#magnetic flux density

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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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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 3.5 \, \text{A m}^2 \) is at \( 0.6 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**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) (2m/r³) . Given: m = 3.5 A m² , r = 0.6 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 3.5/(0.6)³) = 10⁻⁷ × (7.0/0.216) ≈ 3.24 × 10⁻⁶ T . Substituting values gives 3.24 × 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.

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material with \( B = 0.45 \, \text{T} \) and \( H = 3000 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**Magnetic properties** μ_r = 400 indicates 400 times vacuum permeability, so B enhanced 400 times for same nI. H = nI (A/m) for solenoid, M = χ H, B = μ₀(H+M) links microscopic magnetization to macroscopic field. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.45 T , H = 3000 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.45/4π × 10⁻⁷) ≈ 3.581 × 10⁵ A m⁻¹ . M = 3.581 × 10⁵ - 3000 ≈ 3.551 × 10⁵ A m⁻¹ . Substituting values gives 3.551 × 10⁵ A m⁻¹, which matches

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

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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.

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 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