 In the last lecture, we had seen how the method of scalar potential can be used to find out the magnetization and the magnetic field due to a uniformly magnetized sphere. What we do today is continue in the same line and try to solve the same problem namely when we have a uniformly magnetized sphere, what is the corresponding field B and the field H due to it, but this time we will be using the method of vector potential. Of course, both of them being equally good in this case will lead to the same result and first we will do that. After that in the second half of this lecture, we will be introducing, we will be till now we have been talking about static phenomena. We started with electrostatics and then we went to magnetostatics, we will be for the first half we will still be with magnetostatics, but in the later half we will bring in some time varying phenomena and introduce some other aspects of the electromagnetic theory. .. Let me begin by reminding you that we had seen that if I have magnetization m, the corresponding magnetic vector potential due to it A was given by mu 0 by 4 pi well basically m divided by r cube. Now, what we did is to say that if I have a magnetization distribution, then this is actually m cross r divided by r cube. Now, if I have a distribution of magnetization, then we were able to write it in two parts. One was a surface integral part and another was a volume integral part like here and we had seen that the vector potential A can be written as a surface integral of n cross m by r minus r r prime cube. So, that tells me that there is an equivalent surface magnetizing surface current if you like, which is equal to minus n cross m. Remember inside the integral all my quantities have primes. Now, and there is a volume density which is given by del cross m, but in this case I have got a uniform magnetization. So, as a result my del cross m term is 0. So, I need to simply concentrate on the term which is surface integral term which is written as A of r equal to minus mu 0 by 4 pi integral over the surface of n prime cross m of r prime over r minus r prime n d s. Now, this is the quantity that we will like to calculate and this n cross m or with a minus sign that is my equivalent surface current due to magnetization. So, J s of r in this case r prime is given by minus n prime cross m of r prime and what we will do is if you refer back to this picture we will take the direction of the magnetization as the z axis. So, that the angle between the direction of magnetization and the radial vector will be taken as theta. So, therefore, this quantity is nothing, but m sin theta and the direction of that is clearly the azimuthal direction E phi. This is a direction on the surface of the sphere which is perpendicular to both the radial direction and the z direction. Now, if you look at how the phi is defined I will just recall for you I am not drawing a 3 dimensional picture, but I am drawing a 2 dimensional picture. Remember that the radial vector is along the radial direction is the r vector and what I do is to drop a perpendicular onto the x y plane and it is the foot of the perpendicular which is. So, this p is the foot of the perpendicular from the tip of the radius vector and this angle which we make with the x axis with the x axis that is my azimuthal angle phi. Now, we had seen earlier that whenever we define a unit vector it is defined in the direction of the increasing quantity. So, in this case obviously my unit vector will be like this. So, this will be my unit vector which I wrote as E phi. So, as a result this E phi is can be resolved into a component along the x direction which is minus. So, this angle is being phi this angle is phi. So, this is 90 minus phi and a component along the y direction. So, therefore, let us write down E of phi to be equal to m sin E of phi is minus sin phi well I am using phi prime. So, minus sin phi prime unit vector i minus because that is that direction plus cos phi prime unit vector j. So, this is my E phi and we had seen that j s of r prime which is what I am going to write down is given by m sin theta prime times this quantity E phi there which is minus sin phi prime i plus cos phi prime j and I need to now calculate the integral which is involved in the expression for the a of r. So, let us go back to a of r. . So, a of r is given by mu 0 by 4 pi I have taken care of the minus sign which is outside and this will be equal to m sin theta prime into minus sin phi prime i plus cos phi prime j divided by r minus r prime and d s of cos. So, the question is how do I evaluate this integral? Now, what I will do is I will calculate the part which has which is along the y direction and the reason is as I go along I will make a comment which will show that this term can be take shown to be equal to 0 and we will see why, but at the moment I will be just evaluating this term with a cos phi prime j. Now, notice that in doing so what I do is to use the property of the spherical harmonics the orthogonal property of spherical harmonics. So, we know that 1 over r minus r prime can be expanded in spherical harmonics. So, what I will do is this that since I told you that I am going to do only the y component. So, we will we have in y component in the numerator we have a quantity which is sin theta prime cos phi prime. Our idea is to express this quantity as in terms of spherical harmonics. Now, you can check that they from the spherical harmonics table that I can write y 1 1 theta phi in this case theta prime phi prime it does not matter is equal to minus square root of 3 by 8 pi sin theta prime e to the power i phi prime. And y 1 minus 1 theta prime phi prime is plus square root of 3 by 8 pi sin theta prime e to the power minus i phi prime. So, what I do is this I can use the definition of sin and cosine that well sin is of course, already there and cosine I know is e to the power i phi plus e to the power minus i phi by 2. So, therefore, sin theta prime cos phi prime can be written as square root of 2 pi by 3 there is a root 3 by 8 pi, but I need also a factor of 2 in my definition of cos phi and sin phi. This times it turns out to be y 1 1 theta prime phi prime minus actually there is overall minus sin also minus y 1 minus 1 theta prime phi prime. Well sin theta prime cos theta phi prime is this sin theta prime sin phi prime which is there which is I am writing it down. So, that you can see why that term turns out to be 0. So, sin theta prime sin phi prime you can easily show that it is given by root 2 pi by 3 there is an i there because sin is defined as e to the power i phi minus e to the power minus i phi by 2 i. So, this into this is y 1 1 theta prime phi prime plus y 1 minus 1 theta prime phi prime. So, basic idea is this that I am going to express 1 over r minus r prime in terms of spherical harmonics recognize the fact that in the numerator also I have spherical harmonics. So, as a result what I am going to do is to write down the entire thing in terms of spherical harmonics. I recall that 1 over r minus r prime can be written as 4 pi sum over l sum over m 1 over 2 l plus 1. This is something which you did last time also r lesser raise to the power l by r greater raise to the power l plus 1 y l m star theta prime phi prime which are the angles corresponding to the vector r prime into y l m theta phi which are the angles corresponding to the vector l. Now what we do is this I define my point of observation as in the x z plane. So, you remember that I can always do that because the only direction which is given to me is the z direction along which the magnetization direction is there. So, I have the liberty of choosing the x y plane. So, therefore, I choose my observation point to be in the x z plane which means I choose phi to be equal to 0. This is an important point which simplifies the problem substantially. With that a remember I told you that the x component will go away which I will make a comment on later. But let me say what is a y which is which comes from sin theta cos phi prime part. So, I had this mu 0 over 4 pi. I have root up 2 pi by 3 m r square came out of r square d omega which is the angle integration 4 pi from here and then I have an integral to be performed integral over l m 1 over 2 l plus 1. This is a rather long expression, but turns out to be reasonably easy r lesser raised to the power l r greater raised to the power l plus 1. Then I have actual integration will be over. Well you could recall I had shown this to be equal to y 1 minus 1 minus y 1 1 I had taken care of this minus sign by interchanging the y's. Then I have the 2 y l m's which came here y l m star theta prime phi prime y l m theta and 0 and of course d omega prime. Remember that these also have theta prime phi prime as their arguments. Why because it came from sin theta prime cos phi prime. So, if you use the orthogonality condition of this then integral say for example, when you integrate this one with that one then it tells you that l must be equal to 1 and m must be equal to minus 1 and similarly in this case l is equal to 1 and m is equal to 1. So, I get 2 terms and so what I am left with will be something like this. I have of course m r square the I had this root 2 pi by 3 I will retain because this term here. So, I have got still root 2 pi by 3 4 pi has cancelled out with that. So, I have a mu 0 since l is equal to 1 I get 2 l plus 1 which is equal to 3 then of course I have r lesser raise to the power l and l is equal to 1. So, therefore, it is simply r lesser by r greater to the power l plus 1 which means r square times this time it is y 1 minus 1 of theta not theta prime 0 minus y 1 1 and theta 0. Now, remember y both y 1 1 or y 1 minus 1 they had e to the power i phi and e to the power minus i phi respectively, but since phi is equal to 0 each one of them each one of them turns out to be a sin theta, but y 1 1 had a minus sign in front of it. So, therefore, this quantity turns out to be 2 sin theta. So, this along with this factor of 2 pi by 3 this turns out to be 2 sin theta. Well at this stage I will also point out why the x component would vanish in the x component I will have a very similar integral, but I will have y 1 minus 1 plus y y y 1 1 and. So, that would be sin theta minus sin theta which will be equal to 0. So, other than that this is essentially identical. So, let me then with that write down what I get for my A y. So, A y so vector potential only has y component which is equal to m r square mu 0 by 3 well actually 2 mu 0 by 3 sin theta and r lesser divided by r greater whole square. So, this is this is the expression for the y component of the vector potential and. So, what I now do is that if I am inside the sphere I know r greater than is capital R. So, the potential inside. So, I get a r square which will cancel with this r square there. So, I will be left with m r square sorry I will be left with m 2 mu 0 by 3 and r sin theta. So, this is what I would have inside the sphere and outside the sphere I will get A y again is given by m outside this is r r lesser is r. So, I get a m r cube then 2 mu 0 by 3 and r greater is smaller. So, therefore, I get sin theta over r square. Now, if you remember that since phi is equal to 0 my r sin theta is nothing but my x. So, it is m 2 mu 0 inside only by 3 x I am not doing this, but it because it has slightly more complicated, but you could do it as an exercise. Now, so therefore, inside the sphere my vector potential has only y component and it is proportional the y component of the vector potential is proportional to x. If you recall we had shown earlier that if y component of the vector potential is linear in x it means a constant magnetic field. So, I can immediately evaluate the magnetic field inside and obviously, it will only have z component because B z will be then d A y by d x minus d A y d A x by d y which is equal to 0 and since the vector potential is proportional to x this simply gives me m times 2 mu 0 by 3. Now, this is this is identical to the expression that we had obtained using the scalar potential method of course, it has to because it is the same thing. Before we conclude the subject of magneto statics let us do something else. Let me put this magnetized sphere in an external magnetic field of strength B 0. Now, since the equations are linear I can simply add. So, we had in the absence of the external field. So, I am doing magnetized sphere in an external field. So, in the absence of the external field I had this as 2 by 3 mu 0 m, m is the magnetization and if I have an external field I of course, add B 0 to it. Now, I will define I will get the corresponding value of H in terms of 1 over mu 0. So, I have got B 0 plus 2 by 3 which is simply the value of B minus m. This is simply 1 over mu 0 B minus m this expression and if you simplify it it turns out to be simply equal to one third of m. So, this is this is what I would get if I put it in an external field. Now, couple of comments that suppose this material of the material of the sphere happens to be paramagnetic. Now, we had said that the just as in the absence of a magnetic material my relationship between H and B is B by mu is H. Now, if I have a paramagnetic material then my B is related to H through mu H. So, this is for paramagnetic where mu is the permittivity of the material. So, let us see what does it actually mean. So, this means in that case what I will get is this B which is 2 by 3 mu 0 m plus B 0 will be equal to mu times B 0 by mu 0 minus one third m. This is what it would be. Now, you can solve this equation and you get that magnetization m is written as 3 by mu 0 mu minus mu 0 by mu plus 2 mu 0 times B 0 and corresponding value of the magnetic field B will be 3 mu by mu plus 2 mu 0 times B 0. I urge you to look up the expression that we had derived when we had put a polarized electrically polarized material into an electric field and you will find that there is almost a direct similarity between these two expressions. The last comment that we want to make is that what happens if I have water known as ferromagnetic substances. So, we had seen that in case of paramagnet we have a relationship where B and H are parallel. The if you reduce the external field to 0 the B field also goes away. Now, there are materials which are known as ferromagnets and this is a picture of what happens in case of a ferromagnet. So, I am as I apply external field the initially B rises, but for a certain strength of the external field the value of B saturates. Now, the reason is that the ferromagnetism appears due to the magnetic moments being aligned. Now, once in a particular material all the magnetic moments have been aligned I have nothing more to aligned it and that then is the limit after this if you increase the electromagnetic field nothing will happen because saturation has been reached. Supposing you saturate a ferromagnet by applying a magnetic field and then you start reducing the strength of the magnetic field. Now, something interesting happens unlike in a case of a paramagnet the as you reduce the external field the magnetization or the B direction does not retrace its own path. In fact, it picks up a different path and when it picks up a different path it will still have some memory even after you have reduced the external field to be equal to 0 the magnetization is not lost. So, this is what is known as retentivity or the process of hysteresis the material in some sense remembers having been subjected to a magnetic field in the past. So, you prepare a magnet by applying a magnetic field then gradually bring the strength of the magnetic field to 0 you will find the magnetization will still exist as is shown here. In fact, you will need a magnetic field to be applied in the negative direction if you want the magnet magnetism to vanish and this is actually this value is called the coercive field and of course, there is then a symmetry of this this loop is known as a hysteresis loop. Now, so therefore, we have now obtained two expressions. So, there is an expression which we derived namely B the total magnetic induction field is B 0 external field plus 2 by 3 mu 0 m, but this expression this curve gives me a dependence of B on H which is some functional relationship. So, what I do is this that the when I have this let me write this equation for H you go back you can find that H is written as B 0 by mu 0 minus one third m and if you solve these two equations together I by eliminating m you can get this expression which is B equal to 3 B 0 minus 2 mu 0 H. This tells me that the relationship between B and H is a straight line with the slope being equal to minus 2 mu if you are plotting it against mu 0 H. Now, so therefore, the I have a relationship which is given by the hysteresis and I have derived a relationship between B and H. So, what I can do is this that I have once I have subjected the ferromagnet to saturation and reduced then I find out where is the intersection of this with B versus H relationship and this is that point P there which will sort of tell me how much is the value of the magnetization at that point. So, this is the way one determines the magnetization for a ferromagnetic sample that brings us to an end of the magneto statics and what I am going to do now is to switch over to the phenomena time dependent phenomena. So, but before I do that let me start with a few definition. So, one of them which we have talked about in case of electric field very similarly one can define magnetic flux. The magnetic flux will be defined as this B dot d s. Now, look at this situation here I have a surface S 1 which is this oval shaped plate and on the same rim of S 1 I can have a Fisserman's net type of a surface as well I call this S 1 I call this S 2. Now, since the direction of the surface is always defined as an outward normal the normal to the surface S 1 is like this and the normal to the surface S 2 is like that. Now, if you now use the fact that del dot of B equal to 0 which means the surface integral of B B dot d s is equal to 0. Then you can easily show that as long as the boundary is fixed it does not matter what surface you take to define flux. So, phi n of S 1 equal to phi n of S 2. The second point that I want to talk to you is what is known as electro motive force. Remember or let me alert to you that what I am going to talk about is not a force at all it is called a electro motive force because of historical reason. But let us see what happens supposing you have an electric circuit and the electric circuit connected to a battery and in the external circuit you have a current. Now, you can assume something like a Ohm's law so that you know that the current is nothing but the electric field divided by the resistance. So, what I know is the integral of E dot d l is not equal to 0 because there is a current in fact flowing. However, we have seen in electrostatics the integral of E dot d l must be equal to 0. So, how does one understand this apparent contradiction. So, one of the things that I would like you to realize is that integral E dot d l equal to 0 alternatively del cross of E equal to 0 was a consequence of the fact that I was talking about electrostatic fields which are conservative. Now, what happens in a circuit is that for instance let us say that the current is flowing because of a battery connected to the circuit. So, I have an external loop where I can calculate E dot d l going from let us say the positive terminal to the negative terminal and I get some value. Now, when I complete the circuit through the battery the battery provides the motive force by usually in case of a battery converting chemical energy to electrical energy and that is a non-conservative field. So, therefore, the this will provide let us say an integral E prime dot d l which inside the battery will exactly cancel the E dot d l that I have calculated on the outer circuit. So, therefore, my electro motive force is actually integral if you like inside the battery from terminal to terminal of integral E dot d l. However, this non-conservative field is 0 outside the battery. So, I can actually complete the loop and redefine my electro motive force through this relationship here that is electro motive force which I use script E for it is loop integral of E prime dot d l where E prime is the electric field that the battery has produced. Now, as I said we are talking about time dependent phenomena. Salida has had enunciated a law from observation and what he had said is that he found by a set of experiments that whenever there is either a relative motion between the source of magnetic field and a circuit or if by some reason even if there is no relative motion the magnetic field is changing with time. In both the cases it gives rise to an equivalent or gives rise to an EMF and what Faraday had said is that the EMF is equal to the rate of change of magnetic flux through circuit. There is a another law which goes along with the Faraday's law and that is why you will find when we write down the Faraday's law there will be a minus sign that it electro motive force is equal to minus d phi by dt. This minus is not to be treated as an algebraic quantity, but it is a reminder of the fact that there is another law called the Lange's law which tells us that if this is connected to a circuit conducting circuit the direction of the current will be such that that this current would generate magnetic field of its own which will tend to oppose the cause which generated namely if the flux was increasing as a result of our action the induced current will be in such a direction that the flux would tend to decrease as a result. So let us first talk about couple of cases connected with this production of EMF. Now if my circuit is moving if there is motion in the circuit or relative motion I find it easy to explain the Faraday's law and the explanation comes by the Lorentz force that acts on the charged particle. Remember that if a circuit is moving the circuit has charges these charges themselves then are physically moving along with the circuit. Now so therefore when a magnetic field is applied on a circuit the charges that exist inside the circuit are now subjected to a Lorentz force and this Lorentz force can make the charges move. So that would be a source of my electromotive force. Now let us look at that situation here look at this circuit which I have indicated in orange pattern and let us suppose that this is my D L that is this circuit will be moving like this. Now when this happens the circuit is for example instantaneously moving by an amount D L. So the charges in that will be subjected to a force suppose a charge Q in that will be subjected to a force equal to Q times V cross B. So electromotive the EMF which is the integral of electric field dotted with D L which is F by Q dotted with D L which is simply V cross B dotted with D L. Now remember that what we have is the charges are subjected to a transverse force. So the velocity direction is shown like this and supposing as a result this portion of the circuit tends to stretch like this then the idea is that this D L would sweep an area this D L will sweep an area and so this expression here is integral of V cross B dot D L that came from the Lorentz force. I use the fact A dot B cross C is B dot C cross A etcetera and rewrite it as equal to integral B dot D L cross V. Now so D L cross V this is D L this is V is the rate at which the area changes because D L times the distance along this would be the area. So therefore this quantity can be shown to be there is a minus sign you can see it from the way D L cross V is done and so therefore this is B dot D S by D T where S is the area and this is nothing but the rate of change of flux with time. So therefore this tells me the way Faraday's law was stated that the EMF is minus d phi by D T. Now this is a case which we used the fact that I have a loop which is being stretched. Now supposing the loop is not getting stretched but I have just a simple motion. In this picture what I am showing is that I have a magnetic field which is non uniform I have taken the magnetic field along the z direction but let us suppose that I have a so the magnetic field is outward along the z direction perpendicular to the screen I have a rectangle which is moving with a velocity V along the x direction and let us say at a particular instant the left edge of the rectangle is at the position x and the right edge is at the position x plus dx. Now what I am going to do is this that let me calculate then how much is the electromotive force. So this is done remember that electromotive force is integral of V cross B dot d l I am still with what I called as the motional EMF only in this case the velocity is the physical velocity of the loop which is moving with a velocity V along the x direction. Now in this case what I have said is that if you refer to let me draw this picture here again so this is a rectangle its instantaneous position is x here and an x plus dx there and supposing I am taking the loop direction like this. So the magnetic field strength on this edge is B at x the magnetic field strength here is B at x plus delta x. Now what I am going to do is to calculate how much is V cross B velocity is in the x direction the magnetic field is in the z direction. So V cross B will be in the y direction x cross z. So if it is x cross z its minus y direction. So I have contribution to the integral only from these two sides because these are parallel to the y direction. So this EMF can then be written as integral V cross B at point x dotted with now this d l is because I am going like this. So it is minus j if you like delta y plus integral V cross B of B at x plus delta x dot in this case it is along the plus j direction. So plus j delta y now if delta x is small I can write B at x plus delta x is equal to I can use a Taylor series B at x plus d B by dx the inhomogeneous field times delta x. Here I have got two terms one is minus another is plus and what I will now do is I will also give explicit form to the direction of V cross B V is along the x direction B is along the z direction. So V cross B's direction is minus y direction. So this is minus j dot minus j which is plus 1 but this term will be negative because this is minus j dot plus j. So if you now substitute it here what I am going to get is EMF is equal to integral minus V cross delta B x by delta x multiplied by delta x delta y you know V is dx by dt. So therefore if I write this as dx by dt then this gives me d B by dt. So this will become integral of d B by dt and delta x delta y is my area. So this gives me a d s. So which is again minus d by dt of the flux. But you see in when we do experiments the charged particles that are there they do not realize they experience a force. They do not realize whether this is happening this effect is happening because that with respect to another observer the charges happen to be moving that they are a part of a moving circuit or just the magnetic field is changing in time. So whatever be the case there would be an EMF if there is a change in the magnetic flux and this change in the magnetic flux can come from either a physical change like stretching of a circuit actual movement of the circuit or a changing magnetic field. In the last case where there is a time dependent magnetic field I still have Faraday's law valid. But I cannot give you an explanation in terms of Lorentz forces. So for reasons that we do not understand at this moment it seems that if there is a changing magnetic flux for whatever reason that is equivalent to an induced electric field because I need an induced electric field to define my electromotive force. My electromotive force was integral loop integral of E dot d l. So the cause could be physical movement in which case we give it a special name we call it emotional EMF or it could be simply a time dependent magnetic field. With this let us summarize the laws as the Faraday's law as it stands. So we had seen integral of E dot d l which is my EMF is minus d phi by d t and the flux is by definition integral B dot d s. This line integral I am going to convert using Stokes theorem to del cross E dot d s. So this is equal to minus d by d t of B dot d s and this integral is over arbitrary surface. Since the integral is over the arbitrary surface well I can also write this as partial d B by d t dot d s. What I get is del cross of E is equal to minus d B by d t. So Faraday's law which tells us which is the first time varying phenomena that we have introduced tells us that if there is a changing magnetic flux this produces an EMF in the circuit. And this is so in addition to for example del dot of E equal to rho by epsilon 0 del dot of B equal to 0 these were the two Gauss's law I have this equation. Remember in the absence of time varying field del cross of E was equal to 0. I am still left with one equation which I have not yet talked about and that is del cross of B equal to mu 0 j the Ampere's law. What we will do in the next lecture is to find out if there is time varying electric field how does Ampere's law change and once we have done that we would get a set of four complete Maxwell's equations. And that is that will be done next time that is talking about just as time varying magnetic field gave us an equivalent induced electric field. We will see that a time varying electric field can be shown to be equivalent or give rise to a an induced magnetic field as to why that is a rather weak thing or a difficult thing to observe and things like that this will be the content of our next lecture.