 Hello, welcome to the session of magnetic field intensity for infinite and finite length filament. An example, here we are going to see how to solve an example based on magnetic field intensity for finite as well as for infinite length filament. Myself Rohini Mirgu working as an associate professor in electronics and telecommunication engineering department at Vulture Institute of Technology, Sulapur. Learning outcomes, at the end of this video you will be able to calculate the magnetic field intensity for given infinite length filament. You will be able to calculate magnetic field intensity for a given finite length filament. Before we proceed further, as we are finding out the magnetic field intensity for finite and infinite length filament, let us look at first for the infinite length filament. So, recall the equation of magnetic field intensity for infinite length filament. H for infinite length filament can be given as H bar is equal to I upon 2 pi r A5 bar unit is amperes per meter, where I is current, r is the perpendicular distance between the filament and the point P, where magnetic field is desired. And A5 bar is the unit vector indicating the direction of the magnetic field. Now, example, let us consider the example of filament carrying current of 0.4 amperes is in carrying current in AZ bar direction placed in free space at x is equal to 2 and y is equal to minus 4 parallel to z axis. And H bar in the interval z is between minus infinity to plus infinity that is the A part of the question, B part of the question is z is between minus 3 to plus 3. So, when I am saying that z is between minus infinity to plus infinity, so this question comes out to be of infinite length filament and part B is of finite length filament. Let us look one by one, let us first solve for infinite length filament. The question will be like this, first consider H bar for infinite length filament as we already recall the equation which is I by 2 pi r A5 bar and infinite length filament ranging from minus infinity to plus infinity placed at x is equal to 2, y is equal to minus 4. And this filament is parallel to z axis we assume that and this filament is carrying current I in positive AZ by direction. Let us consider the point P 0 1 0, why because we want to find H bar at point 0 1 0, join this point P to the filament and join in such a way that this will be a perpendicular on the infinite length filament and the distance between these two is nothing but the r. Let us move further, here I have I highlighted the things which we want like I which is point 4 amperes and carrying current in AZ bar direction, filament placed at x equal to 2, y is equal to minus 4 and we are right now solving for infinite length filament. So this is the filament carrying current I and the distance that we said r is at 0 1 0 and the perpendicular is r. But what is this point of intersection when I draw a perpendicular from 0 1 0 onto the filament what is this point of intersection. So the point of intersection is q which is 2 minus 4 0, how the coordinates have come as 2 minus 4 0 because x is equal to 2 y is equal to minus 4 is already given as it is perpendicular on z axis or parallel to z axis the z coordinate will remain common between P and q. So the z coordinate is 0 here. Now remember I have shown the arrow head towards point P because I want the field at point P, if I want to find a vector I will be subtracting P minus q arrow head minus del. So h bar is I by 2 pi r a phi as the equation is for infinite length filament as I am solving first for a part of the question. First let us find out r bar or r. So for that the procedure is r bar is head coordinate minus del coordinate. So r bar is 0 minus 2 a x bar 1 minus of minus 4 a y bar and 0 minus 0 a z bar. So r bar comes out to be minus 2 a x plus 5 a y. As I already calculated r bar now let us determine the a phi bar. So moving ahead for determining a phi bar, a phi bar is nothing but indicating the direction of h bar. So it can be given as a l bar cross a r where a l is the direction in which the filament is carrying current. The filament is carrying current in a z bar direction positive a z bar direction. So a phi bar will be given as a z bar cross a r bar. So a r bar I need to find out so a r bar can be given as r bar upon modulus of r bar. So this is the value of a r bar. Now a phi bar to calculate a phi bar I need the cross product a z cross this. So a z cross the a x and this a z part will give me the result like this. A z cross a x is equal to a y and a z cross a y is equal to a x. So h bar if I substitute the values now I get i is of course as we already seen it is 0.4 r is modulus of r which is under root 29 and this part is nothing but a phi and when I calculate it comes out to be minus 10.98 a x minus 4.39 a y unit of h bar is amperes per meter. Now we will proceed further for determining the h bar for finite length filament. So the equation of magnetic field intensity for finite length filament let us recall it the equation for h for finite length filament can be given as h bar is equal to i by 4 r sin alpha 2 minus sin alpha 1 into a phi bar. A phi bar is again indicating the direction of h bar unit is amperes per meter where i is the current r is the perpendicular distance between the point where we want the field and between the filament alpha 1 and alpha 2 are the angle made from point p to the upper end of the filament and to the lower end of the filament and a phi bar is the unit vector indicating the direction of the magnetic field. Again the same example is here current is 0.4 filament is placed at x equal to 2 y is equal to minus 4 and we are determining now not for this but we are determining now for finite length filament that is z is minus 3 to plus 3. So consider the filament now this is not of infinite length now this is of finite length. So finite length filament placed at x equal to 2 y is equal to minus 4 and carrying current i and this filament is ranging from z is equal to minus 3 to z equal to plus 3. If I consider this point this is z is equal to minus 3 this is z is equal to plus 3 because the range is given here and I can name the ends of the filament this as a lower end and this as a upper end how looking at the direction of the current the arrow head I will show to the upper end and arrow tail I will say the lower end. Now consider the point p where we want to find the field draw this perpendicular from point p onto the filament which comes to be r. Now from this point p join to the upper end of the filament like this. This upper end of the filament makes an angle alpha 2. Now while making an angle alpha 2 remember start from considering from this perpendicular to the upper end like this. So from perpendicular to the upper end and arrow should be towards upper end. Similarly join the lower end to the filament and the angle midi is alpha 1. It is from perpendicular to the lower end the angle midi is alpha 1. So the direction I have shown as downwards. So start from perpendicular then towards upper end and start from perpendicular then towards lower end. This is needed for deciding the science. So this is the science of alpha 1 and alpha 2. H for finite length filament can be given as I by 4 by r sin alpha 2 minus sin alpha 1 into a phi. So in that let us first find out r or r bar. So r bar can be given as by the same as we have calculated previously head coordinate minus tail coordinate perpendicular coordinates we already discussed z coordinate will remain common as it is placed along z axis or parallel to z axis rather. And modulus of r bar is under root 29. So r is under root 29. Now let us determine a phi. So a phi bar can be calculated like this a phi bar is a l bar cross a r bar a r bar is r bar upon modulus of r bar again we have already calculated this the same way we get here and let us calculate this cross product a z cross a x a z cross a y for that follow this triangle for calculating the cross product and get the results like this. So I get the result as minus 2 a y minus phi u a x. So the again considering this diagram I by 4 pi r this is the equation in that I have calculated a phi I have calculated r I is known to me only the thing is calculating alpha 1 and alpha 2. So let us calculate alpha 1 and alpha 2. For that first of all decide the signs of alpha 1 and alpha 2. Alpha 1 is considered as negative alpha 2 is considered as positive why? Because alpha 2 is in the same direction of the current whereas alpha 1 is in the opposite direction of the current. So that alpha 2 in the same direction of the current is positive alpha 1 in the opposite direction of the current is negative values of alpha 1 alpha 2 are given like this alpha 2 is tan inverse of the front side that is this side is 3 why? Because the z here is 3 and z here is minus 3. So this coordinate is 0. So 0 to 3 is 3 divided by r, r is under root 29, 29.12 degrees, alpha 1 is also same it is only with minus sign minus 29.12 degrees. So again write the equation of h bar and if I substitute the values I is 0.4, alpha 2, alpha 1 and I have substituted here as a phi which we have calculated. So I get h bar as minus 5.34 ax minus 12.14 ay bar milli amperes per meter. These are the references used for preparing the video thank you.