 Hello, Namaste. Welcome to the session of magnetic field intensity for infinite length filament. Learning outcomes, at the end of this video you will be able to apply Biot-Sivert's law to derive magnetic field intensity for infinite length filament. You will be able to determine the direction of magnetic field for given filament. Now, before I start I want to ask you to recall the equation of Biot-Sivert's law. As we want to apply the Biot-Sivert's law, please recall the Biot-Sivert's law which we have already seen in the previous video. I hope you got the equation of Biot-Sivert's law. Biot-Sivert's law can be given as h bar is equal to closed line integral of i dL bar cross ar bar upon 4 pi r square. Of course, the unit of magnetic field intensity is amperes per meter, where i stands for current, dL is a small section of the filament, small length of the filament and ar bar is the unit vector in the direction of ar bar. And what is r? r is the distance between the filament and the point P where we want the magnetic field. And in between this symbol is not a simple multiplication, this is a cross product. H for infinite length filament we want to find out. For that I will consider the x, y, z axis three dimensional coordinate system and let us assume an infinite length filament like this. And I will consider that the current i is flowing through this infinite length filament. And at the same time I want to tell you that here our intention is to find out the magnetic field at point P in the x, y plane. The point P has coordinates r, phi, 0 in cylindrical coordinates. Why r, phi, 0? Because point P is at certain radius from the origin, so r, as point P passes through some plane, so that is phi and as it is the x, y plane here assume that z equal to 0, so this coordinate is 0. So, the coordinates of point P are r, phi, 0. Let us assume a small section of the filament having length dL bar. Why small section? Because when the filament is of infinite length, for this infinite length I cannot find out h bar instantaneously or rather I need to find out small section dh bar that means the magnetic field intensity for this small section. And then I will come up or integrate of all these sections so that I get h bar for infinite length filament. Let us consider the point where I have considered the small length of the filament that point is 0, phi, z. How the coordinates have come as 0, phi, z? Because radius at this point is 0 because it is the point is on the z axis, phi because the filament passes through some plane phi and z as it is from origin at certain height so that the coordinate is z. Let us join this point with the point P. So, after joining these two points I should put the arrow head towards point P. Why towards point P? Because I need to find the field at point P where we want to find the field at that point arrow head should be there. The distance can be denoted as r bar. I will consider the same diagram here and let us start deriving the equation for h bar. I have written a Biot Savard's law, dh bar is equal to ideal bar cross ar bar upon 4 pi r square. The same equation is repeated only the thing is as I am finding out for this small length I have omitted the integral sign and beyond that I have written as a dh bar because it is for a differential length of the filament. dl bar, here in this case I need to find out the unknowns from this equation. In this equation the unknown is dl bar, i is current as we know some value is given but what is dl bar? What is ar bar? 4 pi is a constant and what is r? These things we need to find out. So out of that dl bar here is dz az bar. Why it is dz az bar? Because the filament is placed along z axis which is ranging from minus infinity to plus infinity. But when I consider a small section of the filament which is dl bar it comes to be dz az because it is along z axis. When I multiply it with i, i dl bar can be given as i dz az bar. Ar bar is the head coordinate minus tail coordinate. Ar bar is this vector. For this vector how I can find out ar bar? It is the head coordinate minus tail coordinate. Head means which head? The arrow head where the arrow is pointing. The arrow is pointing towards point p these are the coordinates minus this coordinate that way we can find out a vector ar bar. So head coordinates here are r minus the tail coordinate is 0. So r minus 0 unit vector ar bar is 5. Head coordinate is 5, tail coordinate is 5 into a phi bar then 0 minus z az bar. So r bar comes out to be r ar bar minus z az bar. Now I want to find out only magnitude as it is needed here also for ar bar I need the magnitude. So the magnitude of ar bar can be given as under root of r square plus z square. Squaring and adding the magnitudes of r bar. So this is r bar modulus of r bar. Now ar bar can be given as a vector upon its magnitude. The vector is this one which I have written here, upon its magnitude the magnitude is here which is written here. So I have found out dl bar. I know now ar I know r. Now my job is to substitute all this in the equation of dh bar that is going to be out. So dh bar is idz az this is nothing but idl bar. r ar minus z az upon under root of r square plus z square this is nothing but ar bar divided by 4 pi r square. r square is under root of r square plus z square whole bracket square and this is in between is a cross product. So let us simplify this equation. When I simplify this equation I get idz cross this bracket r ar minus z az upon 4 pi r square plus z square raise to 3 by 2. How it has come as 3 by 2? This is a square and this part is 1 half. So when I sum up it comes out to be 3 by 2. Now my job is to find out this cross product. Which cross product? Az cross ar. When I am finding out the cross product both should be vectors. So first vector is az another vector is ar. Then az cross az. This I need to find out. To find out the cross product follow the spangle. Like if I want the cross product for example I want the cross product of ar and afi. Ar cross afi will give me az. But if I do reverse afi cross ar then in that case the answer will be minus of az. So in the same way here I want az cross ar. So az cross ar as I am going in the same direction of the arrow my result will be afi. So az cross ar is afi and az cross az is 0. Remember self cross product is always 0. So substitute these values in above equation. So dh bar is ir cross dz afi. And this term has been to 0. So I got ir dz afi upon the denominator is same. 4 pi r square plus z square raise to 3 by 2. Now fine I have found out dh bar. But my intention is to find out h bar for the infinite length filament. So that h bar can be given as integral of dh bar. And the integration limits are from minus infinity to plus infinity. Because it is a infinite length filament. So let us substitute this dh bar here and let us find out h bar. So h bar can be given as integration minus infinity to plus infinity. This is the dh bar which we have calculated. Now I will take out the constants outside and I will get the equation like this. h bar is ir afi divided by 4 pi into the integration sign dz upon r square plus z square raise to 3 by 2. So this term I need to integrate. As integration with dz, integration is with respect to z. So that this part is inside the integral. All others will be the constants which are taken outside the integral. Now this term I cannot directly find out the its answer. So let us try to find out the integral of this part. And this can be done like this. I can solve this integral by substitution method. Substitute z equal to r tan theta. dz comes out to be, when I differentiate dz is r tan differentiation is sec square r sec square theta d theta. At the same time let us find out what will be the limits if I substitute. So limits will be like this. When z is standing towards infinity theta will be tending towards pi by 2. When z is tending towards minus infinity theta is tending towards minus pi by 2. So let us substitute this inside this bracket or in this integral. But before that let us calculate this what is r square plus z square raise to 3 by 2. If I substitute this r square plus z as r tan theta r square tan square theta raise to 3 by 2. This value comes out to be r square comes out as a constant outside raise to 3 by 2. So that it comes out to be r cube. And inside the bracket there will be 1 plus tan square theta. 1 plus tan square is nothing but sec square. So sec square raise to 3 by 2 gives me sec cube theta. So that I can substitute this value and I can substitute dz which is this one at this place. So dh bar this constant part as it is instead of dz there comes r sec square theta d theta. Instead of this bracket it comes to be r cube sec cube theta r and r cancels out. Here it remains r square sec square sec square cancels out. Here it remains sec theta. So when I solve this integral so my answer comes out to be h bar is equal to i by 2 pi r a phi unit of magnetic field intensity is amperes per meter. So this is the equation of h or h bar for infinite length filament. Let us look at this equation again and see what are the terminologies. So h bar is i by 2 pi r a phi bar amperes per meter where i is nothing but a current which is flowing through the filament r is the perpendicular distance between the filament and the point P where we desire the magnetic field. And here it is a scalar quantity and a phi bar is the unit vector indicating the direction of magnetic field. Let us decide how to find the direction. So the direction of magnetic field intensity is given by right handed thumb rule like this. Like if I hold the filament with my right hand and my thumb indicating the direction of the current this thumb is indicating direction of current then my curve fingers will be indicating the direction of magnetic field intensity. So this is the way we can decide the direction of the magnetic field. So this is the method by which we can decide the direction. These are the references used for preparing this video. Thank you.