enttEH2tJs2H,uetO?Hit018Due to D, =o , we findDzn = Psore. -D=PsSince thereexists surface current Json the surface of perfectelectric conductor,consideringthe direction ofthe surface currentdensity and the integral path complying with the right hand rule,H=o,wefinde, ×H= JsH2t = JsorV
Due to 0 ，we find D1n = D2n = S or = S en D Since there exists surface current JS on the surface of perfect electric conductor,considering the direction of the surface current density and the integral path complying with the right hand rule, H1t = 0 , we find H2t = JS H JS or en = E , H → en et ① ② H1t H2t JS
Example.The components of the time-varying electromagneticfieldin a rectangularmetal waveguide ofinterior cross-section a x b are元E, = Eyo sincos(o t - k,z)xa元H,= Hxo sincos(o t -k.2)a元XH. = H.o cossin( o t -k.zaFind the density of thedisplacementcurrentinthe waveguideand thecharges and the currentsLon the interior walls of the鞋waveguide.The inside isvacuum.Electric field lines---Magneticfieldlines
Example. The components of the time-varying electromagnetic field in a rectangular metal waveguide of interior cross-section a b are sin( ) π cos 0 x t k z a Hz Hz − z = cos( ) π 0 sin x t k z a Hx Hx − z = cos( ) π 0 sin x t k z a Ey Ey − z = Find the density of the displacement current in the waveguide and the charges and the currents on the interior walls of the waveguide. The inside is vacuum. a z y x b x z y x y z g b a Electric field lines Magnetic field lines
Solution: (a) We find the displacement current asaD刀18sin( t -k.-)--e,EvoのsinXata(b)At the interior wall y = O, we have福Ps=e,(εE,)=εE)Js =e,x(H +H.)=-e.H,+eHAt the interior wall y = b, we havePs =-e, -(c E,) =-s E,Js=-e,x(H+H)=e.H -e,H
Solution:(a) We find the displacement current as t = D Jd sin( ) π 0 sin x t k z a y Ey − z = −e (b) At the interior wall y = 0，we have S y y Ey = e ( E ) = S y x z z Hx x Hz J = e (H + H ) = −e + e At the interior wall y = b, we have S y y Ey = −e ( E ) = − S y x z z Hx x Hz J = −e (H + H ) = e − e
At the lateral wallx = O, H = O, we haveJs =e ×e,H-o sin(ot -k.z) = -e,Ho sin(o t-k,z)At the lateral wallx = a, H = O , we findJs = -e, ×e.(-H_o sin(o t -k.=)) = -e,H-o sin(o t -k,z)At the lateral wallsx = O and x = a, due to E, = O, and p, = O .Thecurrentsontheinteriorwalls
At the lateral wall x = 0 , , we have = 0 H x sin( ) sin( ) 0 0 H t k z H t k z S = x z z − z = − y z − z J e e e ( sin( )) sin( ) 0 0 H t k z H t k z S = − x z − z − z = − y z − z J e e e At the lateral wall x = a , , we find Hx = 0 At the lateral walls x = 0 and x = a, due to , and . = 0 Ey = 0 S z y x The currents on the interior walls
4.ScalarandVectorPotentialsSuppose the medium is linear, homogeneous, and isotropic, fromMaxwell's eguation we find02HVXJVxVxH+eat?02EaJVxVxE+us-μatat?Using V×V×A= VV.A- V2 A, and consideringV. B=0 andV. D = p ,thentheaboveequationsbecomeaj0E2E-eVpuOt?at802HV2H-e-V×JOt?V
4. Scalar and Vector Potentials J H H = + 2 2 t Suppose the medium is linear, homogeneous, and isotropic, from Maxwell’s equation we find t t = − + E J E 2 2 Using , and considering and ， then the above equations become A A A B = 0 2 = − D = J H H = − − 2 2 2 t + = − 1 2 2 2 t t E J E