 Namaste. Welcome to the session of Magnetic Field Intensity on the Axis of Circular Loop. Myself Rohini Mirgu working as an associate professor at Walshian Institute of Technology in the department of Electronics and Telecommunication Engineering. Learning outcomes at the end of this video you will be able to apply Biod's Havards Law to derive Magnetic Field Intensity on the Axis of Circular Loop. Determine the direction of Magnetic Field Intensity for a given circular loop. Now, before we proceed further as we said we are able to apply the Biod's Havards Law. So, let us recall the equation of Biod's Havards Law. So, this is the equation of Biod's Havards Law. H bar is equal to closed line integral of i dL bar cross ar bar upon 4 pi r square where i stands for current. dL bar is a small section of the filament which is a vector quantity we consider is direction as the direction of the current. r is the distance between the filament and the point p where we want to find the magnetic field. ar bar is the unit vector in the direction of r and x is not just it is a x it is not a simple multiplication. This is a cross product. H on the axis of the circular loop that we want to find out. So, let us consider the x-y system. In this let us assume a circular current carrying loop is placed in x-y plane at z equal to 0 centered at origin. So, consider this loop which is centered at origin. Consider that this loop is carrying current in this direction in anticlockwise direction having the current as i. Let us assume point q at 0 phi z where we want to find the magnetic field intensity due to this circular current carrying conductor. So, for that let us assume a small section of the filament dL bar. Consider the point on this dL bar as r phi 0. The coordinates here has come as r phi 0 because point p is at certain radius and it lies on the plane phi constant plane. So, r phi as I said that the loop is centered at origin and placed in x-y plane with z equal to 0. So, z is coming to be 0. Now, as I want to find the field due to this loop at this point at point q join the point p to point q. So, this is the line which is joined here we want to find out the dH bar due to this section of the filament and then we will find for the complete loop. Consider the line which is joining from p to q is r but show the arrow head towards point q where we want to find the field. So, this vector comes to be r bar. Start deriving. According to Biot-Severts law h bar is equal to i dL bar cross ar bar upon 4 pi r square all this is inside the closed line integral. But recall the previous diagram. In this diagram we have drawn everything this loop is there but we are first trying to find out dH bar for this small section of dL bar. So, that let us modify the equation remove the integral sign and write dH bar which is i dL bar cross ar bar upon 4 pi r square. Now, dL bar here can be given as r d5 a5 why r d5 a5 because dL bar is this circular section circular part. Here I am considering the cylindrical coordinates of steam in cylindrical coordinates of steam the coordinates are r, phi and z. And this circular section all the time corresponds to phi. So, this part is the length is r d5 and unit vector is a5 bar. When I multiply it with i I get i r d5 a5 bar and r bar. Now, I want to find r bar because I want to substitute these values in the equation. I want to find r I want to find ar i dL bar already I have calculated. So, r bar is head minus tail this arrow head is q and arrow tail is p. So, q coordinates minus p coordinates will give me r bar. So, it is 0 minus r ar phi minus phi goes to be 0 next plus z minus 0 a z bar. So, this is my r bar. Now, find here I just want the magnitude. So, let us find the magnitude of r bar which is under root of r square plus z square squaring and adding the terms. Ar bar is nothing but a vector upon its magnitude. So, this vector divided by this magnitude is ar bar. Now, substitute ar bar, idl bar as well as r bar in this equation of dh bar. So, again recall idl bar is this, ar bar is this, r bar is this or modulus of r bar is given by this equation. So, substitute all this in the equation of dh bar. So, first let us substitute idl bar then substitute ar bar then substitute r bar so that I will get dh bar like this. Now, I can simplify this equation I can write this rest to half and this is a square I can combine this together and that goes to be rest to 3 by 2. So, I can write dh bar as i r d phi a phi cross minus r ar plus z a z divided by 4 pi r square plus z square rest to 3 by 2. So, dh bar is this. Now, this is not a simple cross it is a cross product. So, to find the cross product I need to follow this triangle. Here I want the cross product of a phi with r. So, a phi with r I am going in the opposite direction of the arrow of course the result is a z but it is either positive or negative that depends on the direction of the arrow. As I am finding out a phi cross ar I am going in the opposite direction of the arrow so my result is minus a z. Again I also want the cross product of a phi cross a z. So, a phi cross a z I am going in the same direction of the arrow so my result is ar. So, these are the results which I need to substitute in the equation. So, this is minus a z already minus sign is there so it comes to be r a z and here a phi cross a z will give me this ar. So, this is my equation fine this is dh bar. Now, to find h bar I need to integrate all the elements from 0 to 2 pi of dh bar all the dl bars together from 0 to 2 pi will give me h bar. So, while integrating let us think a bit on this. In dh bar there are two terms one is a z bar term another is ar bar term. What I mean by this is this dl bar has two components ar bar and a z bar. So, when I am trying to integrate what happens is there is a diametrically opposite point to the point p which is also having the two components ar and a z like this. Now, if you if you observe carefully this ar and this ar are exactly in the opposite direction. So, they cancel each other and this a z and this a z both are in the same direction so they sum up when we integrate. So, for every dl bar there is another dl bar diametrically opposite to it with which all the a z bar components gets added and and ar bar components get omitted. So, that while integrating omit ar bar components and consider only a z bar components. So, again let us see this equation where you can see I have omitted ar bar component and I have taken only r a z in the bracket. So, this comes to be I r square ar upon 4 pi r square plus z square raise to 3 by 2 integral of d phi. So, d phi is phi is from 0 to 2 pi. So, integration gives me the result as 2 pi. So, finally my answer is I r square upon 2 r square plus z square raise to 3 by 2 a z bar. Unit of magnetic field intensity is amperes per meter. This is the magnetic field intensity due to circular current carrying conductor when I find it on the z axis. So, the unit vector has come as a z. Now, let us look at the quantities. In this equation I is the current, r is the radius of the circular current carrying conductor or filament. z is nothing but the perpendicular distance between the center of the circular filament to the point p. a z bar is the unit vector indicating the direction of the magnetic field. Now, let us think about the direction. Now, as I said in the previous derivation the direction is a z bar. But whether it is all the time a z bar let us look at this. The direction of magnetic field intensity is given by right handed thumb rule like this. If thumb is indicating the direction of the field then chord finger will be indicating direction of current or chord finger will be indicating direction of field and thumb will be indicating the direction of current. But here I am considering a circular loop. So, current will be flowing with, can be shown with my chord fingers and thumb will be indicating the, so the field direction is decided by the direction of the current. knowledge I want to ask you a question what will be the direction of H if point Q is below the circular loop if point Q is below the circular loop what I mean is this is the point Q but if I consider it below here. So whether there will be any change in this equation so let us look at the answer if point Q is below the circular loop still the direction of H is AZ only because the direction of H bar is decided by right handed thumb rule which is dependent on the current direction and it is independent on the position of point Q. Let me ask you one more question if point Q is at the center of the circular loop what will be the change in the below equation in this equation so if point Q is here instead of here if the point Q is here what change will be there so naturally the Z is going to be 0. So the answer is if Q is at the center of the circular loop Z is equal to 0 so equation will be modified as I by 2R AZR these are the references used for preparing this video. Thank you.