00:02
In the given problem three wires current carrying wires are kept perpendicular to the plane of paper and parallel to each other such that their cross sections taken together make an equilateral triangle as shown here so this is the first wire which is carrying current in a direction out of the paper, out of the plain of paper.
00:37
Let it be wire a.
00:38
Then there is another wire b.
00:42
This is also carrying the current outward, means out of the plan of paper.
00:47
And the third wire at the third vertex of this equilateral triangle at c.
00:54
This is also carrying current in the same direction means outward.
00:58
All the three currents ia equals to and here this is c sorry the naming this current is given as c that was c and this is d so this is c this is d and there is one more fourth current carrying conductor that is also perpendicular to the plenopaper but the direction of current is missing here the mass magnitude as well as the direction both are missing here.
01:31
Now the magnitudes of these known currents are ia is equal to ic is equal to id and that is equal to 10 ampere.
01:44
Now as currents in all three wires, a, c and d, are in the same direction.
02:11
So these wires will attract each other.
02:25
Means if you look at wire a, it will be attracted equally by the two wires c and d.
02:34
So here this is the force of attraction exerted at a by c...