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19 public questions tagged with this topic.

Hysteresis, Retentivity, Coercivity and Permanent Magnets

The magnetic potential energy of a dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.2 \, \text{T} \) at \(

**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. U_m = -m B cosθ . Given: m = 0.7 A m² , B = 0.2 T , θ = 0° , cos 0° = 1 . Substitute: U_m = -0.7 × 0.2 × 1 = -0.14 J . Substituting values gives -0.14 J, 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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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A bar magnet with \( m = 2.5 \, \text{A m}^2 \) is at \( 0.7 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m

**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 = (μ₀/4π) (2m/r³) . Given: m = 2.5 A m² , r = 0.7 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 2.5/(0.7)³) = 10⁻⁷ × (5.0/0.343) ≈ 1.46 × 10⁻⁶ T . Substituting values gives 1.46 × 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 > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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Diamagnetism, Paramagnetism and Ferromagnetism

A paramagnetic material has \( \chi = 5 \times 10^{-4} \) and \( H = 2 \times 10^3 \, \text{A m}^{-1} \). What is its ma

**Diamagnetism** exhibits small negative susceptibility χ ≈ -10⁻⁵ to -10⁻⁶, weakly repelled from stronger to weaker field regions, no permanent moment, induced moment opposite to B, present in all materials but dominated by other effects. Superconductor perfect diamagnet with χ = -1, complete field expulsion. M = chi H . Given: chi = 5 × 10⁻⁴ , H = 2 × 10³ A m⁻¹ . Substitute: M = 5 × 10⁻⁴ × 2 × 10³ = 1 A m⁻¹ . Substituting values gives 1 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 > Diamagnetism, Paramagnetism and Ferromagnetism

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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 dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 180^\circ \) has potential energy

**Diamagnetism** exhibits small negative susceptibility χ ≈ -10⁻⁵ to -10⁻⁶, weakly repelled from stronger to weaker field regions, no permanent moment, induced moment opposite to B, present in all materials but dominated by other effects. Superconductor perfect diamagnet with χ = -1, complete field expulsion. U_m = -m B cosθ . Given: m = 0.7 A m² , B = 0.4 T , θ = 180° , cos 180° = -1 . U_m = -0.7 × 0.4 × (-1) = 0.28 J . Substituting values gives 0.28 J, 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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Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 800 turns per meter and current \( 1.5 \, \text{A} \) has a core with \( \mu_r = 300 \). 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 = 800 m⁻¹ , I = 1.5 A , μ_r = 300 , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 300 × 800 × 1.5 = 0.452 T ≈ 0.45 T . Substituting values gives 0.45 T, which matches expected magnitude

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

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Coulomb's Law and Force Between Point Charges

A point charge \( q = 5 \, \mu\text{C} \) is placed at the origin. What is the electric field magnitude at a point 2 m a

**Inverse-square law** for charges states F ∝ 1/r² while increasing with charge product. Using k = 9×10⁹ N·m²/C², force at distance r follows F = k q₁q₂/r², forming basis for pairwise force calculation. Electric field: E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 5 × 10⁻⁶ C , r = 2 m . E = 9 × 10⁹ × (5 × 10⁻⁶/(2)²) = 9 × 10⁹ × (5 × 10⁻⁶/4) = 1.125 × 10⁴ N/C . Substituting values gives 1.125 × 10⁴ N/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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Collision Frequency and Mean Free Path Variation

A gas occupies 44.8 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**Collision frequency** Z = n σ v_rel, σ = π d² cross-section, v_rel = √2 v_avg, so Z ∝ n, for air at STP n≈2.5×10²⁵ m⁻³ d≈3×10⁻¹⁰ m λ≈68 nm, collision frequency ~10⁹ s⁻¹, illustrating frequent collisions at atmospheric pressure. Number of moles (μ) = VolumeMolar volume.μ = (44.8)/(22.4) = 2.0 mol. Substituting values gives 2.0 mol, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

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PHYSICS

A pure Si crystal is doped with 1 ppm of pentavalent impurity. If the Si atom density is 5 × 10²⁸ m^{-3, the number

Given: A pure Si crystal is doped with 1 ppm of pentavalent impurity. If the Si atom density is 5 × 10²⁸ m^{-3, the number of donor atoms per cubic meter is: These values define the system as per NCERT data. Formula: 1 ppm = 1 part per million = 10⁻⁶. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Number of donor atoms = 10⁻⁶ × 5 × 10²⁸= 5 × 10²² m^{-3 . These contribute electrons, assuming full ionization at room temperature. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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PHYSICS

A Si crystal with 5 × 10²⁸ atoms m^{-3 is doped with 0.5 ppm of pentavalent impurity. The number of donor electrons

Given: A Si crystal with 5 × 10²⁸ atoms m^{-3 is doped with 0.5 ppm of pentavalent impurity. The number of donor electrons per cubic meter is: These values define the system as per NCERT data. Formula: 0.5 ppm = 0.5 × 10⁻⁶. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Number of donor atoms = 0.5 × 10⁻⁶ × 5 × 10²⁸= 2.5 × 10²² m^{-3, each contributing one electron. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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Work, Energy and Orbital Mechanics

What is the minimum speed required to escape from a point RE above Earth’s surface? (RE\=6.4×106m,g\=9.8m/s2)

ve = 2GMERE+h = 2gRE22RE = gRE2. ve = 9.8×6.4×1062 = 3.136×107. ve≈5.6×103m/s = 5.6km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.6 km/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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