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Homework Help: Magnetic Force Experienced by a Loop of Wire

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data
    The problem is shown in the link below:

    http://postimage.org/image/do2b7xz0b/ [Broken]

    2. Relevant equations

    Biot-Savart Law's and dF=Idl x B

    3. The attempt at a solution

    My attempt is really just a mess. I essentially want to break it up into 2 segments. (1) Being the straight horizontal segment from -w to w and (2) the semi-oval segment. For the straight segment, I get dl x B to be ydx*z_hat and when I take the integrals in the equation dF=Idl x B, I get a Force of 2wIy*z_hat and since y=0 all along this point I get a Force of 0 due to the line segment at the origin. As for the oval segment I'm probably only correct as far as y=h-x^2 is what needs to be integrated. Any help would be appreciated!
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 10, 2012 #2

    rude man

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    Well, you're right about dividing the problem into two sections, and you figured out the straight (base of the inverted parabola) right.

    So the next step is to write the equation for the inverted parabola: y = y(x).

    I also suggest "going advanced" and using bold lower-case letters for the unit vectors, thus i_hat = i etc. In fact, use bold for all vectors, e.g dF = i*ds x B etc.
    Last edited: Mar 10, 2012
  4. Mar 10, 2012 #3
    Yeah, isn't that what I have? y=h-x2
  5. Mar 11, 2012 #4

    rude man

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    That ca'nt be the right equation, can it? What happened to w?
    Let x = +w or -w, what do you get for y?
  6. Mar 11, 2012 #5
    well in that case just y=h-w^2

    but that doesn't make much sense to me, I was assuming something along the lines:

    y=-a(w+x)(w-x) ??
  7. Mar 11, 2012 #6
    Ouf, nevermind, x=w and y=h-w2 make sense to me now. I reviewed a little bit of my highschool quadratic forms! :P
  8. Mar 12, 2012 #7

    rude man

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    Well, no, you need y as a function of x.

    Hint: what's the standard formula for a parabola with the base at (0,0)?
  9. Mar 12, 2012 #8
  10. Mar 12, 2012 #9

    rude man

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    Make it y = ax2 and you got a deal.

    Now, llok at that graph y = ax2. Picture what you have to do to that graph to make it look like your loop (minus the straight section).
  11. Mar 12, 2012 #10
    This is what led me to believe that it was y=h-x2. Do you want me to say y=-ax2+h ? or am I completely approaching this incorrectly?
  12. Mar 12, 2012 #11

    rude man

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    You're getting warm. Of course, you realize that it can't be right:
    plug x = +w or x = -w into it, and what do you get for y?

    Look at your last formula: y = h - ax2. a is not part of your figure, so how about solving for a by forcing y = 0 when x = + or -w?

    I'm wondering if nowadays they don't teach enough analytic geometry before embarking on calculus. In my day, analyt came first, and I'm talking ivy league college, not high school. You have to master analyt first!
  13. Mar 12, 2012 #12

    y=h(1-x2/w2) ??
  14. Mar 12, 2012 #13

    rude man

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    Now, you will need to integrate the magnetic force along the entire parabola. You have already stated that dF = i*ds x B, which is right. You know i and B, what do you do about ds?
  15. Mar 12, 2012 #14
    I'm not sure, I feel as if I need to express ds in such a way that it takes into account both the change in y and the change in x. Possibly a ratio?
  16. Mar 12, 2012 #15
    I take the derivative of y with respect to x and voila?!
  17. Mar 12, 2012 #16

    rude man

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    You're on the right track.

    Draw an element of ds on your parabola. What would ds be in terms of x and y, and, remembering it's a vector, using unit vectors i and j?

    Is ds different on the left vs. the right-hand part of the parabola?

    BTW let's call the current I instead of i so as not to confuse it with the unit vector.

    Also, note that I is defined couterclockwise.
    Last edited: Mar 12, 2012
  18. Mar 12, 2012 #17
    I wanna say di/dj = ds and so:


    di/dj = dy/dx = -2hx/w2
  19. Mar 12, 2012 #18
    hmm actually,

    ds = dxi + dyj

    dy=-2hx/w2 * dx


    dx= -w2/2hx *dy


    ds = (-w2/2hx*dy)i + (-2hx/w2*dx)j
  20. Mar 12, 2012 #19

    rude man

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    Almost. Draw a small piece ds along the rigth-hand part of the parabola. Which way does it point? (Remeber the direction of I).


    I have to leave for a few hrs. Try to get ds for both sides, remembering the direction of I. Also realize that you will eventually need to do an integration, so you need ds x B to be a function of one variable only.
  21. Mar 13, 2012 #20
    How about. ds x B = (wkydx - 2khx^2/w^2)k

    should i sub in my equation for y in for y?
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