#magnetic torque

Torque on Current Loop, Magnetic Moment and Galvanometer

A rectangular loop of area \( 0.02 \, \text{m}^2 \) with 10 turns carries \( 3 \, \text{A} \) in a field of \( 0.5 \, \t

**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 the plane means sin 30° with the normal. tau = 10 × 3 × 0.02 × 0.5 × sin 60° = 0.3 × 0.866 = 0.2598 ≈ 0.26 N m . Using F = q v B sinθ,

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 Force on Moving Charge - Lorentz Force and Motion

A rectangular loop of area \( 0.07 \, \text{m}^2 \) with 10 turns carries \( 5 \, \text{A} \) in a field of \( 0.8 \, \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 × 5 × 0.07 × 0.8 × 1 = 2.8 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 \( 0.09 \, \text{m}^2 \) with 18 turns carries \( 2 \, \text{A} \) in a field of \( 0.6 \, \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 = 18 × 2 × 0.09 × 0.6 × 1 = 1.944 ≈ 1.94 N m . Using F = q v B sinθ, F = I l B sinθ, 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 \( 0.05 \, \text{m}^2 \) with 16 turns carries \( 2.2 \, \text{A} \) in a 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. tau = N I A B sin θ , where θ = 45° to plane means sin 45° with normal. tau = 16 × 2.2 × 0.05 × 0.7 × sin 45° = 1.232 × 0.707 = 0.871 ≈ 0.87 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 \( 0.04 \, \text{m}^2 \) 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 \( 0.04 \, \text{m}^2 \) 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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Earth's Magnetism and Magnetic Declination

A magnetic dipole of moment \( 0.2 \, \text{A m}^2 \) is in a uniform field of \( 0.9 \, \text{T} \) at \( 30^\circ \).

**Earth's magnetism** approximated as dipole inclined to rotation axis, magnetic declination is angle between geographic north and magnetic north, inclination or dip angle is angle between total field and horizontal. Horizontal component B_H = B cosδ, vertical B_V = B sinδ, δ dip angle, B_H ≈ 3-4×10⁻⁵ T in India. Torque is tau = m B sinθ . Given: m = 0.2 A m² , B = 0.9 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.2 × 0.9 × 0.5 = 0.09 N m . Substituting values gives 0.09 N m, which matches expected magnitude for this magnetic

Ref: NCERT > Physics Book > Magnetism and Matter > Earth's Magnetism and Magnetic Declination

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A dipole with \( m = 0.5 \, \text{A m}^2 \) in \( B = 0.1 \, \text{T} \) at \( 60^\circ \) has torque:

**Soft ferromagnetic materials** have low coercivity and retentivity, narrow hysteresis loop, lose magnetism when external field removed, ideal for electromagnets and transformer cores. Energy loss per cycle proportional to loop area, explaining why soft materials minimize loss. tau = m B sinθ . Given: m = 0.5 A m² , B = 0.1 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . tau = 0.5 × 0.1 × 0.866 = 0.0433 N m ≈ 0.043 N m . Substituting values gives 0.043 N m, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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