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 ..... .

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Answer & Solution

Correct Answer

(A) \(3\)

Step-by-step Solution

Detailed explanation

\(\vec{\tau}=\vec{M} \times \vec{B}=M B \sin 90^{\circ}\)

\(=M B=\frac{i \sqrt{3} \ell^{2}}{4} B\)

\(=\sqrt{3} \times 10^{-5} {N}-{m}\)

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