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

  1. Oct 30, 2007 #1
    Hello all, I thought I was starting to understand these magnetic field problems but this one is driving me crazy. I've inputted a good 6 or 7 answers and they all failed. Help would be awesome as always.

    1. The problem statement, all variables and given/known data
    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?
    Express the force in terms of [tex]I_1, I_2, a, d, mu_0[/tex]

    2. Relevant equations
    None are given, but my guess is this guy will be involved:

    F_parallel_wire = I_x L B_y ; x is one wire, y is another wire
    F_p_w = (u_0 * L * I_1 * I_2) / 2(pi)d

    Ampere's Law as well:
    circle integeral of B * ds = BL = B(2(pi)d = u_0 * I_through

    3. The attempt at a solution
    Most recent answer.
    just to make this easier, mu_0 = u and pi = p

    F = I L B
    F = I_1 * [(d+.5a) + (d-.5a)] * [(u * I_2)/(2 * p * a)]

    not right.
    So anyway, there's 4 sides to the loop so there should be 4 forces.
    F_top_loop is equal and opposite to F_bottom_loop so those can be tossed aside.
    F_left has the same direction of current as the wire so they attract. Negative sign.
    F_right has the opposite direction of current as wire so they repel. Positive sign.
    F_left than F_right to the wire so it'll have a stronger force and the direction will go left.

    The distance between the closer ones is d - .5a and the distance between the distant ones is d + .5a

    B = u_0(I_2)/(2(pi)d)
    since the length of the wire isn't provided, I figure d will be the amount in parallel with the part of the loop, in this case a. d = a.
  2. jcsd
  3. Oct 30, 2007 #2


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    Homework Helper

    First off... what is the magnetic field due to the I2 current at the left segment of the loop... u0*2pi/distance... what is the distance between the I2 current and the left segment of the loop?

    The force acting on the left segment of the loop is I1*L*B. What is L here?
  4. Oct 30, 2007 #3
    The distance between the i2 current and the left current is (d-(a/2)). (d+(a/2)) for the right current.

    I thought L was 'a,' the amount of wire running directly parallel of of l_loop. But that didn't work either.
  5. Oct 30, 2007 #4


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    yes, so B = u0*I2/[2pi(d-0.5a)]

    so the force on the left segment is:

    -I1*distance*B =

    same way the force on the right segment is:

    so add these 2 forces.

    is this what you did? the question asks for the magnitude... so at the end after getting the net force, leave of the minus sign...
  6. Oct 31, 2007 #5
    Thanks you helped a lot
  7. Jul 13, 2008 #6
    But wait if you add those two forces you would end up with zero... or are you suppoed to add the abs of the two values?
    Sorry i'm kind of confused.
  8. Jul 13, 2008 #7
    Note that one of the forces has (d-0.5a) in the denominator while the other has (d+0.5a) due to the different distance from I2, therefore having a different magnetic field.
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