Earth's Magnetism and Magnetic Declination

Earth's Magnetism and Magnetic Declination

A solenoid produces \( B = 1.5 \, \text{T} \) with a core of \( \mu_r = 500 \) and \( n = 2000 \, \text{m}^{-1} \). What

**Geomagnetic field** arises from outer core dynamo, field lines emerge near geographic south pole. Understanding D and I allows conversion between geographic and magnetic coordinates, with B_H = B cos(inclination) used in experiments with tangent galvanometer. B = μ₀ μ_r n I , so I = (B/μ₀ μ_r n) . Given: B = 1.5 T , μ_r = 500 , n = 2000 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (1.5/4π × 10⁻⁷ × 500 × 2000) = (1.5/1.256 × 10⁰) ≈ 1.194 A ≈ 1.2 A . Substituting values gives 1.2 A, which matches expected magnitude for this magnetic configuration, confirming

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Earth's Magnetism and Magnetic Declination

A material with \( B = 0.55 \, \text{T} \) and \( H = 3500 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**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. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.55 T , H = 3500 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.55/4π × 10⁻⁷) ≈ 4.375 × 10⁵ A m⁻¹ . M = 4.375 × 10⁵ - 3500 ≈ 4.34 × 10⁵ A m⁻¹ . Substituting values gives 4.34 × 10⁵ A m⁻¹, which

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Earth's Magnetism and Magnetic Declination

A magnetic needle with magnetic moment \( 0.1 \, \text{A m}^2 \) is placed in a uniform magnetic field of \( 0.2 \, \tex

**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. Torque on a magnetic dipole is tau = m B sinθ . Given: m = 0.1 A m² , B = 0.2 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . Substitute: tau = 0.1 × 0.2 × 0.866 = 0.01732 N m ≈ 0.017 N m . Substituting values gives

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Earth's Magnetism and Magnetic Declination

A material with \( B = 0.2 \, \text{T} \) and \( H = 1000 \, \text{A m}^{-1} \) has magnetization \( M \). What is \( M

**Geomagnetic field** arises from outer core dynamo, field lines emerge near geographic south pole. Understanding D and I allows conversion between geographic and magnetic coordinates, with B_H = B cos(inclination) used in experiments with tangent galvanometer. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.2 T , H = 1000 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.2/4π × 10⁻⁷) ≈ 1.59 × 10⁵ A m⁻¹ . M = 1.59 × 10⁵ - 1000 = 1.58 × 10⁵ A m⁻¹ . Substituting values gives 1.58 × 10⁵ A m⁻¹, which matches expected

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Earth's Magnetism and Magnetic Declination

A bar magnet with magnetic moment \( 1.0 \, \text{A m}^2 \) is placed at a distance of \( 0.5 \, \text{m} \) along its a

**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. The magnetic field along the axis is B = (μ₀/4π) (2m/r³) . Given: m = 1.0 A m² , r = 0.5 m , (μ₀/4π) = 10⁻⁷ . Substitute: B = 10⁻⁷ × (2 × 1.0/(0.5)³) = 10⁻⁷ × (2.0/0.125) = 1.6 × 10⁻⁶ T . Substituting values gives 1.6 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³

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Earth's Magnetism and Magnetic Declination

Which material retains magnetization after the external field is removed and is used in permanent magnets?

**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. Ferromagnetic materials, particularly hard ferromagnets, retain magnetization after the external field is removed and are used in permanent magnets. Substituting values gives Ferromagnetic, 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 paramagnetic material with \( \chi = 3 \times 10^{-3} \) in \( H = 400 \, \text{A m}^{-1} \) has magnetization \( M \)

**Geomagnetic field** arises from outer core dynamo, field lines emerge near geographic south pole. Understanding D and I allows conversion between geographic and magnetic coordinates, with B_H = B cos(inclination) used in experiments with tangent galvanometer. M = chi H . Given: chi = 3 × 10⁻³ , H = 400 A m⁻¹ . M = 3 × 10⁻³ × 400 = 1.2 A m⁻¹ . Substituting values gives 1.2 A m⁻¹, 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 material’s susceptibility becomes negative and large when:

**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. A superconductor has a susceptibility of chi = -1 (large and negative) when it transitions to the superconducting state below its critical temperature, expelling all magnetic fields via the Meissner effect, a unique property among materials. Substituting values gives It becomes superconducting, 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 \( m = 0.2 \, \text{A m}^2 \) 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

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Earth's Magnetism and Magnetic Declination

A material with susceptibility \( \chi = 5 \times 10^{-3} \) has a relative permeability \( \mu_r \) of:

**Geomagnetic field** arises from outer core dynamo, field lines emerge near geographic south pole. Understanding D and I allows conversion between geographic and magnetic coordinates, with B_H = B cos(inclination) used in experiments with tangent galvanometer. μ_r = 1 + chi . Given: chi = 5 × 10⁻³ . Substitute: μ_r = 1 + 5 × 10⁻³ = 1.005 . Substituting values gives 1.005, 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

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 magnetic dipole of moment \( 0.2 \, \text{A m}^2 \) is in a uniform field of \( 0.9 \, \text{T} \) at \( 30^\circ \).

**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. Torque is tau = m B sinθ . Given: m = 0.2 A m² , B = 0.9 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.2 × 0.9 × 0.5 = 0.09 N m . Substituting values gives 0.09 N m, which matches expected magnitude for this magnetic

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