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

2024
Views:

301

Pages:

6
Author:

Jabin F.

Examples: (i) B = B₀x: Possible and uniform (ii) B = αx²: Impossible (∇ x B ≠ 0) (iii) B = By²: Possible and non-uniform (iv) B = Bz²: Possible and non-uniform Here: Fnet ≠ 0, τ ≠ 0 B₀A = B(1/2) * (-2) F₀A = B(1/2) * (-2) FBC = B(1/2) * 2 B = B(1/2) * 2 FBC + FDA ≠ 0 FAB: FAB = ∫(I * dl x B) FAB = ∫(-l/2 to l/2) [I * dy * (B * y * ẑ)] FAB = IB * ∫(-l/2 to l/2) [y * dy] * ẑ FAB = IB * [(y²/2)] from -l/2 to l/2 * ẑ FAB = IB * [(l²/4) - (-l²/4)] * ẑ FAB = IB * (l²/2) * ẑ Similar calculations for FAO and FCD: FAO = 0 FCD = (I * B(l/2) * l) * ŷ Conclusion: FBC ≠ 0 FAO = 0 FCD = (I * B(l/2) * l) * ŷ Vector Potential (A): A(s, φ, τ) = (μ₀nI / 2) * s * φ̂ Magnetic Field (B): B = ∇ x A Calculation: B = (μ₀nI / 2) * [∂/∂s(s * sinφ) * ẑ - ∂/∂φ(s * sinφ) * ŝ] B = (μ₀nI / 2) * [sinφ * ẑ - s * cosφ * ŝ] Conclusion: The calculated magnetic field (B) is the same as the given magnetic field. Therefore, the vector potential A(s, φ, τ) is a valid solution. Equation: 2x - β = 0 x = β / 2 Direction of m₂ under equilibrium conditions: The direction of m₂ under equilibrium conditions would be the same as the direction of B₁. Magnetic Field (B₁): B₁ = (μ₀ * m₁ / 4π * r³) * [2 * cos(α) * r̂ + sin(α) * θ̂] α = 45° r = √(1² + 1²) = √2 B₁ = (μ₀ * m₁ / 4π * 2√2) * [(2 * cos(45°) * r̂ + sin(45°) * θ̂)] B₁ = (μ₀ * m₁ / 8π√2) * (r̂ + θ̂) Top Section m₁ * r₁ = m₁ * (l + 45°) = l / 2 m₂ * r₂ = m₂ * cos(θ - 45°) Middle Section U = (μ₀ * m₁ * m₂ / 4π * r³) * [cos(θ) - 3 * (1/√2) * (cos(θ) + sin(θ))] Bottom Section Under equilibrium conditions, U = minimum. F = -∇U = 0 ∂U/∂θ = 0 -sin(θ) - (3/2) * (-sin(θ) + cos(θ)) = 0 (1/2) * sin(θ) - (3/2) * cos(θ) = 0 tan(θ) = 3