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#magnetic dipole

33 public questions tagged with this topic.

Earth's Magnetism and Magnetic Declination

A dipole with \( m = 0.2 \, \text{A m}^2 \) in \( B = 0.8 \, \text{T} \) at \( 45^\circ \) has torque:

**Earth's magnetism** approximated as dipole inclined to rotation axis, magnetic declination is angle between geographic north and magnetic north, inclination or dip angle is angle between total field and horizontal. Horizontal component B_H = B cosδ, vertical B_V = B sinδ, δ dip angle, B_H ≈ 3-4×10⁻⁵ T in India. tau = m B sinθ . Given: m = 0.2 A m² , B = 0.8 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . tau = 0.2 × 0.8 × 0.707 ≈ 0.11312 N m ≈ 0.11 N m . Substituting values gives 0.11 N m, which matches expected magnitude

Ref: NCERT > Physics Book > Magnetism and Matter > Earth's Magnetism and Magnetic Declination

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Earth's Magnetism and Magnetic Declination

The magnetic potential energy of a dipole with \( m = 0.8 \, \text{A m}^2 \) in a field \( B = 0.25 \, \text{T} \) at \(

**Elements of Earth's field** include declination D, inclination I, horizontal component B_H, total field B = √(B_H² + B_V²). B_H provides compass direction, declination varies with location, important for navigation, inclination 0° at magnetic equator, 90° at poles. U_m = -m B cosθ . Given: m = 0.8 A m² , B = 0.25 T , θ = 180° , cos 180° = -1 . Substitute: U_m = -0.8 × 0.25 × (-1) = 0.2 J . Substituting values gives 0.2 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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Earth's Magnetism and Magnetic Declination

A dipole with \( m = 0.5 \, \text{A m}^2 \) in a field \( B = 0.6 \, \text{T} \) at \( 180^\circ \) has potential energy

**Earth's magnetism** approximated as dipole inclined to rotation axis, magnetic declination is angle between geographic north and magnetic north, inclination or dip angle is angle between total field and horizontal. Horizontal component B_H = B cosδ, vertical B_V = B sinδ, δ dip angle, B_H ≈ 3-4×10⁻⁵ T in India. U_m = -m B cosθ . Given: m = 0.5 A m² , B = 0.6 T , θ = 180° , cos 180° = -1 . U_m = -0.5 × 0.6 × (-1) = 0.3 J . Substituting values gives 0.3 J, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence

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Solenoid with Magnetic Core and Magnetic Properties

A dipole with \( m = 0.7 \, \text{A m}^2 \) in \( B = 0.6 \, \text{T} \) at \( 30^\circ \) has torque:

**Core magnetization** M = (μ_r -1)nI, so B = μ₀(nI + M). High μ_r materials like soft iron increase B dramatically for same nI, used in electromagnets, with μ_r up to 5000, enabling strong fields with low current. tau = m B sinθ . Given: m = 0.7 A m² , B = 0.6 T , θ = 30° , sin 30° = 0.5 . tau = 0.7 × 0.6 × 0.5 = 0.21 N m . Substituting values gives 0.21 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 > Solenoid with Magnetic Core and Magnetic Properties

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Solenoid with Magnetic Core and Magnetic Properties

A dipole with \( m = 0.4 \, \text{A m}^2 \) in \( B = 0.8 \, \text{T} \) at \( 60^\circ \) has torque:

**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. tau = m B sinθ . Given: m = 0.4 A m² , B = 0.8 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . tau = 0.4 × 0.8 × 0.866 ≈ 0.277 N m ≈ 0.28 N m . Substituting values gives 0.28 N m, which

Ref: NCERT > Physics Book > Magnetism and Matter > Solenoid with Magnetic Core and Magnetic Properties

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

The magnetic potential energy of a dipole with moment \( 0.4 \, \text{A m}^2 \) in a uniform magnetic field of \( 0.5 \,

**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. Magnetic potential energy is U_m = -m B cosθ . Given: m = 0.4 A m² , B = 0.5 T , θ = 30° , cos 30° = (√(3)/2) ≈ 0.866 . Substitute: U_m = -0.4 × 0.5 × 0.866 = -0.1732 J ≈ -0.17 J . Substituting values gives -0.17 J, 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

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A dipole with \( m = 0.8 \, \text{A m}^2 \) in a field \( B = 0.3 \, \text{T} \) at \( 90^\circ \) 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 = 0.8 A m² , B = 0.3 T , θ = 90° , cos 90° = 0 . U_m = -0.8 × 0.3 × 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 > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

The torque on a magnetic dipole in a uniform magnetic field is zero when:

**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 torque on a magnetic dipole is given by tau = m B sinθ . It becomes zero when sinθ = 0 , which occurs when the dipole is aligned with the field ( θ = 0° ) or anti-aligned ( θ = 180° ), as the cross product m × B vanishes. Substituting values gives The dipole is parallel or anti-parallel to the field, which matches

Ref: NCERT > Physics Book > Magnetism and Matter > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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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.

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. Substitu

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

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

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