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Electromagnetic Theory and Electrodyp,, , , , , , 8.12. Plane Electromagnetic Waves In Free Space. ;, , well’s equaitons are, i, Max divD = V-D=p, =0, , , , , , , , , , , , divB = V+ o B = pH, curl. E = - = and D=eE, ot J = 0E ~(), and curlH = J+ » :, Free space is characterised by :, p= 0, 0 ='0, p= Hp and € = & ro), Therefore Maxwell's equations reduce to ; ., divE =0 +(a), divH =0 s(b), oH, curlE = —po =- 3 wa) : 14(3), ; and curl H = & x. wd) |°, Taking curl of equation 3(c), we get, i , : curl curlE = - po 2 (curl H), ii ‘it Substituting curl H from [3(d)], we get, ii ms “- eurl curl E = [ 2 (e (« 2), iil . ie. curl curl E = — [Up &9 He : w(4), t, Now: curl curl E = grad divE-V’E :, ie. curlcurllE = —V’E [since div E = 0 from 3(a)], Making this substitution equation (4) becomes, VE ~ Woe 2 “ =0, (6), , Now taking curl of equation [3(d)], we get, , curl curl H = e 2 (curl E)., Substituting curl from [3(c)], we get ,, vH (6), 2 Sar, , = 0 from [3(b)], we obtain, , a oH, 1 &, curl curl H = e) = = (- Uo Gr) =~ Hoes 2H, = grad div H- V’H and Noting that div H, , curl curl H = — yxy:, Making this substitution in equation ©, we.get, , Again using identity curl curl H

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\ or, Equations (5) and (7) Tepresent wave equations : aff), oted that these i Boverning electroma ., a ea @ na May be obtaj Y using (2) j a eg E and H in free Space, It, section. enh ee are vector Equations of identical fo ‘uations (9) and (10) of Precedin,, components of E and Hi separately satisfies the same scalar way aie ones that each of the sir, a Bok rm, v = ou, . 4 Woe ye =0 @), where-u is a scalar and can Stand for one o th ; ‘ ‘, resembles with the general was oie of the components of E and H, It is obvious that equation (8), . 2, Vu = i ou, 3 vor 20), where v is the velocity of wave. F ;, Comparing (8) and (9), we see that the field vectors E and H are propagated in free space as waves at, * a speed equal to, , v= te Since fy = 4x x 1077 weber/Amp-m, , & = 8-542 x 10°” farad/m, , a 4n, = ( Hore, So that = 9 x 10° m/farad., , 4n x 10°7, = 3x 10° m/sec = c, the speed of light., , x (estat * 9 x w), , 5 ; = i the, : x ight i f 7——= ; so equations (5) and (7) take, Therefore it is reasonable to write c the speed of light in place o q), , form Fk . ‘ oo, ra, 17H_ “ay, Varo ye”, 1 fu (12), Vu - 2 aP = 0., and, , tic waves. A plane wave is defined, js ations for plane electromagnet a mene, Now let us find the. solution of eet any point in a plane ae reed to ae directio, i ssthe i en, as a wave whose amp litude sat above equations in gee ernie aoe (13), : The plane wave solution ies Boe! © ", , , , ee

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‘term of moduli, , Ug ange aS, , ge pi B_since Vice c}, , - ratio of magnitude of E to the magnitude of H is, ised 98 20, , Fil lel), , ia Fe ; [snes = 7) :, 4n x 1077 oy, , es) = 376-6 ohms +(21), , were the units of Zo are most easily seen from the-fact, it measures a.ratio of £ in volt/m to H in amp-turn/m, therefore must equal volt/amp or Ohms. Because the, writs of E/H are the same as those of impedance, the value, f Jy is often referred to as the wave impedance of free, , ation in, , , , , , spice, Further since the ratio Z) = s is real and, , , , , , Fig. 8.2, , positive ; this implies that field vectors E and H are in the same Phase i.e. they have the same relative, , magnitude at all points at all times (fig. 8.2)., , The Poynting vector (i.e. energy flow per unit area per unit time) for a plane electromagnetic, , waveis given by, , S=ExH=E x 2XE, Uy, , E, En, , n, , ga, Hoc, BE, Zo, , using (20), , x (nx E) = i (E-E)n-G-n)E], , (since E+ n = 0, E being prependicular to n), , [refer equation (21 )], , mie electromagnetic wave of angular frequency ©, the.average value of S over a complete cycle is, , egseterPst, , oon, XH HH Hl- B, , EG, 27, , , , : 2, ik-r—-iot, <(B é } real, , Ee <cos’(@t-k-r)>n, , : 1, [since (cos”< wt—k+r) > = Fa

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: ‘Electromagnetic Theory and Electrodyng, mics, , 1. 72, => Enns 0, ao™ (22), , (since Ems = #), It is obvious that the direction of Poynting vector is along the direction of propagation of electromagneg, , wave. This means that the flow of energy in a plane electromagnetic wave in free space is along ty, direction of wave. ;, , ‘Ratio of Electrostatic and maguetic energy densities is given by, , 4122 eh eK Kal, Um Epp? Ho HP Ho & ts), , ss Z -V(e)|, z ®%, ie. the electromagnetic energy density is equal to magnetostatic energy density. Total electromagnetic, , energy density, ti = ug tg = 2, = 2X LEE’ = OE, Time average of energy density, , , , <u> = <&E > = &< (Eye! IO all, = eq Ee < cos” (wt-k+4)> = 38 Ea = € Ems (24), Dividing (22) by (24), we obtain / ;, : -<S> 1 il 3 n, = 57 n= n= =cn (25), _<up Zoe " fH) Voto, , Thus we obtain /, : : . <S>=<u>cn (26a), ie. 1 WM energy flux = energy density x c. ..(26b), , " ‘This’ equation implies that the energy density associated with an electromagnetic wave in free space, _ propagates with the speed of light with which the field vectors do., , ” ‘Summarising we may say for electromagnetic waves in free space that:, , 1. In free space the electromagietic waves travel with the speed qflight: ee, 2. ‘The electromagnetic field vectors E and Hi are mutually perpendicular and they are., , ; indicating the, , perpendicular to the direction of propagation of electromagnetic waves. Thereby indicating, , electromagnetic waves are transverse in nature., 3. The field vectors E and H are in same phase. son and the, 4, The direction of flow of electromagnetic energy is along the direction of wave propagation, energy flow per unit area per second is represented by, , <S> = n= <u>cn., % ity, nergy dens, , 5. The electrostatic energy density is equal to the magnetic energy density and the e, associated with the electromagnetic wave in free space propagates with the speed of light.