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!

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

    berkeman

    User Avatar

    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

    fluidistic

    User Avatar
    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:

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

    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?

    [tex]

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

    [/tex]

    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

    and

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



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

    [tex]


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

    [/tex]

    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

    rl.bhat

    User Avatar
    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

    berkeman

    User Avatar

    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

    berkeman

    User Avatar

    Staff: Mentor

    Based on your initial drawing, yes.
     
  20. Nov 5, 2009 #19
    thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Magnetic field in the open loop
Loading...