# Using Maxwell Stress Tensor to get force between two current-carrying wires

I’m trying to find the force per unit length between two parallel wires carrying the same current in the same direction and a distance of 2a apart. I need to use the Maxwell stress tensor and am working in CGS units.

If I set the wires parallel to the z-axis and have both lying in the y-z plane I see the magnetic field on the left wire flows in the positive x direction.

I see the tensor is only non-zero on the diagonal with values:

$$T_{xx} = \frac{-B^2}{8\pi}$$ $$T_{yy} = \frac{B^2}{8\pi}$$ $$T_{zz} = \frac{B^2}{8\pi}$$

My confusion comes from when I go to integrate it. I don’t know what the area should be. So for

$$\vec{F} = \int T\cdot dA$$

What should the area be? Since the solution needs to be force per unit length, it would seem one direction is the length of the wire, but I have no reason to pick anything for the other direction except knowing what the answer should be. Since the solution, I think, should be

$$\frac{F}{l} \sim \frac{I^2}{a}$$

and since

$$B = \frac{I}{ca}$$

it would seem that the force should be something like

$$F_y = \frac{1}{8\pi}\left(\frac{I}{ca}\right)^2\int_0^l\int_0^{2a}dydz$$

in order to get the right units. Unfortunately I don’t have a good reason why this would be chosen, if it’s even close to correct.

Your stress tensor should have a coordinate dependence $$T_{yy} \propto \frac{x^2}{x^2 + a^2}$$, which, when integrated, gives you the proportionality you wanted:
$$I^2\int\limits_{-\infty}^\infty dx \frac{x^2}{x^2+a^2} = I^2\frac{\pi}{2 a}$$