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#potential energy

45 public questions tagged with this topic.

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.

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

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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

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

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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 m/r³ and

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

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

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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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

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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Torque on Magnetic Dipole and Potential Energy

A magnetic dipole with \( m = 0.3 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 90^\circ \) has potentia

**Magnetic dipole in uniform field** experiences torque τ = m B sinθ and potential energy U = -m·B = -m B cosθ, minimum -mB when aligned (θ=0°), maximum +mB at anti-alignment (θ=180°). Work done rotating from θ₁ to θ₂ equals ΔU = mB(cosθ₁ - cosθ₂). U_m = -m B cosθ . Given: m = 0.3 A m² , B = 0.4 T , θ = 90° , cos 90° = 0 . U_m = -0.3 × 0.4 × 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 > Torque on Magnetic Dipole and Potential Energy

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Torque on Magnetic Dipole and Potential Energy

A dipole with \( m = 0.25 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 90^\circ \) has potential energy

**Magnetic dipole in uniform field** experiences torque τ = m B sinθ and potential energy U = -m·B = -m B cosθ, minimum -mB when aligned (θ=0°), maximum +mB at anti-alignment (θ=180°). Work done rotating from θ₁ to θ₂ equals ΔU = mB(cosθ₁ - cosθ₂). U_m = -m B cosθ . Given: m = 0.25 A m² , B = 0.4 T , θ = 90° , cos 90° = 0 . U_m = -0.25 × 0.4 × 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 > Torque on Magnetic Dipole and Potential Energy

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Torque on Magnetic Dipole and Potential Energy

A dipole with \( m = 0.8 \, \text{A m}^2 \) in a field \( B = 0.5 \, \text{T} \) at \( 45^\circ \) has potential energy:

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. U_m = -m B cosθ . Given: m = 0.8 A m² , B = 0.5 T , θ = 45° , cos 45° = (1/√(2)) ≈ 0.707 . U_m = -0.8 × 0.5 × 0.707 ≈ -0.2828 J ≈ -0.28 J . Substituting values gives -0.28 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 > Torque on Magnetic Dipole and Potential Energy

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