#equatorial line

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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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 bar magnet with \( m = 1.5 \, \text{A m}^2 \) produces a field at \( 0.5 \, \text{m} \) on its equatorial line. What i

**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) (m/r³) . Given: m = 1.5 A m² , r = 0.5 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (1.5/(0.5)³) = 10⁻⁷ × (1.5/0.125) = 1.2 × 10⁻⁶ T . Substituting values gives 1.2 × 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

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet produces a field of \( 8 \times 10^{-6} \, \text{T} \) at \( 0.4 \, \text{m} \) on its equatorial line. Wha

**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) , so m = (B r³/(μ₀/4π)) . Given: B = 8 × 10⁻⁶ T , r = 0.4 m , (μ₀/4π) = 10⁻⁷ . m = (8 × 10⁻⁶ × (0.4)³/10⁻⁷) = (8 × 10⁻⁶ × 0.064/10⁻⁷) = 5.12 A m² . Substituting values gives 5.12 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

What is the magnetic field at a point on the equatorial line of a bar magnet with magnetic moment \( 2 \, \text{A m}^2 \

**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. The magnetic field on the equatorial line is B = (μ₀/4π) (m/r³) . Given: m = 2 A m² , r = 0.1 m , (μ₀/4π) = 10⁻⁷ T m A⁻¹ . Substitute: B = 10⁻⁷ × (2/(0.1)³) = 10⁻⁷ × (2/0.001) = 2 × 10⁻⁴ T . Substituting values gives 2 × 10⁻⁴ T, which matches expected magnitude for this

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet produces a field of \( 4 \times 10^{-6} \, \text{T} \) at \( 0.2 \, \text{m} \) on its equatorial line. Wha

**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) , so m = (B r³/(μ₀/4π)) . Given: B = 4 × 10⁻⁶ T , r = 0.2 m , (μ₀/4π) = 10⁻⁷ . m = (4 × 10⁻⁶ × (0.2)³/10⁻⁷) = (4 × 10⁻⁶ × 0.008/10⁻⁷) = 0.32 A m² . Substituting values gives 0.32 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet produces a field of \( 5 \times 10^{-6} \, \text{T} \) at \( 0.4 \, \text{m} \) on its equatorial line. Wha

**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) , so m = (B r³/(μ₀/4π)) . Given: B = 5 × 10⁻⁶ T , r = 0.4 m , (μ₀/4π) = 10⁻⁷ . m = (5 × 10⁻⁶ × (0.4)³/10⁻⁷) = (5 × 10⁻⁶ × 0.064/10⁻⁷) = 3.2 A m² . Substituting values gives 3.2 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet produces a field of \( 6 \times 10^{-6} \, \text{T} \) at \( 0.3 \, \text{m} \) on its equatorial line. Wha

**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) (m/r³) , so m = (B r³/(μ₀/4π)) . Given: B = 6 × 10⁻⁶ T , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . m = (6 × 10⁻⁶ × (0.3)³/10⁻⁷) = (6 × 10⁻⁶ × 0.027/10⁻⁷) = 1.62 A m² . Substituting values gives 1.62 A m², which matches expected magnitude for this magnetic configuration, confirming dipole field

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 0.6 \, \text{A m}^2 \) 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 = 0.6 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (0.6/(0.3)³) = 10⁻⁷ × (0.6/0.027) ≈ 2.222 × 10⁻⁶ T ≈ 2.22 × 10⁻⁶ T . Substituting values gives 2.22 × 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

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 1.0 \, \text{A m}^2 \) produces a field at \( 0.25 \, \text{m} \) on its equatorial line. What

**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) (m/r³) . Given: m = 1.0 A m² , r = 0.25 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (1.0/(0.25)³) = 10⁻⁷ × (1.0/0.015625) = 6.4 × 10⁻⁶ T . Substituting values gives 6.4 × 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

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