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Square loop carrying current near a wire

  1. Feb 12, 2006 #1
    This one is again from Griffiths.
    a) Find the force on a square loop of side 'a' placed at a distance 's' from an infinite wire. Both the loop and the wire carry a current 'I'.

    I found the magnitude of the magentic field using Biot-Savart's law:
    [tex]B = \frac{\mu_0 I}{2\pi s}[/tex]

    The force is given by:
    [tex]\vec F_{mag} = I\int\left(d\vec l \times \vec B\right)[/tex]
    "dl" is a wire element.

    So, when I consider only the magnitude:
    [tex]F = \frac{\mu_0 I^2}{2\pi s}\int dl[/tex]
    Here the wire is infinite, so how is it possible to integrate over the length of the wire?
     
  2. jcsd
  3. Feb 12, 2006 #2

    siddharth

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    The B you first find is due to the infinite wire. So, when you integrate to find the force acting on the loop, dl will be an element on the loop. So you should integrate over the square loop.
     
  4. Feb 13, 2006 #3
    I'm sorry, I don't get it. Here we have to take the direction into consideration. But the current flows in the wire as well as the loop. So if we take the current on the square loop, what will [itex]\int dl[/itex] look like?
     
  5. Feb 14, 2006 #4
    Help me out here someone!!
     
  6. Feb 14, 2006 #5

    siddharth

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    I have my exams going on, so I can't reply immediately.

    Anyway, you find the magnetic field B due to the infinite wire as,
    [tex]B = \frac{\mu_0 I}{2\pi s}[/tex].

    The force on a small element 'dl' of conductor in a magnetic field is
    [tex] F = i \vec{dl} \times \vec{B}[/tex]
    where, B is the external magnetic field, 'i' is the current on the conductor.

    So, in this case, the external magnetic field is due to the infinte wire and 'dl' and i are with respect to the square loop. Therefore, when you integrate, you do so along the square loop, and not the infinite wire.

    Did you get the answer in the book?
     
    Last edited: Feb 14, 2006
  7. Feb 14, 2006 #6

    Gokul43201

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    Since 's' is given as a specific distance in the question, you should use the more general equation :

    [tex]B = \frac{\mu_0 I}{2\pi d}[/tex]
    where d can be either s or s+a or s+x (0<x<a), depending on where in the square you take your elemental length.
     
  8. Feb 15, 2006 #7
    Thanks. I suppose the forces at the sides of the square should cancel. Then effective magnetic force operates at the bottom and at the top of the square loop.
    At the bottom: d=s
    [tex]F = \frac{\mu_0 I^2}{2\pi s}\int dl[/tex]

    [tex]F = \frac{\mu_0 I^2}{2\pi s}s[/tex]

    [tex]F = \frac{\mu_0 I^2}{2\pi }[/tex]

    At the top: d=s+a
    [tex]B = \frac{\mu_0 I}{2\pi (s+a)}[/tex]

    [tex]F = \frac{\mu_0 I^2}{2\pi (s+a)}\int_0^{s+a} dl[/tex]

    Are my boundary conditions for "dl" correct? If so the force obtained will be the same as for the bottom segment.
     
  9. Feb 16, 2006 #8

    Gokul43201

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    Everything is right except for the limits on that last integral. dl refers to the length of an element on the bottom segment. Let this element be at some general position 'x' along the segment. What are the limits for x (ie : what values can x take as you move the element through various positions along the bottom segment) ?
     
  10. Feb 17, 2006 #9
    Limits for x would be 0 to a, right?
     
  11. Feb 17, 2006 #10

    Gokul43201

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    Yes. And make sure you understand why.
     
  12. Feb 20, 2006 #11
    I stumbled upon a glitch here. :bugeye:
    I should have taken the limits on 'dl' as 0 to a here too, right?
     
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