A) \[2mA\]
B) \[1mA\]
C) \[40\mu A\]
D) \[20\mu A\]
Correct Answer: B
Solution :
\[N=1000\,\,turns/an\] \[H=\frac{Nl}{l}\] \[100\frac{A}{M}=\frac{1000\times l}{{{10}^{-2}}}\] \[100A={{10}^{5}}\times I\] \[{{10}^{-3}}A=I\] \[I=1mA\]
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