#torque calculation

Magnetization, Magnetic Intensity, Susceptibility and Permeability

A dipole with \( m = 0.2 \, \text{A m}^2 \) in \( B = 0.5 \, \text{T} \) at \( 90^\circ \) has torque:

**Relation between B, H, M** is B = μ₀(H+M) = μ₀(1+χ)H. Susceptibility χ = μ_r -1 quantifies material response. Given B, μ_r, n, current I = B/(μ₀ μ_r n), with μ₀ = 4π×10⁻⁷ T·m/A, enabling current calculation for desired B with magnetic core. tau = m B sinθ . Given: m = 0.2 A m² , B = 0.5 T , θ = 90° , sin 90° = 1 . tau = 0.2 × 0.5 × 1 = 0.1 N m . Substituting values gives 0.1 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 > Magnetization, Magnetic Intensity, Susceptibility and Permeability

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A dipole with \( m = 0.35 \, \text{A m}^2 \) in \( B = 0.9 \, \text{T} \) at \( 30^\circ \) has torque:

**Magnetic properties** μ_r = 400 indicates 400 times vacuum permeability, so B enhanced 400 times for same nI. H = nI (A/m) for solenoid, M = χ H, B = μ₀(H+M) links microscopic magnetization to macroscopic field. tau = m B sinθ . Given: m = 0.35 A m² , B = 0.9 T , θ = 30° , sin 30° = 0.5 . tau = 0.35 × 0.9 × 0.5 = 0.1575 N m . Substituting values gives 0.1575 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 > Magnetization, Magnetic Intensity, Susceptibility and Permeability

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A magnetic dipole of moment \( 0.25 \, \text{A m}^2 \) is in a uniform field of \( 0.5 \, \text{T} \) at \( 30^\circ \).

**Field due to bar magnet** on axial line is B_axial = (μ₀/4π)·2m/r³, equatorial B_eq = (μ₀/4π)·m/r³, where μ₀/4π = 10⁻⁷ T·m/A, m magnetic moment (A·m²), r distance (m). Axial field twice equatorial at same distance and parallel to moment, equatorial opposite to moment direction. Torque is tau = m B sinθ . Given: m = 0.25 A m² , B = 0.5 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.25 × 0.5 × 0.5 = 0.0625 N m . Substituting values gives 0.0625 N m, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A magnetic dipole of moment \( 0.8 \, \text{A m}^2 \) is in a uniform field of \( 0.5 \, \text{T} \) at \( 45^\circ \).

**Axial versus equatorial** field comparison shows B_axial = 2 B_eq for same r. Using μ₀ = 4π×10⁻⁷ T·m/A, calculation involves r³ = (0.3)³ = 0.027 m³, so B = 10⁻⁷·m/r³ yields moment estimation. Torque is tau = m B sinθ . Given: m = 0.8 A m² , B = 0.5 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . Substitute: tau = 0.8 × 0.5 × 0.707 ≈ 0.2828 N m ≈ 0.28 N m . Substituting values gives 0.28 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A magnetic dipole of moment \( 0.2 \, \text{A m}^2 \) is placed in a uniform field of \( 0.3 \, \text{T} \) at \( 45^\ci

**Axial versus equatorial** field comparison shows B_axial = 2 B_eq for same r. Using μ₀ = 4π×10⁻⁷ T·m/A, calculation involves r³ = (0.3)³ = 0.027 m³, so B = 10⁻⁷·m/r³ yields moment estimation. Torque is tau = m B sinθ . Given: m = 0.2 A m² , B = 0.3 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . Substitute: tau = 0.2 × 0.3 × 0.707 = 0.04242 N m ≈ 0.042 N m . Substituting values gives 0.042 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Electric Dipole - Moment, Field and Torque

A dipole with charges \( +6 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) separated by 4 mm is in a field \( 5 \times 10

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. Dipole moment: p = q × 2a = 6 × 10⁻⁶ × 4 × 10⁻³ = 2.4 × 10⁻⁸ C m . Torque: tau = p E sin θ = 2.4 × 10⁻⁸ × 5 × 10⁴ × sin 60° = 1.2 × 10⁻³ × (√(3)/2) = 1.04 × 10⁻³ N m . Substituting values gives 1.04 × 10⁻³ N m, which matches expected

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

A dipole with \( p = 4 \times 10^{-9} \, \text{C m} \) is at 30° to a field \( E = 6 \times 10^4 \, \text{N/C} \). What

**Interaction of dipole with uniform field** produces pure couple without net force, equal opposite forces forming torque. Potential energy minimum -pE at alignment, maximum +pE at anti-alignment, governing orientation dynamics. tau = p E sin θ . tau = 4 × 10⁻⁹ × 6 × 10⁴ × sin 30° = 24 × 10⁻⁵ × 0.5 = 1.2 × 10⁻⁴ N m . Substituting values gives 1.2 × 10⁻⁴ N m, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

A dipole with charges \( +5 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) separated by 3 mm is in a field of \( 2 \times

**Interaction of dipole with uniform field** produces pure couple without net force, equal opposite forces forming torque. Potential energy minimum -pE at alignment, maximum +pE at anti-alignment, governing orientation dynamics. Dipole moment: p = q × 2a = 5 × 10⁻⁶ × 3 × 10⁻³ = 1.5 × 10⁻⁸ C m . Torque: tau = p E sin θ = 1.5 × 10⁻⁸ × 2 × 10⁴ × sin 45° = 3 × 10⁻⁴ × (√(2)/2) = 2.12 × 10⁻⁴ N m . Substituting values gives 2.12 × 10⁻⁴ N m, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

A dipole with charges \( +3 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) separated by 5 mm is in a field \( 8 \times 10

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. Dipole moment: p = q × 2a = 3 × 10⁻⁶ × 5 × 10⁻³ = 1.5 × 10⁻⁸ C m . Torque: tau = p E sin θ = 1.5 × 10⁻⁸ × 8 × 10⁴ × sin 45° = 1.2 × 10⁻³ × (√(2)/2) = 8.48 × 10⁻⁴ N m . Substituting values gives 8.48 × 10⁻⁴ N m, which matches expected

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) is at 60° to a field \( E = 3 \times 10^4 \, \text{N/C} \). What

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. tau = p E sin θ . tau = 6 × 10⁻⁹ × 3 × 10⁴ × sin 60° = 18 × 10⁻⁵ × (√(3)/2) = 1.56 × 10⁻⁴ N m . Substituting values gives 1.56 × 10⁻⁴ N m, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

A dipole with charges \( +7 \, \mu\text{C} \) and \( -7 \, \mu\text{C} \) separated by 3 mm is in a field \( 4 \times 10

**Electric dipole** consists of charges +q and -q separated by 2a, dipole moment p = q·2a, vector from negative to positive, unit C·m. In uniform field E, torque τ = p × E, magnitude τ = p E sinθ, tending to align p with E, potential energy U = -p·E = -p E cosθ. Dipole moment: p = q × 2a = 7 × 10⁻⁶ × 3 × 10⁻³ = 2.1 × 10⁻⁸ C m . Torque: tau = p E sin θ = 2.1 × 10⁻⁸ × 4 × 10⁴ × sin 60° = 8.4 × 10⁻⁴ × (√(3)/2) = 7.28 × 10⁻⁴ N

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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PHYSICS

A magnetic dipole experiences a torque of 0.02 N m in a field of 0.4 T at 90° . What is its magnetic moment?

Given: A magnetic dipole experiences a torque of 0.02 N m in a field of 0.4 T at 90° . What is its magnetic moment? These values define the system as per NCERT data. Formula: tau = m B sinθ, so m = tau/B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: tau = 0.02 N m, B = 0.4 T, θ = 90°, sin 90° = 1 . m = 0.02/0.4 × 1 = 0.05 A m² . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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