#magnetic moment

Hysteresis, Retentivity, Coercivity and Permanent Magnets

A bar magnet with magnetic moment \( 1.5 \, \text{A m}^2 \) is placed at a distance of \( 0.3 \, \text{m} \) along its a

**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. The magnetic field along the axis is B = (μ₀/4π) (2m/r³) . Given: m = 1.5 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . Substitute: B = 10⁻⁷ × (2 × 1.5/(0.3)³) = 10⁻⁷ × (3/0.027) = 1.11 × 10⁻⁵ T ≈ 1.1 × 10⁻⁵ T . Substituting values gives 1.1 × 10⁻⁵ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on

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 \( m = 1.8 \, \text{A m}^2 \) produces a field at \( 0.6 \, \text{m} \) on its equatorial line. What i

**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 = (μ₀/4π) (m/r³) . Given: m = 1.8 A m² , r = 0.6 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (1.8/(0.6)³) = 10⁻⁷ × (1.8/0.216) ≈ 8.333 × 10⁻⁷ T ≈ 8.33 × 10⁻⁷ T . Substituting values gives 8.33 × 10⁻⁷ T, which matches expected magnitude for this magnetic configuration, confirming dipole field

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

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A dipole with \( m = 0.5 \, \text{A m}^2 \) in \( B = 0.1 \, \text{T} \) at \( 60^\circ \) has torque:

**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. tau = m B sinθ . Given: m = 0.5 A m² , B = 0.1 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . tau = 0.5 × 0.1 × 0.866 = 0.0433 N m ≈ 0.043 N m . Substituting values gives 0.043 N m, 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 dipole with \( m = 0.4 \, \text{A m}^2 \) in a field \( B = 0.8 \, \text{T} \) at \( 0^\circ \) has potential energy:

**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 = 0.4 A m² , B = 0.8 T , θ = 0° , cos 0° = 1 . U_m = -0.4 × 0.8 × 1 = -0.32 J . Substituting values gives -0.32 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 magnetic moment \( 1.5 \, \text{A m}^2 \) is placed at a distance of \( 0.6 \, \text{m} \) along its a

**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. The magnetic field along the axis is B = (μ₀/4π) (2m/r³) . Given: m = 1.5 A m² , r = 0.6 m , (μ₀/4π) = 10⁻⁷ . Substitute: B = 10⁻⁷ × (2 × 1.5/(0.6)³) = 10⁻⁷ × (3.0/0.216) ≈ 1.389 × 10⁻⁶ T ≈ 1.39 × 10⁻⁶ T . Substituting values gives 1.39 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming dipole

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

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Diamagnetism, Paramagnetism and Ferromagnetism

A bar magnet with original \( m = 2.0 \, \text{A m}^2 \) is cut transversely into two equal parts. What is \( m \) of ea

**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. When cut transversely, each part has half the original magnetic moment. Given: m = 2.0 A m² . Each part: m' = (2.0/2) = 1.0 A m² . Substituting values gives 1.0 A m², 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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 3.0 \, \text{A m}^2 \) is at \( 0.4 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**Magnetic field of bar magnet** follows inverse cube law B ∝ m/r³, unlike inverse square for electric dipole. Given B at distance r, moment m = B r³/(μ₀/4π) for equatorial, m = B r³/(2·μ₀/4π) for axial, enabling moment extraction from measured field. B = (μ₀/4π) (2m/r³) . Given: m = 3.0 A m² , r = 0.4 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 3.0/(0.4)³) = 10⁻⁷ × (6.0/0.064) = 9.375 × 10⁻⁶ T ≈ 9.38 × 10⁻⁶ T . Substituting values gives 9.38 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on