(02,E2) Figure 3.4: Refraction of steady current at a material interface ather than along the surface, and a new charge will replace it, supplied by the current for finite conducting regions (3. 38)becomes n12·(J1-J2)=0 (339) a boundary condition on the tangential component of current can also be found Substituting E=J/o into(3. 32) we have JI J2 = We can also write this as Ju J2e (3.40) he We may combine the boundary conditions for the normal components of current electric field to better understand the behavior of current at a material boundary stituting E=J/o into ( 3. 34)we hay EIJun-EJ2n=p (341) where Jin= f12 JI and J2n =112. J2. Combining (3.41)with(3. 39), we have E1n(∈1 where Unless E102-01E2=0, a surface charge will exist on the interface between dissimilar urrent-carrying conductor We may also combine the vector components of current on each side of the boundary to determine the effects of the boundary on current direction(Figure 3. 4). Let 81.2 denote en J1.2 and f12 so that Jin = Ji cos 01, Jir= Ju sin 81 J2= J2 sin g. ②2001 by CRC Press LLCFigure 3.4: Refraction of steady current at a material interface. rather than along the surface, and a new charge will replace it, supplied by the current. Thus, for finite conducting regions (3.38)becomes nˆ 12 · (J1 − J2) = 0. (3.39) A boundary condition on the tangential component of current can also be found. Substituting E = J/σ into (3.32)we have nˆ 12 × J1 σ1 − J2 σ2  = 0. We can also write this as J1t σ1 = J2t σ2 (3.40) where J1t = nˆ 12 × J1, J2t = nˆ 12 × J2. We may combine the boundary conditions for the normal components of current and electric field to better understand the behavior of current at a material boundary. Sub￾stituting E = J/σ into (3.34)we have 1 σ1 J1n − 2 σ2 J2n = ρs (3.41) where J1n = nˆ 12 · J1 and J2n = nˆ 12 · J2. Combining (3.41)with (3.39), we have ρs = J1n  1 σ1 − 2 σ2  = E1n  1 − σ1 σ2 2  = J2n  1 σ1 − 2 σ2  = E2n  1 σ2 σ1 − 2  where E1n = nˆ 12 · E1, E2n = nˆ 12 · E2. Unless 1σ2 − σ1 2 = 0, a surface charge will exist on the interface between dissimilar current-carrying conductors. We may also combine the vector components of current on each side of the boundary to determine the effects of the boundary on current direction (Figure 3.4). Let θ1,2 denote the angle between J1,2 and nˆ 12 so that J1n = J1 cos θ1, J1t = J1 sin θ1 J2n = J2 cos θ2, J2t = J2 sin θ2