#B field

Diamagnetism, Paramagnetism and Ferromagnetism

A material has \( B = 0.42 \, \text{T} \) and \( M = 3.0 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \m

**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 H = (B/μ₀) - M . Given: B = 0.42 T , M = 3.0 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.42/4π × 10⁻⁷) ≈ 3.338 × 10⁵ A m⁻¹ . H = 3.338 × 10⁵ - 3.0 × 10⁵ = 3.38 × 10⁴ A m⁻¹ . Substituting values gives 3.38 × 10⁴

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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

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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

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 2.2 \, \text{A m}^2 \) produces a field at \( 0.3 \, \text{m} \) on its equatorial line. What i

**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. B = (μ₀/4π) (m/r³) . Given: m = 2.2 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2.2/(0.3)³) = 10⁻⁷ × (2.2/0.027) ≈ 8.15 × 10⁻⁶ T . Substituting values gives 8.15 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material has \( B = 0.28 \, \text{T} \) and \( M = 2.0 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \m

**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. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.28 T , M = 2.0 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.28/4π × 10⁻⁷) ≈ 2.228 × 10⁵ A m⁻¹ . H = 2.228 × 10⁵ - 2.0 × 10⁵ = 2.28 × 10⁴ A m⁻¹ .

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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

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material with \( \mu_r = 300 \) and \( H = 600 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**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. B = μ₀ μ_r H . Given: μ_r = 300 , H = 600 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 300 × 600 = 0.22608 T ≈ 0.23 T . Substituting values gives 0.23 T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

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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

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Torque on Magnetic Dipole and Potential Energy

A material has \( B = 0.5 \, \text{T} \) and \( M = 2 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \mu_0

**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). B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.5 T , M = 2 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.5/4π × 10⁻⁷) ≈ 3.978 × 10⁵ A m⁻¹ . H = 3.978 × 10⁵ - 2 × 10⁵ = 1.978 × 10⁵ A

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Torque on Magnetic Dipole and Potential Energy

A material has \( B = 0.54 \, \text{T} \) and \( M = 4.0 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \m

**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 H = (B/μ₀) - M . Given: B = 0.54 T , M = 4.0 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.54/4π × 10⁻⁷) ≈ 4.297 × 10⁵ A m⁻¹ . H = 4.297 × 10⁵

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