#magnetic intensity

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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Solenoid with Magnetic Core and Magnetic Properties

A solenoid has 500 turns per meter and carries a current of \( 3 \, \text{A} \). What is the magnetic intensity \( H \)

**Solenoid with magnetic core** produces field B = μ₀ μ_r n I inside, μ₀ = 4π×10⁻⁷ T·m/A, μ_r relative permeability, n = N/L turns per meter, I current. Core enhances field μ_r times, so given B, μ_r, n, current I = B/(μ₀ μ_r n) can be found, illustrating core effect on field strength. Magnetic intensity H = n I , where n is turns per unit length and I is current. Given: n = 500 m⁻¹ , I = 3 A . Substitute: H = 500 × 3 = 1500 A m⁻¹ . Substituting values gives 1500 A m⁻¹, which matches expected magnitude for this

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 800 turns per meter carries a current of \( 2.5 \, \text{A} \). What is the magnetic intensity \( H \) i

**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. Magnetic intensity H = n I . Given: n = 800 m⁻¹ , I = 2.5 A . Substitute: H = 800 × 2.5 = 2000 A m⁻¹ . Substituting values gives 2000 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.

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

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 600 turns per meter carries a current of \( 3.5 \, \text{A} \). What is the magnetic intensity \( H \) i

**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. Magnetic intensity H = n I . Given: n = 600 m⁻¹ , I = 3.5 A . Substitute: H = 600 × 3.5 = 2100 A m⁻¹ . Substituting values gives 2100 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.

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

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A material has \( B = 0.25 \, \text{T} \) and \( M = 1.8 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \m

**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. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.25 T , M = 1.8 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.25/4π × 10⁻⁷) ≈ 1.989 × 10⁵ A m⁻¹ . H = 1.989 × 10⁵ - 1.8 × 10⁵ = 1.89 × 10⁴ A m⁻¹

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

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 500 turns per meter carries a current of \( 4.5 \, \text{A} \). What is the magnetic intensity \( H \) i

**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. Magnetic intensity H = n I . Given: n = 500 m⁻¹ , I = 4.5 A . Substitute: H = 500 × 4.5 = 2250 A m⁻¹ . Substituting values gives 2250 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.

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

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 400 turns per meter carries a current of \( 5 \, \text{A} \). What is the magnetic intensity \( H \) ins

**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. Magnetic intensity H = n I . Given: n = 400 m⁻¹ , I = 5 A . Substitute: H = 400 × 5 = 2000 A m⁻¹ . Substituting values gives 2000 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.

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

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 300 turns per meter carries a current of \( 6 \, \text{A} \). What is the magnetic intensity \( H \) ins

**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. Magnetic intensity H = n I . Given: n = 300 m⁻¹ , I = 6 A . Substitute: H = 300 × 6 = 1800 A m⁻¹ . Substituting values gives 1800 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.

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

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 700 turns per meter carries a current of \( 2 \, \text{A} \). What is the magnetic intensity \( H \) ins

**Solenoid with magnetic core** produces field B = μ₀ μ_r n I inside, μ₀ = 4π×10⁻⁷ T·m/A, μ_r relative permeability, n = N/L turns per meter, I current. Core enhances field μ_r times, so given B, μ_r, n, current I = B/(μ₀ μ_r n) can be found, illustrating core effect on field strength. Magnetic intensity H = n I . Given: n = 700 m⁻¹ , I = 2 A . Substitute: H = 700 × 2 = 1400 A m⁻¹ . Substituting values gives 1400 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.

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

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Torque on Magnetic Dipole and Potential Energy

A material has \( B = 0.5 \, \text{T} \) and \( M = 2 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \mu_0

**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). B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.5 T , M = 2 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.5/4π × 10⁻⁷) ≈ 3.978 × 10⁵ A m⁻¹ . H = 3.978 × 10⁵ - 2 × 10⁵ = 1.978 × 10⁵ A

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 700 turns per meter carries a current of \( 3 \, \text{A} \). What is the magnetic intensity \( H \) ins

**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. Magnetic intensity H = n I . Given: n = 700 m⁻¹ , I = 3 A . Substitute: H = 700 × 3 = 2100 A m⁻¹ . Substituting values gives 2100 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.

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

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 400 turns per meter carries a current of \( 6 \, \text{A} \). What is the magnetic intensity \( H \) ins

**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. Magnetic intensity H = n I . Given: n = 400 m⁻¹ , I = 6 A . Substitute: H = 400 × 6 = 2400 A m⁻¹ . Substituting values gives 2400 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.

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

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