#square loop

Torque on Current Loop, Magnetic Moment and Galvanometer

A square loop of side \( 0.15 \, \text{m} \) with 50 turns carries \( 1 \, \text{A} \) in a magnetic field of \( 0.6 \,

**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. Torque tau = N I A B sin θ , where A = 0.15 × 0.15 = 0.0225 m² . tau = 50 × 1 × 0.0225 × 0.6 × sin 30° = 0.675 × 0.5 =

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 \( 0.25 \, \text{m} \) with 20 turns carries \( 2.5 \, \text{A} \) in a magnetic field of \( 0.3 \

**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 = 0.25 × 0.25 = 0.0625 m² . tau = 20 × 2.5 × 0.0625 × 0.3 × sin 45° = 0.9375 × 0.707 = 0.6633 ≈ 0.66 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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Magnetic Field Due to Current - Straight Wire and Circular Loop

A square loop of side \( 0.12 \, \text{m} \) with 35 turns carries \( 1.8 \, \text{A} \) in a magnetic field of \( 0.5 \

**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. Torque tau = N I A B sin θ , where A = 0.12 × 0.12 = 0.0144 m² . tau = 35 × 1.8 × 0.0144 × 0.5 × sin 30° = 0.9072 × 0.5 = 0.4536 ≈ 0.45 N m . 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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Magnetic Field Due to Current - Straight Wire and Circular Loop

A square loop of side \( 0.16 \, \text{m} \) with 25 turns carries \( 3 \, \text{A} \) in a magnetic field of \( 0.7 \,

**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 = 0.16 × 0.16 = 0.0256 m² . tau = 25 × 3 × 0.0256 × 0.7 × sin 45° = 1.344 × 0.707 = 0.9502 ≈ 0.95 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 \( 0.2 \, \text{m} \) with 30 turns carries \( 1.5 \, \text{A} \) in a magnetic field of \( 0.4 \,

**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. Torque tau = N I A B sin θ , where A = 0.2 × 0.2 = 0.04 m² . tau = 30 × 1.5 × 0.04 × 0.4 × sin 60° = 1.8 × 0.4 × 0.866 = 0.6235 ≈ 0.62 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 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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