#turns per meter

Torque on Current Loop, Magnetic Moment and Galvanometer

A solenoid with 1400 turns per meter carries a current of \( 2.5 \, \text{A} \). What is the magnetic field inside it? (

**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 . B = 4 π × 10⁻⁷ × 1400 × 2.5 = 14 π × 10⁻⁴ ≈ 4.40 × 10⁻³ T . Using F = q v B sinθ, F

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

View Details

Torque on Current Loop, Magnetic Moment and Galvanometer

A solenoid with 1100 turns per meter carries \( 1.8 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi

**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. B = μ₀ n I . B = 4 π × 10⁻⁷ × 1100 × 1.8 = 7.92 π × 10⁻⁴ ≈ 2.49 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

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

View Details

Torque on Current Loop, Magnetic Moment and Galvanometer

A solenoid has 1250 turns per meter and carries a current of \( 1.5 \, \text{A} \). What is the magnetic field inside it

**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. Magnetic field B = μ₀ n I . B = 4 π × 10⁻⁷ × 1250 × 1.5 = 7.5 π × 10⁻⁴ ≈ 2.36 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

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

View Details

Torque on Current Loop, Magnetic Moment and Galvanometer

A solenoid has 600 turns per meter and carries a current of \( 3 \, \text{A} \). What is the magnetic field inside it? (

**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 . B = 4 π × 10⁻⁷ × 600 × 3 = 7.2 π × 10⁻⁴ ≈ 2.26 × 10⁻³ T . Using F = q v B sinθ, F

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

View Details

Solenoid, Toroid and Ampere's Law

A solenoid with 1300 turns per meter carries \( 1.2 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi

**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. B = μ₀ n I . B = 4 π × 10⁻⁷ × 1300 × 1.2 = 6.24 π × 10⁻⁴ ≈ 1.96 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =