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.


1 Answer 1


If you want to use the stress tensor to calculate the force on the wire, then the volume whose surface you are integrating over must contain that wire. The surface you propose does not contain either of the wires.

You could do something like integrate over a cylinder containing one of the wires, but it would be a nightmare to figure the magnetic field on the surface of such a cylinder.

Instead we could exploit the symmetry of the problem. The setup is symmetric about the xz plane between the two wires. As such, the magnetic field along this plane points only the y direction. This plane is also a surface containing one of the wires. This is perhaps easier to integrate and should lead to the correct solution.

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:

$$I^2\int\limits_{-\infty}^\infty dx \frac{x^2}{x^2+a^2} = I^2\frac{\pi}{2 a}$$

  • $\begingroup$ Technically the surface has to be closed, but we can consider the plane to be "closed at infinity". (Or more accurately, we could take the limit of a half-cylinder as the length and radius go to infinity.) $\endgroup$ Nov 1, 2021 at 23:22

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