 up until now we have been talking about phenomena of electrostatics. Electrostatics basically implies the effect of fields that is created by static electric charges. What we are going to do now is to talk about a different phenomena and that is the production and the properties of the magnetic field for which the source is a steady current that is a charge in motion. Basically there is one major difference between electrostatics and what we will be calling as the field of magnetostatics. We know that electric charges can be isolated namely I can have a single positive charge or a negative charge. However, it has never been possible to isolate a magnetic charge namely either a magnetic north pole or a south pole individually. Whenever a magnetic north pole is there accompanying it will be a magnetic south pole that is the net magnetic charge is always 0. Therefore, that is one of the major difference what we say is magnetic monopoles do not exist. However, I must tell you that there is really no theoretical reason why magnetic monopoles do not exist. However, in spite of several attempts to isolate such things it has not yet been possible. The source of a magnetic field is an electric current that is a steady current which is flowing let us say in a wire or on any other conducting loop. Now, in such a field the a moving electric charge experiences a side wise force. Remember that an electric field can exert a force on a charge whether it is moving or not. However, the magnetic field is characteristic in the sense that it only exerts force on a moving charge and this force which is given by the expression v cross b. So, if I have electric and magnetic field I have the net force is given by q which is the charge times the electric field which is the result of or of the electric field plus v cross b. Now, notice that this force due to the magnetic field is expressed as a cross product of the velocity with that magnetic field which means if the charges were static it will only experience an electric field, but in order that it experiences a force in a magnetic field the charge must be moving. This is known as the Lorentz force expression. So, what we have said is the current is the source of the magnetic field. Now, let us define current in a proper way current by itself is a scalar quantity. This is basically the amount of charge that is crossing the boundary of a surface s of a volume v. So, whatever is the amount of charge that is crossing an unit area per second per unit time that is what I call as the a current. So, in other words this is supposing I take a volume which is let us say defined by a boundary s the rate. So, if certain amount of charges flow into that volume the amount of charge in that volume will increase. Now, in steady magnetic phenomena or in a steady state phenomena I do not expect an accumulation of charge and this gives rise to what is known as the equation of continuity. Now, let us look at what is equation of continuity. Now, firstly let me assume that this is a volume and this s is a surface which separates two parts of that volume the volume to the left and the volume to the right. So, therefore, if electric charges are crossing this boundary the from the left I will expect a depletion of the amount of charge and of course, to the right there will be an increase in the amount of charge. Now, so therefore, what happens is this that this current that I have got which is there is a minus sign wrong here, but which is the dot product of a quantity which I call as the current density and this is the current density which is crossing a surface and it is that surface should be normal to the direction of the flow J dot d s. So, therefore, the rate of change of the charge in that volume which is d q by d t is given by well q is nothing but the charge density integrated over the volume. So, therefore, this quantity since this is the amount of charge that is increasing in the left part of the volume that must be the decrease from this side and which is nothing but the minus the J dot d s which is minus of the current. So, and I know that by using the divergence theorem I can rewrite this as del dot divergence of J over the volume. So, therefore, I have got d rho by d t d v integral is minus integral of del dot J d v and if I bring it to one side and use the fact that this is independent of what volume we take it is true for any volume what I get is this equation which is d rho by d t plus del dot of J is equal to 0. This is known as the equation of continuity basically it tells me talks about conservation of charge. So, if the left side of the volume is losing some charge because of the current flowing into through the surface. So, I must have that increasing a corresponding amount of charge on the right hand side. Since, we will be interested in primarily magnetostatics namely my currents are steady. So, therefore, I must have d rho by d t equal to 0 because there is no accumulation of charge in the volume and as a result the corresponding equation then becomes del dot of J equal to 0. So, this is the steady state form of the equation of continuity one point which of course will not be explaining in any detail is to tell you that this equation is also relativistically valid that is if you use what is known as Lorentz contraction then you will find this equation will transform co-variantly under it. So, this is just to tell you that the equation this equation is lot more rigorous and powerful than we thought it was while doing electrostatics we have been familiar with the basic law which is the Coulomb's law which essentially told you that the force between charges is given by an inverse square law namely q 1 q 2 divided by r square and there was a multiplicative constant dimensional constant 1 over 4 pi epsilon 0. Now, if I have had a charge distribution what I would do is that electric field due to this charge distribution at a point I build up by finding out what is the electric field due to a volume element containing the charge in that distribution at that point P where I am looking for the electric field. Now, since we have said that the source of magnetic field is a current distribution. So, what we do is we look at when we look at a circuit we look at individual current element that is a small length of the wire let us say carrying current and I define the direction of the length element along the direction in which the current happens to be flowing at a particular point on that circuit. Now, if this d l so I times d l is the length element regarded as a vector in the direction of the current. So, I times d l is my current element. Now, what we do is this that we calculate the magnetic field at a point P this picture illustrates this that this red thing is a current carrying wire and here I have shown a small section of it and this is the direction in which the current is being carried and I am interested in calculating the magnetic field due to this current carrying system at this point P. Now, similar to Coulomb's law we have a law here which is known as the Byte-Savart's law and Byte-Savart's law tells you that if you take a current element I d l the magnetic field d b due to that current element is given by a first like 1 over 4 pi epsilon 0 I have a constant here which I will come back to which is mu 0 by 4 pi. Now, current element since I is a scalar I will write it separately I d l. Now, the d l cross the position of the point where I am looking for the magnetic field with respect to the position of this d l. So, if you refer back to this picture again you find that if I am looking for the magnetic field at a point P which is with respect to my arbitrary origin at a vector position r and the current element is at a position r prime vector. So, with respect to that current element the point P is at r minus r prime. So, coming back to this formula again I have d l cross r minus r prime which is the vector distance from that current element and divided by the distance r minus r prime cube and you can see that this is also essentially an inverse square law because there is a power in the top and a power in the bottom. This constant mu 0 is known as the permeability of the vacuum or permeability of the free space. So, at this moment we will leave it at that that this is the constant which appears in the Biot-Cybert's law just as a constant 1 over 4 pi epsilon 0 with epsilon 0 being known as the permeability of the free space given. So, this has something to do with the magnetic properties of the material if there is a material then the mu 0 will change, but we will come back to it at a later stage. So, this is the Biot-Cybert's law. So, when you have a current carrying circuit and you are interested in finding out what is the magnetic field at that point at a particular point then you split your current distribution into a large number of infinitesimal current element and the Biot-Cybert's law gives you the magnetic field and at that point and essentially what you have to do is to sum it up namely integrate it. Now, what I will do is this that I will try to get a slightly different form of this Biot-Cybert's law by doing a little bit of algebra. So, let me let me look at what we are looking for. So, let us look at this. So, this is my I d l cross r minus r prime by r minus r prime cube is my structure. Now, remember what I had, supposing I wanted instead to write it in the form of a current density. So, remember I had I times d l. So, I is a current and d l is of course, a length and I know that what I could do is to write this I as j dot d s, where j is the current density. So, therefore, that area element which goes with the current density to define my current and the length d l taken together gives me a volume. So, therefore, my magnetic field at the point p can be expressed as B let us say at the point r is given by mu 0 by 4 pi integral. Now, remember this r is a fixed position with respect to an arbitrary origin. This is the point at which I am determining the magnetic field. Now, this is not a quantity which is varying, but when I have quantities belonging to the source I am going to write it as given by r prime vector. So, current density at r prime cross r minus r prime divided by r minus r prime cube and d cube r prime. So, this is my magnetic field expression. Now, I am going to do a bit of a algebraic manipulation. Now, recall that gradient of 1 over r minus r prime is minus r minus r prime by r minus r prime cube. So, what I am going to do is this I am going to write this expression in a slightly different way. I will write it as a mu 0 by 4 pi. I have a minus sign if I write it as a gradient of 1 over r minus r prime. Now, in order to take care of that minus sign I am going to interchange the order of the cross product which is j of r prime d cube r. Now, so what I am going to do now is this I am going to rewrite this. So, I have got gradient 1 over r minus r prime cross j r prime. I will write it as equal to mu 0 by 4 pi integral del cross I will come back to the explanation of this of 1 over r minus r prime times j of r prime. Now, let us see why remember that 1 over r minus r prime is a scalar. So, if I have a del cross of a scalar times a vector this is equivalent to this scalar times del cross of that vector which in this case is 0 because this gradient operator is with respect to variable r del which of course, we have said in this particular case is not varying. So, del cross of j r prime is 0. So, I have got del cross scalar times a vector is a scalar times del cross of vector which is 0 plus gradient of the scalar cross product with vector. So, this is the result that I have got. So, I have written it this way and this integrated with d cube r prime. Now, of course, this del cross can be taken out that is because this has nothing to do with the integration variable. So, I will write this as mu 0 by 4 pi del cross of integral of j r prime divided by r minus r prime. Now, this is a rather important form in which the Bayer-Sawart law is expressed. So, this is expressed in terms of a current density distribution in principle. Now, since the magnetic field is written as a curl of a quantity and we know that divergence of a curl is always 0. So, this tells you the magnetic field at the point b is such that its divergence is equal to 0. So, that is del dot of b is equal to 0. So, the magnetic field is such that its divergence is equal to 0. Now, this is actually a consequence of the fact that there are no magnetic monopoles. How do I say that? Recall that in electrostatics the corresponding equation was del dot of e was equal to rho by epsilon 0 where rho was the charge density. So, as a result since in this case I do not have magnetic charges free magnetic charges my that is I do not have monopoles the magnetic field is such that its divergence is 0. Now, we have said several times that if you have a vector field a vector field is uniquely specified by specifying its divergence and curl. Actually, there is something else to be done that is specify also the normal component of the field at some surface, but let us leave it at that. So, basically what we are saying is that a vector field can have a divergence and it can have a different curl different vectors field can have same divergence, but different values of the curl. So, in order to specify it completely I am going to now calculate how much is the curl of the vector v. So, let us look at what is the curl. Now, curl of b now this expression now is what has been obtained written by me sometime back as the Byatt-Sawart's law of expression. So, I have got mu 0 by 4 pi del cross of del cross of this quantity. Now, we have several times seen that a del cross del cross of a vector field is given by gradient of the divergence of that field del of del dot of a minus del square a. So, my curl of b since b itself is written as a curl of something. So, I got mu 0 over 4 pi curl of this is my Byatt-Sawart's law of expression curl of j r prime by r minus r prime d cube r prime and I am going to use this formula that I wrote down that is del cross del cross of this quantity is the constant mu 0 by 4 pi gradient of del dot of this quantity j of r prime by r minus r prime d cube r prime minus del square of j r prime divided by r minus r prime. This looks like a rather horrible expression, but let us try to look at simplify this. So, first what I am going to do is this I am going to look at the first term that is del del of del dot of this quantity. Now, look at this first before I calculate the del let us look at what is this quantity here divergence j of r prime by r minus r prime. So, what I am doing first is this since I know that the quantity is inside the variable of integration is r prime and the gradient is through with respect to r I can simply take this del dot inside. So, I get integral del dot j r prime by r minus r prime d cube r. Now, so this is again divergence of a vector times a scalar, scalar is 1 over r minus r prime. So, the same way we write it as the vector dotted with gradient of 1 over r minus r prime that is because scalar times del dot of j prime j which is a function of r prime is 0. So, I write this and now what I do is this since the gradient acts on a function of r minus r prime I will use a minus sign and make that gradient a gradient prime meaning thereby that it acts on the variable r prime and not an r and that is as long as it is a function of r minus r prime I can do that by simply adding a minus sign. So, I got this now this quantity now everything here is with respect to the prime coordinate. Now, what I do is that I write this as del prime dot j r prime by r minus r prime plus there is a minus sign and hence there is a plus 1 over r minus r prime del prime dot j d cube r. So, this is this is a simplification that we have achieved of the first term of that. Now, let us look at what does it mean firstly this term which is an integration of a divergence of a quantity integrated over d cube r prime these integration variables should be d cube r prime and not d cube r. So, I can convert this using the divergence theorem to this quantity dotted with d s prime and as we have done several times I can take the surface to infinite distances and make this integral vanish. So, I will be left with simply this term and we have seen that we are dealing with steady state magnetic phenomena and I had talked earlier about the continuity relation and we had said that for the steady state phenomena del dot of j must be 0. So, therefore, this is also 0. So, therefore, from the expression that I had written down here of the curl of b the first term vanishes and I am simply left with the second term. So, let me look at what the second term is. So, I am left with curl of b is equal to mu 0 by 4 pi del square of j of r prime by r minus r prime d cube r prime. So, this is and there is a minus sign entrant. So, what I will do is this as we have done several times I will take the del square inside, but I know that del square acting on the variable r has no effect on this j of r prime. So, this will be written as minus mu 0 by 4 pi integral j of r prime times del square of 1 over r minus r prime. Several times we have pointed out that del square of 1 over r is minus 4 pi times a delta function of r. So, this quantity is minus 4 pi times delta function which is a 3 dimensional delta function. So, I write a delta cube of r minus r prime these are all modulus of vectors. Now, because of this delta function I can do this integration there is a minus sign here minus sign here 4 pi cancels out and I will be simply left with j r prime delta r minus r prime d cube r prime which is nothing, but j of r and 4 pi has cancel out I am left with a mu 0 j of r. So, therefore combining the two I have got divergence of b equal to 0 absence of magnetic monopoles and curl of b is equal to mu 0 j due to compare these expressions with the corresponding expression for the electrostatic field. Remember del dot of b equal to 0 I had del dot of e equal to rho over epsilon 0 that was my electrostatic Gauss's law. So, of course we have already understood that why this is 0 and not something del cross e electric field was a conservative field. So, therefore the del cross of that was equal to 0, but you notice that del cross of b is given by mu 0 times j. Let me do an integral formulations of this equation. So, del cross of b equal to mu 0 j now suppose I convert this by taking so this is my circuit and I define a surface and I say that look in that case if I define a surface then I can write down del cross b dot d s is same as j dot d s. So, this would be a surface through which the current is flowing that is surface perpendicular to the direction of the current. Now, since this is true for an arbitrary surface. So, let me write it here. So, I get integral of del cross b dot n d s is equal to mu 0 times j dot n d s that is true of an arbitrary surface that you could take and because of that I must have integral of b dot d l. Remember that I have used Stokes theorem here which converts the surface integral of a curl to line integral of the vector field itself and j dot n d s is my current. So, therefore, this is mu 0 times i and this is usually called the Ampere's law of magnetism. In an identical way del dot of b equal to 0 can be converted into integral b dot d s is equal to 0 in the same way as we had converted del dot of e equal to rho by epsilon 0 to integral e dot d s equal to q divided by epsilon 0. So, this is the Gauss's law of magnetism. We do next is to use the Byrd-Sawart's law and the Ampere's law illustrate a few problems. I would like to point out that the physical content of Ampere's law is no different from that of the Byrd-Sawart's law. However, frequently if I particularly in cases where I have symmetry because it is expressed as an integral b dot d l and if you want to use a definition like that to extract information about b then I need this integral b to be expressed as a functional form which will be possible only if I have substantial amount of symmetry. So, therefore, Ampere's law has limited validity because it cannot be always used, but the your Byrd-Sawart's law is fairly general. So, let me let me start by taking a long straight wire. So, a long straight wire like this. Now, I am interested in calculating what is the magnetic field at a distance let us say r from the wire. Now, purely by symmetry consideration I can imagine that the magnetic field will be the same at the same distance from the wire. In other words, it can only depend upon the distance r from the wire. So, if I take a circle of radius r with the center at the location of the wire then this magnitude of the magnetic field everywhere must be the same. The direction of the magnetic field is given by the standard right hand rule that is if you hold the wire with your finger pointing in the direction of the current. The direction in which your fingers curl that gives you the direction of the magnetic field. In other words, the direction of the magnetic field is azimuthal. So, it is depends upon 5. So, in that case what I can do is I can calculate what is the integral of b dot d l by taking a circle of radius r. Since, b is the azimuthal and the circle that I take is also in the same direction b dot d l is same as magnitude of b times magnitude of d l and then the integration since magnitude of b is the same and the integration over d l will be nothing but 2 pi times r and that must be equal to mu 0 times the current that is flowing through. In other words, the field b at a distance r from the wire is given by mu 0 i by 2 pi r that is its magnitude times of course, the unit vector in the direction of 5 that was easy because the problem has substantial symmetry. Now, what I am going to do is this. I am going to recalculate the same thing by taking a by using Bayes-Sawart's law. So, if I use Bayes-Sawart's law let us look at how does it work. So, let me define the x y z axis the same way there is my x axis there is my y axis and the z axis is out of the plane of the paper. Now, I will take the wire in the direction of x axis and the point at which I am interested in calculating the magnetic field to be at on the y axis remember that I can always do it because I have an infinite wire. So, whichever point I am interested in calculating the magnetic field I simply that and I use it for defining my axis. So, what I do is this is my point p and this I will take as my this is my origin. So, this is my vector r. So, if I take a current element at a vector position r prime then this is my vector r minus r prime. So, let us look at how does it go if you recall your expression for the Bayes-Sawart's law let me go back a little bit. So, remember that what I had was d l prime cross r minus r prime in this case my d l prime is the same as in the x direction. So, it is nothing but I times d x. So, let me return back to that. So, what we need is I times I cross r minus r prime. Now, notice I cross r prime is of course 0 because r prime is also along the x axis. So, I am left with I cross r and r is taken along the y axis. So, therefore, I cross r is along the z axis. So, the unit vector k unit vector k times r minus r prime magnitude sin of this angle times sin of this angle. What I do there now is this that I can write this as equal to I want to write it in terms of this fixed distance r. So, notice that I have got r minus r prime since this angle is theta I can write this distance in terms of r times tangent of theta minus pi by 2 which is equal to minus r cot theta. I use this that my d x then. So, this is my x is equal to this is x is equal to r tan of this angle. So, therefore, it is minus r cot theta. So, d x is equal to r cos x square theta d theta and I can write down r minus r prime magnitude itself as equal to r cos x theta. So, therefore, my magnetic field which is along the direction k mu 0 i by 4 pi. If you collect all these things you have got magnitude of r minus r prime times sin theta coming from I cross r minus r prime. You had a 1 over r minus r prime cube. So, this one will go left with r minus r prime square which is r square cos x square theta and I have a sin theta there. So, I get it as sin theta divided by cos x square theta r square cos x square theta and of course, r cos x square theta d theta that is that is my d x there. Now, notice this thing that things cancel out there. I am left with simply an integration of sin theta from 0 to pi. So, I am left with simply integration of sin theta from 0 to pi which gives me minus cos theta and when I put the limit I get a factor of 2. So, therefore, I am left with again k mu 0 i divided by 2 pi r which is of course, was obtained in a much simpler way in from the Ampere's law. I will use another application now to calculate the magnetic field on the axis of a circular loop as a an expression which is used several times for calculating other things in particular the field inside a solenoid. .. So, let us refer to this picture here. So, I have taken my circle which is in red in the x y plane. So, this is x axis this is y axis and z axis is along the axis of the circle that is perpendicular to the plane of the circle passing through its center. So, remember that what I am going to do I need my to apply biots of art law. So, I first need an expression for d l remember that i d l cross r minus r prime. So, let us look at the vector relationship first. So, I will find it easy to show it just for the circle let me show it along the x axis and y axis suppose I have a circle here this is essentially recalling some vector algebra that we have done this is in the x y plane. Now, what I am going to do is this I am looking at an element at an angle theta here making an angle theta with the x axis and the direction of that vector is along the tangent at the point p. The magnitude of the vector because I am taking a small element there is a times d theta if I take the element to be at the position theta making an angle d theta. So, I have a vector along the tangent to the circle of magnitude a d theta. Now, how do I resolve it along the x and the y axis. So, notice that I can resolve it like this and that now. So, therefore, since this is the vector which is of magnitude a d theta the x component is minus a sin theta because this angle is theta. So, therefore, this angle is theta and hence this is 90 minus theta. So, therefore, my d l will be minus a d theta minus because the direction of that component is along the negative x direction sin theta times i and of course, a d theta cos theta because cos theta direction the direction of along the y axis is positive j. The point p this is of course, really not the point p, but the point let me call it p prime that is the position at which I have got the current element. Now, the vector r prime itself which is this vector I can resolve it and that simply gives me a cos theta times i because this is in this direction it is outward direction. So, r prime vector itself is a cos theta times i plus a sin theta times j and the vector r returning to the slide the vector r is along the z axis I have rewritten the two relationship that we have proved. So, vector r is z times k. So, therefore and further the magnitude of r minus r prime by Pythagoras theorem is r square plus a square square root. Now, you put all of them in what is d l cross r minus r prime now. So, that is very easily calculated because r minus r prime r prime I have written now r is along the z direction. So, therefore you just write this d l cross r minus r prime this is along the k direction you have got the d l expression there you can easily show that this has i j and the k component important one is the k component because d l has an i and a j and r minus r prime will give you this relationship. So, if I divide this by distance which is a square plus z square if there was a cube there. So, a square plus z square to the power 3 by 2 d theta all these quantities are constant other than theta. So, therefore the integral of cos theta or integral of sin theta they give me 0 and I am simply left with a term which is dependent upon k. So, this gives me mu 0 i a square by 2 a square plus z square to the power 3 by 2 k I would like to point out that the current times pi a square which is the area of the loop is equal is defined to be the magnetic moment of the current loop. So, therefore this is also written as mu 0 by 2 pi m by a square plus z square to the power 3 by 2 if you take z to be very large then you notice that this gives me a 1 over r cube dependence of the magnetic field. We will see that this is nothing, but the way that dipole fields behave.