00:01
In this problem we have been given the two wires which are carrying currents 12 amperors and 24 ampers in the opposite directions and these two wires are parallelly placed at a distance of 5 centimeters apart.
00:16
Considering a point p with respect to these two wires symmetrically we observe that p is at a distance of 8 centimeter with respect to the current element whose width is 1 .5 millimeters so in meters it's 1 .5 into 10 raise to minus 3 meters.
00:35
So we need to determine the magnetic field at this point be because of these two current carrying wires.
00:42
So here we apply bayotte sauts law to compute the magnetic field at a point p with respect to current element.
00:49
And according to that the magnetic field is given by mu not i d l into sine theta divided by 4 pi into r square.
01:00
Theta is the angle between dl vector and r vector.
01:06
R is the distance from this current element and the point p.
01:12
So here we observe that this angle theta, if we observe the value of sine theta, in this triangle we can see this is 2 .5 centimeter because that's half of this distance between two parallel wires.
01:24
And this is 8 centimeter.
01:27
So here we can use that sine theta will be equal to perpendicular upon the hypotenuse.
01:34
So that will be 2 .4, 2 .5 upon 8.
01:39
And we observed that according to right -hand thumb rule, if we place the thumb in the direction of the current with respect to both these wires, the magnetic field will be in the same direction and hence we can say that magnetic field because of these two wires will be added together to get the net magnetic field.
01:58
So let's say b1 is the magnetic field due to this first wire, and b2 is the magnetic field due to the second wire...