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#H calculation

2 public questions tagged with this topic.

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

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

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Torque on Magnetic Dipole and Potential Energy

A material has \( B = 0.65 \, \text{T} \) and \( M = 4.5 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \m

**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 = 0.65 T , M = 4.5 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.65/4π × 10⁻⁷) ≈ 5.167 × 10⁵ A m⁻¹ . H = 5.167 × 10⁵ - 4.5 × 10⁵ = 6.67 × 10⁴ A

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

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