question_answer

A current flows in a conducting wire of length L. If we bent it in a circular form, its magnetic dipole moment would be

A)  \[\frac{l{{L}^{2}}}{4\pi }\]                            

B)  \[\frac{lL}{4\pi }\]

C)  \[\frac{{{l}^{2}}L}{4\pi }\]                            

D)  \[\frac{{{l}^{2}}{{L}^{2}}}{4\pi }\]

Correct Answer: A

Solution :

Let a wire length L is bend in a circular form of radius r. \[2\pi r=L\,\,\,\,\,\Rightarrow \,\,\,\,\,r=\frac{L}{2\pi }\,\,\] The magnetic dipole moment of a circular ring \[\operatorname{M}=IA\] \[\operatorname{M}=l\pi {{r}^{2}}\]                             ... (ii) On putting the value of from Eq. (i), \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,M=l\pi {{\left( \frac{L}{2\pi } \right)}^{2}}=l\pi \frac{{{L}^{2}}}{4{{\pi }^{2}}}=\frac{l{{L}^{2}}}{4\pi }\]