A coil in the shape of an equilateral triangle of side $10\, {cm}$ lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field $20\, {mT}$. The torque acting on the coil when a current of $0.2\, {A}$ is passed through it and its plane becomes parallel to the magnetic field will be $\sqrt{{x}} \times 10^{-5} \,{Nm}$. The value of ${x}$ is ..... .
JEE MAIN 2021, Medium
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$\vec{\tau}=\vec{M} \times \vec{B}=M B \sin 90^{\circ}$
$=M B=\frac{i \sqrt{3} \ell^{2}}{4} B$
$=\sqrt{3} \times 10^{-5} {N}-{m}$
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