A) \[2.4\times {{10}^{-4}}H\]
B) \[3.9\times {{10}^{-4}}H\]
C) \[1.28\times {{10}^{-3}}H\]
D) \[3.14\times {{10}^{-3}}H\]
Correct Answer: B
Solution :
(b) \[3.9\times {{10}^{-4}}H\] Here, l = 2 m diameter = 2 cm \[\therefore \] Radius, \[r=\frac{2}{2}=1\,cm=1\times {{10}^{-2}}m\] \[{{N}_{1}}=2000,{{N}_{2}}=1000\] Area \[=\pi {{r}^{2}}=\pi \times \left( 1\times {{10}^{-2}} \right)=3.14\times {{10}^{-4}}{{m}^{2}}\] Mutual inductance, \[M=\frac{{{\mu }_{0}}{{N}_{1}}{{N}_{2}}A}{l}\] \[=\frac{4\pi \times {{10}^{-7}}\times 2000\times 1000\times 3.14\times {{10}^{-4}}}{2}\] \[=3.9\times {{10}^{-4}}H\]
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