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

2024
Views:

390

Pages:

5
Author:

Jabin F.

Text Content: Average Power: ⟨P⟩ = (μ₀ * ⟨p²⟩) / (6πc) Intensity, Iα: Iα sinθ θ: angle between a and r Larmor formula: P = (μ₀ * q² * a²) / (6πc) a = v²/R P = (μ₀ * q² * v⁴) / (6πc * R²) Visual Elements: The image is divided into two sections. The first section contains the formula for average power. The second section contains a diagram illustrating the relationship between intensity, angle, and Larmor formula. The overall layout is clear and organized. Given: d = 2 mm D = 6 mm Capacitance per unit length: C' = (2π * ε₀) / ln(D/d) = (2π * ε₀) / ln(6/2) = (2π * ε₀) / ln(3) C' = (4π * ε₀) / (9 * ln(3)) = (9 * 10^9 * 2 * 1.098) / 9 C' = 5.05 * 10^(-11) F/m Total length: l = 1 m Total Capacitance: C = C' * l C = 5.05 * 10^(-11) F * 1 m C = 5.05 * 10^(-11) F (option-a) Ques. 55: εr = 3.5 d = 0.02 m D = 0.04 m Capacitance per unit length: C' = (2π * ε₀ * εr) / ln(D/d) = (2π * ε₀ * 3.5) / ln(0.04 / 0.02) C' = (4π * ε₀ * 3.5) / (ln(2)) = (9 * 10^9 * 2 * 0.693) * 3.5 C' = 3.5 * 10^(-9) F/m Equation 1: ∇ · A = 0 → Coulomb gauge ∇ · (B × r / 2) = 0 (option-a) Ques 13: A = (1/2) * (B × r) E = E₀ * r * e^(iωt) Again: ∇ · A = 0 → (Coulomb gauge) Ques 17: Φ(r,t) = 0 A(r,t) = (i * qt) / (4πε₀ * r²) E: E = -∇Φ - ∂A/∂t = (-q) / (4π * ε₀ * r²) (option-c)