#magnetization

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

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

The magnetization \( M \) of a sample is \( 5 \times 10^4 \, \text{A m}^{-1} \) in a magnetic field \( B = 0.1 \, \text{

**Permanent magnet requirement** is high retentivity to maintain field and high coercivity to resist demagnetization. Ability to retain magnetism after field removal is property of hard ferromagnets, related to domain wall pinning and anisotropy. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.1 T , M = 5 × 10⁴ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.1/4π × 10⁻⁷) ≈ 7.96 × 10⁴ A m⁻¹ . H = 7.96 × 10⁴ - 5 × 10⁴ = 2.96 × 10⁴ A m⁻¹ ≈ 3 × 10⁴ A m⁻¹ . Substituting values gives

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

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

The magnetic field contribution \( B_m \) due to a material with \( M = 2 \times 10^5 \, \text{A m}^{-1} \) is: (Take \(

**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_m = μ₀ M . Given: M = 2 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 2 × 10⁵ = 0.2512 T ≈ 0.25 T . Substituting values gives 0.25 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 > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

The magnetic field contribution \( B_m \) due to a material with \( M = 3.2 \times 10^5 \, \text{A m}^{-1} \) is: (Take

**Soft ferromagnetic materials** have low coercivity and retentivity, narrow hysteresis loop, lose magnetism when external field removed, ideal for electromagnets and transformer cores. Energy loss per cycle proportional to loop area, explaining why soft materials minimize loss. B_m = μ₀ M . Given: M = 3.2 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 3.2 × 10⁵ = 0.40192 T ≈ 0.40 T . Substituting values gives 0.40 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 > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

The reason a paramagnetic material’s magnetization increases with a stronger external field is:

**Soft ferromagnetic materials** have low coercivity and retentivity, narrow hysteresis loop, lose magnetism when external field removed, ideal for electromagnets and transformer cores. Energy loss per cycle proportional to loop area, explaining why soft materials minimize loss. In paramagnetic materials, magnetization increases with a stronger external field because more atomic magnetic moments align with the field, overcoming random thermal motion, though the alignment remains partial compared to ferromagnetic materials. Substituting values gives Increased moment alignment, 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 > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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Diamagnetism, Paramagnetism and Ferromagnetism

A paramagnetic material has \( \chi = 5 \times 10^{-4} \) and \( H = 2 \times 10^3 \, \text{A m}^{-1} \). What is its ma

**Diamagnetism** exhibits small negative susceptibility χ ≈ -10⁻⁵ to -10⁻⁶, weakly repelled from stronger to weaker field regions, no permanent moment, induced moment opposite to B, present in all materials but dominated by other effects. Superconductor perfect diamagnet with χ = -1, complete field expulsion. M = chi H . Given: chi = 5 × 10⁻⁴ , H = 2 × 10³ A m⁻¹ . Substitute: M = 5 × 10⁻⁴ × 2 × 10³ = 1 A m⁻¹ . Substituting values gives 1 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.