question_answer

A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that \[\vec{B}\] is in plane of the coil. If due to a current i in the triangle a torque \[\tau \] acts on it, the side l of the triangle is            

A) \[\frac{2}{\sqrt{3}}{{\left( \frac{\tau }{B.i} \right)}^{\frac{1}{2}}}\]

B) \[2{{\left( \frac{\tau }{\sqrt{3}B.i} \right)}^{\frac{1}{2}}}\]

C) \[\frac{2}{\sqrt{3}}\left( \frac{\tau }{B.i} \right)\]

D) \[\frac{1}{\sqrt{3}}\frac{\tau }{B.i}\]

Correct Answer: B

Solution :

[b] \[\tau =MB\,\sin \,\theta ,\,\tau =iAB\,\sin \,{{90}^{o}}\] \[\therefore \,\,\,\,A=\frac{\tau }{iB}=1/2(BC)(AD)\] But \[\frac{1}{2}\,(BC)(AD)\] \[=\frac{1}{2}(l)\sqrt{{{l}^{2}}-{{\left( \frac{l}{2} \right)}^{2}}}=\frac{\sqrt{3}}{4}{{l}^{2}}\] \[\Rightarrow \,\,\,\frac{\sqrt{3}}{4}{{(l)}^{2}}=\frac{\tau }{Bi}\Rightarrow \,\therefore \,\,\,l=2{{\left( \frac{\tau }{\sqrt{3}B.i} \right)}^{\frac{1}{2}}}\]