A) \[4L\]
B) \[2L\]
C) \[\frac{L}{2}\]
D) Zero
Correct Answer: C
Solution :
\[\tau =NlAB=Nl\pi {{r}^{2}}B\] \[r2\pi r=l\] \[r=\frac{l}{2\pi N}\] \[\tau =Nl\pi \times {{\left( \frac{l}{2\pi N} \right)}^{2}}B\] \[=Nl\pi +\frac{{{l}^{2}}}{4{{\pi }^{2}}{{N}^{2}}}B=\frac{l{{L}^{2}}B}{4\pi N}\] \[{{\tau }_{\max }}=\frac{l{{L}^{2}}B}{4\pi }\]
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