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

2024
Views:

76

Pages:

7
Author:

Jabin F.

For Tangential (Parallel) Compor: ∇ x B = μoJ = μo(Jf + Jb) → Babove = Bbelow = μo(Kf + Kb) ∇ x E = H = Jf → If there is no free current on the boundary (Kf = 0) Habove = Hbelow = continuous ∇ x M = Jb Mabove - Mbelow = Kb let K = Kf + Kg B = μo(H + M) B1 = μo(H1 + M1) B2 = μo(H2 + M2) B1 - B2 = 0 μo(H1 + M1) - μo(H2 + M2) = 0 H1 - H2 = -(M1 - M2) H1'' = H2'' = 0 ⇒ B1''/μ1 = B2''/μ2 ⇒ B1sinθ1 = B2sinθ2 eqn(2) eqn(1) tan θ1/tan θ2 = μ1/μ2 Limitation: angle blw B and normal to the plane There is no free current on boundary If there is given μ1, μr1, Xm then tan θ1/tan θ2 = μ1/μ2 = μoμr1/μoμr2 = (1 + Xm1)/(1 + Xm2) Q(3).37 Page-55 free space: μ1 = μo μ2 = μoμr B⊥ = B⊥ B1z = B2z ⇒ B2z = Bz = -μj ⇒ H1'' = H2'' B1'/μ1 = B2'/μ2 B1x/μ1 = B2x/μ2 ⇒ Bx/μo = B2x/(μoμr) ⇒ B2x = μrB1x from (i) and (ii) B2 = B2z + B2x = μrB1x + BzR B1y + B2y = 100 (5x10^-6) + (10x10^-6)B2y = 100 ⇒ B2y = (100 - 5x10^-6)/(10x10^-6) = 10^5 - 5 H1y + H2y = 0 Compare H1y and H2y H1y = 100/(5x10^-6) = 2x10^7 ⇒ B2y = (10^5 - 5)x10^-5 = (10^5 - 100)x10^-5 = -11mT B2 = (8x - 11y + 2z)mT