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Hysteresis, Retentivity, Coercivity and Permanent Magnets

The magnetic field inside a bar magnet is directed from:

**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. Inside a bar magnet, magnetic field lines run from the south pole to the north pole to form closed loops with the external field (north to south), maintaining continuity as there are no magnetic monopoles. Substituting values gives South to north, 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

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

The magnetic potential energy of a dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.2 \, \text{T} \) at \(

**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. U_m = -m B cosθ . Given: m = 0.7 A m² , B = 0.2 T , θ = 0° , cos 0° = 1 . Substitute: U_m = -0.7 × 0.2 × 1 = -0.14 J . Substituting values gives -0.14 J, 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

A bar magnet with \( m = 2.5 \, \text{A m}^2 \) is at \( 0.7 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**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 = (μ₀/4π) (2m/r³) . Given: m = 2.5 A m² , r = 0.7 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 2.5/(0.7)³) = 10⁻⁷ × (5.0/0.343) ≈ 1.46 × 10⁻⁶ T . Substituting values gives 1.46 × 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 > 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.

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

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 900 turns per meter and current \( 3 \, \text{A} \) has a core with \( \mu_r = 250 \). What is \( B \) i

**Core magnetization** M = (μ_r -1)nI, so B = μ₀(nI + M). High μ_r materials like soft iron increase B dramatically for same nI, used in electromagnets, with μ_r up to 5000, enabling strong fields with low current. B = μ₀ μ_r n I . Given: n = 900 m⁻¹ , I = 3 A , μ_r = 250 , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 250 × 900 × 3 = 0.8478 T ≈ 0.85 T . Substituting values gives 0.85 T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque

Ref: NCERT > Physics Book > Magnetism and Matter > Solenoid with Magnetic Core and Magnetic Properties

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Diamagnetism, Paramagnetism and Ferromagnetism

The net magnetic flux through a closed surface surrounding a bar magnet is:

**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. Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero, as magnetic field lines form closed loops with no monopoles. Substituting values gives Zero, 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 > Diamagnetism, Paramagnetism and Ferromagnetism

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Diamagnetism, Paramagnetism and Ferromagnetism

A dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 180^\circ \) has potential energy

**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. U_m = -m B cosθ . Given: m = 0.7 A m² , B = 0.4 T , θ = 180° , cos 180° = -1 . U_m = -0.7 × 0.4 × (-1) = 0.28 J . Substituting values gives 0.28 J, 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 > Diamagnetism, Paramagnetism and Ferromagnetism

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Diamagnetism, Paramagnetism and Ferromagnetism

A material’s magnetic susceptibility becomes zero when:

**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. For a superconductor below its critical temperature, the susceptibility chi = -1 and relative permeability μ_r = 0 , but in a broader context, susceptibility can approach zero for non-magnetic materials or when magnetization is negligible compared to the applied field. However, the question implies a specific state, and superconductors achieve this uniquely with perfect diamagnetism. Substituting values gives It exhibits perfect diamagnetism, which matches expected magnitude for

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

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Diamagnetism, Paramagnetism and Ferromagnetism

The absence of a net force on a magnetic dipole in a uniform field implies:

**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. In a uniform magnetic field, the forces on the dipole’s poles are equal in magnitude and opposite in direction, resulting in no net translational force. This occurs because the field strength does not vary, unlike in a non-uniform field where a gradient would produce a net force. Substituting values gives The field strength is constant, which matches expected magnitude for

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

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Diamagnetism, Paramagnetism and Ferromagnetism

The potential energy of a magnetic dipole in a uniform field is highest when:

**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. The potential energy U = -m B cosθ is highest when cosθ = -1 , i.e., θ = 180° , when the dipole is anti-parallel to the field. This is the least stable position, as energy is maximized. Substituting values gives It is anti-parallel to the field, 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 > Diamagnetism, Paramagnetism and Ferromagnetism

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Diamagnetism, Paramagnetism and Ferromagnetism

A material with \( \mu_r = 0 \) completely expels magnetic fields and is classified as:

**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. A material with μ_r = 0 (and chi = -1 ) exhibits perfect diamagnetism, characteristic of a superconductor due to the Meissner effect. Substituting values gives Superconductor, 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 > Diamagnetism, Paramagnetism and Ferromagnetism

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