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 \[\mathbf{\vec{B}}\] 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) \[\frac{2}{\sqrt{2}}{{\left( \frac{\tau }{Bi} \right)}^{1/2}}\]                           

B) \[\frac{2}{\sqrt{3}}\left( \frac{\tau }{Bi} \right)\]              

C) \[2{{\left( \frac{\tau }{\sqrt{3}Bi} \right)}^{1/2}}\]           

D)        \[\frac{1}{\sqrt{3}}\frac{\tau }{Bi}\]

Correct Answer: C

Solution :

Torque acting on equilateral triangle in a magnetic field \[\overrightarrow{B}\] is \[\tau =i\,AB\sin \theta \] Area of triangle LMN \[A=\frac{\sqrt{3}}{4}{{l}^{2}}\]  and \[\theta =90{}^\circ \] Substituting the given values in the expression for torque, we have \[\tau =i\times \frac{\sqrt{3}}{4}{{l}^{2}}B\sin 90{}^\circ \]   \[=\frac{\sqrt{3}}{4}i\,{{l}^{2}}B\]                           \[(\because \sin 90{}^\circ =1)\] Hence,  \[l=2{{\left( \frac{\tau }{\sqrt{3}Bi} \right)}^{1/2}}\]