#magnetic dipole

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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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

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

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Diamagnetism, Paramagnetism and Ferromagnetism

The absence of a net force on a magnetic dipole in a uniform field implies:

**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. In a uniform magnetic field, the forces on the dipole’s poles are equal in magnitude and opposite in direction, resulting in no net translational force. This occurs because the field strength does not vary, unlike in a non-uniform field where a gradient would produce a net force. Substituting values gives The field strength is constant, which matches expected magnitude for

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

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Diamagnetism, Paramagnetism and Ferromagnetism

The potential energy of a magnetic dipole in a uniform field is highest when:

**Ferromagnetism** shows large positive χ ≈ 10³ to 10⁵, strong attraction, domain structure with spontaneous magnetization, hysteresis, retentivity. Distinction based on sign and magnitude of χ and behaviour in non-uniform field, explaining attraction versus repulsion. The potential energy U = -m B cosθ is highest when cosθ = -1 , i.e., θ = 180° , when the dipole is anti-parallel to the field. This is the least stable position, as energy is maximized. Substituting values gives It is anti-parallel to the field, 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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Diamagnetism, Paramagnetism and Ferromagnetism

A magnetic dipole placed in a uniform magnetic field experiences no net force but a torque. This is because:

**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. In a uniform magnetic field, the forces on the north and south poles of a dipole are equal and opposite, canceling out to produce no net force. However, these forces act at different points, creating a torque that tends to align the dipole with the field. Substituting values gives Forces on the poles cancel out but produce a couple, which

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

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Diamagnetism, Paramagnetism and Ferromagnetism

A magnetic dipole of moment \( 0.15 \, \text{A m}^2 \) is in a uniform field of \( 0.8 \, \text{T} \) at \( 60^\circ \).

**Ferromagnetism** shows large positive χ ≈ 10³ to 10⁵, strong attraction, domain structure with spontaneous magnetization, hysteresis, retentivity. Distinction based on sign and magnitude of χ and behaviour in non-uniform field, explaining attraction versus repulsion. Torque is tau = m B sinθ . Given: m = 0.15 A m² , B = 0.8 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . Substitute: tau = 0.15 × 0.8 × 0.866 ≈ 0.1039 N m ≈ 0.104 N m . Substituting values gives 0.104 N m, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque

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 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.

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

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A magnetic dipole in a non-uniform field experiences a net force because:

**Magnetization M** is magnetic moment per unit volume (A/m), magnetic intensity H = B/μ₀ - M, susceptibility χ = M/H dimensionless, permeability μ = B/H = μ₀(1+χ), relative permeability μ_r = μ/μ₀ = 1+χ. For solenoid with core, B = μ₀ μ_r n I, n turns per meter (m⁻¹), I current (A). In a non-uniform field, the field strength varies across the dipole, causing the forces on its poles to differ in magnitude. This imbalance results in a net force, unlike in a uniform field where the forces cancel out. Substituting values gives The field strength varies spatially, which matches expected magnitude for this magnetic

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetization, Magnetic Intensity, Susceptibility and Permeability

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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.

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

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

The magnetic potential energy of a dipole with \( m = 0.9 \, \text{A m}^2 \) in a field \( B = 0.2 \, \text{T} \) at \(

**Magnetic properties** μ_r = 400 indicates 400 times vacuum permeability, so B enhanced 400 times for same nI. H = nI (A/m) for solenoid, M = χ H, B = μ₀(H+M) links microscopic magnetization to macroscopic field. U_m = -m B cosθ . Given: m = 0.9 A m² , B = 0.2 T , θ = 90° , cos 90° = 0 . Substitute: U_m = -0.9 × 0.2 × 0 = 0 J . Substituting values gives 0 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 > Magnetization, Magnetic Intensity, Susceptibility and Permeability

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A dipole with \( m = 0.4 \, \text{A m}^2 \) in a field \( B = 0.3 \, \text{T} \) at \( 0^\circ \) has potential energy:

**Magnetization M** is magnetic moment per unit volume (A/m), magnetic intensity H = B/μ₀ - M, susceptibility χ = M/H dimensionless, permeability μ = B/H = μ₀(1+χ), relative permeability μ_r = μ/μ₀ = 1+χ. For solenoid with core, B = μ₀ μ_r n I, n turns per meter (m⁻¹), I current (A). U_m = -m B cosθ . Given: m = 0.4 A m² , B = 0.3 T , θ = 0° , cos 0° = 1 . U_m = -0.4 × 0.3 × 1 = -0.12 J . Substituting values gives -0.12 J, which matches expected magnitude for this magnetic configuration, confirming dipole

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetization, Magnetic Intensity, Susceptibility and Permeability

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A magnetic dipole of moment \( 0.8 \, \text{A m}^2 \) is in a uniform field of \( 0.5 \, \text{T} \) at \( 45^\circ \).

**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. Torque is tau = m B sinθ . Given: m = 0.8 A m² , B = 0.5 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . Substitute: tau = 0.8 × 0.5 × 0.707 ≈ 0.2828 N m ≈ 0.28 N m . Substituting values gives 0.28 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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