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?

  • $\begingroup$ 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. $\endgroup$ Commented Dec 14, 2015 at 11:26
  • $\begingroup$ @Floris You are correct, sorry for the confusion. $\endgroup$
    – KSPR
    Commented Dec 14, 2015 at 11:30
  • $\begingroup$ @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? $\endgroup$
    – KSPR
    Commented Dec 14, 2015 at 11:33
  • $\begingroup$ 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. $\endgroup$ Commented Dec 14, 2015 at 17:18

1 Answer 1


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 $a=\ell$.

This is worked out in detail in the answer to this earlier question.

You should be able to ignore the force due to to short elements. Force scales with length and is inversely proportional to distance; there will be a small component of magnetic field due to the end wires that intersects the currents in the long wire at right angles, and this will result in a small amount of additional force at the ends - but compared to the total force due to the opposing wire, it can safely be neglected.

  • $\begingroup$ Thank you for answering. But how come that the long wire does not feel the force from the short wire? Is it because they are perpendicular which means that one short segment gives F=BIL and one gives F=-BIL? $\endgroup$
    – KSPR
    Commented Dec 14, 2015 at 13:33
  • $\begingroup$ @KSPR - I actually had to correct my statement... I don't think what I said originally was quite true $\endgroup$
    – Floris
    Commented Dec 14, 2015 at 18:30

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