A) \[\frac{B{{A}^{2}}}{{{\mu }_{0}}\pi }\]
B) \[\frac{B{{A}^{3/2}}}{{{\mu }_{0}}\pi }\]
C) \[\frac{B{{A}^{3/2}}}{{{\mu }_{0}}{{\pi }^{1/2}}}\]
D) \[\frac{2B{{A}^{3/2}}}{{{\mu }_{0}}{{\pi }^{1/2}}}\]
Correct Answer: D
Solution :
\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi I}{r}=\frac{{{\mu }_{0}}I}{2r}\] Or \[I=\frac{2Br}{{{\mu }_{0}}}\] Also area, \[A=\pi {{r}^{2}}\] or \[r={{\left( \frac{A}{\pi } \right)}^{1/2}}\] Magnetic moment of the loop \[M=IA=2\frac{Br}{{{\mu }_{0}}}A\] \[=2\frac{BA}{{{\mu }_{0}}}{{\left( \frac{A}{\pi } \right)}^{1/2}}=2\frac{B{{A}^{3/2}}}{{{\mu }_{0}}{{\pi }^{1/2}}}\]
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