A) 1.5 H
B) 2.5 H
C) 3.5 H
D) 0.5 H
Correct Answer: A
Solution :
\[L=\frac{{{\mu }_{r}}{{\mu }_{0}}{{N}^{2}}A}{l}\] \[=\frac{600\times 4\times {{10}^{-7}}\times {{(2000)}^{2}}\times (1.5\times {{10}^{-4}})}{0.3}\] \[=1.5\,H\]
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