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Homework Help: Force current between an Infinitely Long Wire and a Square Loop

  1. Apr 1, 2006 #1
    Im having some major trouble on this problem for my physics HW: A square loop of wire with side length a carries a current I_1. The center of the loop is located a distance d from an infinite wire carrying a current I_2. The infinite wire and loop are in the same plane; two sides of the square loop are parallel to the wire and two are perpendicular as shown. 20743A.jpg

    What is the magnitude, F, of the net force on the loop?
  2. jcsd
  3. Apr 1, 2006 #2
    The force between two perpendicular current carrying wires is zero. Only the sides of the square parallel to the inifinite wire contribute to the force. The force on current carrying wire in a magnetic field is F = (length of wire)*IxB = (lenght of wire)*I*B*sin(theta). If the wire is perpendicular to the magnetic field (meaning parllel to the wire creating the mag. field) then theta = pi/2 and the force becomes (length)*I*B.

    Anyways, you gotta find the magntic field due to the infinte wire using the law of Biot and Savart (I think the answer is someting proportional to the inverse of two times the distance fromt the wire). It is probably given in your book. Once you know the mag. field due to the inf wire as a function of distance from the wire you use the formula given in the first paragraph for parallel wires to get F= -a*I*B(d-a/2) + a*I*B(d+a/2) as the total force acting on the two parallel sides of the square. The net force should be negative (meaning directed to the left and in the plane containing the square and inf. line). You could calculate the magnetic field due to the square then find the force on the infinite wire from that magnetic field, but I don't recommed it. By the way, the side of the square closest to the wire is attracted, the side furthest is repelled (currents in the same direction attract, opposite directions repel.

    Oh, and here is the mag field for an infintely long wire: B= u*I/(2*Pi*a) where 'a' is the distance from the wire and 'u' is a constant.
    Last edited: Apr 1, 2006
  4. Apr 1, 2006 #3
    I inputed your answer in and it still gave it to me as wrong..im not sure why though....
  5. Apr 1, 2006 #4


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    The B fields acting on each section are different, right? One is [itex] {\mu_0 I \over 2 \pi (d-a/2)} [/itex] and the other is [itex]{\mu_0 I \over 2 \pi (d+a/2)} [/itex]. Then the force on each wire has a magnitude I a B with the B field evaluated at the two positions, and the two forces are in opposite directions.
  6. Feb 23, 2008 #5
    this question is confusing me to no end. how would i go about solving this?

    i know that i need to find the total force as the sum of the forces on each straight segment of the wire loop and i know that my first step is to calculate the magnitude of the force F_1 on the section of the loop closest to the wire, that is, a distance d - a/2 from it ... but I am still lost.

    BTW, the Formula for the force on a current-carrying conductor:
    The magnetic force on a straight wire segment of length l, carrying a current I with a uniform magnetic field B along its length, is

    F = Il x B where l is the length of the loop and B is the magnetic field and I is the current
  7. Oct 20, 2010 #6

    i had an inkling. the mellon shaped dwarf inside me head told me this was correct
  8. Apr 29, 2012 #7
    How would you find the magnetic flux through the loop?
  9. Apr 29, 2012 #8


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    Since this was originally posted 6 years ago, and subsequent requests for help by others have gone unanswered, I am locking this thread.

    People still wanting help with the problem are welcome to start a new thread -- be sure to fill out the homework template if you do so.
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