#axial field

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

View Details

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

View Details

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

View Details

Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 3.0 \, \text{A m}^2 \) is at \( 0.5 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**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 = (μ₀/4π) (2m/r³) . Given: m = 3.0 A m² , r = 0.5 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 3.0/(0.5)³) = 10⁻⁷ × (6.0/0.125) = 4.8 × 10⁻⁶ T . Substituting values gives 4.8 × 10⁻⁶ 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

View Details