InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Consider an infinitely long wire carrying a current I (t), With `(dI)/(dt) = lambda` = constant . Find the current produced in the rectangular loop of wire ABCD if its resistance is R, Fig |
|
Answer» In Fig. XY is infinitley long wise carrying current I (t), such that `(dI)/(dt) = lambda =` constant. At any distance r from the wire, strength of manetic field, `B (r ) = (mu_(0) I)/(2 pu r)`. It is directed out of the pape. Total magnetic flux linked with the rectangular loop `phi = int_(x_(0))^(x) B (r ). dA = int_(x_(0))^(x) (mu_(0) I)/(2 pi r) ldr` `phi = (mu_(0) Il)/(2 pi) (log_(e)) (x)/(x_(0)) = LI`, where L is self-inductance of the loop `L = (mu_(0) l)/(2 pi) (log_(e)) (x)/(x_(0))` Induced current, `I = (e)/(R ) = (L)/(R ) (dI)/(dt) = (L)/(R ) lambda = (mu_(0) l lambda)/(2 pi R) (log_(e)) (x)/(x_(0))` |
|