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Homework Help: Magnetic Field

  1. Nov 15, 2005 #1
    By direct integration show that for any arbitrary loop carry a current i
    [tex] F = \oint idL \cross B = 0 [/tex]
    Note that an arbitrary current loop doesn ot need to lie in a plane
    WEll since that is true then both dL and d(theta) are integration varaibles here
    should something like this be formed?
    [tex] B \int_{0}^{L} \int_{\theta_{1}}^{\theta_{2}} \sin(\theta)dL + L \cos(\theta) d\theta [/tex]
    am i right? I m not sure about the limits of integration tho...

    please advise

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

    Physics Monkey

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    Is the magnetic field constant? If so, doesn't the force on a closed loop equal
    [tex]
    \vec{F} = I \oint d\vec{\ell} \times \vec{B},
    [/tex]
    which reduces to
    [tex]
    \vec{F} =\left( I \oint d\vec{\ell} \right) \times \vec{B},
    [/tex]
    since the magnetic field is constant. What is
    [tex]
    \oint d\vec{\ell} \,\,?
    [/tex]
     
    Last edited: Nov 15, 2005
  4. Nov 15, 2005 #3
    [tex] \oint dL = 0 [/tex] because of Green's Theorem??

    In my calc text book it says that [tex]\int_{C} \nabla f = o [/tex] for any piecewise differentiable curve?
     
    Last edited: Nov 16, 2005
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