Skip to content

Torque on Magnetic Dipole and Potential Energy

Latest questions in this category.

30 questions

A magnetic dipole with \( m = 0.3 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 90^\circ \) has potentia

**Magnetic dipole in uniform field** experiences torque τ = m B sinθ and potential energy U = -m·B = -m B cosθ, minimum -mB when aligned (θ=0°), maximum +mB at anti-alignment (θ=180°). Work done rotating from θ₁ to θ₂ equals ΔU = mB(cosθ₁ - cosθ₂). U_m = -m B cosθ . Given: m = 0.3 A m² , B = 0.4 T , θ = 90° , cos 90° = 0 . U_m = -0.3 × 0.4 × 0 = 0 J . Substituting values gives 0 J, 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 > Torque on Magnetic Dipole and Potential Energy

When a magnetic dipole is placed perpendicular to a uniform magnetic field, the torque acting on it is maximum because:

**Torque on magnetic dipole** in uniform field B is τ = m × B, magnitude τ = m B sinθ, m moment (A·m²), B field (T), θ angle between m and B (degrees). Torque tends to align moment with field, zero at θ = 0°, maximum mB at 90°, direction given by right-hand rule. The torque on a magnetic dipole is given by tau = m B sinθ . It reaches its maximum value when sinθ = 1 , which occurs at θ = 90° (perpendicular orientation), as the cross product m × B is greatest when the angle between the dipole moment and field

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

The alignment of a magnetic dipole in a uniform field results in:

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. In a uniform field, a magnetic dipole experiences a torque that aligns it with the field to minimize potential energy ( U = -m B cosθ ), reaching a stable equilibrium when parallel ( θ = 0° ), with no net force due to field uniformity. Substituting values gives A stable equilibrium position, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

A dipole with \( m = 0.25 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 90^\circ \) has potential energy

**Magnetic dipole in uniform field** experiences torque τ = m B sinθ and potential energy U = -m·B = -m B cosθ, minimum -mB when aligned (θ=0°), maximum +mB at anti-alignment (θ=180°). Work done rotating from θ₁ to θ₂ equals ΔU = mB(cosθ₁ - cosθ₂). U_m = -m B cosθ . Given: m = 0.25 A m² , B = 0.4 T , θ = 90° , cos 90° = 0 . U_m = -0.25 × 0.4 × 0 = 0 J . Substituting values gives 0 J, 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 > Torque on Magnetic Dipole and Potential Energy

The magnetic field contribution \( B_m \) due to a material with \( M = 3.5 \times 10^5 \, \text{A m}^{-1} \) is: (Take

**Torque on magnetic dipole** in uniform field B is τ = m × B, magnitude τ = m B sinθ, m moment (A·m²), B field (T), θ angle between m and B (degrees). Torque tends to align moment with field, zero at θ = 0°, maximum mB at 90°, direction given by right-hand rule. B_m = μ₀ M . Given: M = 3.5 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 3.5 × 10⁵ = 0.4396 T ≈ 0.44 T . Substituting values gives 0.44 T, which matches expected magnitude for this magnetic configuration, confirming dipole

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

A magnetic dipole oscillates in a uniform field when displaced from its equilibrium position because:

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. The torque on a magnetic dipole ( tau = m B sinθ ) acts as a restoring force when displaced from its equilibrium position (aligned with the field). This torque causes oscillatory motion, similar to a pendulum, as it seeks to return to the stable alignment. Substituting values gives The torque acts as a restoring force, which matches expected magnitude for this magnetic configuration, confirming

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

A magnetic dipole in a uniform field is in unstable equilibrium when:

**Magnetic dipole in uniform field** experiences torque τ = m B sinθ and potential energy U = -m·B = -m B cosθ, minimum -mB when aligned (θ=0°), maximum +mB at anti-alignment (θ=180°). Work done rotating from θ₁ to θ₂ equals ΔU = mB(cosθ₁ - cosθ₂). The potential energy U = -m B cosθ is maximized when θ = 180° (anti-parallel), making it an unstable equilibrium position. Any small perturbation causes the dipole to rotate toward the stable position ( θ = 0° ), as the energy decreases in that direction. Substituting values gives It is anti-parallel to the field, which matches expected magnitude for this

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

A dipole with \( m = 0.7 \, \text{A m}^2 \) in \( B = 0.2 \, \text{T} \) at \( 45^\circ \) has torque:

**Magnetic dipole in uniform field** experiences torque τ = m B sinθ and potential energy U = -m·B = -m B cosθ, minimum -mB when aligned (θ=0°), maximum +mB at anti-alignment (θ=180°). Work done rotating from θ₁ to θ₂ equals ΔU = mB(cosθ₁ - cosθ₂). tau = m B sinθ . Given: m = 0.7 A m² , B = 0.2 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . tau = 0.7 × 0.2 × 0.707 = 0.09898 N m ≈ 0.1 N m . Substituting values gives 0.1 N m, which matches expected magnitude for this magnetic configuration, confirming

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

A material with \( B = 0.25 \, \text{T} \) and \( H = 1500 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**Torque on magnetic dipole** in uniform field B is τ = m × B, magnitude τ = m B sinθ, m moment (A·m²), B field (T), θ angle between m and B (degrees). Torque tends to align moment with field, zero at θ = 0°, maximum mB at 90°, direction given by right-hand rule. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.25 T , H = 1500 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.25/4π × 10⁻⁷) ≈ 1.99 × 10⁵ A m⁻¹ . M = 1.99 × 10⁵ - 1500

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

A material with \( \mu_r = 400 \) and \( H = 500 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**Torque on magnetic dipole** in uniform field B is τ = m × B, magnitude τ = m B sinθ, m moment (A·m²), B field (T), θ angle between m and B (degrees). Torque tends to align moment with field, zero at θ = 0°, maximum mB at 90°, direction given by right-hand rule. B = μ₀ μ_r H . Given: μ_r = 400 , H = 500 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 400 × 500 = 0.2512 T ≈ 0.25 T . Substituting values gives 0.25 T, which matches expected magnitude for this magnetic

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

A material has \( B = 0.3 \, \text{T} \) and \( M = 1.5 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \mu

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.3 T , M = 1.5 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.3/4π × 10⁻⁷) ≈ 2.387 × 10⁵ A m⁻¹ . H = 2.387 × 10⁵ - 1.5 × 10⁵ = 8.87 × 10⁴ A m⁻¹ ≈ 8.9 × 10⁴

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

A material with \( \mu_r = 150 \) and \( H = 800 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**Torque on magnetic dipole** in uniform field B is τ = m × B, magnitude τ = m B sinθ, m moment (A·m²), B field (T), θ angle between m and B (degrees). Torque tends to align moment with field, zero at θ = 0°, maximum mB at 90°, direction given by right-hand rule. B = μ₀ μ_r H . Given: μ_r = 150 , H = 800 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 150 × 800 = 0.15072 T ≈ 0.15 T . Substituting values gives 0.15 T, which matches expected magnitude for this magnetic

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy