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

2024
Views:

256

Pages:

6
Author:

Jabin F.

Electric force per unit area: (electrostatic) F₁ = q₁E₂ ⇒ f₁e = F₁/A = q₁/A * E₂ = σ₁E₂ E₂ = -σ₂/(2ε₀) (-ẑ) f₁e = -σ₁σ₂/(2ε₀) (-ẑ) attractive type force. Equation: ∮cB.dl = μ₀Ienc = μ₀(+I + I) = 2μ₀I (b) BL = (μ₀I)/(2πr₂) * [I(x̂ + ŷ)] ⇒ BL = (μ₀I)/(2πr₂) * (cos(45°)x̂ + sin(45°)ŷ) ⇒ BL = (μ₀I)/(2πr₂) * (1/√2)(x̂ + ŷ) ⇒ BL = (μ₀I)/(2√2πr₂) * (x̂ + ŷ) Bm = (μ₀I)/(2πr₂) * (ŷ - x̂) ⇒ Bm = (μ₀I)/(2πr₂) * (-x̂ + ŷ) ⇒ Bm = (μ₀I)/(4π) * (-x̂ + ŷ) ⇒ BLm = BL + Bm = (μ₀I)/(2√2πr₂) * (x̂ + ŷ) + (μ₀I)/(4π) * (-x̂ + ŷ) ⇒ BLm = (μ₀I)/(4π) * ((2/√2)x̂ + (2/√2 + 1)ŷ) ⇒ BLm = (μ₀I)/(4π) * ((√2 + 1)x̂ + (√2 + 1)ŷ) ⇒ BLm = (μ₀I)/(4π) * (√2 + 1)(x̂ + ŷ) For doubling the magnetic field: BN = μ₀I/(2πr) * ŷ However, in the question: BN = μ₀I/(2πr) * (ŷ + ẑ) From equation (6): x = 1 Therefore: BN = μ₀I/(2πr) * [I(x̂ × ẑ)] ⇒ BN = μ₀I/(2πr) * (-ŷ) ... (-ẑ) × (-x̂) = +ŷ Dimension of I should be (-2) so that equation (6) is satisfied. ⇒ I = I(-2) ⇒ x = 1