#magnetic field

Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 900 turns per meter and current \( 3.5 \, \text{A} \) has a core with \( \mu_r = 150 \). What is \( B \)

**Magnetic field inside solenoid** with core B = μ₀ μ_r n I is uniform, direction along axis given by right-hand grip rule. For n = 2000 m⁻¹, μ_r = 400, B = 1.2 T, I = 1.2/(4π×10⁻⁷×400×2000) ≈ 1.19 A, showing modest current produces tesla-level field with high μ_r core. B = μ₀ μ_r n I . Given: n = 900 m⁻¹ , I = 3.5 A , μ_r = 150 , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 150 × 900 × 3.5 = 0.59346 T ≈ 0.59 T . Substituting values gives 0.59 T, which matches expected magnitude

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

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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

The magnetization \( M \) of a sample is \( 5 \times 10^4 \, \text{A m}^{-1} \) in a magnetic field \( B = 0.1 \, \text{

**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. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.1 T , M = 5 × 10⁴ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.1/4π × 10⁻⁷) ≈ 7.96 × 10⁴ A m⁻¹ . H = 7.96 × 10⁴ - 5 × 10⁴ = 2.96 × 10⁴ A m⁻¹ ≈ 3 × 10⁴ A m⁻¹ . Substituting values gives

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 contribution \( B_m \) due to a material with \( M = 3.2 \times 10^5 \, \text{A m}^{-1} \) is: (Take

**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_m = μ₀ M . Given: M = 3.2 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 3.2 × 10⁵ = 0.40192 T ≈ 0.40 T . Substituting values gives 0.40 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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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

The reason a paramagnetic material’s magnetization increases with a stronger external field is:

**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. In paramagnetic materials, magnetization increases with a stronger external field because more atomic magnetic moments align with the field, overcoming random thermal motion, though the alignment remains partial compared to ferromagnetic materials. Substituting values gives Increased moment alignment, 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 potential energy of a dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.2 \, \text{T} \) at \(

**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 = 0.7 A m² , B = 0.2 T , θ = 0° , cos 0° = 1 . Substitute: U_m = -0.7 × 0.2 × 1 = -0.14 J . Substituting values gives -0.14 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 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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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 900 turns per meter and current \( 3 \, \text{A} \) has a core with \( \mu_r = 250 \). What is \( B \) i

**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. B = μ₀ μ_r n I . Given: n = 900 m⁻¹ , I = 3 A , μ_r = 250 , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 250 × 900 × 3 = 0.8478 T ≈ 0.85 T . Substituting values gives 0.85 T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque

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

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Diamagnetism, Paramagnetism and Ferromagnetism

A dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 180^\circ \) has potential energy

**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. U_m = -m B cosθ . Given: m = 0.7 A m² , B = 0.4 T , θ = 180° , cos 180° = -1 . U_m = -0.7 × 0.4 × (-1) = 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 > 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 material has \( B = 0.42 \, \text{T} \) and \( M = 3.0 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \m

**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. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.42 T , M = 3.0 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.42/4π × 10⁻⁷) ≈ 3.338 × 10⁵ A m⁻¹ . H = 3.338 × 10⁵ - 3.0 × 10⁵ = 3.38 × 10⁴ A m⁻¹ . Substituting values gives 3.38 × 10⁴

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

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