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

2024
Views:

423

Pages:

5
Author:

Jabin F.

# Interaction Energy b/w two magnetic dipoles :- U = (μ₀ / 4πr³) [m₁⋅m₂ - 3 (m₁⋅r̂) (m₂⋅r̂)] U = (μ₀ / 4πr³) [m₁⋅m₂ - 3 (m₁⋅cosθ₁) (m₂⋅cosθ₂)] (r>>a,b,ϕ) For a point dipole τ = m × B magnetic vector potential (A') for dipole Adipole = (μ₀ / 4πr³) (mx) Adipole = (μ₀Msinθ) / (4πr²) Bdipole = ∇×A = (μ₀M / 4πr³) [2cosθr̂ + sinθφ̂] # In Electrostatic Edipole = (P / 4πε₀r³) [2cosθr̂ + sinθφ̂] Vdipole = (Pcosθ) / (4πε₀r²) τ₂ = m₂×B₁ B₁ = (μ₀M₁ / 4πr³) [2cosθ₁r̂ + sinθ₁φ̂] B₁ = (μ₀M₁) / (4πr³) (-2) m₂ = I₂y = m₂y τ₂ = m₂y × (μ₀M₁ / 4πr³) (-2) τ₂ = (μ₀M₁m₂) / (4πr³) (-2) τ₂ = (μ₀M₁m₂) / (4πr³) (-y) τ = (μ₀M₁m₂) / (4πr³) (-y) τ = (2μ₀(7πa²)(7b²)) / (4πr³) (-y) τ = (2μ₀I²a²b²) / (4r³) (-y) τ = (μ₀I²a²b²) / (4r³) (-y) Case-I If m₁ is fixed and m₂ is free to rotate then under equilibrium, what orientation m₂ take → U = -m₂⋅B₁ B₁ = (μ₀m₁) / (4πr³) (-z) m₂ takes (-z) direction under equilibrium (II) (opposite of Case I) [m₁ fixed, m₂ free] Here m₁ take (+y) direction under equilibrium.