How can we clarify that the relation between linkage flux an flux in a cylinder shaped wire is \$ d\lambda=\frac{x^2}{a^2}\ d\phi \$?can we illustrate this by electromagnetic formula?I think I would be able to explain this if I knew the exact definition of linkage flux,I think linkage flux is not a real flux that exists(like flux itself),any help would be appreciated.

  • \$\begingroup\$ The flux linkage is the product of the flux \$ \Phi \$ and the number of coils, N, exist in the wire. Now are you able to explain that? \$\endgroup\$
    – dirac16
    Oct 11 '17 at 14:43
  • \$\begingroup\$ I know this ,but here we don't have a coil!it is a cylinder! \$\endgroup\$
    – MAh2014
    Oct 11 '17 at 15:06
  • \$\begingroup\$ Here you should think of N as the fractional volumes. \$\endgroup\$
    – dirac16
    Oct 11 '17 at 16:08
  • \$\begingroup\$ see physics.stackexchange.com/questions/176783/… \$\endgroup\$
    – dirac16
    Oct 11 '17 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.