Electromagnetic Waves PDF

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 107 106 Blectromagnetic Waves Manvell's Eguations 0 E characteristic parameter of the medium surrounding the The quantity e is a medium. Even no medium like the permittivity of the charge Q. and is called is generally denoted by Eo. The permittivity vacuum has a finite permittivity, and D is of the free-space (vacuum) (b) Anisotropic medium (a) Isotropic medium 10° 367 F/m (3.3) Fig. 3.2 Electric field and electric displacement vectors lor isotroplc and of vaccuum. The anisotroplc media. Generally, other media have pemittivities e higher than that ratio of e and eg is called the relative permittivity or the dielectric constant of the 3.2 shows the E and D for medium and is denoted by e,. Equation (3.8) is same as Eqn (3.5). Figure inside the magnetized plasma and anisotropic media. Inside crystals and (3.4) isotropic is a tensor. (ionosphere) the dielectric constant We observe here that, since, the permittivity is different for different media, the 3.1.2 Electric Scalar Potential charge Q produces different electric fields at the same distance from it. A one point to another, r electric field if we move a charge from same In the presence of an quantity which remains independent of the medium characteristics is called the work done is essenually stored in the charge in the electric displacement density, D, and is defined as the product of the electric field we have to do a work. The called the form of potential energy. This energy a is characterized by parameter and the medium permittivity. Hence, is called done is a scalar quantity, this potential electric potential. Since the work P with respect to some electric potential V of a point D= ¬E = f 47r2 (3.5) electric scalar potential. The the work done in carying a unit positive charge reference point O is defined as reference point is zero defination the potential of the f the medium is isotropic, meaning properties of the medium are not direction from point O to P. Then by zero. a unit charge from O to itself is identically dependent, then the permittivity is a scalar quantity. For this medium it can be e as the work done in carrying distance 'dl' from P infinitesimally small seen from Eqn (3.5)that the direction of D is same as that of E. On the other Let us say that the point P is at an hand, for an anisotropic medium (a medium for which the properties are direction (Fig. 3.3). dependent) the permittivity in general is a tensor and then Eqn (3.5) has to be re-written as D=:E (3.6) dz where is a 3 x 3 matrix. In the cartesian co-ordinate system this can be written as dy (3.7) For an anisotripic medium, the direction of D is not same as be seen from that of E. As can Eqn (3.7) D, is not produced Same is true for D, and by only E, but also by E, and Ez. D,. Depending upon the elements of the X tensor, D can have any is special case of Eqn (3.7) where a arbitrary direction with respect to E. The permittivity Work done in moving charge elemenis ¬xy, Ex=¬yy = isotropic case Fig. 3.3 Ezz = 6 and other fz fyx. ¬yz.fa¬zy = 0. For off diagonal isotropic medium Eqn (3.7) reduces LO P is E the work required to move a unit charge Then if the electric field at -EE from P to Pi is (3.9) dV=-E dl (3.8)  Electromagnetic Waves 109 108 Maxwell's Equations the ongin of a co-ordinas. take the point P as we can Newton/coulomb sincc the clectric ficld is force per unit charge. Without losing generality, locaton with respect tothe electric ficld as P can have any arbitrary and Newton/coulomb cquivalent units. The commonly used system. Since, the point are has three components along So, Vol/meter the length vector dl general in he D will be derived later after we haveC co-ordinate axes, unit of the electric displaceinent density three principal axes. covered nmore laws of electromagnetics. dl dxî+dy9 +dzâ (3.10) Flux Density 3.1.3 Magnetic Field and Magnetic asS field the electric field can be written in its components the stationary charges give rise to clectric Similarly, As we have mentioned earlier, in motion field. Since the charges E E,+ E,ý+ E,2 (3.11) whereas the charges in motion produce magnetic current as the starting point for the generation constitute a current, we can take the Substituting for E and dl in Eqn (3.9) we get field. As the presence of a charge is felt by the clectric field of the magnetic of a current can be felt the magnetic field generated by dV= -(E,dr + E,dy + E,dt) (3.12) produced by it, the presence field could have been written on the lines similar of Now, if the potential of the point P is V, the change in potential over infinitesimal by it. The definition magnetic a unit i.e. the magnetic field is the force experienced by to that of the electric field, like distance dl (using differential calculus) is monopoles in reality magnetic monopole. However, there are no magnetic be defined differently. 8V av the electric charges. magnetic field therefore has to The dv d +dy+d (3.13) Due to a current magnetic element the field is defined by the Biot-Savart law. The of a wire carrying current I in it. Consider an infinitesimally small piece Realizing that is then defined by the product dl. The current moment length of he piece is say direction the length dl wire can be oriented in any Idl. Since, the piece of the We can write Eqn (3.13) as w-++ (3.14) is a vector quantity and therefore should be denoted by dl. scalar quantity making the current moment Idl a vector The current I is a quantity. Without loss of located at the origin of the that the current element is generality, let us assume dV = VV d co-ordinate system as shown in Fig. 3.4. (3.15) Substiuting for dV in Eqn (3.9) we get dl (VV+E)dl = 0 (3.16) Now, since, the charge has been moved P over a finite distance PP the PI dl can not be zero. Therefore Eqn (3.16) demands that length dd VV +E=0 E= -VV The electric fheld (3.17) ldlx at any point is negative of the point. gradient of the potential at that Equation (3.17) is very useful in distribution of finding electric field due to charges. If one direcly goes one has to carry out for calculation of complex vector addition of the electric neld due to a current element Magnetic field which is quite cumbersome. Instead, oneelectrical then can find its can add fields due to individual charges Flg. 3.4 their addition is gradient to give the electric field. potentials due to all charges and field due to intinitesimally much simpler Since'the potentials According to Biot-Savart the law agnetic fielddH ( Equation (3.17) is also usedcompared to the addition of the electric are scalas element) at a point P is given as potential has a V commonly in defining the unit of, the electric fields. used unit of volt. Small current dl x f and therefore has The operator V is field.:The (3.18) the unit of dH = or Vm'. Instead of m-'.The Eqn. (3.17) ifwe Had a electric field hence should have a spatial deriyaliv 4Tr? element the line joining the current used Eqn (3.2) we unit V/ in the direction of would get the where f is the unit vector unito  Eectromagnetie Wave 111 E 110 Marvell's Bquations vectors dl and F for this point (and also for B and the observation point P. Since, the has the cross-product of the two a direction point P) lie in the plane of the paper, the direction of the current is perpendicular tothe plane of the paper. Since, of dl is upwards and the direction of the nmagnetic assumed upwards, the direction at point P. If we take B field dH will be into the paper as indicated by ® going the lett of the current element, by right hand some other observation point Pa on Isotropic medium Anisotropic medium come out of the paper as indicated by O. If one screw rule, the magnetic field will vectors in isotropic and streches his imagination a littue, he can then see that the magnetic field forms a Fig. 3.5 Magnetic feid and magnetic flux density element. anisotropic media. circular loop around the axis of the current From Eqn (3.18) it is clear that the magnitude of the magnetic field is directly proportional to the current moment and is inversely proportional to the square As shown in Fig. 3.5, in an isotropic medium the directions of B and H are of the distance between the source and the observation point. The cross-product same, whereas they are different in an anisotropic medium. dl x f also suggests that the strength of the magnetic field is niaximum when d It should be emphasized here that an anisotropic medium does not mean that and f are pependicular to cach other, that is, for equal to. As 6 reduces, the and A medium can be electrically anisotropic but it has both e j as tensors. angle between dl and f reduces and consequenly the strength of uhe magnetic and vice-versa. magnetically isotropic On the right hand field reduces. For extreme case of 6 = 0 the vectors dl and f are colinear making The unit of magnetic field H can be derived from Eqn (3.18). the unit of dl is meter(m). Cross product and hence the magnetic field identically zero. The magnetic field, side of Eqn (3.18) the unit of curent is Ampere(A), f is unitless. The unit of the therefore, is identically zero at the axis of the current element. the unit of distancer is meter and the unit vector flux density is It should also be noted that in Eqn (3.18) there is no quanity which depends magnetic field therefore is Amnpere/meter. The unit of magnetic upon the medium property. The magnetic field induced by a current element is, either Tesla' or Webers/m* therefore, independent of the medium surrounding it. Another quantity which characterizes the magnetic field is the magnetic flux 3.1.4 Conduction Current Density J density B. The magnetic flux density is defined as the total magnetic lines of forces it is rather inadequate to deîne In situations where the current has spatial varnation passing through a unit area placed perpendicular to them. The flow in the system. One can then use the current density as is related to B magnetic field H just the total curent through a medium characteristic parameter called as the curTent flow unit area per of the medium as 'penneability the primary parameter which is deñned B B = uH.19) The permeability of vacuum is denoted by lo and its value is Henry/meter. A ratio of permeability of a medium 47 x 10 the relative to that of the vacuum is called pemeability 4, giving As we have seen in case of (3.20) medium and it is a permittivity here also u is a scalar for a isotropic medium, tensor for an herefore, we have in anisotropic medium. For anmagnetically anisoropic general B:H where is a 3 x 3 matrix. (3.21) In the cartesian co-ordinate Fig. 3.6 system this can be written as conducting matenal with an arbitrary cylinder made of a =yyy L y 4zy J Hy (3.22) Consider a infinitesimal cross-section S (see Fig. 3.6). Let us a s s u m e t the conductüvity of the that the curment density cylinder is o H: Jexists in the cylinder over a length 7.  Blectromagnetic Waves 112 well's Equations 113 S between its two ends A The resistance R of the cylinder nd and 3.2 BASIC LAWS OF ELECTROMAGNETICS its resistivity is 1/a. B is then given by Resistivityx length I In the previous section, we got introduced to some of the fundamental quantities (3.23) of electromagnetics ike electric ficld E, magnetic field H, etc. In the following area of cross section sections, we state physical lawS which establish relationship between various is constant all over the cross. For simplicity assuming that the current density electromagnetic quantities. We will see that the laws can be mathematically section, the total curent in the conductor is written in integral or differential form. Depending upon the suitability one can choose the integral form or the differential form. I=JS (3.24) there is a voltage drop V between point Gauss's Law Due to the current flow in the conductor, 3.2.1 to be of infinitesimal size we can A and B. 1f we assume the conducting cylinder The Gauss's law states that, the total ourward electric displacement through any assume the electnc field associated with the voltage drop to be constant between closed surface surrounding charges is equal to the total charge enclosed. A and B. Then one can wriie the voltage V to be Consider a charge at some location in space (Fig. 3.7). The electric V= El (3.25) displacement density D at some point P at a distance r from Q is given by Eqn (3.5) as Now from Ohm's law we have V= IR (3.26) D= (3.31) Substituting for V. I and R from Eqns (3.23), (3.24) and (3.25) we get El=Is J=aE (3.27) The magnitude of the conduction curent density J is proportional to the electric field D strength E. Now, from the Ohm's law we also know that the current flows in the direction in which the potential drop is maximum. That is, the curent flows in the direction in which the potential V has maximum change. This direction is opposite to the direction of the gradient of V. From Eqn (3.17) the direction opposite to gradient of Vis same as that of the electric field E. Therefore, we can conclude that not only the Fig. 3.7 Electric displacement vector and area norma is also same as that magnitude of J is proportional to E but its direction of E. The Eqn (3.27) for vector J and E can be written as electnic J=GE Suppose, place a frame enclosing area A at location P. Then the we (3.28) the area of For anisotropic materials the displacement passing through the frame indeed will depend upon conductivity also is a tensor and then the Cross-section A of the frame but it will also depend upon the orientation of the has to be re-wniten as Eqn (3.28) frame with respect to D.Ifthe direction ofD is nomal the area A, the electric to displacement through the frame will be DA. On the other hand, if the direction J=:E (3.29) of D is tangential to the area A, the electric displacement through the frame will where the tensor conductivity given as is be zero. In general we can see that the électric displacement through the frame proportional to the of the area A perpendicular to D. So if the angle projection s î is e, the projected area will be A cos 4. between D and the normal to the area (3.30) The electric through the frame therefore will be DA cos 6. In vector displacement of D and the area A, where the direction of an area For an anisotropic material the algebra this is the dot product as that of E as is direction of the conduction current is not IS defined by its normal. Therefore, evident from San ionosphere the conducüvity is Eqns (3.29) and (3.30). For a medium like ne Electric displacement = D. A a tensor. Let us now consider a closed surface S surrounding the charges Q1. Q2, Q3 on the surface, the as shown in Fig. 3.8. Then, through an incremental area da  Blectromagnetic Waves Mawell's Equations 115 114 clectric displacement de. 1he total Can utwand electric displacemcnt will be D Then according to the Gauss's la law ouation (3.35) has to be valid for any arbitrary volumc. This can happen surface S. provided the integrand is identically zero making over the be obtuinced by integrating +0t+ en (3.32) V.D =p (3.36) D-da= Q1 This is the differential fornm of the Gauss's law. From Eqn (3.36) the physical meaning of divergence becomes more clear. The divergence of D represents the netoutward flow of D per unit volume The Gauss's law can be effectively used in finding charge distribution in a region having clectric field., One can find divergence of D to obtain charge density at that location. If there is no accumalation ofcharges in the given region the charge density p wil be zero and the divergence of D will be identically zero. The total da inward electric displacement for this region then will be equal to the total outward clectric displacement. aQ Q S Fig. 3.8 Region containing charges Insiead of discrete charges Q1. Q3. 0, if there is a continious distribuion of (a) (b) eharge inside the closed surface, we have to find total charge by integrating over Flg. 3.9 displacement vectors for (a) a single charge Q (b) two charges Electric the enclosed volume. The distributed charges can be corectly represented bya Q anda Q2 charge densitly p which in general is a function of space. The p is the charge per unit vohume huving units Coulomb/m. The charge in a small volume dv will which is One can also make an important observation here that, any charge be of electric displacement on pdu."The tolal charge cnclosed by the volume can be obtained by integrating over the volume V, For distributed Outside the surface S, although affects the distribution Consider a charge Q at charges Eqn (3.32) then becomes S, does not alter the net outward electric displacement. the center of a spherical surface a shown in Fig. 3.9(a). The electric displacement D da Js (3.33) IS radially outward, and hence is normal to the spherical surface at every point, The total outward electric displacementtimesQ: is This is the Guuss law the integral form. outside in Now, suppose we place a charge a Q (a is much greater than l1) The equivalent diflerential form can be obtained by of the electric displacement thcorem to Eqn (3.33). Using Eqn (C.28), the LHS of applying the Divergence the spherical surface. The direction and magnitude but the net Eqn (3.33) can be written as on the surface of the spherewill be completely altered (Fig. 3.9(6) the oulward displacement will still be equal to Q because he charge enclosed by D da.Dd (3.34) surface is still Q only. Substituting in Eqn (3.33) and bringing all the terms on the left hand side of e inside a hollow-sphere is p = cquality sign we get EXAMPLE 3. A volume charge density enclosed within the sphere. Also find the 10eA C/m'. Find the total chargeof the sphere. I(V D) Ciectric flux on the surface - pl dv = 0 (3.35) density  Electromagnetic Waves Marwell's Equations 117 116 within 2he surtace of the sphere of im radius is Solution: The D just enclosed 20T The total charge D = 5 C/m 4T(1)2 p r sin 6drdbdo - density of the sphere is p,, the D just out side is S+ P, = If the surface charge 5/2 (given.) dosiaBde10-20dr We, therefore,get p, = - 2. 5 C/m2 407 Find electric fux density is given as D =x'i+xy|. =40rdr= 4000 100 EXAMPLE 3.3 The at the origin with inside a cube of side 2m placed centered the charge density Since the distribution of charge is spherically symmetric, the electric its sides along the coordinate axes. density is also uniform on the surface of the sphere. By Gauss law the to electric flux from the surface of a sphere is Q. Solution: C for expression of divergence in The volume charge density is (see appendix 47rD Q2 the spherical coordinate systemn) /100 p=VD OD8D D4t 4T (2)2 ay z = 6.25 x 10 C/m2 ()+0+ ) 3x3 EXAMPLE 3.2 A volume charge density of 20r C/m3 lies within the cube is a spherical surface of Im radius. Find the electric displacement densiy The charge enclosed by everywhere in the space. Also find the surface charge density which should b& placed on the surface of a sphere of 1m radius so that the electric o- dzdyd: | - sddyd density just inside the sphere is double of that just outside the displacemen sphere. Solution: - 12xds= 12=8C The charge enclosed within a sphere of radius r < 1mis 100 cos 28 Q)= 20r sin 6dedodr is given by D= EXAMPLE 3.4 The electric flux density I r 0, 4,2 = 20 and e;2 1. If = = 2, and for for the time = the magnetic ficlds in the two fundamental phenomenon of electromagnetics. H = 6 +5 +82 regions are wave is, therefore, an important unbound media or ideal homogeneous media in A/m have either Although we do not as it provides the basic wave is very usetul H = 2-3 +22 A/m practice, the concept of uniform plane in a medium. Moreover, Find the linear curent wave propagation density in the current sheet. understanding of the electromagnetic dimensions much larger than the wavelength. lor media which have physical In practice. solution. the uniform plane wave the solution closely approximates to solution is adequate where uniform plane wave herefore, there are many problems and is quite useful. in the understanding otf unitorm plaue wave n this chapter, we develop of a unilorm plane investigale s o m e interesting properties Carlesian geometry and wave.  Electromagnetic Waves rm Plane Wa 144 145 R MEDIUM 4.1 HOMOGENEOUS UNBOUND On the lines similar to that used for thetransmission line analysis, the two unbound medium without any electi cd cquations for E and H are decoupled by differentiating the cquations Let us consider a homogeneous, isotropic, e and permeability couore with respcct to the space coordinate. Since, now, we are dealing with medium permittivity or magnetic source. For this case the entire m e d i u m. T h e source f oncedimensional space, the space derivatives are the vector space derivative constants over the the t h r e scalar quantities which are in tha Taking curl of Eqn (4.11) we get are are no free charges and currents m e a n s that there o p e r a t o r s. nature of the medium then reduce to medium making, p =0 and J = 0. The Maxwell's equations VxV xE=-4Vx OH (4.13) V B=0 (4.1) OE V.D=0 (4.2) VxVxH=¬V x (4.14) aB (4.3) nterchanging space and time derivatives on the RHS, we get VxE=ar aD a( x H) (4.4) VxVx E=-4- (4.15) VxH=a VxVxH=¢ x E) From constitutive relations we have.16) B uH (4.5) for (V x H) in Eqn (4.15) from Eqn (4.12) and for (V E) in Substituting x D ¬E (4.6) Eqn (4.16) from Eqn (4.11) we get and e are constants 3E Since the medium is homogeneous and non-time varying, (4.17) as a function of space and time. The Maxwell's equations, therefore, reduce to V B = V (uH) = uV H = 0 4.7) V x V x H=- H 4.18) V- H 0 (4.8) V D=V (eE) = ¬V.E = 0 Using the vector identityV x V A =V(V A) - x VA, where A is any and (4.9) be written as arbitrary vector, the LHS of Eqns (4.17) and (4.18) can V.E=0 (4.10) 3E (4.19) xE=- (4.11) V.E)-V*E=-HE 2 a2H 3(¬E) V(V H)- V'H=-uE2 (4.20) Vx H E (4.12) 0 and V E 0, Eqns (4.19) and Since from Eqns (4.8) and (4.10), V.H = = It can be seen from Eqns (4.11) and (4.12) that the time derivative of the magnetic field is related to the space derivative of the electric field and the time (4.20) reduce to derivative of the electric field is related to the 3E space derivative of the magnetic (4.21) field. These two equations suggest that a time without corresponding electric field and vice a varying magnetic field can not exist E=a versa. It is, therefore, clear that if the fields are time a H varying, both electric and magnetic fields have to co-exist and 2H E a? (4.22) we can not get only electric field or only Then this is magnetic field which varies with time. their solutions represent conceptually different than the electrostatic or Equations (4.21) and (4.22) are the wave equations and which are non-time varying and therefore can exist without each magnetostatic fields the wave phenomenon in three dimensional space. other. fields have to exist in Equations (4.11) and (4.12) are similar to the transmission line Eqns (2.5) We, therefore, observe that if at all the time varying and (2.6) for the form of a wave. voltage and current respectively. The a homogeneous, unbound medium, they have to.exist in in nature. These equations, however, are more general two coupledequations are exist together. Hence this compared to that for the oreover, both magnetic fields have electric and to transmission lines as they the transmission line is a represent the fields in three dimensional space whereas phenomenon is called 'Electromagnetic Wave" one dimensional geometry.  Electromagnetic Waves 146 Uniform Plane Wave HARMONIC FIELDS 147 4.2 WAVE EQUATION FOR TIME the wave cquation for each component is a scalar equation as vectors are functions of the space (. For time varying fields the E and H v = -uev that the ficlds are sinusoidal fun and also functions of time. If we assume (4.34) of time (time harmonic fields) and if we take the time variation explicitly out could be E,, Ey, E, or H, Hy, H. F where F could be cither E or H) as then, we can write a field vector ( Fx, y, z, 1) = Fx, y, z)e 4.2) 4.3 sOLUTION OF THE WAVE EQUATION (4.23) field. Now onwards we wi Herc angular frequency of the time varying is the will As mentioneda the beginning of the chapter, our philosophy is to find the simplest assume that F is only a function of space (x. y, z) and its time variatin a vector ation ssiblesolution which satisfies the given constraints, in the ero is implicit. Ption. Let us, therefore, ask a question "Can we have an present case, the wave electric field which is For time harmonic fields we observe that me varying but is conslant through out the space?" That is, can an electric field E(r, y, z, 1) = Kejon Fla, y.z.)= Fr, y, z)eja) (4.24) (4.35) = jwF(x. y. z)elo= jwF(r, y, z, 1) olution of the wave equation, where K is a vector which has same orientation 4.25) and magnitude throughout the space. Similarly, Substituting from Eqn (4.35) into Eqn (4.31) we get vK= - ' peK Flx. y. 2, di= | Flx. y.z)eJ"d (4.26) (4.36) Since K is constant over the space its space, derivatives are zero, therefore vK= 0. It implies that F(x, y, z)el + C (4.27) jw opeK =0 (4.37) F.y,2.) jw +C (4.28) Since u and e for a medium cannot be zero, either o should be zero or K should where C is a constant function of time. Since we are interested here in only time be zero. If o=0, the field E represents a DC field or an electrostauc field. Here since we are analysing time varying fields, o #0and the only option we have is varying quantities, we take C = 0. From Eqns (4.25) and (4.28) we see that a time derivative is equivalent to multiplication by jw and a time integration is that K=0 making E E0. We, therefore, conclude that a uniform electric field in the three-dimensional cquivalent to division by ja. We, therefore, have varying, and since the magnetic field is related to space can not exist if it is time the electric field, even the magnetic field cannot exist. We also understand that jo (4.29) no time varying field can exist without varying as a function of space. 32 Having seen that the simplest uniform solution for the wave equation does not ar2Jo jw = -w (4.30) let us see whether the level of complexity that is the uniformity of the CXISt, next and so on. neld in a plane leads us anywhere! Without losing generality let us orient our r-axis. Since now COordinate system such that the electric field E aligns along the Replacing 8-/81 by -w in Eqns (4.21) and (4.22) we get the wave equations it could be constant in ry or for the time harmonic fields as Cae assuming the field to be constant in a plane, Iz or yz plane. field is constant in a plane E =-w peE (4.31) Let us first consider a case where the electric take an r-directed E field which field. Let us vH= -peH pendicular to the direction of the (4.32) onstant in the yz-plane ( it means the field varies only as a function of x). The Each of the Eqns (4.31) and (4.32), in fact, field can then be written as is a set of three each vector component of E and H. In other words, each equations, one 1o component of E ana Eo(r)eJo (4.38) satisfies the wave equalion. Since the V E(r, y, z, 1) = operator is a scalar operator, (4.33)  niform Plane Wave 149 Electromagnetic Waues , =0 (4 48) 148 is a conslant function of Eqn (4,11). Since Enlz) z, 9E(2)/8z 40 and consequently Let us substitute E in one ofthe Maxwell's cquations, exists, E also cxists. 3H H,an H, note that in the above discussion, although we have taken x-oriented VxE a -joH (4.39 One can eld constant in xy-plane, the arguments are valid for any plane which contains the field vector AUmmarize the above discussion as follows dx ay -jouH (4.40) time varying clecric or magnetic field which in uniforin A in the three E E, E dimensional space, cannot exist. = 0 and a/3z 0. Punh. her A túme varying held which is conslant in a plane perpendicular to the field Since E is not a function of y and z, we have 3/3y hence becomes also cannot cxist. E, 0. Eqn (4.40) direction, E has onlyx-component so E, = = can cxist is the field which The simplest form of held which in a planc containing the field vector, and consequently has spatial variation isconstant 0 0 -jouH (441) along the dircction perpendicular to the constant field plane Eplx) 0 0 UNIFORM PLANE WAVE The determinant in Egn (4.41) is identically zero giving 44 As discusscd above, the time varying fields which can exist in an unbound, -jwuH =0 (4.42) are constant in a plane containing the field vector and homogencous medium, field H has to be zero. Since no time to the plane. This phenomenon is then called the Again, except when =0, the magnetic w