#rectangular coil

Torque on Current Loop, Magnetic Moment and Galvanometer

A rectangular loop of area \( 0.06 \, \text{m}^2 \) with 15 turns carries \( 2.5 \, \text{A} \) in a field of \( 0.8 \,

**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. tau = N I A B sin θ , where θ = 60° to plane means sin 30° with normal. tau = 15 × 2.5 × 0.06 × 0.8 × sin 60° = 1.8 × 0.866 = 1.5588 ≈ 1.56 N m . Using F = q v B sinθ, F =

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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 rectangular loop of area \( 0.05 \, \text{m}^2 \) with 15 turns carries \( 2 \, \text{A} \) in a field of \( 1 \, \tex

**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. tau = N I A B sin θ , θ = 90° to plane means sin 0° = 1 with normal. tau = 15 × 2 × 0.05 × 1 × 1 = 1.5 N m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ

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 rectangular loop of area \( 0.08 \, \text{m}^2 \) with 15 turns carries \( 2.5 \, \text{A} \) in a field of \( 0.6 \,

**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. tau = N I A B sin θ , where θ = 30° to plane means sin 60° with normal. tau = 15 × 2.5 × 0.08 × 0.6 × sin 30° = 1.8 × 0.5 = 0.9 N m . 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 > 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 rectangular loop of area \( 0.04 \, \text{m}^2 \) with 18 turns carries \( 2.8 \, \text{A} \) in a field of \( 0.9 \,

**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. tau = N I A B sin θ , where θ = 60° to plane means sin 30° with normal. tau = 18 × 2.8 × 0.04 × 0.9 × sin 60° = 1.8144 × 0.866 = 1.5712 ≈ 1.57 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 rectangular loop of area \( 0.03 \, \text{m}^2 \) with 20 turns carries \( 2 \, \text{A} \) in a field of \( 0.7 \, \t

**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. tau = N I A B sin θ , where θ = 45° to plane means sin 45° with normal. tau = 20 × 2 × 0.03 × 0.7 × sin 45° = 0.84 × 0.707 = 0.5939 ≈ 0.59 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 Force on Moving Charge - Lorentz Force and Motion

A rectangular loop of area \( 0.06 \, \text{m}^2 \) with 22 turns carries \( 3 \, \text{A} \) in a field of \( 0.5 \, \t

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. tau = N I A B sin θ , θ = 90° to plane means sin 0° = 1 with normal. tau = 22 × 3 × 0.06 × 0.5 × 1 = 1.98 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 Force on Moving Charge - Lorentz Force and Motion

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A rectangular loop of area \( 0.08 \, \text{m}^2 \) with 10 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 = 10 × 4 × 0.08 × 0.9 × 1 = 2.88 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 Force on Moving Charge - Lorentz Force and Motion

A rectangular loop of area \( 0.04 \, \text{m}^2 \) with 25 turns carries \( 3 \, \text{A} \) in a field of \( 1.2 \, \t

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. tau = N I A B sin θ , θ = 90° to plane means sin 0° = 1 with normal. tau = 25 × 3 × 0.04 × 1.2 × 1 = 3.6 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 Force on Moving Charge - Lorentz Force and Motion

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A rectangular loop of area \( 0.07 \, \text{m}^2 \) with 12 turns carries \( 3.5 \, \text{A} \) in a field of \( 0.5 \,

**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. tau = N I A B sin θ , where θ = 60° to plane means sin 30° with normal. tau = 12 × 3.5 × 0.07 × 0.5 × sin 60° = 1.47 × 0.866 = 1.273 ≈ 1.27 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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