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, \[l=2{{\left( \frac{\tau }{\sqrt{3}Bi} \right)}^{1/2}}\]You need to login to perform this action.
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