Net force on square loop from parallel current-carrying wire

[vsage]

In the figure below, a long straight wire carries a current of 8.5 Amps.
(Attached)
Find the magnitude of the net force (in N) on a rigid square loop of wire of side, L, equal to 5.35 cm. The center of the loop is d = 10.40 cm from the wire and the loop carries a current of 10.0 Amperes.

I don't really know where to start on this one. I realize that the segments perpendicular to the length of wire with current I1 can be discarded as not contributing any force right? I don't know where to proceed from there. Your help is greatly appreciated.

Well I sort of figured out how to do it but I think I'm screwing the sign up somewhere:

$$F = iL \cross B$$
$$B = \frac{\mu I_1}{2 \pi r}$$
And sum up the force on the top and bottom of the square from the long wire? Do I add the force being exerted on the top of the loop from the bottom of the loop and vice versa?

Edit: Here is my work (let subscript t be top and b be bottom)

$$F = \frac {u_0I_1I_t L_t}{2 \pi r_t} + \frac {u_0I_1I_bL_b}{2 \pi r_b}$$

$$F = \frac {2 \times 10^{-7} \times 8.5 \times -10 \times 0.0535}{0.07725} + \frac {2 \times 10^{-7} \times 8.5 \times 10 \times 0.0535}{0.13075}<br /> <br /> = about 9 \times 10^-6N$$

which is wrong

 

[vsage]

Nevermind I got it. I have been plugging in the wrong numbers for hours :\

 

[wais]

.

To start solving this problem, we can use the formula F = iL x B, where F is the force, i is the current, L is the length of the wire, and B is the magnetic field. In this case, we have a long straight wire carrying a current of 8.5 Amps, so we can calculate the magnetic field at a distance of d = 10.40 cm using the formula B = (μ0I)/(2πr), where μ0 is the permeability of free space, I is the current, and r is the distance from the wire.

B = (4π x 10^-7)(8.5)/(2π x 0.1040) = 0.00205 T

Now, we can use this value of B to calculate the force on each segment of the square loop. Since the loop is rigid, we can treat each segment as a separate wire and use the same formula F = iL x B for each segment.

For the top and bottom segments, the force will be directed towards the left, as the current in the loop is opposite to the current in the long wire. So we can use the formula F = iL x B with i = 10.0 A, L = 0.0535 m, and B = 0.00205 T.

Ft = (10.0)(0.0535)(0.00205) = 1.09 x 10^-3 N

For the left and right segments, the force will be directed towards the right, as the current in the loop is in the same direction as the current in the long wire. So we can use the same formula with i = 10.0 A, L = 0.0535 m, and B = 0.00205 T.

Fr = (10.0)(0.0535)(0.00205) = 1.09 x 10^-3 N

Now, to find the net force on the loop, we need to take the vector sum of the forces on each segment. Since the forces on the top and bottom segments are in opposite directions, they will cancel out, leaving us with only the forces on the left and right segments.

Fnet = Fr + Fr = 2(1.09 x 10^-3) = 2.18 x 10^-3 N

Therefore, the magnitude