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

2024
Views:

384

Pages:

5
Author:

Jabin F.

# maxwell eqn in Conducting media from page-242 (Continue)2 ∇⋅E=0 ∇⋅B=0 ∇×E=−∂B∂t ∇×B=μσE+με∂E∂t These are first order diff. egh. Now. we weill make II order ditt eqh. from eqh take curd => ∇×(∇×E)=−∂∂t(∇×B) ∇⋅(∇⋅E)−∇2E=−∂∂t[μσE+με∂E∂t] ∇2E−μσ∂E∂t−με∂2E∂t2=0 Damping Terms. wave equation of electric field. Similarly or wave on for ragne th ∇2B=μσ∂B∂t−με∂2B∂t2 ∇×E=−∂B∂t−0 at 4 =let Soln equation (5) and (6) are, respectively. E=Eoe^(i(K*r-wt)) B=Boe^(i(K*r-wt)) K* → Complex wave vector from eqn (5) (iK*)(iK*)E=μσ(-iw)E-με(-iw)(-iw)E=0 Since E≠0 -(K*)^2+iμσω+μϵω^2=0 (K*)^2=μϵω^2+iμσω let K*=α+iβ Here α= propagation coefficient β= attenuation coefficient. Since expression of E & B are soln of wave eqn. Hence these Satisfy the maxwell eqn.- ∇⋅E=0 → (i(K*⋅E))=0 K*⋅E=0 → K*⊥E ∇⋅B=0 → (i(K*⋅B))=0 K*⋅B=0 → K*⊥B ∇×E=−iωB → (i(K*×E))=-i(-iw)B and, ∇×B=i(K*×E) → B=K*×E/ω In a Conductor, E, B and K* are mutually ⊥ to each other. # if Ē=E^0ñ, K*=K^0z, B=B^0ŷ then E=E₀e^(i(K*z-wt)) = E₀ñe^(i(K^0z-wt)) Similarly B=B₀ŷe^(i(K*z-wt)) ∴ K*z=K^0z from eqn (8) K*=α+iβ E=E₀ne^[i(α+iβ)z-wt] E=E₀ne^(-βz)e^(iαz-wt) Similarly B=B₀ye^(-βz)e^(iαz-wt) α → Propagation Coefficient β → attenuation Coefficient. free space E₀ Conductor E=E₀e^(-βz) E=E₀e^(-z/δ) oscillator soln Damping soln z=0