A) 0.1256 mH
B) 12.56 mH
C) 1.256 mH
D) 125.6 mH
Correct Answer: C
Solution :
\[L=\frac{{{\mu }_{0}}{{N}^{2}}A}{l}=\frac{4\pi \times {{10}^{-7}}\times {{(1000)}^{2}}\times 10\times {{10}^{-4}}}{1}\] =1.256 mHYou need to login to perform this action.
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