1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determining the force on a loop cause by an infinite line

  1. Jul 3, 2016 #1
    1. The problem statement, all variables and given/known data
    An infinitely long line of current $I_1=6[A]$ is following along the positive z-axis in the direction of +$\hat{a_z}$. Another current is following a triangular loop counter clockwise from the points A(0,2,2), B(0,6,2) and C(0,6,6).

    2. Relevant equations
    To start I applied Ampere's Law since an amperian path could be applied to current $I_1$ which results in
    $$\vec{H}=\frac{I_1}{2\pi\rho} \hat{a_{\phi}}$$

    Applying the force on two conductors formula $$F=\int{I_2 \vec{dl_2}} \times{\vec{B_1}}$$

    3. The attempt at a solution
    Where I am stuck is at segment AB, I have determined that the following for this segment $$dl=dy\,\hat{a_y}$$

    The problem that I am having is determining the direction of the magnetic field vector, is it it directed in the positive $ \hat{a_z} $ because of how the problem is described or is it another direction all together? The teacher provided a solution but did not outline why they did certain steps.
  2. jcsd
  3. Jul 3, 2016 #2


    User Avatar
    Homework Helper
    Gold Member

    Welcome to PF!
    You know ##\vec{H}=\frac{I_1}{2\pi\rho} \hat{a_{\phi}}##. Thus, ##\hat{a_{\phi}}## gives you the direction.

    Are you familiar with the magnetic field pattern of a long straight current? (Right hand rule)
  4. Jul 3, 2016 #3
    Do you mean that since the current is moving in the positive [itex] \hat{a_x} [/itex] that I apply the right hand rule? If so then the result is that the force is directed in the negative [itex] \hat{a_x} [/itex] direction. Therefore, [tex] \vec{B}=\frac{-{\mu}_0\,I_1}{2\pi\,y} \hat{a_x}[/tex]

    At this point I integrate the and solve for segments AB and BC which results in $$\vec{F_{1,AB}}= 6{\mu}_0\,ln(3) \,\hat{a_z}$$ $$\vec{F_{1,BC}}= -4\mu \,\hat{a_y}$$

    Up to this point I understand and did correctly according to the given solution, however, the final integral confounds me...After the cross product is applied and simplified my result is this $$6{\mu}_0[\int\frac{-dz}{y}\,\hat{a_y} + \int\frac{dy}{y}\,\hat{a_z} ]$$

    I believe the second integration should result in [itex] 6{\mu}_0\,ln(1/3)\,\hat{a_z} [/itex] but I dont know about the about the first integral since there is the is "y" term in the denominator. Do I make that y=6 because that is where the integration started at or y=2?
  5. Jul 3, 2016 #4


    User Avatar
    Homework Helper
    Gold Member

    Typo? The current is in the direction of [itex] \hat{a_z} [/itex]. The field is in the negative x direction.


    I can't check the above, since ##I_2## was not specified.

    For the first integral you'll need to relate ##y## and ##z## along the path CA.
  6. Jul 3, 2016 #5

    Ok I will see what I can do, sorry I forgot but [itex] I_2 = 2 [A] and I_1=6\pi [A][/itex]. I am still trying to get a handle on latex
  7. Jul 3, 2016 #6


    User Avatar
    Homework Helper
    Gold Member

    OK. Your results so far look good.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted