 Namaste. Welcome to the session of Magnetic Field Intensity for Finite Length Filament. Myself, Dr. R. R. Mergo, Associate Professor, working in the ANTC department at Walsham Institute of Technology, Sulapu. Learning outcomes. At the end of this video, you will be able to apply Biodiversity Law to derive Magnetic Field Intensity for Finite Length Filament. You will be able to determine the direction of Magnetic Field for the given filament. Before moving further, let us recall the equation of H bar for infinite length filament. In the previous video, we have discussed the H bar. Here, even though I have shown as H, we have discussed about H bar because H is the magnetic field which is a vector quantity. So, H bar for infinite length filament. So, can you recall the equation? Thank you. I hope you are able to write the equation of H bar for infinite length filament. H bar is equal to I upon 2 pi r A phi bar unit is amperes per meter, where I is nothing but the current which is flowing through the filament. R is nothing but the perpendicular distance between the filament and the point P where magnetic field is desired and whereas A phi is the unit vector indicating the direction of magnetic field. When I am saying I want to find H or H bar for finite length filament, let us consider the finite length filament which is placed in XYZ sustain. Consider this is a finite length filament ranging from Z1 to Z2. So, finite length filament placed along Z axis ranging from Z1 to Z2. The current I is flowing through this finite length filament in the upward direction. Now, consider the point P which is placed in Z equal to 0 plane. This point P has the coordinates r phi 0. Why r phi 0? Because point P here as having certain radius from origin point P lies on some plane which is phi constant plane. So, phi as I am assuming that this point P is in XY plane but Z equal to 0 plane. So, Z is 0. Now, consider a small length of this filament like this which is dL bar and let us consider the point P on this dL bar which is 0 phi Z. Why the coordinates are 0 phi Z? Because this point lies on some phi constant plane. This point is at certain height Z from origin and this point as it is at on the Z axis the radius will be 0. Now, join this point to the point P in order to proceed further to find h bar for finite length filament at point P. Join this point to the point P. Now, show the arrow head towards point P. Why towards point P? Because I want the field at point P. So, arrow head is towards point P as I have shown the arrow here. So, this is a vector. So, that the distance which is named here is r bar. Bar indicates it is a vector. Now, join the upper end of the filament to point P. This upper end makes an angle alpha 2 with Z equal to 0 plane and this join the lower end of the filament to the point P which makes an angle alpha 1 and the point which is which I have considered on the small length of the filament making an angle alpha. So, once this assumption and these terminologies are known to us, let us move on for deriving h bar for finite length filament. h bar for finite length filament for that I have again the same diagram is shown here. This is a finite length filament from ranging from Z 1 to Z 2 current i is flowing in upward direction. I have considered small section of this length dL bar at 0 phi Z. I have considered point P at r phi 0 where I want to find the field and then we have joined all this the upper end and lower end to point P as well as the 0 phi Z point to point P making angles alpha 1 alpha 2 and alpha and this distance is or this vector is r bar. Now, let us start deriving for that start from Biot-Savart's law. According to Biot-Savart's law dh bar is equal to i dL bar cross ar bar upon 4 pi r square. Now, here dL bar is dz a z bar here dL bar is dz a z bar why because it is placed along the Z axis. The filament is placed along Z axis. So, when I multiply it with i i dL bar will give me i dz a z bar. So, I have calculated this term. Now, let us move on for calculating r modulus of r and ar. So, r or r bar is can be given as head coordinate minus tail coordinate. Head coordinate is this arrow head coordinates which are r phi 0 and tail coordinate is 0 phi Z. So, r bar comes out to be r minus 0 into ar bar phi minus phi into a phi bar and 0 minus Z into a z bar. So, r bar is r ar minus z a z. Modulus of r bar is under root of r square plus z square. Why modulus I want? I want this here the magnitude of r. At the same time also I want to find ar bar. Ar bar unit vector is also needed r bar upon modulus of r bar. So, that I need our modulus of r bar also in the calculation. Ar bar is r bar upon modulus of r bar. So, r bar comes here this modulus of r bar comes here. Now, moving further substitute the values in the equation of Bayard Savart's law. What was Bayard Savart's law? dh bar is idl bar cross ar bar upon 4 pi r square. So, idl bar comes here as idz a z bar cross ar bar is this term upon 4 pi r square. So, modulus of r bar is substituted here. Now, let us simplify this equation. So, I get it as idz a z cross r ar minus z a z. This under root term comes at the ground. So, here instead of square it comes out to be raised to 3 by 2. To find this cross product I need to follow this triangle. The cross product of a z cross ar as well as a z cross a z I need to find out. Now, it is simple because a z cross a z is giving me 0 because self cross product is 0. Now, only the thing is I need to find out a z cross a r. This is also simple how because I just need to follow this triangle a z cross a r. I am going in the same direction of the arrow. So, thus cross product will give me the third term which is a phi which is positive. If I go in the opposite direction of the arrow the answer will be minus a phi. But here as you see a z cross a r I am following the same direction of the arrow. So, answer is plus a phi. So, I can substitute this which is a z cross a r is a phi and a z cross a z is equal to 0 in the above equation. So, I get i r dz a phi upon 4 pi r square plus z square raise to 3 by 2. This is fine that I have found out for this small length of the filament. But my intention here to find for the finite length filament which is ranging from z 1 to z 2. So, h bar here will be integral of dh bar z 1 to z 2. Now, you can recall the previous video of infinite length filament in which we have considered the limit minus infinity to plus infinity because that was an infinite length filament. But here the filament is of finite length with the range as z 1 to z 2. So, the limits come as z 1 z 2. Now, substitute the value which we have calculated of dh bar here the limits are z 1 to z 2. So, this is h bar. Let us simplify this as this is the integration with respect to z i r a phi as well as 4 pi. These terms are the constants can be taken outside the integral h bar is equal to i r a phi upon 4 pi integration z 1 to z 2 dz upon r square plus z square raise to 3 by 2. Now, the thing is I need to find this integral. This integration can be found out by using substitution method like this. Substitute z is equal to r tan alpha dz when I differentiate this gives me r instead of tan alpha it goes to be 6 square because the differentiation of tan is 6 square 6 square alpha d alpha. Let us look at the limits when z is tending towards z 2 alpha is tending towards alpha 2 when z is tending towards z 1 alpha is tending towards alpha 1. From where I have shown these we can recall from the previous diagram. There we have shown these angles alpha 1 and alpha 2 made with the lower end and upper end of the filament. Also we can calculate this term r square plus z square raise to 3 by 2. So, here r square instead of z square it is r square tan square alpha raise to 3 by 2. So, r cube sec cube alpha. I can substitute here. I get i r a phi upon 4 pi integration is from z 1 to z 2 no it is not z 1 to z 2 it is alpha 1 to alpha 2 r sec square alpha d alpha upon r cube sec cube alpha. Now r and r cancels out sec square cancels out. So, what remains inside the integral is like this i r a phi 4 pi as it is limits alpha 1 to alpha 2 d alpha upon r square sec alpha. So, 1 upon sec is nothing but cos. So, cos alpha d alpha. So, that integration will give me as h bar h bar is equal to i r a phi upon 4 pi r square sin alpha alpha 1 to alpha 2. So, finally I get the equation as h bar is equal to i by 4 pi r sin alpha 2 minus sin alpha 1 a phi bar in amperes per meter. So, this is the equation which is obtained for finite length filament. So, let us see what are the terminologies inside this equation. The terminologies are i is the current r is the distance between the filament and the point p where magnetic field is desired and a phi bar is nothing but the unit vector indicating the direction of the magnetic field alpha 2 and alpha 1 are the angle made with respect to the point p with the upper end of the filament and the lower end of the filament respectively. To decide the direction for that consider the filament is in our right hand thumb indicating the direction of the current. So, my god the fingers will be indicating the direction of the magnetic field like this. So, these are this is the direction of the magnetic field. So, this direction can be decided by right handed thumb. These are the references used for preparing this video. Thank you.