00:01
Here we have two wires interacting magnetically by their currents.
00:07
And we have an equilibrium situation in which we'll need to use duton's second law that the sum of the x components of the forces on a wire has to add up to zero and the same with the y.
00:30
In addition, we'll need the magnetic force on a current carrying.
00:34
Wire is a product of the current in the wire times the length of the wire times the magnetic field in which it sits.
00:46
So we're going to be thinking about one of the wires carrying a current in the magnetic field of the other.
00:53
So we will also need amper's law and the magnetic field of a long straight wire, which is mu -0 times i over 2 pi times the distance from the wire.
01:12
So i will pick the wire on the left.
01:15
Now both the currents are the same and the masses are the same of the wires.
01:22
So for that reason, you can expect them to be hanging vertically from the same angle.
01:29
So there is some symmetry in this situation that we can make some use.
01:34
Use of.
01:35
So i will choose to draw a force diagram on the left wire, sorry, the right wire.
01:45
We'll choose the right wire, rather arbitrary.
01:50
So what i'll show is that there is mg downwards.
01:57
There is the magnetic force, and i will label that number one because it's coming from wire number one.
02:09
And this is on wire number two.
02:18
They are identical, so those are just for labeling.
02:23
And then there's the tension that's holding the wire up, and we'll pretend that we know the length of the wire, which we will eventually.
02:40
So we can see that it's a fairly simple equilibrium.
02:46
What we'll need to take components on are the tension, so it has a y component equal to tension times cosine of theta over two.
03:04
And it has an x component tension sine of theta over two.
03:20
And let's see.
03:21
So by writing our balance equations, we can come up fairly rapidly with our equations that the force magnetic, in the x direction, the force magnetic on wire number 2 due to number 1 is equal to t sine of theta over 2.
03:48
And in the y direction, we have m g equals t cosine then theta over 2.
04:05
And if we take a ratio between these, we have take a ratio.
04:11
It's always a good thing to do to get rid of the tension.
04:20
And we get fmag 1 over m g is equal to tangent of theta over 2.
04:34
Now furthermore, we're going to have to substitute in for the magnetic force as well as the magnetic field of the wire.
04:46
And here i'm going to be very careful about whose current i'm talking about, but we can simplify it a little bit later.
04:56
So our force magnetic one is equal to the force on wire two due to number one.
05:11
And what goes into the current is the current in wire number two times the length of wire number two, and i'm going to call that h, so i don't confuse it with the string.
05:27
It's not the length of the string...