 In the previous lecture, we had started discussing the time dependent phenomena and I had enunciated what is known as Faraday's law. What Faraday had proposed based on certain experiments is that, whenever there is a relative motion between a magnet and a circuit or if the magnetic field strength is changing with time, that is if any of these effects results in the flux through a circuit changing, then it leads to an EMF in this circuit. And as we had seen EMF, we defined EMF as line integral of electric field and so what we had seen is that the time dependent phenomena led to a non-conservative type of electric field because del cross of E did not become 0, but had a value. So, this is what we did that the EMF which is the line integral of the electric field is rate of change of flux. So, EMF is given by minus d by d t of b dot d s and EMF by definition is the line integral of the electric field which by Stokes law, we know that we can write it as a surface integral of the curl of the electric field and by comparing this expression with this expression, since this is valid for any surface, we can write down a relationship which is the differential form of Faraday's law that is del cross of E is equal to minus d b by d t. So, notice that for the static phenomena my del cross of E was equal to 0, but now I write del cross of E as time derivative of the magnetic field. So, I repeat again that whatever is the cause for a flux through a circuit change, the we talked about motion of EMF which arises when there is a relative motion between the magnetic field and the circuit. For instance either a magnet is moved towards a circuit or if a circuit moves towards or away from a magnetic field which is of course, we know by relative motion it is the same effect. The other problem is that while this emotional EMF could be explained by invoking the Lorentz force that is as the circuit moves the charge on the circuit also experiences a force due to the magnetic field, but what was surprising is that even if there is no relative motion, if the magnetic field strength changes with time even then there is an EMF this is borne out by the experiment. So, therefore, we postulated that whenever there is a changing magnetic field that is a change flux changes through a circuit for whatever reason then there is an induced electric field produced and that is what is contained in this law that I have written out. With this let me introduce the concept of an inductance. So, notice that supposing I have two circuits I have called this loop 1 and this I have called as loop 2. Now, let us suppose that there is a magnetic there is a current which is flowing through loop 1 and this current is changing with time as a result the magnetic field that it produces changes with time and the loop 2 is a loop which intersects these magnetic field lines. So, therefore, as the magnetic field strength changes magnetic field strength produced due to the first loop changes it results in a change of flux in the second loop and let us now look at what it actually means. Now, if you recall your Biot-Saberts law it tells me B 1 because I have a loop 1 B 1 is the magnetic field due to the loop 1 that is the lower loop in that picture that is given by mu 0 by 4 pi then the integral over the loop of D L 1 by index 1 subscript 1 is supposed to be for the loop 1 cross R divided by R cube. So, this is the magnetic field produced by the current in the first loop. So, therefore, what I get is this mu 0 I must put it there. So, the flux through the second circuit this circuit is given by of course, integral B 1 dot d s if you like I will write it as d s 2. So, B 1 is the magnetic field produced by first loop and d s 2 is because I am integrating over the surface area of the second loop. So, you notice since B 1 is proportional to the current let me write it as I 1. So, phi 2 which is the flux through the second circuit is proportional to the current in the first circuit. So, this tells me that the flux I can write the flux through the second circuit I can write as equal to some constant let me write it as m 2 1 multiplied by I 1. So, therefore, flux in the second circuit is proportional to the current in the first circuit and that is what it is. And now the E m f in the second circuit which is minus d phi 2 by d t then is simply equal to minus m 2 1 times d i 1 by d t. So, you notice that this proportionality constant which comes when we express the E m f in the second circuit with the rate of change of current in the first circuit this is what I call as a mutual inductance. The index the indices have been subscripts have been written. So, that this m 2 1 stands for the mutual inductance when I am talking about the E m f in the second circuit due to the cause is in the first circuit the effect is in the first index now we can do something interesting though this formula that I will derive is of no great use, but on the 100th establishes two things. Firstly, it will establish that the mutual inductance is a property which depends upon the geometry of the loops relative position of the two loops and also I will show that m 2 1 is equal to m 1 2 that is whether I am changing in the changing current in the flux circuit and looking at the effect in the second circuit that is measure the flux in the second circuit. This gives me the same proportionality constant if I do the reverse namely if I change the current in the second circuit and look for the flux in the second circuit. . So, therefore, let us look at what we did. So, for example, we said that phi 2 is integral b 1 dot d s 2 b 1 because the magnetic field is produced by the first loop. Now, let me express this in terms of the vector potential produced by the current in the first circuit. So, I will write this as del cross a I will write as a 1 again all the quantities dot d s 2 and by Stokes law this would become line integral of a 1 dot d l 2. Notice that the vector potential is produced due to the first circuit the loop integral is taken over the second circuit, but I know that a 1 that is the vector potential produced by the current I 1 in the first circuit is given by mu 0 I 1 over 4 pi line integral of d l 1 over r. This is something which we had discussed while we defined the vector potential. Now, if I plug this into this expression. So, this tells me that phi 2 is given by mu 0 I 1 by 4 pi integral d l 1 dot d l 2 over r. So, this tells me that m 2 1 which is what we wrote down which is proportional the proportionality constant of phi 2 with I 1 is given by mu 0 by 4 pi double loop integral of d l 1 dot d l 2 divided by r. Now, you can see that this expression is symmetric in 1 and 2 because this r is actually r 1 2. So, therefore, this is identical to what you would have if you reverse the procedure namely m 1 2 is equal to m 2 1. Thus, this is what is known as the Neumann's formula. Neumann's formula establishes that the number 1 the structure tells us it depends only on geometry and the relative position. The fact that m 2 1 is equal to m 1 2 tells us that the mutual inductance is symmetric in the indices of the loops. Now, now I come to a rather curious things. If you change the current in loop number 1 the flux through loop number 2 it cuts the flux through loop number 2 which changes and establishes an Neumann. But let us now ask the following question supposing I do not have 2 loops I just had 1 loop. Now, this loop which I have when I change the current through this loop itself the magnetic field produced by this current changes and as a result flux through this circuit also changes. So, in some sense it is a self effect we are talking about and since the when there is a changing magnetic flux through a loop the loop does not care about what caused this changing flux as long as there is a changing in flux it will generate a Neumann. So, what will happen that you change the current through a loop it leads to changing magnetic field produced by that loop and which results in changing magnetic flux through the loop itself. So, this would mean that you have what is called a self effect that is the magnetic field when it changes and leads to changing magnetic flux there will be an emf generated in the same circuit. So, I will write that this emf this emf is minus by definition minus d phi by d t and let me write this as minus delta phi by delta i and of course delta i by d i by d t. Now, this proportionality constant that you have got here rate of change of flux through a circuit with respect to the changing current in that circuit itself is known as a self inductance and it is denoted by the letter L. So, therefore, the definition of self inductance is d phi by d t where phi stands for changing flux in a given circuit. So, we have talked about what is self inductance and what is mutual inductance, but let me first calculate just as illustrative example let me talk about calculation of some simple cases of self and mutual inductance. So, here in this picture you find that there is a current through a solenoid and this current is going to be changed. Now, I know if there is a current in a solenoid current i then this leads to a magnetic field B given by mu 0 n i. So, magnetic field B in a solenoid is mu 0 n i and actually its direction I should specify that direction is along the axis let me take it along the z axis. Now, so basically what is happening is this. So, I have this solenoid now when the current changes through the circuit of course, the magnetic field changes, but then there is a concept of how much of flux is linked with each loop. So, we say it is flux linked with each loop with each turn if you like. So, this is equal to simply the pi times r square which is the area of that loop multiplied by the magnetic field which is mu 0 n i. So, if you take a length l of the loop let me write it as a small l. So, that I do not confuse it with the notation for the self inductance itself. Now, I know that this many this length since the number of turns per unit length is small m the number of turns is equal to n times l. So, total flux linked this is equal to pi r square mu 0 n I have another n from there. So, I have a n square l i and now of course, the self inductance is simply d 5 i d i and this is linear n i. So, this works out to pi r square mu 0 n square times l. So, this is the expression and you can see it depends upon only the dimensionality is the geometry of the radius the number of turns the length of that loop. So, this is an example of calculation of self inductance of a solenoid. Now, let us do the following let us calculate the mutual inductance of the solenoid. So, here what I have done is this simple I made it somewhat simple I have put one loop over the other you can see this in this picture there is this blue colored loop blue or green which I was talking about earlier and now I have super post on it rather tightly the another solenoid which let us say we will assume that the first solenoid has let us say n 1 number of turns the second solenoid will have let us say n 2 number of turns and they are tightly bound. So, that we assume all the magnetic field that is produced by one of the solenoids passes or will be linked with the flux will be totally linked with the other one. So, let us look at how much flux will be linked. So, I am talking about now phi 2 I am changing I am passing a current i through the first loop and which we had seen gives me a magnetic field mu 0 n i is the current. So, the first loop and the total area is the area of each turn is pi r square then total number of turns that I have in the second loop is n 2 into n. So, since m 1 2 or m 2 1 is nothing but phi 2 divided by i that is simply given by mu 0 this is n 1 mu 0 n 1 n 2 pi r square l very interesting relationship that can come out of here. Remember that we showed that the self inductance of a solenoid is given by pi r square mu 0 n square l. So, therefore, self inductance of first solenoid is given by pi r square mu 0 n 1 square into l self inductance of the second solenoid will be pi r square mu 0 n 2 square into l. Remember I made an assumption that the two solenoids are tightly bound to each other. So, therefore, whenever flux changes because of change in current the entire flux is linked with the turns. So, if you compare this with this you find m 1 2 can be written as m 1 2 which we had seen as same as m 2 1 can be written as square root of l 1 into l 2. This is not really a general expression because we have assumed that the two solenoids are tightly bound to each other. But what we find is that the relationship between m 1 2 and the self inductances of the two loops can be generally written as some constant kappa times root of l 1 into l 2 and this coefficient kappa will depend upon relative orientation of the loop and this I can call it as a coefficient of coupling. Let us do another exercise. I am talking about I am taking two circular loops and again I made the problem somewhat simple. There is two co-planers are co-planer and concentric circular loop that is their centers are the same and they are in the same plane. I have assumed that the current is changing through the second circle the bigger circle. Now if I assume that the current is changing through the bigger circle and there is a smaller circle inside I have magnified it, but let us assume that r 1 r 1 is much smaller than r 2. Now when that happens I know how to find out the magnetic field due to a circular loop at its center and that expression the magnetic field due to the bigger circle is given by mu 0 I 2 because I 2 is the current in that divided by 2 I 2. Once again I could give it a direction which is let us say z axis because I could change the make the current flow that way. So, that if I use the right handed rule and so therefore, that could be this direction. Now it is reasonable to assume that if r 1 is much less than r 2 then the magnetic field strength at the center of the bigger circle is roughly the magnetic field strength over the smaller circle. So, once I make this statement that then the flux through the first circle will be simply b 2 into pi r 1 square and b 2 we have seen is mu 0 I 2 by 2 r 2 into pi r 1 square. So therefore, m 1 2 is divided phi 1 divided by I 2 is given by pi r 1 square mu 0 divided by 2 r 2. Once again it shows that mutual inductance is a property of the geometry and mostly the dimensionality of the things. So, with these examples let me now go over to a slightly different idea. Let us talk about how to find the energy of a current distribution. Recall that when we did electrostatics we discussed how to find the energy of a charge distribution and what we did at that time was assume first initially all the charges are away at infinity and that I took as the situation of 0 energy. Then I brought one charge put it in its place no work is done because there is no field, but once the first charge comes there is an electric field and then of course, I as I bring in the charges I would require it would be necessary for me to do work against the electric field that have been established and we calculated how much is the energy. Now, I have a problem now I cannot remove the current distribution to infinity. So, I cannot do that, but so this technique of starting with a 0 potential energy and then establishing the charges will not work here. So, let us see what is the way in which we will do it. Now, so these are the basic principle when we establish a current in a circuit we have just now seen an induced EMF is developed. Now, I we must do work to overcome this induced EMF. This induced EMF we have seen by Lange's law opposes the cause. So, therefore, if you are moving for example, if you are moving a magnet towards a circuit this circuit will generate an induced EMF circuit will have an induced EMF and if that happens to be a conducting circuit then of course, there will be an induced current. So, therefore, if you want to move for example, this magnet with a uniform velocity you will have to do work on it. Now, unlike the case of resistance this work is completely recoverable because the you know you get if you are moving something you need to do some work, but then you can always get it back in the reverse situation. That is situation number 1 and even if there is one circuit there will be an EMF in that circuit and we will have to overcome do work to overcome it. If there are more than one conductors then we have seen there are mutual effects that is supposing I concentrate on a particular loop let us call it loop number 1. Now, loop number 1 will have an EMF generated because of changing current in loop number 1 itself. Secondly when there are changing currents in other conductors loop number 2, 3, 4 etcetera then also there will be an EMF generated in loop number 1 and we must do work for that also. And so, in order to calculate the total energy we have to take into account both this situations. So, let us see how it works. So, let me I am going to calculate how much is the flux through ith circuit supposing I have got very large number of loops what is the flux change through the ith circuit ith loop let us say. So, we did that just now. So, let me call it phi i. Firstly, we notice that there is a there is a flux generated if there is a current i is in the circle loop ith loop itself. And we had seen that the proportionality constant was the self inductance since it is ith loop I will call it L i times I i. The second part is that we have seen that when there are currents in the other circuits j naught equal to i then I have a flux generated in the ith circuit which is given by M i j i j where M i j is the mutual inductance between i and j and I am assuming that M i j is equal to M j i. So, the E M f through the first circuit ith circuit which is minus d phi i by d t that is equal to L i d i i by d t plus sum over j naught equal to i M i j d i j by d t. So, this is the E M f and this is the E M f established in ith circuit. So, the rate at which work must be done to establish this E M f or to overcome this E M f. So, rate at which work is done that is obviously given by E i times i i E M f times i. So, this is simply multiplying these with i i all over let us do that L i i i d i by d t plus sum over j naught equal to i M i j i i i by d t plus sum over j naught by d i j by d t. Now, I will write it in a little more symmetric fashion. Firstly, you notice i i d i by d t can be written as a half of L i of course, is a constant d by d t of i square i i square. So, that d by d t of i i square is 2 i i d i i by d t 2 and a half cancels out. The second term requires a little bit of explanation, but it turns out I can write it like let me first write it down and then we will see why half sum over j naught equal to i M i j d by d t of i i i j. Now, this is really nothing great because what we have done is to say this is d i i by d t multiplied by i j plus i i times d i j by d t. Now, because M i j is equal to M j i I could interchange like this that is write it as you know d i by d t i times d j by d t i j times d i by d t. So, you notice so that these are written very symmetrically. So, with this we can now calculate how much is the total work done. So, this is the rate at which the work is being done. So, all that you need is to integrate it over d t. So, that this will give me the work done which will convert to the potential energy which is integral sum over i e i i i strictly speaking this last line that I did it is valid only if I sum over i. So, that you can sum over this and there is a sum over i here and then I use M i j equal to M j i. So, then this once I integrate over time I find I get half sum over i l i i i square and plus half sum over i sum over j i not equal to j of M i j i i i j. Now, if you bring back the sum over the expression for the flux this is the expression for the flux you notice this expression is nothing, but sum over this is nothing, but sum over i of i i phi i the multiply i i with i i you get i i square and multiply here you get i i i j and sum. So, this is equal to half sum over i i i times phi i now I will write it in a slightly different fashion now this is half sum over i i i this is flux. So, flux by definition b dot d s. So, I write it as b dot d s i this is the flux through the i th circuit and I will write this as again write b as del cross of a. So, del cross of a dot d s i and then I use stokes law. So, that this becomes loop integral of a dot d l i this is for discrete. Now, how will it change if I have a continuous distribution now see if I have continuous distribution then of course, all that I need is this that this i i with a surface will give me the current. And so therefore, this will go to half of loop integral of a dot j current this is current density and d cube r the current density because this is per area I mean this is we had current by area. So, here since there is just a length. So, therefore, I have got a d cube r there because I have divided by this thing. So, this is the expression for the work done by deep by establishing while establishing currents in various circuits and this is the corresponding continuous version with this much of introduction to Faraday's law. Let me go over to a slightly different story now. Now, let us recall where are we with respect to the electromagnetism as of now. So, we had del dot of e equal to rho by epsilon 0 del dot of b equal to 0 this actually will not change generally then we have just now seen that del cross of e is minus d b by d t. We have not yet talked about what happens to this del cross b equal to mu 0 j or the corresponding expression del cross h equal to just j free. Similarly, here if you want to talk in terms of the displacement field then this is simply equal to rho free. This is where we stand with respect to Maxwell's equation, but you notice that we understand the asymmetry between del dot of e and del dot of b because this tells us there are charges this tells us that there are no magnetic monopoles. Whereas, there seems to be some difference between the way these appear also. Once again I understand why the term corresponding to j free is not there because there are no magnetic monopoles which are moving around, but what is not clear is why is not there a time dependence here. Now, this was actually Maxwell's contribution to the electrodynamics and that is why these set of equations as modified later are known as Maxwell's equation. So, let me go through that. Firstly, let us look at a simple example of charging of a capacitor. Now, remember that when I pass a current through a circuit which contains a capacitor, there is no current which passes through the gap of the capacitor. But, however there is an electric field established in the capacitor. We will take the simple example of parallel plate capacitor and we know that as I am charging a capacitor the electric field through this changes because electric field will depend upon what is the instantaneous charge on the plates of the capacitor. And as you connect a battery to a circuit even though it is a small time, during that time the charge goes from 0 to whatever its final values. So, during that time there is a changing electric field through a capacitor. There is a problem and the problem comes up this. Let us look at the charging of a capacitor. There is a there is a capacitor with a current I flowing in and this current I is let us say momentarily changing because of my having connected a battery to the circuit and the electric field is of course, given like this. Now, consider a loop which I have called as C like this. Now, I know that I could calculate I could calculate the magnetic field by using the standard Ampere's law which says integral b dot d l equal to mu 0 i. Now, during the charging of the capacitor the charge accumulating in the plates of the capacitor is changing. As a result there is a current in the external circuit. Now, if there is a current in the external circuit and I look at this loop I can calculate the loop integral and say it is equal to mu 0 times i. i is the instantaneous current that instant at which I am calculating the magnetic field. Now, let us look at what is the problem. Now, if I have this loop this is the artificial loop mathematical loop. Now, let us suppose that I put a surface on the plane of the loop. So, let me do that. So, this is what we said just now integral b dot d l equal to mu 0 i. Now, let me put a surface now on that loop which I have called as S 1. So, through that surface a current is flowing and the magnetic flux line are intersecting that surface. So, so far as this surface is concerned I understand that del cross of b equal to mu 0 j because b dot d l which I had written down integral b dot d l equal to mu 0 i. That b dot d l could be converted to curl b dot d s and over this surface which I have called as S 1. The I can calculate the magnetic field at every point and. So, therefore, del cross b dot d s is my mu 0 i and if you since this will be true for any surface of this type I expect del cross b equal to mu 0 j. This is something which you have done in the earlier, but let us pause for a moment. We have said several times if I have a loop which I showed earlier these result del cross b equal to mu 0 j is independent of what surface I take as long as this is the boundary of the surface that has been defined. So, suppose instead of the surface S 1 through which the current passes and as a result the magnetic field lines also passes. Let us look at slightly different situation. Again I have the same situation I have however a slightly different surface. You notice this is a pot like surface I have this same loop C, but this time what has happened is instead of choosing the surface on the plane of the loop I had taken it like this. So, therefore, on this surface the through this surface there is no current. Now, if there is no current through that surface I expect del cross the I dot d l to be equal to 0 as a result del cross b should be equal to 0, but there is a dichotomy. We had said earlier the result should not depend upon which surface you choose because both of them represent the current passing through the loop C 1. So, how can I get one result for this type of a surface and another result for the surface S 1 which was on the plane of the loop. Now, what is the problem? The problem is that there is no magnetic field here. So, that b dot d s should be equal to 0, but is there something inside that? Now, notice there is an electric field inside that and that electric field is changing. So, as a result this is a dichotomy which Maxwell had realized that I get two different answers for the same problem depending upon surface I choose. So, what he did is to conjecture that just as by Faraday's law whenever there is a changing magnetic flux, whenever there is a changing heat changing magnetic flux through a circuit there is an emf generator. He wanted alternatively a changing magnetic field a time dependent magnetic field gives rise to an induced electric field. So, Maxwell now postulated going by parallelism that if a time dependent magnetic field gives rise to an induced electric field. It is reasonable to assume that a time dependent electric field of the type that we have here the as the charging is taking place the electric field through the capacitor is changing. So, a time dependent electric field is equivalent to a magnetic field the effects must be symmetric and this indeed was sort of proved or verified by experiment. So, how does one fix this dilemma let us look at that. So, just to understand let us assume that the space between the dielectric is filled with space between the capacitor plates is filled with dielectric. Now, if there are dielectric there are charges and when you change the current through it the electric field changes that of course pushes this charges. So, what will say is this that this is a space in which the electric field is changing with time and let us calculate how much is the what is the effect. So, the electric flux in this region is given by surface integral I have this time I am taking surface s 2 d dot d s I have taken the flux of the d field because I have said it is filled with direct. Now, so therefore d phi by d t actually I should write full. So, this is equal to d by d t of integral of d dot d s this I will write this in a using the divergence theorem as d by d t of del dot of d d q r voluminative, but del dot of d is nothing but rho rho free d q r which is nothing but the total charge q and. So, therefore this is d q by d t notice d q by d t that is rate of change of charge on the capacitor plate is same as that of the current that is flowing in. So, therefore this is this has the dimension of correct. So, what we have said is I d the name came from displacement current this was Maxwell gave it a name displacement current that is rate of change of the electric flux with time. So, how did Maxwell handle it? So, Maxwell said that let us then modify the last law that is the Ampere's law which was del cross h equal to j by adding a term d d by d t. You notice that del dot del cross h is identically 0 because it is divergence of a curl. So, therefore I get from here del dot of j plus d by d t of del dot of d this is equal to 0, but del dot of d is rho. So, that gave me del dot of j plus d rho by d t equal to 0. You identify recall this is nothing but our continuity relation. Thus this is the equation which should replace del cross of h instead of j should now be added with a term d d by d t. That completes our establishment of the 4 Maxwell's equation and we will discuss them all together in a more coherent way from the next lecture.