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Homework Help: Force between a square loop and a long wire

  1. Nov 15, 2008 #1
    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. What is the magnitude, F, of the net force on the loop? (See picture below)

    Now I know that F= IL X B
    and the force of two wires is F= (u0 * L * I1 * I2)/(2*pi*d)

    but I am at a loss as to how to start this problem.
    Please help.


    Attached Files:

  2. jcsd
  3. Nov 15, 2008 #2


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    Hi StephenDoty,

    Right, and remember the cross product is general, while the second formula is only used for parallel wires.

    Try thinking of the loop as four separate wires, and then the force you are looking for is the net force.
  4. Nov 22, 2008 #3
    The total force equals the force of the left parallel bar of the loop - the force of the right parallel bar of the loop.

    F= (u0 * a * I1 * I2)/(2*pi)*(1/.5d - 2/3d)?

    I am having trouble figuring out the distance from the long wire to the parallel bar for both the left and right parallel bars of the loop. Please see the figure in the first post of this thread.

    Any help would be appreciated. Thank you.
    Last edited: Nov 22, 2008
  5. Nov 22, 2008 #4


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    It looks like you are thinking that (1/2) d = a, but I don't see where that is said.

    Think about actual numbers: for example, if d was 30 cm and a was 10cm, what would the two distances be for the left and right sides? Remember that d is measured from the straight wire to the middle of the loop.

    Once you have that, what are the two distances in general, in terms of d and a?
  6. Nov 22, 2008 #5
    a is the length

    and I used fractions of d for the distance between the long wire and the two parallel parts of the loop. And the distance has nothing to do with the length a.

    I do not know how to find the distance using d. I really need an explanation of how you would use d to find the distance from the long wire to the parallel parts of the loop.

    In the formula:F= (u0 * a * I1 * I2)/(2*pi)*(1/.5d - 2/3d)
    a is the length. and this formula is the formula for the force between two parallel wires. I do not see how you thought that .5d=a. The (1/.5d - 2/3d) part of the formula was the (1/r) part of the formula for the force between two parallel wires where r is the distance from one parallel wire to the long wire.
  7. Nov 22, 2008 #6


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    I definitely was not thinking that 0.5 d = a; I was just trying to determine where you got the values in your denominators.

    Is this equation saying that one loop side is 0.5 d away, and the other is (3/2) d away? That does not look right to me.

    Here is how you determine you two distances:

    If you look at your diagram, if you start at the straight wire and move a distance d to the right, you get to the midpoint of the loop. How much farther to the right do you have to go to get to the rightmost side of the loop? Add those together to get the distance to the right side of the loop.

    Then repeat the process. Starting at the straight wire and moving to the right a distance d gets you to the center of the loop; how much do you then have to move to the left to get to the left side of the loop? Subtract those values to get the total distance between the straight wire and the left sides.

    Those two values (each of them should have d and a in them) will be the two r values for the denominator.

    (I was looking back at your earlier post; you mentioned that [itex]a[/itex] is the length. That is true, the distance [itex]a[/itex] will be the wire length L in the formula, but since the loop is a square, the length [itex]a[/itex] will also be part of the r value in the formula.)
    Last edited: Nov 22, 2008
  8. Nov 23, 2008 #7
    Is it d+.5a for the left
    and d-.5a for the right?
  9. Nov 23, 2008 #8


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    I think you have the left and right switched (the on on the right is farther away), but those look like the correct distances to me.
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