# A conductor shaped like a square, what's the force between the two long wire segments? [duplicate]

In a conductor shaped like a rectangle, a current I=0.3A flows. The lenght is l=4.5m and the width is b=9cm.

a) Determine the absolute value of the force between the two long wire segments.

b) Determine the direction of the force.

I am only used to problems such as "calculate the force between two infinite long wires" and am a bit lost on how to set up the integrals when it is shaped like a square. I know that I am supposed to use $$F= \int (I\vec{d}l \times \vec{B}) dS$$

Since $b\ll l$, can you neglect the force from the short sides of the rectangle?

Otherwise, how do you set up the integral?

• Think about the direction of force due to current flowing in a wire. You will see, that the short wires will not exert force on the long wire because its perpendicular. SO the force field of short wires will be parallel to the long ones. Commented Dec 14, 2015 at 11:26
• @Floris You are correct, sorry for the confusion.
– KSPR
Commented Dec 14, 2015 at 11:30
• @seeking_infinity Thank you for answering! But if they are perpendicular, then the force F = BIL =\= 0. Is it because the other short wire is F=-BIL and therefore they cancel each other out?
– KSPR
Commented Dec 14, 2015 at 11:33
• Sorry, I was not clear enough. Try to visualize. Consider a wire at x=-l to x=+l. Now, its force field will span the region [{l to l}, y,z]. Force experienced by anything with x coordinates, x<-l and x>l will be zero. At long wires are at -l and l only in your problem. Commented Dec 14, 2015 at 17:18

You need a double integral to solve this. First find the force on an infinitesimal element due to a finite wire from $a$ to $b$. Next integrate that expression by setting $b=a+\ell$ and integrating from $a=-\ell$ to $a=\ell$.