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Calculating magnetic force on semicircle conductor

  1. Oct 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Given the figure below

    aaGm11v.png
    I need to calculate the total magnetic force on the semicircle section of the conductor.
    Current is I, Radius is R, and the Magnetic Field is B.

    2. Relevant equations
    [tex]d\vec{F} = Id\vec{l} \times \vec{B}[/tex]


    3. The attempt at a solution

    [itex]dl [/itex] is equal to [itex]Rdθ[/itex]

    Since [itex]d\vec{l}[/itex] and [itex]\vec{B}[/itex] are perpendicular, the magnitude of the force on the segment [itex]d\vec{l}[/itex] is equal to [itex]I dl B = I(Rdθ)B[/itex], and the components of these forces is [itex]I(Rdθ)Bcosθ[/itex] and [itex]I(Rdθ)Bsinθ[/itex]. Integrating these separately from 0 to π and adding the result gives [itex]IB(2R+L)j[/itex].

    However, I am trying to solve it using the cross product to obtain [itex]d\vec{F}[/itex] and then solve it from there.

    The magnetic field vector is [tex]0i + 0j + Bk[/tex]
    And the vector for [itex]d\vec{l}[/itex] I got was [tex]Rdθcosθ i + Rdθsinθ j + 0k[/tex]

    Calculating the cross product yields
    [tex]-IBRdθsinθi + IBRdθcosθj + 0k[/tex]

    Integrating to from 0 to π to obtain [itex]\vec{F}[/itex] yields
    [tex]-2IBRi + IBRj + 0k[/tex]

    This vector is obviously different than what was obtained before, and I am wondering where I went wrong.
    I think that a place I could have gone wrong is when I found [itex]d\vec{l}[/itex] as [tex]Rdθcosθ i + Rdθsinθ j + 0k[/tex], but I don't know of any other way I could get the components for the vector.

    I'd appreciate if somebody could enlighten me as to where I went wrong. Thank you.
     
  2. jcsd
  3. Oct 22, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    If j is the vertical axis and your angle goes from 0 to pi, the vector dl needs the cosine for the j component (and sine for i): the cable starts alone the j direction.

    The integration over the cosine should lead to 0 due to symmetry.
     
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