 We have come to the end of this course and last time we were talking about radiation theory. So, we will bring that to some conclusion today and then we will sort of summarize whatever we have learnt in this course. So, if you recall this course has been primarily for understanding the physical implications and the mathematical importance of electromagnetic theory. And we have introduced the concept of a vector and a scalar potential. So, last time what we are doing was to talk about a localized oscillating source and we had seen that the these are the general expression for the vector potential and the scalar potential corresponding to a current distribution and the and a charge distribution. And so, what we said is that the if you have oscillating source we have essentially fields going as e to the power i omega t and we divide our region of interest into three. The first one is what we call as the near field where the dimensions of the source which is given by d here is much less than the distance from the source where we make the observation which is still less than the wave length of the radiating field. There is an intermediate zone where we do not do much because it requires lot more rigorous solutions where this r is of the order of lambda and our main interest as we said is on the radius and zone where the dimensions of the source is much less than the wave length which in turn is much less than the distances at which we make our observation. And we have seen that the near field is quasi stationary this statement is understood in the following way that if you look at the field other than for the time variation like e to the power i omega t the solutions will be the same as what you would get in case of static sources or steady currents. So, this is our near field and as we said we are interested primarily in the radiation field for which d is much less than lambda which in turn is much less than r. So, in other words r is the largest scale in the problem. .. Now, let us say see we have been using vector x to represent x y z and let the unit vector corresponding to that be n and the magnitude is x which we will also indicate as r occasionally that is the magnitude of the vector x will be taken as r. Now, if you do that then x minus x prime modulus which we have seen can be written as x square which is r square plus x prime square remember prime is the variable corresponding to the source and minus two times r times vector n unit vector n multiplied with x prime and that raise to the power half. So, if you expand this out we had seen that this gave me r minus sorry this is not a vector r minus the unit vector n dotted with x prime. I also require now this will as we know will appear as the exponent of the exponential function and I have a 1 over x minus x prime there which is essentially the same quantity, but with the raise to the power minus half and what we had seen is that this is given by 1 over r plus n dot x prime divided by r square plus there are other terms like this, but we will not really be interested in these terms these are higher order terms which are there, but just for to indicate. So, we write this term now last time we had sort of seen what was the electric dipole approximation. So, just to recall for you what we did is to put this quantity as in the exponent. So, I have e to the power x minus x prime divided by x minus x prime. So, this was written as equal to or of the e to the power i k r by r then I have a 1 plus 1 by r minus i k times n dot x n dot x prime and plus terms which we have said we neglect. Now, what we did last time is to ignore this term and replace e to the power x minus x prime by x minus x prime as e to the power i k r by r and that gave us the dipole approximation to the problem. The terms that we have here we will take this up and these we will see give us two things one is what is called as the magnetic dipole approximation and the next one which we will not really have much of a time to do is essentially electric quadrupole approximation. So, let us look at how does it go. So, the radiation field that I have got is. So, first I will look at how this works out. So, my radiation field A the vector potential A of r or x I use it interchangeably mu 0 over 4 pi e to the power i k r by r this is just the first approximation integral j of x prime d q x prime. This is the lowest order approximation that we have got. Now, so what we need to do is this that notice one thing that if I now calculate del dot of x i times j. Now, we know that this is gradient of the scalar dotted with the vector. So, this is del x i as to why I am doing it will become obvious in a second del x i dotted with j vector j. So, this is of course, a vector x i is the ith component of the vector x plus x i times del dot of j. Now, by equation of continuity the second term I have got del dot of j. So, therefore, this is same as d rho by d t and since I have said that the variation with respect to time is e to the power minus i omega t. So, I get this as del gradient of x i dotted with j minus minus plus i omega x i times rho. So, notice this is gradient of x i. So, since this is gradient of x i. So, this is essentially i depending upon what i is this is just a unit vector in whichever direction i is and when it is dotted with j I simply get a j i plus i omega x i times rho. So, the reason why I am doing this is that if I look at this integral I have integral j of x prime d cube x prime. Now, what I do is let us look at the ith component of this. So, this ith component of this this is a d cube x. So, I will write this down as equal to the del dot of x i prime j d cube x prime minus i omega integral rho. So, basically what I am doing is I am replacing for this j i del dot of x i j minus i omega x i rho. So, this is rho x i prime d cube x prime. Now, so this is this quantity of course, as expected by divergence theorem can be converted into a surface term and since I know my current is confined within a small region the as if I take this surface go to infinity this term will drop out and what I will be left with minus i omega rho x i prime d cube x prime. If you recall this is nothing, but the definition of the dipole moment actually I have component on the dipole moment. So, this is what I have got from this j x prime integration. So, therefore, if I substitute this into the expression for a I get a of x is equal to mu 0 by 4 pi as before e to the power i k r by r which was there times minus i omega p. So, minus i omega p. So, this is what I get and if we could sort of use that omega by c is equal to k and this can be written as 1 over 4 pi epsilon 0 times i k by c well basically omega is being written as c k and then mu 0 epsilon 0 is 1 over c square. So, this just works out then I have got e to the power i k r by r times b. Now, this is incidentally just the dipole approximation. Having got the vector potential a I calculate the magnetic field b by simply working out what is del cross of this quantity. So, which is equal to minus 1 over 4 pi epsilon 0 skip this terms i k by c which are common which is constant times del cross of p del cross of this quantity times p, but p is a dipole moment which is a fixed factor. So, therefore, what I get is gradient of e to the power i k r over r cross p. Now, this gradient is easily calculated. So, it is minus 1 over 4 pi epsilon 0 i k by c. So, calculate this gradient e to the power i k r gives you a i k then I have got I have to differentiate 1 over r. So, that I get 1 over r square actually minus 1 over r square. So, let me just take out e to the power i k r by r outside this times a unit vector r cross p this unit vector came from the gradient. So, this is my expression for the magnetic field b and the corresponding edge field which is of course, simply obtained by dividing b by mu 0 and realizing that 1 over mu 0 epsilon 0 is c square. So, what I get is c k square by 4 pi e to the power i k r by r times 1 plus I have just changed the order of these two terms to take care of this minus sign i by k r times r cross p. So, this is the expression for the magnetic field. Now, I can obtain the electric field by realizing that electric field is nothing but epsilon d e by d t sorry the epsilon d e by d t is equal to del cross h. So, I can calculate the del cross h of this quantity and d by d t is minus i omega. So, therefore, using this I can get an expression for the electric field as well. So, that is a rather long expression, but here it is on the screen this is there is really nothing that you need to do what we realize is e is basically i c mu by k del cross of h and this vector. So, this is little complicated because it has a scalar it has vectors only p is a constant vector. So, you have to have the gradient of the scalar cross and things like that. So, if you do that you get a huge expression like this. Now, at this stage let us let us look at for example, what is the radiation field like. Now, I have this is the electric field and I am sort of looking at. So, these are the final expressions for the electric and the magnetic field. I will just put this screen for some time. So, that you can have a look at how the algebra works out I have actually sort of in detail worked out the whole thing. So, look at that and so the final expression as we have seen is given by let me write it down again is h equal to c k square by 4 pi e to the power i k r by r times 1 plus i by k r times unit vector r cross p and the electric field expression finally works out to which I have not worked out here, but I have shown it how to do it e to the power i k r by r 4 pi epsilon 0 r. If you do it correctly it will become k square unit vector r cross p cross r minus i k by r 1 plus i by k r into 3 times r dot p actually it is a unit vector r dot p minus r dot p times unit vector r minus p. So, these are the two expressions that I have. Now, let us look at what are the points that are coming out of this expression. Firstly you notice that the magnetic field is transverse to the direction of r perpendicular to the direction of r, but electric field has a longitudinal components because this is r cross r cross p or r cross p cross r. Both these fields e and b at large distances very large r go as 1 over r this is that 1 times this this is the major field expression and this is also true of the electric field here there is a i k by r. We at small distances however the magnetic field is dominated by 1 over r square which is this term whereas, the electric field is dominated by r cube term and this is this term there is a 1 over r there i k by r there and i by k r there. So, this is 1 over r cube this is what dominates the electric field. So, therefore, we write down for the radiation zone where r is very large I neglect this term which gives me a rather simple expression for the magnetic field. So, in the radiation zone I get magnetic field h is equal to C k square by 4 pi e to the power i k r by r times unit vector r cross p and if you look at the same approximation you will find that the electric field can be written in terms of the magnetic field as C mu h cross r. So, this also tells you that the electric field is perpendicular to the direction of the magnetic field and the magnetic field is of course, perpendicular to direction of the unit vector r. The once we have written these down this is incidentally true only in the radiation approximation. Now, that is 1 over r is the dominating factor for both electric field and the magnetic field. The what we are interested in is how does the emitted power go. If you remember the the emitted power is given by your average value of the .pointing vector. So, since we are using complex quantities. So, it is half of real part of e cross h star and by plugging in these two expression remember that e is also written in terms of h. So, therefore, e cross h star I will have 2 h there. So, I have to simply square this. So, if you plug this in you are going to get omega k C mu by 8 pi this times real part of well e is given by h cross r and of course, cross h star. So, this quantity is well I have made probably some slight error there what I have done is to write these numbers down earlier. So, does matter really. So, this clearly because we have said the magnetic field is perpendicular direction of r. So, therefore, this would be in the direction of r itself. So, this quantity will be r omega k C mu by 8 pi times h square and that is equal to well r I have got a here I have got C k square there. So, I write this down. So, omega k whole square C mu divided by there was a 4 pi there there is a 8 pi there. So, I got 32 pi square 1 over r and 1 over r gives me 1 over r square and of course, r cross p whole square which is there in the structure of the magnetic field. So, if you look at this you find that this is given by unidirector r omega k square C mu over 32 pi square 1 over r square r cross p. So, therefore, this gives me p times p square because this is square there times sin square theta. So, notice that the pointing vector is proportional to sin square theta and what we can do is we can find out how much is the power flowing through a unit solid angle in the direction of theta phi. So, which is like doing simply find out how much d p by d omega this is the power is given by well r square r dot the pointing vector and if you calculate this you get omega raise to the power 4 times mu divided by 32 pi square C times p square sin square theta and if you want to calculate the total power that is emitted you can simply integrate this over theta remember that there is no azimuthal dependence. So, you get a 2 pi there you have a sin square theta sin theta d theta which you can integrate out sin cube theta d theta and. So, you get omega 4th mu divided by 12 pi C times p square this is sin square theta has been integrated out. So, if you look at now these 2 structures you notice that this is the vector p the dipole moment vector p and the way the power vector s is being radiated that is symmetric with respect to the direction of p. And this is where we are showing that at an angle theta the you know there is a solid angle d omega and the length of the you know if you join from here to the circumference then that would give you the magnitude of s. So, this is the way the radiation pattern looks like for a dipole it is perfectly symmetric about the direction of p. So, that was dipole approximation where we took the simplest possible thing. So, let us look at the next approximation the in the next approximation there are basically 2 terms. So, this is I recall back A x is given by this expression and it is e to the power i k x minus x prime by x minus x prime I had shown to leading order this gives you these terms. And so, as a result what we did till now is to take only ignore this factor and take only e to the power i k r by r and that gave us the electric dipole approximation. So, that was the dipole radiation field now what we now want to do is to go over and take account of this term I am not writing the full A x now I am only writing the term in A which corresponds to this. So, this is A x equal to mu 0 by 4 pi e to the power i k r by r these are all there and I have 1 over r minus i k well I have once in a while the unit vector n is same as the magnitude of x I have occasionally I have been writing it as unit vector r small r. So, this is the term which is in the my next order of approximation the first term I have not written down that is the electric dipole term. So, let us let us write it down. So, I got A of x is equal to mu 0 by 4 pi e to the power i k r by r integral j x prime 1 over r minus i k n dot x prime d cube x prime I will now do a bit of a complicated algebra because this is a rather complicated expression, but nevertheless let us try to work this out. Notice the this term does not come into the integration at all because it is independent of x prime this is equal to mu 0 by 4 pi e to the power i k r by r 1 over r minus i k integral of j of x prime n dot x prime d cube x prime. Notice this left hand side is a vector and the vector character comes from because this is scalar. So, vector character comes from j. So, let us look at the component wise what what do we get from this integration in other words let me look at let us say the you know I mean what do I get for example, from the i s component or things like that. So, let us write down this expression here well I will for the moment skip this constant term. So, let me just plug it in here like this and n dot x prime what I can do is this that let me let me evaluate instead of n dot x prime let me take out a 1 over r there this is this part. So, that let me call this star. So, this is this star there on the screen you can see the full expression 1 over r integral j of x prime x dot x prime d cube x prime remind you once again that vector x is the position at which we are evaluating the vector potential and x prime is of course, an integration variable. So, what we do is this that since we are interested in this part only this part of the expression is given by since it is x dot x prime which is summation over 3 components. So, let us call this sum over j equal to 1 to 3 integral j of x prime well x j times x prime of j d cube x prime. So, this is this is what this part looks like this is what we are interested in the remaining part at this moment. So, far as the calculation of a of x is concerned the remaining part remains unaltered. So, therefore, we will need it while calculating the magnetic field which will bring them at that time. So, let me then do this try to evaluate this quantity. So, let us look at what we are trying to do. So, we are trying to evaluate sum over j equal to 1 to 3 integral j of x prime remember j is a source quantity x j x j prime d cube x prime. Now, before I do this I want to do and subsidiary algebra we know that if we took the divergence of the whole thing. Suppose I wrote del dot of this quantity which is written there x j x j prime times capital J d cube x prime using divergence theorem and extending the surface to infinity since j is localized I get this quantity to give me 0. Now, what we are trying to say is this if this is 0 this is del dot of a scalar times a vector. So, let us look at what it gives me. So, this tells me gradient of a scalar. So, let us put gradient of x j x j prime dotted with j the vector plus the scalar times x j x j prime times del dot of j. So, this quantity is equal to 0. Now, let us look at what does this give me. Now, remember that this is gradient of a product of two things. So, therefore, this gradient of x j x j prime is given by well I did do a bit of a mess up, but let me sort of redo this what I want to do is the following this is what I want to calculate, but instead what I want to say is this that let me just pull it aside let me pull it aside. So, this is the quantity which we are interested in calculating j equal to 1 to 3 integral j of x prime x j x j prime d cube x prime. So, what I am saying is this that consider divergence of x i prime x j prime j this is equal to 0 for because this is divergence means I can make it a surface integral and j will vanish on the surface. And this del dot I know can be written as gradient of gradient of x i prime x j prime dotted with j plus x i prime x j prime del dot of j d cube x prime is equal to 0. Now, this gradient that I have got I got gradient of x i prime x j prime is equal to a dot with j. So, let us look at what does it give me. Firstly, I can take x j prime this is just a chain rule. So, I can take x j prime out I get gradient of x i prime dotted with j which is nothing but j i. So, this is x j prime j i and like that when I take the x i prime out gradient of x j prime dotted with j gives me j j. So, I get x i prime j j. So, if I now substitute it here I get integral this is x i prime what I have got is this I get x i prime or rather x j prime j j x j prime j i plus x i prime j j. And if you remember that I had this term i omega rho prime the second term which I was not bothered about x i prime this is this comes from the fact that I have got del dot of j which is equal to minus d rho by d t and which goes as omega rho prime. So, therefore, I get i omega rho prime x i prime x j prime d cube x prime is equal to 0. Now, this is the expression which I have to evaluate j x prime x j x j prime sum over j equal to 1 to 3. Now, once I have said that this quantity is equal to 0 what I can do is to rewrite sum over j equal to 1 to 3 integral suppose I am talking about the i th component. So, I have got x i j i x j x j prime this is what I want to talk about and that is equal to sum over j equal to 1 to 3. Now, I bring the x j out and I have got j i x j prime d cube x prime. Now, this is where I use this identity that I had obtained. So, I had said x j prime j i and so this is what I have got x j prime j i. So, I will write this as negative of these two terms which will go to the other side. So, this quantity will then be equal to minus sum over j equal to 1 to 3 x j and integral of j i rather j j x i prime plus i omega rho prime x i prime x j prime d cube x prime. And this is the one which I will replace with that what I now do is this the this is this is equal to this. So, I am going to do take half of this expression and half of that expression. So, that I write it in a slightly symmetric fashion. So, it is half of sum over j equal to 1 to 3 I get x j integral of j i x j prime minus j j sorry x i prime minus i omega rho minus because there is overall minus sign x i prime x j prime sorry there is a prime which was missing here d cube x prime. So, my expression is this and if I now want to integrate this out if you look at the screen this is what I had written it down. Now, notice this term that I have got here j i x j prime minus j j x i prime. You notice this is nothing but the k th component of x prime cross j and of course, I need a epsilon i j k because I do not know what whether i j k what is the natural order there. So, this will be minus epsilon i j k x prime cross j component k and the other one I have not done anything about it. Now, if you now plug this sum over j equal to 1 to 3 x j and bring it inside and you notice that there is a epsilon i j k there with the minus sign there minus sign I can bring it out the this is nothing but x cross x prime cross j i component because this is j th component this is k th component with epsilon i j k. So, therefore, it is x cross x prime cross j i th component and the other term other than this minus sign which is there outside now I have taken down this is remained exactly the same. So, this is what we have proved so far this is what we have proved so far because this is what I had and I just now showed that this sum over j equal to 1 to 3 this is what we have proved there and. So, therefore, if you write this j vector j as i j i x etcetera then this is what you get this was i th component. So, therefore, this is a vector relationship now it is this term it is this term which is my magnetic dipole term and let us see why. So, this is what we have proved now notice that the magnetic moment m which we know is the current times the area vector and the area vector can be written as half of loop integral of r cross d l and if I take i times d l remember i is a current. So, i times d l is goes to if the current exists in a medium rather than in a wire then i times d l is essentially j d v because i is j dot d s and this is d l. So, therefore, this definition of magnetic moment which is given by this can also be written as equal to half of r cross j d v. So, once I have done that that this is r cross or x cross j. So, you notice this that what we are trying to say is here I got x cross x prime cross j and I have the integration of course the remember the integration variables could be prime. So, this term gives me minus x cross m the this is the magnetic moment term and this is the term which will give me the quadrupole moment term, but that we will see little later. So, therefore, if I want to now simply restrict myself to the magnetic field to the magnetic dipole then I need to only worry about this term. I bring back the constants that were there in the problem mu 0 by 4 pi e to the power i k r by r 1 over r minus i k and integral j x prime n dot x prime d cube x prime and I write this because I have shown this integral to be given by n cross m. So, this is the complete expression for the magnetic field for the magnetic vector potential. Now, from this my job is exactly the same as before. So, what I do is I write down calculate the del cross of this a m a m is magnetic dipole term I have neglected all other terms. So, I calculate del cross m this is bit of an algebra, but let us go through this the I have got mu 0 by 4 pi this numbers n cross m I want del cross of this quantity. So, first thing that I do is realize that this is all scalar i k I will take out because it does not really depend on anything. So, this is a scalar. So, I have got instead of dealing with n cross m let me divide it by r and write it as r cross m and because I have divided by r and there is another r there I have written as gradient of e to the power i k r i k by r square this r and that extra 1 over r that I have pulled out minus 1 over r cube into r cross m this is what I have got. So, I got gradient of this quantity times this vector because I am calculating del cross m and. So, therefore, well actually it is del of this quantity cross r cross m there is a cross missing here and of course, this quantity which is a scalar itself times del cross of r cross m. So, these are the two terms there is a cross missing here this gradient can be easily calculated. So, you can see it that e to the power i k r take for example, this term when you take the gradient of e to the power i k r I get i k. So, this is i k e to the power i k r and of course, the same thing plus I keep e to the power i k r differentiate this gives me minus 2 by r cube because it is 1 over r square this gives me minus 3 by r fourth, but there is a minus here. So, it is plus 3 by r fourth times this n cross r cross m because it should be along the unit vector and this other term I have not distributed. So, this is this is the full expression, but then I should be able to simplify them just add them up properly because there are similar terms coming out of here there is a 1 over r cube term here there is 1 over r cube term here add them up properly this gives you a 3. So, you notice that I would get this. So, here I had got i k into i k is minus k square. So, I got minus k square by r minus 3 i k by r square and plus 3 by r cube which came from there from this term and this term which I said is also there. Now, what I do is this it is this second term which I am simplifying del cross r cross m is del dot m r this is the standard vector algebra when you have del cross of 2 vectors you get del dot the second vector r minus del dot r m etcetera. Notice that m is a constant vector. So, this term is 0 and similarly this term is also 0 because it is gradient of m. So, I am left with these 2 del dot of r is known to be equal to 3 and this is m dot del r you know what does it mean it means you have to take the gradient of each 1 of the components and then dot it with the i times 1 component plus j times the other component. So, that gives you just m because gradient of x for example, is 1. So, this gives me minus 3 m this gives you plus m. So, I get a minus 2 m. So, instead of del cross r cross m put in a minus 2 m there and simplify these things further. So, if you do that what you find is like this the this is your h m and we have written this down this is already n cross n cross m simplify this and we have rewritten what these things are n what I have done is this I have simplified n cross n cross m using a cross b cross c formula namely b a dot c minus c a dot b and I have written it in this fashion and this is that term which I just now showed to be equal to minus 2 m. If you plug in all these things you get e to the power i k r by 4 pi k square by r n cross m cross n plus this thing. The reason I am not really concentrating so much on these is I am mostly interested in the radiation zone where only 1 over r term becomes important. So, therefore, in the radiation approximation I am only interested in this term which goes as e to the power i k r by 4 pi k square by r n cross m cross n this is in the radiation zone. The magnet the electric field can be found by directly calculating del cross h or by showing that this is equal to minus d a by d t, but but basically the radiation zone in the radiation zone this is the expression for the electric field these are the only the magnetic dipole term. Now, we do exactly what we did earlier namely we calculate the pointing vector and calculate the total radiated power. So, if I do that I find that this is the this is the radiation zone and this is the electric field in the radiation zone and the corresponding power expression works out to mu 0 c k to the power 4th divided by 12 pi. Notice this omega to the power 4 dependence is very very common because k to the power 4th is also proportional to the omega to the power 4th this times m square and the I have not drawn the power pattern, but power pattern is very similar to the electric dipole pattern. I will not be talking about the electric quadruple moment and that comes from this term that comes from this term and you can see I have not yet worked it out that the correspondingly the magnetic field expression can be written like this where q i j is the quadruple moment tension and is given like this. The quadruple radiation pattern shows the maximum at angle equal to pi by 4 and this is the way the quadruple moment looks like this is something which you will do well to take it as an exercise. So, this is basically what we want to talk about in the radiation basically talking about how the power you know is transported the radiation zone is 1 over r and how much is the power which is sent from the transmitting antenna and which is later on received by a receiving antenna.