 Welcome to the 14th lecture in the course Engineering Electromagnetics. The topics for discussion in this lecture are we continue with the concepts of wave propagation and introduce the important topics of polarization and pointing vector. Taking up the first topic that of polarization first define what we mean by polarization. What is meant by polarization? Polarization refers to the behavior of the electric field, electric field vector as a function of time at a fixed point in space. And when we consider the behavior of the electric field vector as a function of time at a fixed point in space a number of interesting possibilities arise. The first possibility we call as linear polarization as the name suggests in this case what we expect or what happens is that the electric field vector or the tip of the electric field vector moves along a straight line and that case is known as the case of linear polarization. To explain this further we consider a wave propagating in the z direction. Let us consider uniform plane wave propagating in the positive z direction. This is the departure from the previous lecture we are considering propagation in the positive z direction. One can draw the coordinate system let us say like this. So x, y and z and the wave is propagating in the positive z direction. As a result recalling the properties of the uniform plane waves such a wave will have no z field components no field components along the z direction that is the direction of propagation. And it will have field components only in the x y plane. And therefore, since we are considering the electric field vector the behavior of the electric field vector the electric field vector can have an either an x component or a y component. Then therefore, there arise a number of possibilities some of which would correspond to the case of linear polarization. For example, if we have the time varying x component of the electric field vector 0 then only the time varying y component of the electric field vector will exist. And as the vector amplitude varies sinusoidally with time it will have an orientation like this or like this and then depending on the phase it will have a different value of the magnitude. But all along the tip of the electric field vector will move along the y axis positive y axis or negative y axis. And therefore, we see that the tip of the electric field vector will always move along the straight line in this case when only E y is present. Alternatively one could have the situation that E y is 0 so that only E x is present. And then also one can make out that the tip of the electric field vector will move along the x direction and depending on the phase it will have different orientation but always along the positive x axis or along the negative x axis. So once again the tip of the electric field vector will move along a straight line as a function of time let us say at z equal to 0 at a fixed point in space. The third possibility could be that both E x and E y are present. But they are in time phase. Then if we consider the x y plane since E x and E y are in time phase and they are varying sinusoidally with time the electric field vector overall electric field vector will have a movement along a straight line like this making an angle theta with the x axis where theta is the tan inverse of E y by E x. So we could have three possibilities either only one of the components is present or both are present but they are in time phase so that they increase or decrease exactly in synchronism and the overall electric field vector therefore does not change direction. The tip of the electric field vector moves along a straight line which the orientation of which straight line is fixed at a particular location for all time and therefore that is the case of linear polarization. A uniform plane wave with satisfying any of these three requirements or any of these three conditions will be called a linearly polarized plane wave. Now it is easy to see that this balance between E x and E y field components which is required for linear polarization can be disturbed very easily and therefore in general we can have more complicated situations. Going on to the next possibility again a fairly specialized situation and more interesting is the case of circular polarization. How do we introduce circular polarization? Let us say that we have the x component of the electric field varying with time and of course z is equal to some amplitude factor E naught times sin of omega t minus beta z. We are continuing with the assumption that the wave is propagating in the positive z direction. Hence a factor like this appears and it is varying sinusoidally with time. Let us say that the corresponding y component exists and has a form which is having the same magnitude E naught, but is cosine of omega t minus beta z. That is it has a phase difference with respect to the x component of 90 degrees. In particular it is ahead in phase by 90 degrees. A very special situation is being set up. There are two components, they are of equal magnitude and they have a phase difference of 90 degrees. Now, let us say that we focus attention on these field components at a fixed point in space say z equal to 0 being the most convenient location. So, at z equal to 0 we have E x equal to E naught sin omega t and E y equal to E naught sin omega t and E y equal to cosine omega t. And we want to consider the time behavior of the overall electric field vector. And therefore, let us put down the values of the x and the y components at some different time instance say omega t equal to 0 pi by 4 pi by 2 3 pi by 4 and pi at omega t equal to 0 the x component is 0, but the y component is maximum it is E naught. At pi by 4 they are both of equal value E naught by root 2. At pi by 2 E x is maximum, but E y is 0. And similarly, here we have this as E naught by root 2 this is minus E naught by root 2 and here it is 0 and minus E naught. So, that in the x y plane now a plot can be made to obtain the overall electric field vector this is x and this is y and one can make out that the tip of the electric field vector is going to move on a along the circle. So, we draw the circle first and then plot these various points this is the omega t equal to 0 point this is the omega t equal to pi by 4 point this is the pi by 2 point 3 pi by 4 point and the omega t equal to pi point. One could have continued with increasing values of omega t and seen that the electric field vector tip of the overall electric field vector is going to move on a circle like this. The overall magnitude can be estimated from here and it will remain E naught. So, E naught is the radius of the circle and as a function of time the electric field vector tip of the electric field vector is rotating around the circular path at a fixed point in space. Therefore, this is clearly a case of circular polarization, but it is achieved under very special circumstances the magnitudes of the two components which exist are equal and the phase difference is 90 degrees. One could write this in a more compact way by drawing upon the phasor notation for example, this can be written E x can be written as something which is like minus j E naught will understand the basis in a minute and E y can be written as simply E naught E y goes to this kind of notation. So, that the overall electric field in phasor notation can be written as minus j E naught x cap plus E naught y cap and now if you want to go back to the actual time varying components we multiply by e to the power j omega t and take the real part and we see that we will get back E x equal to E naught sin omega t and E y equal to E naught cosine omega t. One could have seen from here that it is a case of circular polarization by processing it in the following manner we can write E x squared plus E y squared is equal to E naught squared which is the equation of a circle centered at the origin with radius E naught. Now, once we realize the essence of the circular polarization it is quite easy to see that many other combinations satisfying this special requirement can be thought of all we need is orthogonal field components specially orthogonal field components which have a phase difference of pi by 2 or 90 degrees. For example, E could be equal to plus j x cap plus y cap E naught or E could be x cap plus minus j y cap E naught in phasor notation all these cases would be corresponding to circular polarization. There will be one difference in these various possibilities of circular polarization and that is the sense of rotation of the electric field vector which is also called the sense of polarization. For example, the illustration that we took up first what is the direction in which the electric field vector will rotate as time advances it will be the clockwise direction. What is the bottom most point corresponds to omega t equal to? Omega t equal to pi. So, well we have written down the component values this is the plot that comes and in 2 pi radians the entire circle is completed. Now, how do we define the sense of polarization or how do we specify it? We rotate let us say right handed screw in the direction in which the electric field vector is rotating and then make out in which direction the screw will advance. Now, when a right handed screw is rotated in the clockwise fashion it will advance into the plane of the blackboard. Which direction is negative z? However, we specified that the wave is propagating in the positive z direction. This direction is not corresponding to that. On the other hand if we rotated a left handed screw in the same fashion that would advance in the direction of propagation of the wave and therefore, this example that we have taken is that of a left circularly polarized wave or left handed circularly polarized wave. It is easy to see that if we change the sign of one of these 2 components it would become a right circularly polarized wave and for these other various possibilities also one can see easily what is the sense of polarization. So, we would have either a left circularly polarized wave or a right circularly polarized wave. Now, as we have already mentioned the circular polarization is a very specific case and requires equal magnitudes and a phase difference of 90 degrees. And if we disturb any of these 2 conditions one would get in general the case of elliptical polarization. For example let us disturb the magnitudes and say that in phasor notation the total electric field vector reach as follows. It is E equal to say x cap A plus j y cap B that is we retain the phase difference of 90 degrees, but the magnitudes are made unequal. Now, we see that E x as a function of time will be A cosine omega t E y as a function of time will be minus B sin omega t as can be made out by multiplying by e to the power j omega t and taking the real part. So, that if we process this these 2 components in this manner which led us to circular polarization conclusion we would get E x squared by A squared plus E y squared by A squared plus E y squared by B squared equal to y implying that now this is a case of elliptical polarization and depending on the values of A and B assuming that A is the greater one here the tip of the electric field vector will move on an elliptical path like this and since y is ahead of x this again will be the tip of the electric field vector will rotate in the clockwise direction and this again will be a case of left circularly polarized way. The ellipse axis major axis and the minor axis are A and B and for this example the major axis is oriented along the x axis the minor axis along the y axis and the axial ratio ratio of the major axis upon minor axis is A by B. This is the axial ratio this is the sense of polarization and there is a third quantity which needs to be put down to specify a particular elliptical polarization that is the orientation of the ellipse. For example what is the orientation of the major axis and what is the orientation of the minor axis. One can visualize more complicated possibilities for example the phase difference could be anything not just 90 degrees and the result again will be elliptically polarized uniform plane wave. Of course one can see that this is the most general case and by putting in special conditions it will result in the simpler cases that we have considered earlier. For example if A is equal to B that will be a circularly polarized wave if either A or B are 0 it will be a linearly polarized wave and if J is absent that is if the two components are in phase then again it will be a case of linear polarization. So this is the most general situation or let us put it down more specifically the most general situation because this is still assuming a phase difference of 90 degrees. In general one can write E as say X cap A plus Y cap E to the power J theta and the case that we considered earlier is the case for theta equal to pi by 2 which will be the most general situation from which all other special cases that we considered can be derived or obtained. Before we consider how to handle that let me specify how the elliptical polarization is specified three quantities need to be specified. One is the sense of polarization according to the test that we described the second is the axial ratio ratio of the major axis to the minor axis and the orientation of the major axis or the minor axis. Once they are specific once one of the axial orientation is specified the other one is automatically found out because the axis are going to be at 90 degrees with respect to each other for an ellipse. So given a general case of elliptical polarization how are we going to obtain these three defining or specifying pieces of information that will be a problem. And just coming to that we will be able to find it and in any case when the electric field vector is along the major axis its magnitude is A. Let me just complete this I think your question will be automatically answered. So considering this general case of elliptical polarization we obtain E x equal to A cosine omega t and E y equal to B cosine omega t plus theta. Then the magnitude squared of the total electric field is going to be E x squared plus E y squared or one could write as a function of time what will be the magnitude. As we have seen the magnitude is going to change with time it is not going to remain constant. Now to obtain the major axis and minor axis information we recognize that these are extreme points maxima or minima and therefore the time derivative here d by d t of E squared is going to be 0 for these instance of time. And therefore this will give us the information about omega t value at these instance of time. And once we have that substituting it back here we will obtain the magnitude of the electric field for these instance of time. And also the orientation with respect to a certain cartesian certain coordinate system. So this way given any general situation it should be possible for us to find out these three defining properties of the given elliptical polarization. The signals that are transmitted from the satellite or that are transmitted to the satellite have circular polarization. The television broadcast signals are linearly polarized just to give you very broadly where these different polarizations are utilized. If you have any questions we can discuss those or else we go on to the next. How these are local maxima extreme points here the first derivative with respect to time must be 0 yes please. The magnitude is available here as a function of omega t. See the magnitude let us say omega t for the major axis is equal to m 1 or m j and for the minor axis it is m n. Now when you substitute back these values here you will get the magnitude of the electric field for the maximum value and the minimum value. Naturally when you find the actual ratio that maximum magnitude and the minimum magnitude is automatically coming into picture. You will be able to find this when you find the magnitudes separately. The steps in finding the actual ratio will be to find the maximum magnitude corresponding to this value of omega t. Find the minimum value of the magnitude corresponding to this value of omega t then take the ratio. Then you will be in a position to specify the actual ratio. The sequence once you understand that there should be no problem. So I suppose we can go on to the next topic which also is very important and that is of point in vector. Let us see if we have time at the end of the lecture then we will consider. We go on to the second topic of the day and that is the point in vector. When a wave propagates from a receiver to a transmitter it will do something at the receiver. For example some information will be transferred or some relay could be actuated. For example we are dealing with remote control of vehicles and therefore it is quite clear that some amount of energy transfer is taking place when a wave is travelling from the transmitter to the receiver. And therefore there is some power flow associated with any kind of thing which power is being supplied by the transmitter and a part of it or whole of it depending on the situation is utilized by the receiver. It is natural to visualize that this power flow should be related to the amplitude of the field vectors, electric field and the magnetic field. And it will be of importance to consider what is the actual quantitative relation between the amplitudes of these field vectors and the power flow. So that is what we consider next and that is given by the pointing vector. We start with the what is called the first equation of Maxwell del cross h equal to del t by del t plus j which we rewrite as j equal to del cross h minus epsilon del e by del t. The way we have written it we are assuming a homogenous isotropic medium but we are keeping the possibility that there is some loss in the medium because the j term is included which would be there for conducting medium. The units of various quantities in this equation are amperes per meter square that can be seen. Now we process this in the following way we take the dot product of all the terms with the electric field vector. So that we write e dot j equal to e dot del cross h minus epsilon e dot del e by del t. Why this is being done will be clear as we proceed. Now we write a utilize a vector identity which reads as del dot a cross c is equal to c dot del cross a minus a dot del cross c and we replace a by e and we replace a by e and we replace a by c by h. So that we get del dot e cross h equal to h dot del cross e minus e dot del cross h. So that we can substitute for e dot del cross h from this relation. When we do that we get h dot del cross e minus del dot e cross h minus epsilon e dot del e by del t. And then we utilize Maxwell's second equation which reads as del cross c. Del cross e equal to minus del b by del t and under the assumption of a homogenous isotropic medium it is minus mu del h by del t which substitution can be made for del cross e. And therefore, we get minus mu h dot del h by del t. And then we get minus mu h dot del h by del t and we put this term a little earlier minus epsilon e dot del e by del t minus del dot e cross h. Now the first two terms can be rewritten as minus mu by 2 del by del t as minus mu by 2 del by del t of h square. Similarly, minus epsilon by 2 del by del t of e square and the third term remains as it is. Now to be able to provide an interpretation to these various terms we consider a certain volume and close by a surface. And we carry out the volume integral of all these terms over that volume. We integrate all these three terms the entire equation over a certain volume. So, that we have e dot j d v equal to minus del by del t integral of half mu h square plus half epsilon e square d v minus the volume integral of the third term. And then using the diverges theorem we say that the volume integral of the divergence is equal to the surface integral of the vector itself. And therefore, it becomes e cross h dot d a where s is the surface bounding this volume. This is as far as one could proceed mathematically. And now the task of interpretation starts. We consider the left hand side. What does it contain? It contains the volume integral of e dot j d v. Now j is equal to sigma e. Therefore, this is equal to sigma e square d v. Now if we consider the units of e and sigma we have the units coming out as volume integral volts per meter squared. And then sigma can be considered to be the reciprocal of the resistivity and therefore, it will be ohm meters. And therefore, these units are equivalent to watts per meter cube. Ohm could be written as v by i. So, that will make it clear that the units are watts per meter cube. What does therefore, this term represent? It represents the power loss in the medium over the volume over which the integration has been carried out. So, this term represents power loss or power dissipated in the medium. Next we consider the right hand side which has two terms. So, we do it term by term. What about the first term? It is minus del by del t integral of half mu h square plus half epsilon e square d v. Now we recognize that these terms are the stored energy densities energy stored per unit volume in the form of the magnetic field. And in the form of the electric field respectively. Therefore, the integral represents the energy stored in the volume that we have considered. The time derivative shows the time rate of change of the stored energy. And the negative sign shows the rate of decrease of the stored energy within this volume that we have considered. And therefore, as far as the interpretation of this first time is concerned, it is equal to the rate of decrease of stored energy. Now comes the second term. Let us look at the left hand side and the right hand side terms that we have already interpreted. Left hand side represents the power dissipated in the medium. There has to be some source supplying this dissipated power. What could be that source? That source could be the stored energy. As a result of power dissipation, the stored energy keeps decreasing. Plus there could be some power flow from the outside. Therefore, the second term is the power flowing in into the volume under consideration. Which is equal to minus E cross H dot d A. And therefore, E cross H dot d A integrated over a surface represents the total power out of a surface S. And therefore, the vector E cross H will give us the power flow per unit area. We had surmised that the power flow will be associated with the strength of the electric and the magnetic fields. And now we find that the power flow per unit area comes out as E cross H. And that takes us to what is called the pointing theorem. It says that the vector E cross H is the pointing vector. And it is a measure of power flow per unit area. Any point. And since it is a vector, it also gives us the direction of power flow. The direction of power flow is E cross H. The direction of power flow will be E cross H divided by its magnitude. So, that is a very important concept. The pointing vector, the pointing theorem says that the pointing vector is a measure of power flow per unit area at any point. And the direction of the pointing vector is the direction of power flow. We could have an isotropic media without any difficulty. And one could see that, one can see that this theorem is going to apply to that. Non-homogeneous media could also be considered. The some problem occurs when we have time varying media. Otherwise, this theorem holds good. Any other question at this point? Yeah, there will have to be some source supplying this power. Yes, actually that is what we have done. We have invoked the conservation of power and whatever power flows into this volume and whatever is the rate of decrease of the stored energy in this volume is the power dissipated. It is that power balance that we have utilized. Any other questions? Then we stop this lecture here. In this lecture, we have introduced the two important concepts. One is that of polarization and we have discussed the specific cases of linear polarization, circular polarization and elliptical polarization. Following that we introduced the concept of pointing vector and we gave the statement of pointing theorem. Thank you.