Magnetic Field Due to Bar Magnet - Axial and Equatorial

Magnetic Field Due to Bar Magnet - Axial and Equatorial

A material’s weak repulsion from a magnetic field is due to:

**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. Diamagnetic materials exhibit weak repulsion from a magnetic field because an external field induces small currents in their atoms that generate an opposing magnetic moment, per Lenz’s law, resulting in a slight reduction of the field inside the material. Substituting values gives Induced opposing moments, 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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

The net magnetic flux through a closed surface surrounding a current-carrying solenoid is:

**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. Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero, as magnetic field lines form closed loops. 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.

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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

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A dipole with \( m = 0.6 \, \text{A m}^2 \) in a field \( B = 0.8 \, \text{T} \) at \( 0^\circ \) has potential energy:

**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. U_m = -m B cosθ . Given: m = 0.6 A m² , B = 0.8 T , θ = 0° , cos 0° = 1 . U_m = -0.6 × 0.8 × 1 = -0.48 J . Substituting values gives -0.48 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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with original \( m = 2.8 \, \text{A m}^2 \) is cut transversely into two equal parts. What is \( m \) of ea

**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. When cut transversely, each part has half the original magnetic moment. Given: m = 2.8 A m² . Each part: m' = (2.8/2) = 1.4 A m² . Substituting values gives 1.4 A m², 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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

In a material where magnetic domains spontaneously align over large regions, the susceptibility is:

**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. Ferromagnetic materials have magnetic domains that spontaneously align, resulting in a large positive susceptibility ( chi gg 1 ). This strong magnetization occurs due to cooperative interactions among atomic magnetic moments, distinguishing them from other materials. Substituting values gives Large and positive, 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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A material has \( B = 0.25 \, \text{T} \) and \( M = 1.8 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (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 (H + M) , so H = (B/μ₀) - M . Given: B = 0.25 T , M = 1.8 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.25/4π × 10⁻⁷) ≈ 1.989 × 10⁵ A m⁻¹ . H = 1.989 × 10⁵ - 1.8 × 10⁵ = 1.89 × 10⁴ A m⁻¹

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 2.2 \, \text{A m}^2 \) produces a field at \( 0.3 \, \text{m} \) on its equatorial line. What i

**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) (m/r³) . Given: m = 2.2 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2.2/(0.3)³) = 10⁻⁷ × (2.2/0.027) ≈ 8.15 × 10⁻⁶ T . Substituting values gives 8.15 × 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.

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A material with \( \mu_r = 500 \) and \( H = 300 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**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 μ_r H . Given: μ_r = 500 , H = 300 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 500 × 300 = 0.1884 T ≈ 0.19 T . Substituting values gives 0.19 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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 1.2 \, \text{A m}^2 \) produces a field at \( 0.3 \, \text{m} \) on its equatorial line. What i

**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) (m/r³) . Given: m = 1.2 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (1.2/(0.3)³) = 10⁻⁷ × (1.2/0.027) ≈ 4.44 × 10⁻⁶ T . Substituting values gives 4.44 × 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.

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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.

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with original \( m = 1.6 \, \text{A m}^2 \) is cut transversely into two equal parts. What is \( m \) of ea

**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. When cut transversely, each part has half the original magnetic moment. Given: m = 1.6 A m² . Each part: m' = (1.6/2) = 0.8 A m² . Substituting values gives 0.8 A m², 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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