A) \[\frac{Li{{B}^{2}}}{2}\]
B) \[\frac{L{{i}^{2}}B}{2}\]
C) \[\frac{{{L}^{2}}iB}{4\pi }\]
D) \[\frac{L{{i}^{2}}B}{4\pi }\]
Correct Answer: C
Solution :
\[{{\tau }_{\max }}=NiAB=1\times i\times (\pi {{r}^{2}})\times B\] \[\left( 2\pi r=L,\,\,\Rightarrow r=\frac{L}{2\pi } \right)\] \[{{\tau }_{\max }}=\pi i{{\left( \frac{L}{2\pi } \right)}^{2}}B=\frac{{{L}^{2}}iB}{4\pi }\]
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