#physics problem

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

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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

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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

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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.

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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.

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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.

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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

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A diamagnetic material has a susceptibility \( \chi = -2 \times 10^{-5} \). What is its magnetic permeability \( \mu \)

**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. Magnetic permeability μ = μ₀ (1 + chi) . Given: chi = -2 × 10⁻⁵ . Substitute: μ = μ₀ (1 - 2 × 10⁻⁵) = μ₀ × 0.99998 . Substituting values gives μ₀ × 0.99998, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A bar magnet with \( m = 1.6 \, \text{A m}^2 \) is at \( 0.8 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**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) (2m/r³) . Given: m = 1.6 A m² , r = 0.8 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 1.6/(0.8)³) = 10⁻⁷ × (3.2/0.512) ≈ 6.25 × 10⁻⁷ T . Substituting values gives 6.25 × 10⁻⁷ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid has 500 turns per meter and carries a current of \( 3 \, \text{A} \). What is the magnetic intensity \( H \)

**Solenoid with magnetic core** produces field B = μ₀ μ_r n I inside, μ₀ = 4π×10⁻⁷ T·m/A, μ_r relative permeability, n = N/L turns per meter, I current. Core enhances field μ_r times, so given B, μ_r, n, current I = B/(μ₀ μ_r n) can be found, illustrating core effect on field strength. Magnetic intensity H = n I , where n is turns per unit length and I is current. Given: n = 500 m⁻¹ , I = 3 A . Substitute: H = 500 × 3 = 1500 A m⁻¹ . Substituting values gives 1500 A m⁻¹, which matches expected magnitude for this

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Diamagnetism, Paramagnetism and Ferromagnetism

A magnetic dipole of moment \( 0.5 \, \text{A m}^2 \) is in a uniform field of \( 0.3 \, \text{T} \) at \( 60^\circ \).

**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. Torque is tau = m B sinθ . Given: m = 0.5 A m² , B = 0.3 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . Substitute: tau = 0.5 × 0.3 × 0.866 ≈ 0.1299 N m ≈ 0.13 N m . Substituting values gives 0.13 N m, which matches expected magnitude for this

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 800 turns per meter carries a current of \( 2.5 \, \text{A} \). What is the magnetic intensity \( H \) i

**Magnetic field inside solenoid** with core B = μ₀ μ_r n I is uniform, direction along axis given by right-hand grip rule. For n = 2000 m⁻¹, μ_r = 400, B = 1.2 T, I = 1.2/(4π×10⁻⁷×400×2000) ≈ 1.19 A, showing modest current produces tesla-level field with high μ_r core. Magnetic intensity H = n I . Given: n = 800 m⁻¹ , I = 2.5 A . Substitute: H = 800 × 2.5 = 2000 A m⁻¹ . Substituting values gives 2000 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.

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