 0.13) = (28 π × 10⁻⁶/0.26) = 1.0769 π × 10⁻⁴ ≈ 3.38 × 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 > Solenoid, Toroid and Ampere's Law

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Solenoid, Toroid and Ampere's Law

A circular loop of radius \( 0.15 \, \text{m} \) with 15 turns carries a current of \( 2 \, \text{A} \). What is the mag

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 15 × 2/2 × 0.15) = (12 π × 10⁻⁶/0.3) = 4 π × 10⁻⁵ ≈ 1.26 × 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 > Solenoid, Toroid and Ampere's Law

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A circular coil of radius \( 0.11 \, \text{m} \) with 25 turns carries \( 3 \, \text{A} \). What is the magnetic field a

**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. B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 25 × 3/2 × 0.11) = (30 π × 10⁻⁶/0.22) = 1.3636 π × 10⁻⁴ ≈ 4.28 × 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 = N