Magnetic Field and direct integration

[stunner5000pt]

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

 

[Physics Monkey]

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]

 

[stunner5000pt]

[tex] \oint dL = 0 [/tex] because of Green's Theorem??

In my calc textbook it says that [tex]\int_{C} \nabla f = o [/tex] for any piecewise differentiable curve?