- #1
stunner5000pt
- 1,461
- 2
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
[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: