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

2024
Views:

141

Pages:

5
Author:

Jabin F.

Heading: #Intensity of radiation Equation: ⃗I = <⃗s> = (μ0 * p0^2 * ω^4 * sin^2(θ)) / (32 * π^2 * r^2 * c) Note: ⃗I ∝ sin^2(θ) / r^2 Conditions: If axis of dipole: θ = 0°, I = 0 If equatorial plane: θ = 90°, I = maximum Section 1 Reference: Question 4.31, Page-97 Given: Initial velocity (vi) = u Final velocity (vf) = u/2 Equations: vf = vi + at vf^2 = vi^2 + 2as a = -3u^2 / 8s Section 2 Equations: Power radiated (P) = (μ0 * e^2 * a^2) / (6πc) Energy radiated (U) = (μ0 * e^2 * a^2 * t) / (6πc) U = (μ0 * e^2 * u^3) / (32 * π * ε0 * c) Section 1 Reference: Question 9, Page-100 Given: |E|^2 - |B|^2 = 29 B = (2√5) * ḱ Calculations: |E|^2 - |B|^2 = 29 |E|^2 - 20 = 29 |E|^2 = 49 |E| = 7 Conclusion: E' = (7/√2) * (î + ĵ) Section 2 Reference: Question 10, Page-100 Given: E = (α, 0, 0) B = (α, 0, 2α) Calculations: E'⋅B' = E⋅B |E'|^2 - |B'|^2 = |E|^2 - |B|^2 E'⋅B' = α * Ex' + α * By' α * Ex' + α * By' = α (Ex')^2 + α^2 - (α^2 + (By')^2 + α^2) = α^2 - 4α^2 (Ex')^2 + By' = α Section 1 Equation: (Ex')^2 - (By')^2 = α^2 = -4α^2 Simplification: (Ex')^2 = (By')^2 = -3α^2 Section 2 Reference: From equation (1) Conclusion: Option-b and option-d satisfy the above equation. Calculation: Option-d: Ex' = -α, By' = αλ α^2 - 4α^2 = -3α^2 Conclusion: Option-d is the answer, with Ex' = -α and By' = 2α.