Wire Suspended By Magnetic Field

[Lancelot59]

I'm given the following situation:

The suspended wire has a radius of 1.20mm (0.0012m), and is being held up by the magnetic fields generated by the bottom two wires which have a current of 50A each.

I started with a simple force setup:

$$F_{g}=2F_{M}sin(60)$$
$$mg=2F_{M}sin(60)$$
$$F_{M}=\frac{mg}{2sin(60)})$$

To get the mass of the wire I assumed it was made of copper, the density being 8.94 g/cm^3. Not being given any information about the length I just decided to ignore it.

$$m=d*v$$
$$m=(89.2 \frac{kg}{m^{3}})*\pir^{2}*length$$

$$m=(89.2 \frac{kg}{m^{3}})*\pi(0.0006)^{2}*length$$
$$m=0.000100883kg$$

So the total weight force pulling the top wire down is: 0.0009896622N

Now to get the magnetic force I treated the wires as infinitely long wires:
$$F=I_{2}lB$$ and then for the magnetic field generated from each wire:$$B=\frac{\mu_{0}I_{1}}{2\pi r}$$

Then sticking everything together, and doubling the magnetic field:
$$F_{M}=I_{2}l2B$$
$$F_{M}=I_{2}l2(\frac{\mu_{0}I_{1}}{2\pi r})$$

Putting that into the net force equation:
$$F_{g}=2(I_{2}l2(\frac{\mu_{0}I_{1}}{2\pi r}))sin(60)$$
$$F_{g}=2(I_{2}l(\frac{\mu_{0}I_{1}}{\pi r}))sin(60)$$

So I substituted in the values and solved for I2. I decided to ignore the length term.
$$0.0009896622=2(I_{2}l(\frac{(4\pi x10^{-7})(50.0A)}{\pi (0.035m)}))sin(60)$$
I wound up with I2 being 0.999918041, which is way off. After guessing a few times I got the correct answer of 199 from the system after feeding it 200. Where did I go wrong?

 

[Elwin.Martin]

To get the mass of the wire I assumed it was made of copper, the density being 8.94 g/cm^3. Not being given any information about the length I just decided to ignore it.

This where I think you went wrong, did you consider doing it in terms of a linear mass density? You know the cross-sectional area of the wire, it might help.

 

[Lancelot59]

This where I think you went wrong, did you consider doing it in terms of a linear mass density? You know the cross-sectional area of the wire, it might help.

Sorry, could you explain that to me?