Example 5.4 Tye electric field in fre space is given by, E = s0cos(10%s + Bx)3, Vim 1. Find the diction of wave propagation, 2. Calculate Band the time ittakes to travel a distance 22 3. Sketch the wave at ¢= 0, 74, 772. Solution: 1. From the positive sign in (or + fi), it is concluded that the wave is propagating in ~ L i, direction, 10! 3x10) As the wave is travelling atthe speed of light, ¢ 2. Here, B 33 radls Atte TIA, £,~ s0coe{ + x= —S0sin Bx (: of 22} ‘The wave is plotted at = 0, 7/4, 7/2 and shown in Fig, 5.3,Example 5.6 in a medium, £=16¢ ° sin x10" -21)4, (W/m). Find (a) the propaga- tion constant, () the wavelength, (c) the speed of the wave, and (d) the skin depth. Solution Here, «=0.05, B=2, 0= 2 108 1, Propagation constant, 7 = (c+ /B)=(0.05 + j2) = 2.000025 288.568° per metre 2, Wavelength, 4 = 3. Speed of the wave 4. Skin depth, “Example 5.7 ‘the clectric field intensity assoviated with a plane wave travelling in a perfect, dielectric medium is given by E,(2.t)= Veos(2n x10" = 0.1 22) Vim. Find: 1, Velocity of propagation. 2. Intrinsic impedance. Solution: 1. Velocity of propagation, v= = = 10 Boog 7 2x 1Ohms 2. Now, Tee Ge fe tr vebstattete meta <1} Mote 377% ft ee 63x10! v 2x10" 2 ax2=2518 @ Intrinsic impedencs, 7*Example 5.8 The electric field intensity of an electromagnetic wave in free space is given by ‘] Detomnine the expression for the components of the magnetic Geld intensity 17 Also, find i Solution: ByMaxvell’s equation, and Thus, the components of the magnetic fields ar: 5 H,=0, H,=0 and H,=Frewalr-4) FLeyye fe- AEX” nen = 3770 y loEo & re I auexi0 NB: This is known as the imvrinste impedance of jree space.Example 5.9 The elecuic field intensity associated with a plane wave wavelling in perfect dickeetric medium having 1 = tis given by 10cox(6x 1071 —-0.49¢2)7 Vim Find the phase velocity, the pemmittvity of the medium and associated magnate feld vector #1 ‘Velocity in five space = 3 x LO¥ ms. Solution Here, = 61x 10" nds B= 0.42 Phase velocity, 13x10 ms Now: By Maxwell’ equation, 1 een 1Osin(in x 10"¢ 0.42) ae « 1 a AF stn 107042 rat ) 4, $F fssyor x10" 0n2yt [coson x10" 0482) 1 oa fi = cont 10"s 0.482) im Hence, the magnetic ld isiven conn x 1071 042)7 AcaExample 50) The x sry components of siculny polar eeelomaynetc wave in fee sccare sini = ey costar - Find the expression for the displacement curront density nd draw a nctskech shaming the ks, Solution Hess, the mogninude ofthe slice fold having erculurpolaisstion and components 3° _ (Eee “The displacement cuneat density is eas (ei Be) +3 si (or fa) Example 5.13 1. Determine a 8... vand 6fora damp soil at | MElz. Given: = 12, 02 x 10" mha/m. 2 Determine the propagation constant 7 for n waerial having fy = 1, &, =¥ and @'=0.25pSim, if the wave frequency = 1.6 MHz '. Find the skin depth 3 ata frequency of 1.6 MHz in aluminium whore @= 38.2MS/m and 4.1 Also, find yand the wave velocity v. Solution: o 2x10- 1, Here, @=—_ 2810" ny ‘oe Da 10H 8NSAKID ENT ence, the moterial is good conductor. =0281 permere (02814 0.281) =03974 245° per mere [FE = PEE 5.959 0 fo _2n x10" PERI 2230x108 mic 559m 02510" 2a K16XIDXB.EST IE «l 2 tics Hence, the material is © good cistectie. a fl 02500? yy MT BPM ref pa ogre 2 an 1.63106 fT RIO =O DHE atin c+ Jf) = 0038 2 Skin dei i i f = 64377 ym FIMO Gx 10 ae xO IBD KIO a= {2 aus =15535.579 pormetre “3 mae Y=Ca+ /p fo _2F x16x 108 Bo Ss3379 6 666x100 (0948 290° per metre 1967.798 245° per met 15588 579 + /15853.578) y 647185 en's“Example 5.14 A uniform plane wave in a lossy medium has a phase constant of 1.6 rad/s at 107 Hz and its magnitude is reduced by 60% for every 2 m travelled, Find the skin depth and speed of the wave, Solution Here, B=1.6ral/s f= 10" Hz Since the reduction in magnitude for travel of x= 2 mis by 60%, we can find the attenuation constant, as follows. 0.916 = a =0438 Ose = Be? = ~20 =I 04)= ‘Thus the skin depth is, 6 = =—1 3.183 m Speed of the wave is, y= B Example 5.15 Find the depth of penetration of a mega-sycle wave into copper which has conductivity ¢~ 5.8% 107 mho/m anda permeability approximately equal to that of fre space. Solution: Given: o=5.8% 10" mholm, f= | MHz, = ly = 4x 107 Him Depth of penetration is 2 L 1 8 ee ous Sime Jixio' xm x 4x x10" Example 5.16 4 uniform plane wave is specified by 7 wave is 2 x 10° mis and the relative permeability is 16, find the frequency, relative permittivity ‘wavelength and intrinsic impedance. 602 x10 m 7 Alm, I the velocity of the Solution Here, 6=0.1, v=2x 10" ms. u.= 1.6 O95 ae On Frequency, Now, v= =—£ Relative permittivity €, (J xh Wavelength 2= 2% = 2% - 209 = 62.83 m 5H nc ingtee 1 [2 =f nes*Example 5.17 An clecric field vector £ of an clectromagnetic wave in fre space is given by ‘his expression E, = E.= 0, pnacl Using Maxwel’s equation for free space condition detemine expressions forthe components of the magnetic field vector J Solution By Maswall’s equation, Comparing both sides. we get and where, 17 = [Me isthe invinsic impedance of fice space ‘Ths, the componen’s ofthe magnetic Fields at: =) and Example 5.18 The clectrc eld intensity associated witha planar travelling wave ina dielectric ‘medium is given by E,(z,t)=1Ocos(2n x10"t = 0.12) Find the wavelength (2), phase shift constant (8) and velocity of propagation (1). Solution Phase shift constant, 6=0.17 Wavelengih, ‘Velocity of propagation, 2x10 ms“Example 5.19 A lossy dielectric has an intrinsic impedance of 200230° © at a particular frequency. If at that frequency, the plane wave propagating through the dielectric has the magnetic field component a, Alm find E and ce Determine the skin depth and direction ofthe wave polarisation H=10.e** cosa - Solution Here, Ho 1 =200 230° Ey Sm => Fy = 2000230 E = Re[2000 £2077] = 20006 coo Since the wave is propagating in the G, direction and the magnetic field is having only @, component, the electric field will have only ~@, component [-. direction of wave propagation is given by @ XG | Hence, the electric field is given as Now, we know that DIR But 1 a= px == 0.2887 npim PRO P dai =3.464m Since the wave has only z-component of the electric field, itis polarised along th ‘The skin depth is given as,Example 5.20 For sea water with = 5 mho/m and ¢, = 80, f= fl, find the distance a radio signal can be transmitted at 25 kcps and 25 Meps, ifthe range is taken to be the distance at which 90% of the wave amplitude is attenuated Solution The wave is attenuated by the factor e~®. In this case, the wave is attenuated 10 90%, EF 0.1 -ax=In(Ol) + x= 2802 @ Substituting the given values, we have the following resus For 25 heps, cr = x 25% 198 [ERI * SOB ESSIO™| fp a (2m x 25% 10°) (8.854 x10") =0.7025 2302 _ Fae $3278 m For 25 Meps, cen dn x 28% 08 [AEXIOT 808.854 10° { ff 5 ~-1 z (2m x 25x10°)7(8.854 x10) =21.960 1,969 = 0.105 mExample 5.21 For a uniform plane wave in fresh lake water ¢= 10? mho/m, e, Calculate o, B, 1 and A for two frequencies 100 MHz and 10 kHz. Solution For 100 Miz frequency, ge _2mfe _ 2 x100 x10" 8854 x10? x80 = aus So, the lake water acs as a good dileste. B ~offe(ie 5s) an x100x10°GeH10 xBBSHRIO = KO {+ tos aa x100 10) 8.854xT For 10 kHz frequency, fe s10110 888400 80g ogs So, the lake water acts as a good co c= [AE «PETIT ag 109 Nepevin B ctor an 10? radianin fo yso.._ [2m x10X10" x4 x10 o 107 2 4228-20 m2 = 8.886 245° 80, = byExample 5.22 1. Earth considered tobe a goo! conductor when <<, Determine the highs eqns for which earth canbe considered a good conduit if <1 means lets than D1. Assune the Following constants o= 510% mhoim, e= 106, Can ar be assumed zero at these frequencies? o 2. Ifthe earth is considered to be a perfect dielecrc for Z< 0.1; then at what ffequencies may 5X 10°mbolm and u, = 1. € earth be considered a perfect dielectric? Assume: Can be assumed zero at these frequencies? Solution 1. tats as, 20. Wid. 015x109 © “g.gsax 10? x10 <5.647x10° Thc highest frequency for which ert canbe considered good conductor ia 5.647 x 10° fe a Fora good conductor, =] 562 = [STRIP ADOT AIO? 9123 pen Hence, ee eannot be 2er0. = 0899 MHz 2. Inthis ease, S 705.8910" ‘The highest frequency for which earth con he considered a good conductor is a, _ 705.89. 108 fase S235 MHz ive 2 8854x101? xB Hence, ais independent of frequency and whatever be the higher frequencies, a will be about 0333 Ap/n,Example 5.25 An clocromagnetic wave impinges on « moalic shoot. Compare the reflection coeficinis for copper and ioa if 8x10 hol for copper, x10 mbom —_ forieon, far copes, for ion, for copper and izon. Take frequency to be | megacyce. Solution Reflection cocffiviens, P= 2— nm where, isthe intrinsic impedance of air and nis the intrinsic impedance of metal For copper Ton jeans 10 Ke 10 = je | SERIO jx 2H 1H E.8SA KIO = 3.680613 10-4 245° = (2.608951 x10" 2.60895 x10") Hence, reflection coefficient far copper is p36 KIO L4S?— 37720" orm 5 69 x IO LATS STII n= {2 PIR KAO" 40x 1000107 BEV a+ jee” Til0® + jx 2 10x 8.854 x10" =0.0889.245° almost perfect reflector For iron: Hence, reflection coefficient for iron is o.08s0. 245° — 377.20" 0 ppegyZasea sTT Zo ~ 8% We Boe relReclor “Example 3.26 f and i waves travelling infos epace, re normally incident onthe interes ‘with a perfect dsteetric with ¢, = 3. Compute the magnitudes of incident, effected and ansmitied E tnd ZF aves tte interface. Take £,~ 1.5 mV'm in medion | For medium 2 (perfect “ neuf =n cen (Eon ‘The transmission coefficient is given as Solution For medium I (five space), 120r=3770 electric) 2217.66 Ett 317+ 217.66 t = 0.732 ‘The reflection coefficient is given as Since the amplitude of the incident electric wave is E, E,_1sx1 A Amplitude ofthe incident magnetic field, Hy= F+= S57 — = 3.98 wala Amplitude of the transmitted electric field, F, =1 x F, = 0.732% 15= 1.098 mV/m fs E,_ 1.098 , Amplitude ofthe transmitted magnetic Held, Hy= T= 77g = £04 Alm Amplitude of the reflected electric field, £, = Ty x E,= —0268x 1S = ~0.402 mV/m Amplitude ofthe reflected magnetic eld, H, = £2 = 5942 ease NeExample 5.27 _Detecmine the normal incidence reflection coefficients for seawater, freshwater, and “good! cart at frequencies of 60 Hz, | MHz and I GHz. Use ¢, = 80, o= 4 mho'm for sea water, €,= 80, 0=5 x 10" for fresh water; and ¢, 5, O= 10 x 10°? for good eatth. ” Solution The reflection coefficients given as, Pg eal where, 1 isthe intrinsic impedance of air and 1 isthe intrinsic impedance of the other medium. Forair, n= | M2 =120n = For ssawater, n= {iat Tele XOOXAEXIO™ — _ going 450 RT Yas iae Yor joese, Ya+ [2x x60x8854x 10x80 For freshwater, n= [2 Foy LV a+ jae Yar joe,e, PRIX IO Ka X10 NOON 3434.224.16° SLO 72 x1 1OX8.854 x10 BO For good water, jou Joy T+ 108 ~\ TF j0e,F, BRIO ARI gs a 7g 540 10x10 + je x 1x 10? x 8.854 x10 18 So, the reflection coefficients are 170 o.o1o9z4s°~ 377 Tnsa wate = Qg109 24°F 377 7 09910” 34224.16"—377 sar T agesh wate) $4420 16°F 377 OT ANS. or.arzo3s3°—377 . T aigonsemts) = 97.37 ZO BEST ITT = 0: AZL10Example 5.28 tn fce-space (2 <0), aplane wave with a cos (1081 — Bz), amAln is incident normally on a lossless medium ( 1, = $44) in region 2 > 0. Determine the reflected For the lossless medium, the intrinsic impedance is, 1: = =2n,=240" Given that: 7, =10 eas (10'r — 2), mAlm <0, p-@-10 1 Fore 0 BE eS Forz>0, Bonne =o me Be From the given form of the incident magnetic field, we may expect that a =Ox4=ap =F 9005 (01 G2}, where, and, Now, z, n=, men 2mm 400 e0s(l0'r +42), mVim simitarly, £ SE Deslioe