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Magnetic force on a rigid semicircular wire

  1. Jun 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A rigid wire carrying a current I, consists of a semicircle of radius R and two straight portions of length L. The wire lies in in a plane perpendicular to a uniform magnetic field B. The straight portions are also within the magnetic field. Determine the net force due to B on the wire.


    2. Relevant equations
    F1 = IL1 x B
    F2 = IL2 x B
    F3 = dF = I dL3 x B





    3. The attempt at a solution

    can somebody help me start this problem off correctly by converting these vectors into a uniform coordinate system so that I cab take this simple integral?

    B = <0, 0, -Zb>
     
    Last edited: Jun 23, 2009
  2. jcsd
  3. Jun 23, 2009 #2

    LowlyPion

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    Welcome to PF.

    Consider that the plane of the loop is perpendicular to the magnetic field.

    Where will the forces from the B field necessarily be, that is in what direction?
     
  4. Jun 23, 2009 #3

    The forces for the two straight portions are easy.

    F1 = <0, IL, 0> x <0, 0, -zB> = -IL(zB) i
    F2 = <0, -IL, 0> x <0, 0, -zB> = IL(zB) i

    F3 = dF3 = I dL x B
    this is where I am stuck!!!
     
  5. Jun 23, 2009 #4
    Also, do magnetic fields of north or south pole attract or repel either a positive or negative charge??
     
  6. Jun 23, 2009 #5

    LowlyPion

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    Is the wire in a closed loop?

    Or is it shaped like the outline of a derby hat?

    I don't have the advantage of seeing any diagram.
     
  7. Jun 23, 2009 #6
    the rigid wire goes like this. straight up...then forms a semi circular curve...then straight down... no closure stated.
     
  8. Jun 23, 2009 #7

    LowlyPion

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    OK. So what direction are the forces on each of the 2 || segments?

    What happens when you add them together?
     
  9. Jun 23, 2009 #8
    the two parallel section forces cancel each other out. I'm not sure about the upward force. I must take the integral, but I'm a bit stuck.
     
  10. Jun 23, 2009 #9

    LowlyPion

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    What upward force are you considering?
     
  11. Jun 23, 2009 #10
    the upward force due to the semi circle that connects the 2 parallel sections of the overall wire. F3
     
  12. Jun 23, 2009 #11

    LowlyPion

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    So long as you mean upward in the y direction and not out of the paper in z , as there is no force in the || direction to the B field. The cross product will only yield x,y forces. (The plane of your wire.)

    Since the semi-circle is symmetrical, then you can ignore any x-directed forces, taking x to be perpendicular to the straight wires.

    So maybe set up an integral of just the y-component recognizing that you can express the Force component in y as R*sinθ, where θ is the angle with the x axis. From 0 to π, since it is a semi-circle should do it don't you think?
     
  13. Jun 23, 2009 #12
    that's what i've been trying to do...

    dF3 = IdL x B how does the integral of dF3 equal to the integral of Rsin@ d@?????
     
  14. Jun 24, 2009 #13

    LowlyPion

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    Note that your F3 can be written as

    dF3(θ) = |dF3|cosθ i + |dF3|sinθ j

    but by symmetry for each positive directed |dF3|cosθ i you have on the other side of the mid-line a negative |dF3|cosθ i component. This resolves itself into

    dF3(θ) = |dF3|sinθ j

    F3(θ) = ∫ |dF3|sinθ dθ , evaluated from 0 to π

    This force is all y-directed, since all x-directed force has canceled with each other.
     
    Last edited: Jun 24, 2009
  15. Jun 24, 2009 #14

    Isn't there another way to set this integral up? Can I convert all this into spherical coordinates?

    I'm asking this because step by step i'm not sure how to get to above after I write down the equation dF3= IdL x B. B = <0, 0, -Bz>. IdL= <? , ?, ?>
     
  16. Jun 24, 2009 #15

    LowlyPion

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    Why would you want to do that?

    Have you evaluated the integral of Sinθ from 0 to π yet?

    Why try to thrash about on it?
     
  17. Jun 24, 2009 #16
    Yes, and I got -2.

    What I'm wondering is this step:

    dF3= dF3 cos@ i + dF3 sin@ j. = IR d@ x B

    IR d@ =<?, ?, ?>
    B = <0, 0, -Bz>
     
  18. Jun 24, 2009 #17

    LowlyPion

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    OK so you know then that the vertically directed force in the j direction is going to be proportional to 2R. (The same as if you had a straight wire 2R long bridging the same gap.)

    The i components, if you were to evaluate that integral will be the integral of cosθ from 0 to π, which looks to me to be 0. I ignored their contribution, recognizing from symmetry that for each wire element with +i directed force, there is an equal and opposite -i directed force element. The sum of all those will be 0.
     
  19. Jun 24, 2009 #18
    I'm not sure how (IR d@ x B) distributes into dF cos@ i + dF sin@ j

    (IRd@ x B) cos@ i + (IRd@ x B) sin@ j

    How would you evaluate this integral?
     
  20. Jun 24, 2009 #19
    Thanks for pointing out the symmetry that cancels out the I component. I see that, but I don't see the technique to evaluate the j component
     
  21. Jun 24, 2009 #20

    LowlyPion

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    Perhaps I see your confusion.

    Note I wrote my force component equation based on the cross product already taken. I was expressing the components of the force on the individual wire elements as a function of θ from intuition. Then taking the integral.

    I think the current-wire elements are originally directed as <I*sinθ, I*cosθ, 0> This gets crossed with <0, 0, B*k>

    I think this results in <B*I*cosθ, B*I*sinθ, 0> which I integrated to give
    <0, 2R*B*I, 0>
     
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