#magnetic field intensity

Hysteresis, Retentivity, Coercivity and Permanent Magnets

A material has \( B = 0.15 \, \text{T} \) and \( M = 8 \times 10^4 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \mu_

**Hysteresis loop** plots B versus H for ferromagnetic material, retentivity (remanence) is residual B at H=0 after saturation, coercivity is reverse H needed to reduce B to zero. Hard ferromagnets have high retentivity and coercivity, retaining strong magnetism after external field removed, suitable for permanent magnets. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.15 T , M = 8 × 10⁴ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.15/4π × 10⁻⁷) ≈ 1.194 × 10⁵ A m⁻¹ . H = 1.194 × 10⁵ - 8 × 10⁴ = 3.94 × 10⁴

Ref: NCERT > Physics Book > Magnetism and Matter > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A material with \( \mu_r = 350 \) and \( H = 400 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**Hysteresis loop** plots B versus H for ferromagnetic material, retentivity (remanence) is residual B at H=0 after saturation, coercivity is reverse H needed to reduce B to zero. Hard ferromagnets have high retentivity and coercivity, retaining strong magnetism after external field removed, suitable for permanent magnets. B = μ₀ μ_r H . Given: μ_r = 350 , H = 400 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 350 × 400 = 0.17584 T ≈ 0.18 T . Substituting values gives 0.18 T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and

Ref: NCERT > Physics Book > Magnetism and Matter > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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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 .

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

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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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Torque on Magnetic Dipole and Potential Energy

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

**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 μ_r H . Given: μ_r = 600 , H = 300 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 600 × 300 = 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.

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

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