A) \[B{{R}^{3}}/2\pi {{\mu }_{0}}\]
B) \[2\pi B{{R}^{3}}/{{\mu }_{0}}\]
C) \[B{{R}^{2}}/2\pi {{\mu }_{0}}\]
D) \[2\pi B{{R}^{2}}/{{\mu }_{0}}\]
Correct Answer: B
Solution :
\[B=\frac{{{\mu }_{0}}i}{2R}\Rightarrow i=\frac{B\times 2R}{{{\mu }_{0}}}\] Now, \[M=i\times A=i\pi {{R}^{2}}=\frac{B\times 2R}{{{\mu }_{0}}}\times \pi {{R}^{2}}=\frac{2\pi B{{R}^{3}}}{{{\mu }_{0}}}\]
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