Magnetic Field and direct integration

[stunner5000pt]

By direct integration show that for any arbitrary loop carry a current i
F = \oint idL \cross B = 0
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?
B \int_{0}^{L} \int_{\theta_{1}}^{\theta_{2}} \sin(\theta)dL + L \cos(\theta) d\theta
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
<br /> \vec{F} = I \oint d\vec{\ell} \times \vec{B},<br />
which reduces to
<br /> \vec{F} =\left( I \oint d\vec{\ell} \right) \times \vec{B},<br />
since the magnetic field is constant. What is
<br /> \oint d\vec{\ell} \,\,?<br />

 

[stunner5000pt]

\oint dL = 0 because of Green's Theorem??

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