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

2024
Views:

221

Pages:

7
Author:

Jabin F.

# Surface current density (K) K = I / dl dl = length ⊥ to direction of current flow K = I / b I ↔ λ K ↔ v → σ λ = q / l σ = q / A ρ = m / V = λ / A J ↔ ρ V = AJ = q / (nA) = λ / n from (1) K = I / b if we make A pole of paper by this surface K = I / 2πR current surface: (φ = z) surface φ fix → θ = z surface θ fix → φ = z surface z fix → θ = φ surface K = I / 2πR page-39 r = mv / 2B r is inversely proportional to q/m and 1/B (option-b) page-39 According to Bohr's theory L = mvrₙ = n(h/2π) Energy En = (1/2)mvₙ² = (q² / 4πε₀rₙ) || = || / 2 || = α₀|| En = K - 2K = -K = -m²vₙ² / 2 = -q² / 8πε₀rₙ mvₙ²/rₙ = q² / 4πε₀rₙ mvrₙ = n(h/2π) m²vₙ² = (nh/2π)² vₙ² = (nh/2π)² * (qB/m²) from eqn (iii) En = (1/2)mvₙ² = (1/2)m(nh/2π)² * (qB/m²) from eqn (iii) En = (1/2)mvₙ² En = (1/2)m(nh/2π)² * (qB/m²) En = (n²h²qB) / (8π²m) (option-d) Ques.9 page-39 (i) Magnetic forces do not work (ii) F = 2(v × B) Both Statement are true. dw = work done = F · dl Page-40 KE = 1880 MeV (Relativistic case) Relativistic KE → mc² - m₀c² m₀c² = 938 MeV (rest mass of proton) m = m₀ / √(1 - v²/c²) K = (m₀c² / √(1 - v²/c²)) - m₀c² = m₀c²[(1 / √(1 - v²/c²)) - 1] K = 1880 MeV m₀c² = 938 MeV α = [(1 / √(1 - v²/c²)) - 1] → √(1 - v²/c²) = 1/3 1 - v²/c² = 1/9 v²/c² = 8/9 → v = (2√2/3)c Equation of Circular motion mv²/r = qvB γ = mv / qB m = m₀ / √(1 - v²/c²) = m₀ / (1/3) γ = m₀ / (2B)