#permeability

Production of EM Waves and Hertz Experiment

Why does the speed of electromagnetic waves in a medium depend on the medium’s properties?

**Hertz experiment** used induction coil connected to two rods with gap, spark produced oscillating charge, emitted EM wave, received by loop with gap sparking when E induced, measured wavelength by standing wave, demonstrated EM wave properties, validating Maxwell. The speed of electromagnetic waves in a medium is determined by the medium’s permittivity ( ε ) and permeability ( μ ), as v = (1/√(μ ε)) , which modifies the wave’s propagation characteristics. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Dependence on permittivity and permeabil

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

View Details

Energy, Intensity and Momentum of EM Waves

According to Maxwell's equations, what is the relationship between the speed of electromagnetic waves in vacuum and the

**Energy in EM wave** equally divided between electric and magnetic fields, energy density u = ½ ε₀ E² + B²/(2μ₀) = ε₀ E² = B²/μ₀, average u_avg = ½ ε₀ E₀², intensity I = c u_avg = ½ c ε₀ E₀² = E₀ B₀/(2μ₀) = c B₀²/(2μ₀), radiation pressure p = I/c for absorption, 2I/c for reflection. The document states that the speed c of electromagnetic waves in vacuum is given by c = (1/√(μ₀ ε₀)) . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields c = (1/√(μ₀ ε₀)), illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

View Details

Displacement Current and Ampere-Maxwell Law

The velocity of light in a medium with permittivity \( \varepsilon \) and permeability \( \mu \) is:

**Displacement current** I_d = ε₀ dΦ_E/dt, Φ_E = ∫ E·dA electric flux (V·m), ε₀=8.85×10⁻¹² F/m, ensures continuity of current in charging capacitor where conduction current stops between plates, I_d equals conduction current in wires, 3 A conduction ⇒ 3 A displacement, maintaining Ampere's law ∮ B·dl = μ₀(I_c+I_d). As per Maxwell's equations, the velocity of light in a medium is v = (1/√(μ ε)) . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields (1/√(μ ε)), illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

View Details

Displacement Current and Ampere-Maxwell Law

According to Maxwell's equations, what is the speed of electromagnetic waves in a medium with permittivity \( \varepsilo

**Charging capacitor** conduction current in wires equals displacement current between plates because dQ/dt = I_c = ε₀ A dE/dt = ε₀ dΦ_E/dt = I_d, preserving charge conservation, magnetic field between plates due to I_d, same as that due to conduction current. The document states that the speed of electromagnetic waves in a medium is v = (1/√(μ ε)) . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields v = (1/√(μ ε)), illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

View Details

Solenoid with Magnetic Core and Magnetic Properties

A solenoid with 1100 turns per meter and current \( 2 \, \text{A} \) has a core with \( \mu_r = 250 \). 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 = 1100 m⁻¹ , I = 2 A , μ_r = 250 , μ₀ = 4π × 10⁻⁷ . B = 4π × 10⁻⁷ × 250 × 1100 × 2 = 0.6908 T ≈ 0.69 T . Substituting values gives 0.69 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

Hysteresis, Retentivity, Coercivity and Permanent Magnets

A diamagnetic material has a susceptibility \( \chi = -2 \times 10^{-5} \). What is its magnetic permeability \( \mu \)

**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. Magnetic permeability μ = μ₀ (1 + chi) . Given: chi = -2 × 10⁻⁵ . Substitute: μ = μ₀ (1 - 2 × 10⁻⁵) = μ₀ × 0.99998 . Substituting values gives μ₀ × 0.99998, 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

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

Solenoid with Magnetic Core and Magnetic Properties

A solenoid produces \( B = 0.96 \, \text{T} \) with a core of \( \mu_r = 300 \) and \( n = 1000 \, \text{m}^{-1} \). Wha

**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 = 0.96 T , μ_r = 300 , n = 1000 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (0.96/4π × 10⁻⁷ × 300 × 1000) = (0.96/3.769 × 10⁻¹) ≈ 2.548 A ≈

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 \( B = 0.6 \, \text{T} \) with a core of \( \mu_r = 200 \) and \( n = 1000 \, \text{m}^{-1} \). 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 = 0.6 T , μ_r = 200 , n = 1000 m⁻¹ , μ₀ = 4π × 10⁻⁷ . I = (0.6/4π × 10⁻⁷ × 200 × 1000) = (0.6/2.513 × 10⁻¹) ≈ 2.39 A ≈ 2.4 A .

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

View Details

Membrane Transportation

What is the correct order of permeability through the lipid bilayer?

Permeability across pure phospholipid bilayer without proteins follows solubility-diffusion model where permeability coefficient P equals oil-water partition K times diffusion within hydrocarbon D divided by thickness delta. Consequently small hydrophobic molecules lacking hydrogen bonds partition efficiently into acyl core exhibiting highest permeability. Gases O2 and CO2 permeability around ten to fifty cm per second diffusing almost freely enabling instantaneous alveolar capillary equilibration and mitochondrial respiration. Water despite small size eighteen Daltons polar forming bonds perm

Ref: Alberts et al., Molecular Biology of the Cell, 6th ed., Chapter 11: Relative Permeability – O2, CO2, H2O, Glucose, Ions, RNA.

View Details

Microbial Physiology

The permeability of the Gram-negative outer membrane is controlled by:

Gram-negative outer membrane serves as molecular sieve preventing entry of large hydrophilic and hydrophobic antibiotics like vancomycin, daptomycin, and bile salts. Permeability is governed primarily by porins, abundant trimeric beta-barrel proteins forming water-filled diffusion channels with constrictions determined by internal loop L3. General porins OmpF and OmpC allow passive diffusion of molecules below about 600 daltons, including nutrients and some beta-lactams, dependent on charge and size. Specific porins like LamB and ScrY facilitate uptake of maltodextrins and sucrose via binding

Ref: Nikaido, Microbiol Mol Biol Rev 2003, Outer Membrane Porins and Resistance; Delcour, BBA 2009, Porin Regulation.

View Details