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#physics calculation

123 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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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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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 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

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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

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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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A bar magnet with \( m = 2.2 \, \text{A m}^2 \) produces a field at \( 0.3 \, \text{m} \) on its equatorial line. What i

**Axial versus equatorial** field comparison shows B_axial = 2 B_eq for same r. Using μ₀ = 4π×10⁻⁷ T·m/A, calculation involves r³ = (0.3)³ = 0.027 m³, so B = 10⁻⁷·m/r³ yields moment estimation. B = (μ₀/4π) (m/r³) . Given: m = 2.2 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2.2/(0.3)³) = 10⁻⁷ × (2.2/0.027) ≈ 8.15 × 10⁻⁶ T . Substituting values gives 8.15 × 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 > 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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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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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material has \( B = 0.28 \, \text{T} \) and \( M = 2.0 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \m

**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 (H + M) , so H = (B/μ₀) - M . Given: B = 0.28 T , M = 2.0 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.28/4π × 10⁻⁷) ≈ 2.228 × 10⁵ A m⁻¹ . H = 2.228 × 10⁵ - 2.0 × 10⁵ = 2.28 × 10⁴ A m⁻¹ .

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

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

The magnetic potential energy of a dipole with \( m = 0.5 \, \text{A m}^2 \) in a field \( B = 0.3 \, \text{T} \) at \(

**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). U_m = -m B cosθ . Given: m = 0.5 A m² , B = 0.3 T , θ = 0° , cos 0° = 1 . Substitute: U_m = -0.5 × 0.3 × 1 = -0.15 J . Substituting values gives -0.15 J, which matches expected magnitude for this magnetic configuration, confirming

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

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