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Force on a wire

  1. Nov 15, 2005 #1
    The figure shows a wire of arbitrary shape carrying a current i between points a and b. The wire lies in a plane at right angles to a uniform magnetic field B. Prove that the force on the wire is teh same as that on a straight wire carrying a current i directly from a to b. (Hint: Replace the wire by a series of "steps" that are parllel and perpendicular to the straight line jkoining a and b.)
    WEll umm
    well the force on the stirahgt wire from a to b is simply
    [tex] F = iL \cross B = iLB \sin( \theta)[/tex]
    for the arbitrary wire
    [tex] F = \int idL \cross B [/tex]
    but i is constant and B is constant so
    [tex] F = iB \int dL = iBL [/tex] whicvh is the same as the stariaght wire. Is this correct?
     

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    Last edited: Nov 15, 2005
  2. jcsd
  3. Nov 15, 2005 #2

    Physics Monkey

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    See my reply to your other post. What is the integral
    [tex]
    \int d\vec{\ell}
    [/tex]
    along the curvy wire from [tex] \vec{a} [/tex] to [tex] \vec{b} [/tex]? Does the integral depend on path? Why?
     
  4. Nov 15, 2005 #3
    The only reason i can think of an integral not depending on path is because a force is conservative. But why is the force due to a magnetic field conservative? Is it because the energy reamins constant in this system?
     
  5. Nov 15, 2005 #4

    Physics Monkey

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    You are integrating a differential, what does that tell you about the result of the integration? If you are uncomfortable with this argument and want to be a bit more formal, then try using the fundamental theorem of line integrals on the integral
    [tex]
    \int_C \vec{v}\cdot d\vec{\ell} = \vec{v} \cdot \int_C d\vec{\ell}
    [/tex]
    where [tex] \vec{v} [/tex] is an arbitrary vector and C is a path from [tex] \vec{a} [/tex] to [tex] \vec{b} [/tex]. Hint: is [tex] \vec{v} [/tex] the gradient of something? What does this result tell you about [tex]
    \int_C d\vec{\ell} \,\, ?
    [/tex]
     
  6. Nov 15, 2005 #5
    ok the result of this integration [tex] v \oint dl = v \int_{a}^{b} dl = v [l]_{a}^{b} = v (b - a) [/tex]

    im not sure what v is the gradient of, however.
     
  7. Nov 16, 2005 #6

    Physics Monkey

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    A couple of things and you should have your answer.

    First, don't forget that [tex] \vec{a} [/tex] and [tex] \vec{b} [/tex] are vectors. I don't know if you're confused about this, or if you just don't bother to notate it, or if you can't get LaTeX to work. I just wanted to emphasize this point either way.

    Second, are you sure you can't find a function that [tex] \vec{v} [/tex] is the gradient of? Hint: the partial derivative with respect to x of your unknown function is [tex] v_x [/tex], a constant.

    Third, your very close to your answer, here and in the other thread. All you need to do is convince yourself (or more properly, the grader) that
    [tex]
    \int_C d\vec{\ell} = \vec{b} - \vec{a}
    [/tex]
    independent of the path C. Once you have this fact, the answers to both your questions are easy to get.
     
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