 Welcome to the session of Biod Savard's Law. Myself Rohini Mirgu, Learning Outcomes. At the end of this video you will be able to state and explain Biod Savard's law. You will be able to determine the direction of magnetic field intensity. You will be able to apply Biod Savard's law to current carrying conductor to calculate magnetic field intensity. These are the contents of the presentation. First of all I would like to ask you a question, how magnetic field is produced? You can think on this question, you can find out the answer, you can write the answer. Of course the initial answer can be, when the magnets are there, there is a magnetic field. But if the permanent magnet is not available, then how magnetic field is produced? So, how magnetic field is produced? I hope you got the answer. The electric current produces a magnetic field. So, when we pass the electric current, then it produces a magnetic field like this. The current is passed through this loop or any current carrying conductor. When the current is passed through any current carrying conductor, around this the flux lines are formed and these are nothing but the magnetic field lines. So, this is what is the magnetic field. Now look at the Biod Savard's law or what is the importance of Biod Savard's law? Biod Savard's law talks about the equation for magnetic field intensity. If magnetic field is produced when current is passed, how much magnetic field is produced? What is the intensity of the magnetic field produced? That is, can be explained using Biod Savard's law. So, let us look at its statement. Before that statement, we need to assume something like, assume a differential DC current carrying conductor producing a magnetic field in free space. Here, I assume that the current carrying conductor is placed in free space. So, like this, this is a current carrying conductor carrying current I in this direction and I assume that this conductor is placed in free space. Before going to a statement, let me tell you what other things I have shown in this diagram. In this diagram, there is a current carrying conductor carrying current I. A small length of this current carrying conductor I considered having the length DL as it is having the direction in the direction of the current, the length is called as DL bar. And let us assume that around this, of course, the field is produced and let us assume that I want to find the field at this point at point P due to this current carrying conductor. So, the line joining this point P to this DL is nothing but R. And the unit vector in the direction of R is nothing but R cap or AR bar. And this line making an angle theta with the filament, filament is one and the same as a current carrying conductor. So, this is a filament, this is a current carrying conductor. Now, statement at point P, the magnetic field or magnitude of magnetic field intensity produced by differential element is directly proportional to the product of current, the magnitude of differential length, the sign of the angle lying between the filament and the line connecting the filament to point P at which field is desired. The magnetic field intensity is inversely proportional to the square of the distance from differential element to point P. Now, when I talk about the direction, the direction of magnetic field intensity is normal to the plane which is containing the differential filament and the line drawn from filament to point P. So, to these two, the direction is normal and the magnitude is directly proportional to the current I, also the product of I and the magnitude of differential length and the sign of the angle and inversely proportional to the distance which is R. So, mathematically I can express the same field as I say directly proportional to the product of current the differential length and sign of the angle inversely proportional to square of the distance and when I combine these two, I get dH bar. H is the magnetic field intensity as I am finding out the magnetic field intensity for differential length I say it is dH bar, it is ideal sin theta upon R square. Now, constant of proportionality is according to Biot-Savart, the constant of proportionality is 1 upon 4 pi. So, dH bar comes out to be ideal sin theta upon 4 pi R square. As we know that A bar cross B bar is equal to modulus of A bar, modulus of B bar and sign of the angle between two. So, that I can replace this ideal sin theta by I is the scalar quantity as it is dL is dL bar cross AR bar I can write instead of this sin theta. Now, this is for a small length dL. So, for this for the part of the filament this is the equation, but if I want to find for a complete length of the filament in that case I need to integrate dH bar I need to integrate this equation. So, H bar is closed line integral of ideal bar cross AR upon 4 pi R square. The unit is amperes per meter and this is what is called as Biot-Savart's law. This is a closed line integral why it is called closed line integral because I need to consider every length of the filament all elements of the filament. Now, Biot-Savart's law is this fine. Now, what are the different quantities or terms in this where I is current dL is a small length of the filament small section of the filament R is a distance between the filament and the point P. What is the point P? Point P is the point where I want to find the magnetic field intensity and then the unit vector in the direction of R is nothing but AR. So, here I use AR in the diagram it is shown AR cap both are one and the same and this in between is not X or it is not just a multiplication it is a cross product. So, there are alternate forms of Biot-Savart's law like this I can also write Biot-Savart's law in terms of J bar J bar is a current density conductor of magnetic rectangular cross section is shown here having the sides B and H. So, current density is given as current upon area it is current upon area means multiplication of two lengths as J is written as a vector. So, some unit vector will be there along with it. So, I can find out the current as surface integral of J bar dot dS bar as it is a current its unit is amperes. Another form is like this K is also called as a current density but it is a surface current density. So, the current is distributed over a surface having the length B. So, it is given as current upon length its unit is amperes per meter K is I upon B current can be given. So, the alternate forms of Biot-Savart's law are H bar is volume integral of J bar dV cross AR upon 4 pi r square why it is a volume integral because it is a integration with respect to volume dV. Also H bar can be given as surface integral of K bar dS cross AR upon 4 pi r square. Why it is a surface integral here? Because the integration is with respect to surface. Surface integral means double integral, volume integral means triple integral. Let me talk about a direction. The direction of magnetic field intensity is normal to the plane containing a differential filament and the line drawn from filament to point B. So, the direction will be like this if my first finger is showing ideal, second finger is showing the direction of R then thumb will be indicating the direction of magnetic field intensity. Now you may ask me here like I have said magnetic field intensity as dH but here I am writing it as a dB. What is dB? dB is differential magnetic flux density. So, magnetic flux density and magnetic field intensity are one and the same with a small difference or what is the relation between B and H that I will show you further. But at this stage we will talk about the direction of magnetic field intensity. The magnetic field intensity is perpendicular to the plane of R and ideal. So, the perpendicular or a normal component in that direction is the direction of magnetic field intensity. The relation between B and H, B bar can be given as mu naught H bar. Now let us apply the Biot-Savart's law to this given example. Assume that there is a point P1 at 1, 2, 3, there is a point P2 at minus 3, 1, 2 and this is the I delta L1. DL can also be written as delta L with small filaments, small part of the filament can be written as delta L. So, delta L1 is given to us and we need to calculate delta H2. So, how we can calculate this? Let us go for the solution. So, first apply the Biot-Savart's law. DL is given in the form of delta L. It is given but it is given in the form of delta L. So, I can modify my equation like this. Now, my job is just to substitute the values of unknowns. I delta 1 is known to me. Only unknown is R and AR. I need to find out R and AR. So, as I said, R bar is arrow head coordinate minus TEL coordinate. So, arrow head coordinates are P2 and TEL is P1. So, minus 3 minus 1 into the unit vector AX bar. As we consider three perpendicular directions AX, AY, AZ. So, unit vector is here AX. Then minus 1 minus 2 into AY. Then 2 minus 3 is AZ. So, this is R bar. The unit vector is nothing but a vector upon its magnitude. I get this as AR bar. Now, substitute AR and AR bar in the equation of delta H2. Continue the example. Then delta H2 is this one. I take under root 26 below so that I get under root 26 raise to 3 by 2. Now, I want to find this cross product. How to find the cross product? So, for that I have written these two brackets here. And then first row comes to be AX, AY, AZ, modulus of that. And the coefficients of first brackets should be written here. Remember, cross product in the cross product I cannot interchange the order. I must first write the coefficients of this minus 1, then 1. Then AZ coefficient is 2. Again, AX coefficient is minus 4. Then minus 3 AZ coefficient is minus 1. Then AX bar into the bracket cross product like this. Minus 1 minus of minus 6 minus 1 minus of minus 6 into AX minus of minus so it comes to be plus. When I am writing the determinant I should take minus AY. Remember, there is a minus sign with AY. So, with AY I leave this row and then I will multiply this minus 1, minus 1 comes to be plus 1. Minus of 4 into sorry minus 4 into 8 is minus 8. So, 1 plus 8 plus AZ bar into the bracket that should leave this column and I should find out like this 3 minus of minus 4. So, 3 plus 4. So, this is my cross product. So, I get the cross product. I have written the cross product here and this is delta H2. So, this is my magnetic field intensity at point 2 due to the current carrying element at point P1 and its unit is in amperes per meter. These are the references used for this video. Thank you.