Skip to content
Login Register

#solenoid

34 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

View Details
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

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 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

View Details
Diamagnetism, Paramagnetism and Ferromagnetism

The net magnetic flux through a closed surface surrounding a solenoid 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

View Details
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

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 magnetic field \( B = 0.8 \, \text{T} \) inside has a core with \( \mu_r = 400 \) and \( n = 800 \, \tex

**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 , so I = (B/μ₀ μ_r n) . Given: B = 0.8 T , μ_r = 400 , n = 800 m⁻¹ , μ₀ = 4π × 10⁻⁷ . Substitute: I = (0.8/4π × 10⁻⁷ × 400 × 800) = (0.8/4π × 3.2 × 10⁻²) ≈

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

The net magnetic flux through a closed surface surrounding a current-carrying solenoid is:

**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. Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero, as magnetic field lines form closed loops. 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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
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
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

View Details
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

View Details