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  • A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that <img alt="\mathrm{\overrightarrow{B}}" src="https://entrancecorner

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

Option: 1

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


Option: 2

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


Option: 3

\frac{2}{\sqrt{3}}\left(\frac{\tau}{\mathrm{Bi}}\right)


Option: 4

\frac{1}{\sqrt{3}} \frac{\tau}{\text { B.i }}


Answers (1)

\mathrm{\tau=M B \sin \theta }

\mathrm{\tau=\mathrm{iAB} \sin 90^{\circ} }

\mathrm{ \therefore \mathrm{A}=\frac{\tau}{\mathrm{iB}} \\ }

\mathrm{ \text { Also, } A=1 / 2(B C)(A D) \\ }

\mathrm{ \text { But } \frac{1}{2}(\mathrm{BC})(\mathrm{AD})=\frac{1}{2}(\mathrm{l}) \sqrt{\tau^2-\left(\frac{1}{2}\right)^2} \\ }
\mathrm{ =\frac{\sqrt{3}}{4} \mathrm{l}^2 \\ }

\mathrm{ \therefore \mathrm{l}=2\left(\frac{\tau}{\sqrt{3} \mathrm{Bj}}\right)^{\frac{1}{2}} \quad-\frac{\sqrt{3}}{4(\mathrm{l})^2}=\frac{\tau}{\mathrm{Bi}} }
 

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Kshitij

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