Magnetic Field

1. Nov 15, 2005

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...

thank you

Last edited: Nov 15, 2005
2. Nov 15, 2005

Physics Monkey

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
$$\oint d\vec{\ell} \,\,?$$

Last edited: Nov 15, 2005
3. Nov 15, 2005

stunner5000pt

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

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

Last edited: Nov 16, 2005