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#relative permeability

26 public questions tagged with this topic.

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 material with \( \mu_r = 350 \) and \( H = 400 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**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 μ_r H . Given: μ_r = 350 , H = 400 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 350 × 400 = 0.17584 T ≈ 0.18 T . Substituting values gives 0.18 T, 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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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 material with susceptibility \( \chi = -5 \times 10^{-5} \) has a relative permeability \( \mu_r \) of:

**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. μ_r = 1 + chi . Given: chi = -5 × 10⁻⁵ . Substitute: μ_r = 1 - 5 × 10⁻⁵ = 0.99995 . Substituting values gives 0.99995, 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 material with \( \mu_r = 500 \) and \( H = 300 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**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 μ_r H . Given: μ_r = 500 , H = 300 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 500 × 300 = 0.1884 T ≈ 0.19 T . Substituting values gives 0.19 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Solenoid with Magnetic Core and Magnetic Properties

A solenoid produces \( B = 1.2 \, \text{T} \) with a core of \( \mu_r = 400 \) and \( n = 1500 \, \text{m}^{-1} \). What

**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 , so I = (B/μ₀ μ_r n) . Given: B = 1.2 T , μ_r = 400 , n = 1500 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (1.2/4π × 10⁻⁷ × 400 × 1500) = (1.2/7.539 × 10⁻¹) ≈ 1.592 A ≈ 1.6 A . Substituting values gives 1.6 A, which matches expected magnitude for this magnetic

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

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material with susceptibility \( \chi = 2 \times 10^{-3} \) has a relative permeability \( \mu_r \) of:

**Relation between B, H, M** is B = μ₀(H+M) = μ₀(1+χ)H. Susceptibility χ = μ_r -1 quantifies material response. Given B, μ_r, n, current I = B/(μ₀ μ_r n), with μ₀ = 4π×10⁻⁷ T·m/A, enabling current calculation for desired B with magnetic core. μ_r = 1 + chi . Given: chi = 2 × 10⁻³ . Substitute: μ_r = 1 + 2 × 10⁻³ = 1.002 . Substituting values gives 1.002, 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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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 1200 turns per meter and current \( 2.5 \, \text{A} \) has a core with \( \mu_r = 200 \). What is \( B \

**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. B = μ₀ μ_r n I . Given: n = 1200 m⁻¹ , I = 2.5 A , μ_r = 200 , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 200 × 1200 × 2.5 = 0.7536 T ≈ 0.75 T . Substituting values gives 0.75 T, which

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 1200 turns per meter and current \( 2.5 \, \text{A} \) has a core with \( \mu_r = 150 \). What is \( B \

**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 = 1200 m⁻¹ , I = 2.5 A , μ_r = 150 , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 150 × 1200 × 2.5 = 0.5652 T ≈ 0.57 T . Substituting values gives 0.57 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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Solenoid with Magnetic Core and Magnetic Properties

A solenoid produces \( B = 0.9 \, \text{T} \) with a core of \( \mu_r = 300 \) and \( n = 1500 \, \text{m}^{-1} \). What

**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 , so I = (B/μ₀ μ_r n) . Given: B = 0.9 T , μ_r = 300 , n = 1500 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (0.9/4π × 10⁻⁷ × 300 × 1500) = (0.9/5.654 × 10⁻¹) ≈ 1.59 A ≈ 1.6 A .

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

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material with \( \mu_r = 300 \) and \( H = 600 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**Relation between B, H, M** is B = μ₀(H+M) = μ₀(1+χ)H. Susceptibility χ = μ_r -1 quantifies material response. Given B, μ_r, n, current I = B/(μ₀ μ_r n), with μ₀ = 4π×10⁻⁷ T·m/A, enabling current calculation for desired B with magnetic core. B = μ₀ μ_r H . Given: μ_r = 300 , H = 600 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 300 × 600 = 0.22608 T ≈ 0.23 T . Substituting values gives 0.23 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 > Magnetization, Magnetic Intensity, Susceptibility and Permeability

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material with \( \chi = -1 \) and \( \mu_r = 0 \) is classified as:

**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). A material with chi = -1 and μ_r = 0 exhibits perfect diamagnetism, characteristic of a superconductor due to the Meissner effect. Substituting values gives Superconductor, 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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