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Homework Help: Magnetic field in the open loop

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data
    I am trying to find the magnetic field at the point in the middle of the picture can someone give a hint on how to approach the problem?

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Nov 3, 2009 #2
    Attachments Pending Approval...
  4. Nov 4, 2009 #3
    any help?
  5. Nov 4, 2009 #4


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    Staff: Mentor

    The hint is to use the Biot-Savart law... (Hint: it often simplifies the integration to make your variable of integration theta...)
  6. Nov 4, 2009 #5
    you should divide the wire into some parts and find m.f created by these parts.
  7. Nov 5, 2009 #6
    would the left and the right side cancel out....

    So I have to do it just for the bottom side???
  8. Nov 5, 2009 #7
    so thus this integral work :

    B= ([tex]\mu[/tex]*I)/(4*pi*x)[tex]\int[/tex] between(-pi/4) and (pi/4) of cos ([tex]\theta) d\theta[/tex]
  9. Nov 5, 2009 #8


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    Gold Member

    I believe so (though I might be wrong). In all cases you should prove it via arithmetic so don't use this shortcut if you are not sure.
  10. Nov 5, 2009 #9
    No. Biot-Savart:

    dB = \frac{\mu_0 I}{4 \pi} \frac{dl sin\theta}{r^2}

    With [tex]\theta[/tex] is angle between dl and r. imo, you can use symmetry property to make life easier.
  11. Nov 5, 2009 #10
    so i stil do not quite get it....does the two side cancel out?


    dB = \frac{\mu_0 I}{4\pi r^2} {dl sin\theta}


    will the angle be (-pi/4) to (pi/4)
  12. Nov 5, 2009 #11
    No. r is changing with theta. Can you find relationship between sin(theta) and r? and between r and x?

    About two sides, you can say that they cancel out.
  13. Nov 5, 2009 #12
    i see how you are looking at the picture.

    basically the

    sin (theta) dl = cos (theta') dl


    1/r^2 = cos^2(theta')/ x^2

    when i put all of them in the integral reduce down to


    dB = \frac{\mu_0 I}{4\pi x} {dl cos\theta}


    i do not know yet what boundary to use for the integral
  14. Nov 5, 2009 #13
    from this, i guess theta and theta' as shown in fig below:

    http://img149.imageshack.us/img149/6568/88285645.jpg [Broken]

    is that right?
    Last edited by a moderator: May 4, 2017
  15. Nov 5, 2009 #14


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    Homework Helper

    dB = μo/4π*I*di*sinθ/r^2
    where sinθ = x/r = x/sqrt( x^2 + l^2)
    Take the limits of integration from -x/2 to +x/2.
  16. Nov 5, 2009 #15
    yes the picture with the angle is exactly what I was thinking....what are the limits for the integral angles???
  17. Nov 5, 2009 #16


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    Staff: Mentor

    The limits come from the geometry of the diagram. You really should be able to figure that part out.

    And on your PM question to me, know, the B-field contributions from the two sides on the left and right do not cancel out.
  18. Nov 5, 2009 #17
    i did to -45 to 45....will that work?
  19. Nov 5, 2009 #18


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    Staff: Mentor

    Based on your initial drawing, yes.
  20. Nov 5, 2009 #19
    thank you
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