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

2024
Views:

329

Pages:

5
Author:

Jabin F.

# Ampere's law in magnetized material: ∇×B = μ₀J ⇒ ∇×B = μ₀(Jfree + Jbound) Jfree = free volume current density Jbound = bound volume current density ∇×B = μ₀[Jf + (∇×M)] ⇒ ∇×(B/μ₀ - M) = Jf ⇒ ∇×H = Jf Where H = B/μ₀ - M B = μ₀(H + M) H = Magnetic field intensity (Amp/meter) M = magnetization (A/m) B = magnetic field or magnetic flux density (Wb/m² or Tesla) ∇×H = Jf → Mf by only free current ∇×M = Jb → Mb by only bound current ∇×H = Jf ⇒ ∮ H·dℓ = If Ampere's law in magnetized material Jf = free volume current density Page-55 (a) For Not Correct? (a) ✓ (b) ✓ (c) ✓ (d) X (wrong statement) ques.40 page-55 1. D₁⊥ - D₂⊥ = σf 2. D₁∥ - D₂∥ = (P₁∥ - P₂∥) 3. H₁∥ - H₂∥ = Kf 4. B₁⊥ - B₂⊥ = 0 (a) ˆn × (D₁ - D₂) = 0 D₁∥ - D₂∥ = 0 (b) ˆn × (H₁ - H₂) = J H₁∥ - H₂∥ = Kf (c) ˆn · (D₁ - D₂) = σ D₁⊥ - D₂⊥ = σ (d) ˆn · (B₁ - B₂) = 0 B₁⊥ - B₂⊥ = 0 option (d) not correct. Magnetic vector potential defined in a region R of space by Ax = ln(xy) Ay = 2 + sin(x) Az = 0 If 'L' be a square loop of wire in the xy-plane with its ends at (0,0), (0,0.25), (0.25,0.25), (0.25,0) in an appropriate unit and it lies entirely in the region R, the numerical value of flux of the above magnetic field passing through L is (a) 0.543 (b) 3.31 (c) -0.75 (d) zero Φ = ∫_S B·dS = ∮ A·dl B = ∇×A = ∂Az/∂y - ∂Ay/∂z i + ∂Ax/∂z - ∂Az/∂x j + ∂Ay/∂x - ∂Ax/∂y k = i(0 - 0) - j(0 - 0) + k(π ln(xy) - (-5π sin(xy))) Φm = ∫∫ B·dS = ∫∫ [π(ln(xy) + 5 sin(xy))] k dxdy = π[(sin(xy) / π + 5 sin(xy)) (y cos(xy) - 5 cos(xy) / π)] = π[(sin(π/4) / π + 5 sin(π/4)) (1/4 cos(π/4) - 5 cos(π/4) / π)]