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329sp09he2soln - ECE 329 Introduction to Electromagnetic...

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Unformatted text preview: ECE 329 Introduction to Electromagnetic Fields Spring 09 University of Illinois Goddard, Peck, Waldrop, Kudeki Exam 2 Thursday, March 19, 2009 —— 7:00-8:15 PM Name: \ Ection: “; 9 AM 12 Noon 1 PM 2 PM J Please clearly PRINT your name in CAPITAL LETTERS and circle your section in the above boxes. This is a closed book exam and calculators are not allowed. You are allowed to bring notes on a 3x5 index card —— both sides of the card may be used. Please Show all your work and make sure to include your reasoning for each answer. All answers should include units wherever appropriate. Problem 1 (25 points) Problem 2 (25 points) Problem 4 (25 points) "TOTAL (100 points)J —’ Problem 3 (25 points) ‘ 1. A monochromatic plane TEM wave in vacuum has an instantaneous Poynting vector 73 = fiPo cos2(wt — ,B-y), wavelength A = % In, and the wave electric field at t = O and r = (0, 0, 0) is E = 0.6:? — 0.82 V/rn. Determine (in proper units): a) (3 pts) Wavenumber ,3, 62% 22’11‘: =q—rgg/ Van b) (4 pts) Wave frequency w, [2771109 46-4 “J--c= 2m"; 9 w :Mnxhw ac/ [b c) (4 pts) P0 in W, 0112‘ lg: Exfi ”HEP. H, &»2(w:~(53.) ) 031044—032; .. d) (3 pts) Direction of wave propagation, A / ‘ e) (4 pts) Electric field vector E at t = 0 and r— — (010) in, W =~ow+ w? M f) (4 pts) Magnetic field intensity vector H at t = 0 and r = (0, $0) In §= W1; [Zo % g) (3 pts) Polarization type of the wave. l/{m—f. #3 ED: ZA/W V g‘? a) (13 pts) Consider an infinite surface current density Js : —.’2sz0 cos(wt) flowing on 2 = 0 surface, where J30 > 0 is real—valued amplitude of the monochromatic surface current measured in A /m units. It is found that J3 injects field energy into propagating tianSVerse electIOmagnetic (TEhl) waves away from the z— — 0 plane at an aVerage rate of 4 W /In2 — that is the magnitude of the average Poynting vectox (73) is 2 W/ n12 for the waves excited by the surface current. Denoting the TEM waves excited by Js (above and below the z = 0 plane) as E = itEo cos(wt 3F fiz) V/m and H = :lzngo cos(wt 2}: [32) A/m, where wavenumber /3 = '— determine the numerical values of wave amplitudes E0 and H0 in VII/111 and A / m units (assuming wave propagation in free space). Hint: equate the expression for the magnitude of (75) for each wave to 2 'VV/mz. /\ <57=éE0H,2: 22 a; Eoufiq a; étsq no §\ 1 N H3399, 1:- ‘4'? mm 1’3 (-H 1 A’ “V t7 . b) (12 pts) We have on 3: = 0 plane a pulse of sheet current J 3(t) = —y2t rect(,:) A/m, where 7‘ = 2 us. Determine and plot Ey(az, t) and Hz(:c,t) vs a: for t = 311.5. Assume that the current sheet is embedded in free space. WA fig. 20 4" p seals): 3. A perfect dielectric slab having thickness W in the :6 direction is surrounded by free space and placed within a constant electric field E, = 181?: V/m. The induced polarization of the bound charges inside the dielectric material reduces the electric field strength inside the slab from 18:2 VI/m to E = 350 V/m. a) (5 pts) What is the displacement field D inside the slab? Hint: since the region, including the dielectric material, contains no free charges, V - D = 0 across the region. a— A ‘ ‘ ‘ pslfio x’énmob. anal m‘l'qu.‘ 1‘44- Sled, ML 1)) (6 pts) What is the polarization—field Einside the slab? A /\ ,_ /\ C :2) PZD"?.E=WE:.X”32.X 3&ng M 5:2,?” c) (5 pts) What is the relative permittivity 67» = i of the slab material? 9:21? a 2:192. = 629 z) 2r = 9/, A 3x 132. x d) (5 pts) What is the electric susceptibility Xe of the slab material? 8 = 659 = (1+7(¢)Ea ,9 Kc ,__ C/ e) (4 pts) If the dielectric slab is removed, leaving a vacuum, what. would be the new values for 67., Xe, P, E in the region of space previously occupied by the slab? £371 A90 an ,— 1: A 6650:?! v 3/. 4. Wave parameters in material media obey 7'17 = jam and ’71]: a +wa. Consider a plane TEM wave propagating in a non—magnetic material (p, = pa) with a. magnetic field intensity H = 1258—32 008(107t — 47,) £. m Determine: 3 Th 1 H 1’ a) ( pts) epiasor . I? 4 ‘9 H 2 X . a 9' 14/ a x "F; $-13 25—h“ {fiber} (4'34?) V%. c) (4 pts) Expression for time—averaged Poynting vector (75) in terms of [771 and 7'. at ’2’. * /‘ A 1. ”C mend?” l" Eé? 14116“ W9 i w/*% 2' W on (t) d) (4 pts) The propagation constant 7' = a + jfl, y: aux” 354%. e) (4 pts) Explicit expressions for In} and 'r 47 igr. «Kama a. mam Mr 54££ ”3"; ——7 kHz-QEIL/ fs'iZ-«q% g f) (6 pts) Explicit values of permittivity 6 and conductivity 0, Ar awe @+j9)1»lb+jl%i-ifl”=fl£} "7 '71:?" '7? ”'74?“ m w 36“! r; . I :«L is d :7 V=‘L‘§%?:% 'SW/ i’dugfli 10*?“ 4“” ...
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