00:01
Our question says that we have three long parallel wires with lengths l of 3 .5 centimeters or 0 .035 meters, each carrying a current i of 8 .0 ampiers.
00:13
So looking along them, they are at the three corners of an equilateral triangle.
00:19
But the directions of wire m is opposite that of wire n and in p.
00:23
So you can look at figure 54 to get an understanding of this.
00:25
And it's asked us to determine the magnetic force per unit length on each wire due to the other two.
00:32
Okay, so the force per unit length between any two wires is calculated by the formula that says that the force over the length is equal to mu not times the current in the two wires, but the current is the same in both.
00:51
So the current in the two wires is just going to get squared, divided by, and this is mu not, magnetic permeability of free space, divided by two times pi times the length l.
01:05
Okay, so now that we have the equation for the force per unit length between any two wires, so we can use this to apply it to all three points, and we can calculate the total magnetic force per unit length.
01:17
So the current, again, we have the current in each wire written, the length of each wire written.
01:22
So if we plug in the values into this expression, remembering that mu not is 4 pi times 10 of the minus 7 newton per meter squared, we plug those into this expression, and we find that this comes out.
01:34
To equal 3 .66 times 10 to the minus 4.
01:42
And of course, since it's force per unit length, the units here are going to be newton's per meter.
01:49
Okay, so now we can find the force acting at the different points, m, n, and p.
01:54
So let's go ahead and find the first one.
01:58
At point m, f of m, is going to be equal to the force of on m from p times the angle at which this force is applying, and that's going to be 30 degrees.
02:09
This is going to be the cosine of 30 degrees.
02:13
And then we're going to add to this the force of n on p and then times the angle, and it's the same angle, cosine of 30 degrees.
02:25
But these forces are not just forces in newton's, but forces per unit length.
02:31
Since these forces are equal and the angles are equal, we can go ahead and add them together.
02:35
So we have two times f over l that we found above because it's the same in each case, right? because each one has the same current that's acting on it.
02:43
So it's 2 times ferville times the cosine of 30 degrees.
02:48
So plugging those values into this expression, we find that the force on m is 6 .33 times 10 to the minus 4.
03:03
And the units here would be newton's per meter.
03:07
And this is at an angle of 90 degrees.
03:12
So we can say this is at theta equals to 90 degrees from the image...