A) \[{{\mu }_{0}}{{N}^{2}}\text{A/}l\]
B) \[{{\mu }_{0}}N\text{A/}l\]
C) \[{{\mu }_{0}}{{N}^{2}}l\text{A}\]
D) \[{{\mu }_{0}}N\text{A}l\]
Correct Answer: A
Solution :
Self-inductance \[L={{\mu }_{0}}{{n}^{2}}lA\]where \[n=\frac{N}{l}\] (turns per unit length) \[\therefore \] \[L=\frac{{{\mu }_{0}}{{N}^{2}}}{{{l}^{2}}}.lA=\frac{{{\mu }_{0}}{{N}^{2}}A}{l}\]
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