#bar magnet

Hysteresis, Retentivity, Coercivity and Permanent Magnets

The strongest magnetic field of a bar magnet is observed:

**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. The magnetic field of a bar magnet is strongest at its poles, where field lines are most concentrated, as opposed to the central region where the field is weaker and less dense. Substituting values gives At its poles, 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 magnetic field inside a bar magnet is directed from:

**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. Inside a bar magnet, magnetic field lines run from the south pole to the north pole to form closed loops with the external field (north to south), maintaining continuity as there are no magnetic monopoles. Substituting values gives South to north, 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 bar magnet with magnetic moment \( 1.5 \, \text{A m}^2 \) is placed at a distance of \( 0.3 \, \text{m} \) along its a

**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. The magnetic field along the axis is B = (μ₀/4π) (2m/r³) . Given: m = 1.5 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . Substitute: B = 10⁻⁷ × (2 × 1.5/(0.3)³) = 10⁻⁷ × (3/0.027) = 1.11 × 10⁻⁵ T ≈ 1.1 × 10⁻⁵ T . Substituting values gives 1.1 × 10⁻⁵ T, 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 bar magnet with \( m = 1.8 \, \text{A m}^2 \) produces a field at \( 0.6 \, \text{m} \) on its equatorial line. What i

**Hysteresis loop** plots B versus H for ferromagnetic material, retentivity (remanence) is residual B at H=0 after saturation, coercivity is reverse H needed to reduce B to zero. Hard ferromagnets have high retentivity and coercivity, retaining strong magnetism after external field removed, suitable for permanent magnets. B = (μ₀/4π) (m/r³) . Given: m = 1.8 A m² , r = 0.6 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (1.8/(0.6)³) = 10⁻⁷ × (1.8/0.216) ≈ 8.333 × 10⁻⁷ T ≈ 8.33 × 10⁻⁷ T . Substituting values gives 8.33 × 10⁻⁷ T, which matches expected magnitude for this magnetic configuration, confirming dipole field

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

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A bar magnet with magnetic moment \( 1.5 \, \text{A m}^2 \) is placed at a distance of \( 0.6 \, \text{m} \) along its a

**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 magnetic field along the axis is B = (μ₀/4π) (2m/r³) . Given: m = 1.5 A m² , r = 0.6 m , (μ₀/4π) = 10⁻⁷ . Substitute: B = 10⁻⁷ × (2 × 1.5/(0.6)³) = 10⁻⁷ × (3.0/0.216) ≈ 1.389 × 10⁻⁶ T ≈ 1.39 × 10⁻⁶ T . Substituting values gives 1.39 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming dipole

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

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A bar magnet with \( m = 1.6 \, \text{A m}^2 \) is at \( 0.8 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**Hysteresis loop** plots B versus H for ferromagnetic material, retentivity (remanence) is residual B at H=0 after saturation, coercivity is reverse H needed to reduce B to zero. Hard ferromagnets have high retentivity and coercivity, retaining strong magnetism after external field removed, suitable for permanent magnets. B = (μ₀/4π) (2m/r³) . Given: m = 1.6 A m² , r = 0.8 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 1.6/(0.8)³) = 10⁻⁷ × (3.2/0.512) ≈ 6.25 × 10⁻⁷ T . Substituting values gives 6.25 × 10⁻⁷ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³

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

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A bar magnet with \( m = 2.5 \, \text{A m}^2 \) is at \( 0.7 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**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. B = (μ₀/4π) (2m/r³) . Given: m = 2.5 A m² , r = 0.7 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 2.5/(0.7)³) = 10⁻⁷ × (5.0/0.343) ≈ 1.46 × 10⁻⁶ T . Substituting values gives 1.46 × 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 > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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Diamagnetism, Paramagnetism and Ferromagnetism

The net magnetic flux through a closed surface surrounding a bar magnet is:

**Paramagnetism** has small positive χ ≈ 10⁻³ to 10⁻⁵, weakly attracted towards stronger field, random moments align partially with B, magnetization decreases with temperature following Curie law χ ∝ 1/T. Materials have unpaired electrons with permanent moments. Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero, as magnetic field lines form closed loops with no monopoles. Substituting values gives Zero, 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 bar magnet’s field strength decreases with distance because:

**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 magnetic field strength of a bar magnet decreases with distance as the field lines spread out, reducing their density and thus the field intensity, following an inverse cube relationship ( B ∝ 1/r³ ) for a dipole at large distances. Substituting values gives The field lines spread out, 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 bar magnet with \( m = 0.9 \, \text{A m}^2 \) is at \( 0.2 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**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. B = (μ₀/4π) (2m/r³) . Given: m = 0.9 A m² , r = 0.2 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 0.9/(0.2)³) = 10⁻⁷ × (1.8/0.008) = 2.25 × 10⁻⁵ T . Substituting values gives 2.25 × 10⁻⁵ 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 > Diamagnetism, Paramagnetism and Ferromagnetism

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Diamagnetism, Paramagnetism and Ferromagnetism

A bar magnet with original \( m = 2.0 \, \text{A m}^2 \) is cut transversely into two equal parts. What is \( m \) of ea

**Paramagnetism** has small positive χ ≈ 10⁻³ to 10⁻⁵, weakly attracted towards stronger field, random moments align partially with B, magnetization decreases with temperature following Curie law χ ∝ 1/T. Materials have unpaired electrons with permanent moments. When cut transversely, each part has half the original magnetic moment. Given: m = 2.0 A m² . Each part: m' = (2.0/2) = 1.0 A m² . Substituting values gives 1.0 A m², 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 bar magnet with \( m = 3.0 \, \text{A m}^2 \) is at \( 0.4 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**Magnetic field of bar magnet** follows inverse cube law B ∝ m/r³, unlike inverse square for electric dipole. Given B at distance r, moment m = B r³/(μ₀/4π) for equatorial, m = B r³/(2·μ₀/4π) for axial, enabling moment extraction from measured field. B = (μ₀/4π) (2m/r³) . Given: m = 3.0 A m² , r = 0.4 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 3.0/(0.4)³) = 10⁻⁷ × (6.0/0.064) = 9.375 × 10⁻⁶ T ≈ 9.38 × 10⁻⁶ T . Substituting values gives 9.38 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on

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

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