University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

31

Pages:

5
Author:

Jabin F.

Ionosphere: Behaves as a mirror for radio waves. ω_p: Plasma frequency ω: Frequency of radio wave N: Electron concentration ω_p = √(Ne²/ε₀m) Conditions: For good conductors, N ≈ 10^28/m³ ω_p is in the ultraviolet (UV) region Conclusion: For visible light, metals are good reflectors because ω << ω_p. Plasma acts as a high pass filter. Problem: Question 62 Page 77 Values: N₁ = 6x10²² cm⁻³ N₂ = 1.8x10²² cm⁻³ Conditions: If ω < ω_p, then metal is opaque. If ω > ω_p, then metal is transparent. Calculations: ω_p = √(Ne²/ε₀m) f_p = ω_p / (2π) f_p₁ = √(N₁) / √(N₂) f_p₂ = √3 * f_p₁ = 1.73x1.88x10¹⁶ Hz = 2.39x10¹⁶ Hz Conclusion: f_p₂ is in the UV region. Problem: Question 12 Page 68 Calculations: J_d = ε₀ * ∂E/∂t B(t) = B₀t^2 * k I_d = ∫ J_d * dS₁ ∫ E * dt = ∫₀ᵗ (-2B₀t) * r² * dr * dθ E_out = -R² * B₀ * φ J = -R² * B₀ * E₀ * φ dS₁ = r * dr * dz I_d = ∫∫ (-R²B₀E₀φ) * r * dr * dz Problem: Question 4 Page 69 Values: σ = 10² S/m ε = 3ε₀ Equations: Conduction current density: J_c = σE Displacement current density: J_d = ε₀ * ∂E/∂t E = E₀sinωt ∂E/∂t = E₀ωcosωt J_c / J_d = σ / (ε₀ω) Condition: If I_c = I_d, then σ = ε₀ω Calculations: f = (σ / (ε₀ω))^(1/2) f = (10⁻² * 2) / (2π * ε₀ * 3 * 2) f = 6x10⁶ Hz = 60 MHz For Good Conductors: σ / (ε₀ω) >> 1 J_c >> J_d For Poor Conductors: σ / (ε₀ω) << 1 J_c << J_d