 Hello everyone, this is Dr. Rupali Shalke working as an associate professor in Department of Electronic Engineering at Walsh and Institute of Technology, Sholapur. In this video, we are going to discuss with the Ampere Circuital Law. Learning outcomes, at the end of this video students are able to state and derive Ampere Circuital Law or which is also known as Ampere's Works Law. These are the contents which are going to be covered in this video that are the, first we will state the Ampere's Law and then we derive the equation for that and we will see some problems related to the Ampere Circuital Law and the last references. The Ampere Circuital Law, it is stated that a line integral of magnetic field intensity around the closed loop path is equal to the current enclosed by that path. Mathematically, when we write this statement, it is a line integral which is nothing but a close integral of h bar. What is h bar? h bar is the magnetic field intensity, magnetic field intensity and dl is the loop covering or the entire loop. Therefore, close integral of h dl is h bar dl is given as a equal to i. i is nothing but a current enclosed by that path. Now, let us derive this statement or this mathematical equation for that we will consider an infinite current carrying filament which is placed along the z axis. In this figure, it is shown that an infinite line filament which is current current is i flowing through this. It is an infinite current filament and it is placed along the z axis. Then the magnetic field is created in a form of circular loop and let us consider that as a radius r it is covering. Now, to verify this, we will state the magnetic field intensity, the magnetic field intensity or magnetic field due to it is given by h bar is equal to i upon 2 pi r a phi bar where i is the current flowing through the filament, r is the path or a circular path radius of the circular path and a phi is the direction. As we are seeing in a circular path, the direction will be a a phi. This derivation, we are already discussed in a previous slides, now previous videos which is a called as a field magnetic field intensity due to the infinite filament. Now, let us consider a path surrounding as it is a circular path for this the d l the magnetic field this magnetic d l is given as a d l bar is equal to d r a r bar plus d r r d phi a phi bar plus d z a z bar. As it is we are considered it is a circular path therefore, we will see the differential component of the length in a cylindrical coordinates. This is a d r in a cylindrical coordinates. Then if r is the radius of the path and it is placed in a horizontal plane, if it is in a horizontal plane then d r and d z because we are only considering in a circular path that is why the d r and d z will be equal to 0. So, when we substitute in the above in a d l equation the d l equation will be reduced to r d phi a phi bar only we are considering for the a phi direction while for a r direction and z direction it will be 0 change in the length therefore, we are considering only r d phi a phi. Now, when we substitute this h d l will take h bar d l the i h is given as a i upon 2 pi r a phi and d l is r d r a phi d phi a phi as a pi the properties of dot matrix the self dot product of the unit vector is always equal to 1 in our case it is a a phi dot a phi therefore, both are the similar therefore, the dot product of this is equal to 1. So, the r r get cancelled out over here and only the equation will remain with the i upon 2 pi r d phi. Now, we will substitute in the mathematical equations of the ampere circular law which is a close integral of h d l is equal to i we will consider the LHS side LHS is equal to close integral of h d l here it is the close integral of h d l is equal to integral we will substitute now ideal from here that is a i upon 2 pi a phi i upon 2 pi a phi as the phi a range of the phi is varying from 0 to 2 pi that is a total 360 degree. So, we will integrate it over the length over the limits 0 to 2 pi as i and 2 pi is constant it will be outside the integral sign and only the we will integrate the t phi integration of d phi is phi. So, after substituting the limits it will be 2 pi by when we cancel out this 2 pi 2 pi and only it remains with the i where i is nothing but the current. So, we can state say that LHS is equal to the RHS by derivation also the RHS value the final value is the i therefore, the LHS is equal to the RHS this is what the Ampere's law states. Now what could be the conditions or what is a equation whether this equation satisfied for the different path now in this figure we are seeing that there are the three different path A path is a circular path as you see B path is a square path and this is a uneven path. So, when the current carrying when it is a this loops are formed across the current carrying conductor i the consider a conductor shown in the figure which is carrying the current i then the line integral h bar across the closed loop path will be as we have marked this path as a b and c for the path a and b it will be a equal to the what the current flowing through it according to the Ampere's circuit law it satisfies the equation because it is a closed integral but for the line integral across the path c is less than i since since the entire current is not enclosed by that path this is a reason why see if the it is a entire current is not enclosed by this path. So, it is a less than the current okay in a figure c we are showing the the line integral at a different condition or different cases in the first case if you see that h bar across the closed loop path to the current carrying conductor enclosed by path 1 and path 2 for the wire 1 and path 2 it will be the same it will be equal to i because if you see that this is a current carrying conductor across this it is covering that is for closed integral of h d l will be equal to i according to the law and this see only the position is being changed but it is covering the circular path for this conductor that is for the closed integral of h d l i d l is equal to i for this condition also but in the third case in a third figure if you observe that the conductor is over here and the magnetic field is shown to the side side then it is a there is no current there will be no magnetic field no magnetic field due to this current carrying conductor that is for the closed in closed current flowing through this will be equal to enclosed by this will be equal to 0 the current enclosed by this path will be equal to 0 therefore according to the equation as i is equal to 0 that is for closed integral of h d l will be equal to 0 and the last one that is the fourth one in this figure if you see that the closed path is not completely completed it is not a closed path but it is breaking over here even if the current carrying conductor is there this path is not entirely covered as a closed path so in this case also the closed integral of h d l is equal to 0 so this only what we can say that when it is a closed path that only the equation is been satisfied so let us see a problems on this now for that we will consider a find i for a square whose sides are 2 meter with the edges coinciding with the plus x and plus y axis and one corner is at the origin now what according to the statement we will draw the diagram we will only consider two axis x axis and y axis as the edges are coinciding so to the sides that's why the sides of the squares were lying along the axis and i is the current flowing through it one edge is at the origin so according to the law i is equal to they ask you to find out the i current so what will the i equation close integral of h d l when you expand this h is given to you h is given to you that is a 2 y square a x now as it is in a Cartesian coordinate the d l will be dx a x bar plus dy a y bar because only two directions we are considered so now we will take a dot product a x into a x it is equal to 1 but a x into a y is equal to 0 when you multiply with this bracket then the equation will reduce to 2 y square dx so after integrating this the value of y is equal to 2 which is a constant and the integration of the dx is x substituting the limits it is a i will be equal to minus 16 ampere so similarly we can take this for the h bar is equal to 2 x square y a z bar h in this case our h is same equation 3 x square y a z bar when you multiply this dot products here there is no z products in this equation this is a dz into a x will be equal to 0 dz into dy will also be equal to 0 because they are not a self dot in this case the current is going to be a 0 these are the references thank you