#H field

Diamagnetism, Paramagnetism and Ferromagnetism

A material with \( B = 0.35 \, \text{T} \) and \( H = 1800 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**Paramagnetism** has small positive χ ≈ 10⁻³ to 10⁻⁵, weakly attracted towards stronger field, random moments align partially with B, magnetization decreases with temperature following Curie law χ ∝ 1/T. Materials have unpaired electrons with permanent moments. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.35 T , H = 1800 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.35/4π × 10⁻⁷) ≈ 2.785 × 10⁵ A m⁻¹ . M = 2.785 × 10⁵ - 1800 ≈ 2.767 × 10⁵ A m⁻¹ . Substituting values gives 2.767 × 10⁵ A m⁻¹, which matches

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

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Diamagnetism, Paramagnetism and Ferromagnetism

A material with \( B = 0.4 \, \text{T} \) and \( H = 2000 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \time

**Ferromagnetism** shows large positive χ ≈ 10³ to 10⁵, strong attraction, domain structure with spontaneous magnetization, hysteresis, retentivity. Distinction based on sign and magnitude of χ and behaviour in non-uniform field, explaining attraction versus repulsion. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.4 T , H = 2000 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.4/4π × 10⁻⁷) ≈ 3.183 × 10⁵ A m⁻¹ . M = 3.183 × 10⁵ - 2000 ≈ 3.163 × 10⁵ A m⁻¹ . Substituting values gives 3.163 × 10⁵ A m⁻¹, which matches expected magnitude

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 600 turns per meter carries a current of \( 3.5 \, \text{A} \). What is the magnetic intensity \( H \) i

**Magnetic field inside solenoid** with core B = μ₀ μ_r n I is uniform, direction along axis given by right-hand grip rule. For n = 2000 m⁻¹, μ_r = 400, B = 1.2 T, I = 1.2/(4π×10⁻⁷×400×2000) ≈ 1.19 A, showing modest current produces tesla-level field with high μ_r core. Magnetic intensity H = n I . Given: n = 600 m⁻¹ , I = 3.5 A . Substitute: H = 600 × 3.5 = 2100 A m⁻¹ . Substituting values gives 2100 A m⁻¹, 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 > Solenoid with Magnetic Core and Magnetic Properties

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 500 turns per meter carries a current of \( 4.5 \, \text{A} \). What is the magnetic intensity \( H \) i

**Magnetic field inside solenoid** with core B = μ₀ μ_r n I is uniform, direction along axis given by right-hand grip rule. For n = 2000 m⁻¹, μ_r = 400, B = 1.2 T, I = 1.2/(4π×10⁻⁷×400×2000) ≈ 1.19 A, showing modest current produces tesla-level field with high μ_r core. Magnetic intensity H = n I . Given: n = 500 m⁻¹ , I = 4.5 A . Substitute: H = 500 × 4.5 = 2250 A m⁻¹ . Substituting values gives 2250 A m⁻¹, 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 > Solenoid with Magnetic Core and Magnetic Properties

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A material with \( \mu_r = 500 \) and \( H = 300 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**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. B = μ₀ μ_r H . Given: μ_r = 500 , H = 300 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 500 × 300 = 0.1884 T ≈ 0.19 T . Substituting values gives 0.19 T, 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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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 400 turns per meter carries a current of \( 5 \, \text{A} \). What is the magnetic intensity \( H \) ins

**Magnetic field inside solenoid** with core B = μ₀ μ_r n I is uniform, direction along axis given by right-hand grip rule. For n = 2000 m⁻¹, μ_r = 400, B = 1.2 T, I = 1.2/(4π×10⁻⁷×400×2000) ≈ 1.19 A, showing modest current produces tesla-level field with high μ_r core. Magnetic intensity H = n I . Given: n = 400 m⁻¹ , I = 5 A . Substitute: H = 400 × 5 = 2000 A m⁻¹ . Substituting values gives 2000 A m⁻¹, 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 > Solenoid with Magnetic Core and Magnetic Properties

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material with \( B = 0.36 \, \text{T} \) and \( H = 2500 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**Magnetization M** is magnetic moment per unit volume (A/m), magnetic intensity H = B/μ₀ - M, susceptibility χ = M/H dimensionless, permeability μ = B/H = μ₀(1+χ), relative permeability μ_r = μ/μ₀ = 1+χ. For solenoid with core, B = μ₀ μ_r n I, n turns per meter (m⁻¹), I current (A). B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.36 T , H = 2500 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.36/4π × 10⁻⁷) ≈ 2.864 × 10⁵ A m⁻¹ . M = 2.864 × 10⁵ - 2500 ≈ 2.839

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetization, Magnetic Intensity, Susceptibility and Permeability

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 300 turns per meter carries a current of \( 6 \, \text{A} \). What is the magnetic intensity \( H \) ins

**Magnetic field inside solenoid** with core B = μ₀ μ_r n I is uniform, direction along axis given by right-hand grip rule. For n = 2000 m⁻¹, μ_r = 400, B = 1.2 T, I = 1.2/(4π×10⁻⁷×400×2000) ≈ 1.19 A, showing modest current produces tesla-level field with high μ_r core. Magnetic intensity H = n I . Given: n = 300 m⁻¹ , I = 6 A . Substitute: H = 300 × 6 = 1800 A m⁻¹ . Substituting values gives 1800 A m⁻¹, 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 > Solenoid with Magnetic Core and Magnetic Properties

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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

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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

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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

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 700 turns per meter carries a current of \( 3 \, \text{A} \). What is the magnetic intensity \( H \) ins

**Core magnetization** M = (μ_r -1)nI, so B = μ₀(nI + M). High μ_r materials like soft iron increase B dramatically for same nI, used in electromagnets, with μ_r up to 5000, enabling strong fields with low current. Magnetic intensity H = n I . Given: n = 700 m⁻¹ , I = 3 A . Substitute: H = 700 × 3 = 2100 A m⁻¹ . Substituting values gives 2100 A m⁻¹, 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 > Solenoid with Magnetic Core and Magnetic Properties

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