A) \[\lambda =\frac{{{n}^{2}}}{2l}\]
B) \[\lambda =\frac{{{I}^{2}}}{2n}\]
C) \[\lambda =\frac{2l}{n}\]
D) \[\lambda =2\ln \]
Correct Answer: C
Solution :
\[\lambda =\frac{2l}{n},\]where \[n=\]number of loops. at \[n=1\] \[\lambda =2l\] at \[n=2\] \[\lambda =\frac{2l}{2}\] at \[n=3\] \[\lambda =\frac{2l}{3}\] at \[n=n\] \[\lambda =\frac{2l}{n}\]You need to login to perform this action.
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