A) \[0.022\ Wb{{m}^{-2}}\]
B) \[0.4\ Wb\ {{m}^{-2}}\]
C) \[0.8\ Wb\ {{m}^{-2}}\]
D) \[1\ Wb\ {{m}^{-2}}\]
Correct Answer: A
Solution :
\[B=\frac{{{\mu }_{0}}Ni}{2r}=\frac{4\pi \times {{10}^{-7}}\times 100\times 2\times \sqrt{\pi }}{2\times {{10}^{-2}}}\]\[=0.022\ wb/{{m}^{2}}\]
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