A) 4\[\pi \]
B) \[\frac{4}{\pi }\]
C) \[\frac{2}{\pi }\]
D) 2\[\pi \]
Correct Answer: B
Solution :
Suppose the length of wire is\[l\]. When wire formed in circular loop then radius of loop \[r=\frac{l}{2\pi }\] Magnetic dipole Mi \[{{M}_{1}}=iA=i\pi {{r}^{2}}\] \[=i\pi \times {{\left( \frac{l}{2\pi } \right)}^{2}}=i\times \frac{{{l}^{2}}}{4\pi }\] When wire formed in square loop then the side of loop \[a=l/4\] Magnetic dipole \[{{M}_{2}}=IA=i\times {{a}^{2}}=i\times \frac{{{l}^{2}}}{16}\] \[\therefore \] \[\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{4}{\pi }\]
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