A) 12.5 mH
B) 1.25 mH
C) 15.0 mH
D) 0.12 mH
Correct Answer: B
Solution :
\[L=\frac{{{\mu }_{0}}{{N}^{2}}A}{l}=\frac{4\pi \times {{10}^{-7}}\times {{\left( 500 \right)}^{2}}\times 20\times {{10}^{-4}}}{0.5}\]=1.25mH
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