question_answer

A coil in the shape of an equilateral triangle of side I 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, then the side I of the triangle is

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

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 \[\vec{B}\]is \[\tau =iAB\sin \theta \] Area of triangle LMN \[A=\frac{\sqrt{3}}{4}{{l}^{2}}\]and \[\theta ={{90}^{o}}\] Substituting the given values in the expression for torque, we have \[\tau =i\times \frac{\sqrt{3}}{4}{{l}^{2}}B\sin {{90}^{o}}\] \[=\frac{\sqrt{3}}{4}i{{l}^{2}}B\]                    \[(\because \,sin{{90}^{o}}=1)\] Hence, \[l=2{{\left( \frac{\tau }{\sqrt{3}Bi} \right)}^{1/2}}\]