Skip to content
Login Register

#H field

18 public questions tagged with this topic.

Earth's Magnetism and Magnetic Declination

A material with \( B = 0.2 \, \text{T} \) and \( H = 1000 \, \text{A m}^{-1} \) has magnetization \( M \). What is \( M

**Geomagnetic field** arises from outer core dynamo, field lines emerge near geographic south pole. Understanding D and I allows conversion between geographic and magnetic coordinates, with B_H = B cos(inclination) used in experiments with tangent galvanometer. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.2 T , H = 1000 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.2/4π × 10⁻⁷) ≈ 1.59 × 10⁵ A m⁻¹ . M = 1.59 × 10⁵ - 1000 = 1.58 × 10⁵ A m⁻¹ . Substituting values gives 1.58 × 10⁵ A m⁻¹, which matches expected

Ref: NCERT > Physics Book > Magnetism and Matter > Earth's Magnetism and Magnetic Declination

View Details
Earth's Magnetism and Magnetic Declination

A paramagnetic material with \( \chi = 3 \times 10^{-3} \) in \( H = 400 \, \text{A m}^{-1} \) has magnetization \( M \)

**Geomagnetic field** arises from outer core dynamo, field lines emerge near geographic south pole. Understanding D and I allows conversion between geographic and magnetic coordinates, with B_H = B cos(inclination) used in experiments with tangent galvanometer. M = chi H . Given: chi = 3 × 10⁻³ , H = 400 A m⁻¹ . M = 3 × 10⁻³ × 400 = 1.2 A m⁻¹ . Substituting values gives 1.2 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 > Earth's Magnetism and Magnetic Declination

View Details
Earth's Magnetism and Magnetic Declination

A paramagnetic material with \( \chi = 4 \times 10^{-4} \) in \( H = 2000 \, \text{A m}^{-1} \) has magnetization \( M \

**Geomagnetic field** arises from outer core dynamo, field lines emerge near geographic south pole. Understanding D and I allows conversion between geographic and magnetic coordinates, with B_H = B cos(inclination) used in experiments with tangent galvanometer. M = chi H . Given: chi = 4 × 10⁻⁴ , H = 2000 A m⁻¹ . M = 4 × 10⁻⁴ × 2000 = 0.8 A m⁻¹ . Substituting values gives 0.8 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 > Earth's Magnetism and Magnetic Declination

View Details
Hysteresis, Retentivity, Coercivity and Permanent Magnets

A paramagnetic material with \( \chi = 8 \times 10^{-4} \) in \( H = 2500 \, \text{A m}^{-1} \) has magnetization \( 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. M = chi H . Given: chi = 8 × 10⁻⁴ , H = 2500 A m⁻¹ . M = 8 × 10⁻⁴ × 2500 = 2 A m⁻¹ . Substituting values gives 2.0 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 > Hysteresis, Retentivity, Coercivity and Permanent Magnets

View Details
Hysteresis, Retentivity, Coercivity and Permanent Magnets

A material with \( \mu_r = 800 \) and \( H = 250 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7}

**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 μ_r H . Given: μ_r = 800 , H = 250 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 800 × 250 = 0.2512 T ≈ 0.25 T . Substituting values gives 0.25 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

View Details
Diamagnetism, Paramagnetism and Ferromagnetism

A material with \( B = 0.35 \, \text{T} \) and \( H = 1800 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**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 M = (B/μ₀) - H . Given: B = 0.35 T , H = 1800 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.35/4π × 10⁻⁷) ≈ 2.785 × 10⁵ A m⁻¹ . M = 2.785 × 10⁵ - 1800 ≈ 2.767 × 10⁵ A m⁻¹ . Substituting values gives 2.767 × 10⁵ A m⁻¹, which matches

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

View Details
Diamagnetism, Paramagnetism and Ferromagnetism

A material with \( B = 0.4 \, \text{T} \) and \( H = 2000 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \time

**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 (H + M) , so M = (B/μ₀) - H . Given: B = 0.4 T , H = 2000 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.4/4π × 10⁻⁷) ≈ 3.183 × 10⁵ A m⁻¹ . M = 3.183 × 10⁵ - 2000 ≈ 3.163 × 10⁵ A m⁻¹ . Substituting values gives 3.163 × 10⁵ A m⁻¹, which matches expected magnitude

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

View Details
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

View Details
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

View Details
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

View Details
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

View Details
Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material with \( B = 0.36 \, \text{T} \) and \( H = 2500 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**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). B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.36 T , H = 2500 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.36/4π × 10⁻⁷) ≈ 2.864 × 10⁵ A m⁻¹ . M = 2.864 × 10⁵ - 2500 ≈ 2.839

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetization, Magnetic Intensity, Susceptibility and Permeability

View Details