I'll state the problem and then explain.
(There's a picture that goes along with this, but I don't think it's necessary, and since I don't know how to get it here, I'll leave it out)
This figure shows wire 1 in a cross section, the wire is long and straight, carries a current of 4.00mA out of the page, and is at distance d1=2.40cm from a surface. (the ground, basically). Wire 2, which is parallel to wire1 and also long, is at horizontal distance d2=5.00cm from wire 1 and carries a current of 6.80mA into the page. What is the x component of the magnetic force per unit length on wire 2 due to wire 1?

Well the main problem I have here is all the equations I know how to use all require me to know L, the length of the wires involved, and this problem doesn't tell me that. Also the "force per unit length" is confusing me, and I'm not clear as to what that means. So I can calculate the magnetic fields of both of these wires, but that doesn't seem to get me anywhere. I have been given the answer, and everything I've tried isn't remotely close to it. Can anyone give me a hint or two about how to get started? Thanks.

I think you can consider the 2 wires to be infinitely long. I'm tempted to use Ampère's law to calculate the magnetic field due to wire 1 on wire 2
Once you have B (sorry Latex doesn't work for now), you can get the force exerted on the wire 2 by the formula F=ILxB where x denotes the cross product and L the length of the considered segment of the wire.

__________________
Isaac
If the problem is too hard just let the Universe solve it.

But what is the value of L if the wire is infinitely long? How can it have any value?
If it helps, the answer to the problem is 88.4 pN/m. Now how to get to that... good luck. haha. This is driving me nuts, I know I'm getting really close but I just can't quite grasp it.

I also don't reach the answer so I'll wait for someone to help us.
By the way, what does p represent in the answer?
L is not important in the formula I gave. As L and B are orthogonal, the cross product reduces into a product. Hence the force they ask should be F=ILB (according to me).
I get a very small answer compared to your 84 pN/m (again, what is p?)

__________________
Isaac
If the problem is too hard just let the Universe solve it.

the p means that it is 88.4*10^-12. so the answer is very small.

Ah ok, I didn't realize p was pico meter.
Unfortunately I still don't reach the answer. I reach 4.02*10^(-10) N/m using the way I described. I wonder if my method is appropriated. Let's wait for someone else to help us. I'm probably taking the same course as you.

__________________
Isaac
If the problem is too hard just let the Universe solve it.

Well the closest I've been able to get is this:

first find the magnetic field generated by wire 1.
B=u0*ia (permiability constant times current through wire 1, sorry for lack of proper formatting)
divded that by 2*pi*.055 (i got .055 from the pythagoream theorem, assuming that to be the distance between the two wires.)
so that gives you about 1.4463*10^-8

then to find the force per unit length on the x component, use the equation
F12=ib*L*Ba*sin90 and since it's the force per unit length, you can divide the L out, so then it becomes a non issue. (ib is the current in wire 2, Ba is the magnetic field of wire 1 that I calculated above).
so you'd get .0068A*1.4463*10^-8 * sin 90 (make sure your calc is set to radians)
this gives you 8.79229626*10^-11. So it's quite close, I think I may have to fool around with the rounding of numbers to get it exact. But the 8.8 is there, just not the 8.84. Hopefully I'm on the right track.

I got it. Instead of using the permiability constant of 1.26*10^-6, use the exact one of 4pi*10^-7. everything else was right, substituting this gives 8.84.... *10^-11, which is equal to 88.4 *10^-12 N/m. woo. thanks for your help man, I really appreciate it. This has been eating away at me all week.

I'm going to bed right now, so I cannot check all the details you've gone through but I think we did it the same way. Except that I misunderstood the distance between the 2 wires. I thought they were on the same horizontal line, but if you used Pythagoras to find the distance between them I guess the 2nd wire was on the ground level.
By the way I'm glad you found out the result. Have confidence in yourself.
A pleasure to help.

__________________
Isaac
If the problem is too hard just let the Universe solve it.