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 8 is in plane of the coil. If due to a current i in the triangle a torque t acts on it, the side I of the triangle is:

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

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

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

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

Correct Answer: C

Solution :

                 Torque acting on equilateral triangle in a magnetic field\[\overrightarrow{B}\]is \[\tau =iAB\sin \theta \]                                          ...(1) Area of triangle LMN \[A=\frac{\sqrt{3}}{4}{{l}^{2}}\] And                        \[\theta =90{}^\circ \] \[\therefore \]                  \[\tau =i\times \frac{\sqrt{3}}{4}{{l}^{2}}B\sin {{90}^{o}}\]                                 \[=\frac{\sqrt{3}}{4}i{{l}^{2}}B\]                \[(\therefore \sin {{90}^{o}}=1)\] Hence,  \[l=2{{\left( \frac{\tau }{\sqrt{3}Bi} \right)}^{1/2}}\]