Question

(30 pts) The magnetic field of a uniform plane wave propagating in a lossless medium with E,=2 is H(x,t) = cos(3x108t+4x)ā, – sin(3x10$t +4x)ā.(Alm). Find: (1) The phasor form expression for the magnetic field (2) The direction of propagation for the wave (3) The velocity Vp (4) The polarization state. If it is a circular polarization, you need to determine it is left or right handed. Provide details for your reasoning. (5) The average power density Sav.

Answer 1

Given magnetic netic field F(x)+) = cos(3x10ft +4xô -sin(3x10ft + z)ố, with relative permitivity of medium (fy) = 2 jwt F169,) e Re keicot elpr@y] + Refilelor.jpy;}} Reſ calcut ope) (dy +jam] وا * Re [cây tiag) elkot ope) escut +Px) ay W = 3x108 p=4 Ħ(,) = (e' (3x10++ 4)) ay From the above expression Ha,t) = ; (3x108 +- (-40%)) â e .. The direction of propagation of Em wave -âs (-ve X-direction) is 3 The polarization is circular polari şation. But there is phase Shift of 90 bw 7-component 4 y-component. ... The polarization is right circular polarization 02-04 1/2 =) Orientation is Cloul wise رم Right circular polarization

phase Velocity (Vp) 3x108 4 ve = 0.75 X10 15 x108 m/sec 01 Sang = È x Direction of Sang Direction of EM wave --* direction - az xây = câ -âyx (-âz) =-ân Myllo Creo - mu no Ver : 84.917 sy - 12011 va Hy=1 Ey = 84.9ti H2 = 1 84.911 fel Ē (2, 1) = -84.91760s (3x108t+44) âg -84.9 IT sin (3x108 +140)ãy E(It) --84-an(es(3x4 vua), tẩn (3x16+47)âu) -84.951 (COS (3X1084+4x)âz + sin (3x10$t*42) ay) Saug Cco3 (3x18+ + Mz)ô - Sơn (3 x 18+ tualay) - -84• qan cos(3x18 +++z) (a, xa - Sin(3x18 tua cây xâz) -84.90[cos (3x10® +42) ¢ +siny(3x1004 +42)] c-ax) 84.90 a Sovey
I hope everything is clear.