Magnetic Field (1 Viewer)

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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
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Physics Monkey

Science Advisor
Homework Helper
Is the magnetic field constant? If so, doesn't the force on a closed loop equal
\vec{F} = I \oint d\vec{\ell} \times \vec{B},
which reduces to
\vec{F} =\left( I \oint d\vec{\ell} \right) \times \vec{B},
since the magnetic field is constant. What is
\oint d\vec{\ell} \,\,?
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[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?
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