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#potential energy

41 public questions tagged with this topic.

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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Damped Oscillations

A spring system has \( m = 1.0 \, \text{kg}, k = 400 \, \text{N/m}, A = 6 \, \text{cm} \). What is the potential energy

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. Potential energy: U = (1/2) k x² . k = 400 N/m, x = 0.03 m . U = 0.5 × 400 × (0.03)² = 0.5 × 400 × 0.0009 = 0.18 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.18 J follows, reflecting SHM

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Equations of SHM, Phase and Angular Frequency

Which condition must be satisfied for the potential energy in an SHM system to be expressible as a quadratic function of

**General equation of SHM** x = A sin(ωt + φ) or A cos(ωt + φ) includes amplitude A (m), angular frequency ω = √(k/m) (rad/s) for spring system, and initial phase φ (rad) setting t=0 position. Phase (ωt + φ) determines instantaneous state, phase difference Δφ governs interference of two SHM motions. Potential energy in SHM ( U = (1/2) k x² ) is quadratic when the force is conservative and linear ( F = -kx ), distinguishing SHM from non-linear systems. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Equations of SHM, Phase and Angular Frequency

A spring system has \( m = 0.3 \, \text{kg}, k = 120 \, \text{N/m}, A = 8 \, \text{cm} \). What is the potential energy

**General equation of SHM** x = A sin(ωt + φ) or A cos(ωt + φ) includes amplitude A (m), angular frequency ω = √(k/m) (rad/s) for spring system, and initial phase φ (rad) setting t=0 position. Phase (ωt + φ) determines instantaneous state, phase difference Δφ governs interference of two SHM motions. Potential energy: U = (1/2) k x² . k = 120 N/m, x = 0.04 m . U = 0.5 × 120 × (0.04)² = 0.5 × 120 × 0.0016 = 0.096 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E =

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Equations of SHM, Phase and Angular Frequency

In SHM, what condition results in the potential energy being zero?

**General equation of SHM** x = A sin(ωt + φ) or A cos(ωt + φ) includes amplitude A (m), angular frequency ω = √(k/m) (rad/s) for spring system, and initial phase φ (rad) setting t=0 position. Phase (ωt + φ) determines instantaneous state, phase difference Δφ governs interference of two SHM motions. Potential energy U = (1/2) k x² is zero when displacement x = 0 , which occurs at the mean position, where all energy is kinetic. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Zero displacement follows, reflecting

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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