A) \[2.96\times {{10}^{-4}}H\]
B) \[5.3\times {{10}^{-5}}H\]
C) \[3.52\times {{10}^{-3}}H\]
D) \[8.3\times {{10}^{-5}}H\]
E) \[2.96\times {{10}^{-3}}H\]
Correct Answer: A
Solution :
\[{{M}_{21}}=\frac{{{\mu }_{0}}{{N}_{1}}{{N}_{2}}{{A}_{2}}}{{{l}_{1}}}\] \[\therefore \]\[{{M}_{21}}=\frac{\begin{align} & (4\times 3.14\times {{10}^{-7}})\times 1500\times 100 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\times \{3.14{{(2\times {{10}^{-2}})}^{2}}\} \\ \end{align}}{80\times {{10}^{-2}}}\] \[{{M}_{21}}=2.96\times {{10}^{-4}}H\] \[\Rightarrow \] \[{{M}_{21}}={{M}_{21}}=2.96\times {{10}^{-4}}H\]
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