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Home » A rectangular coil of length and width having 50 turns of wire is suspended vertically in a uniform magnetic field of strength Weber . The coil carries a current of . if the plane of the coil is inclined at an angle of with the direction of the field, the torque required to keep coil in stable equilibrium will be: (1) (2) (3) (4) .
A rectangular coil of length 0.12 \mathrm{~m} and width 0.1 \mathrm{~m} having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber / \mathrm{m}^{2}. The coil carries a current of 2 \mathrm{~A}. if the plane of the coil is inclined at an angle of 30^{\circ} with the direction of the field, the torque required to keep coil in stable equilibrium will be: (1) 0.12 \mathrm{Nm} (2) 0.15 \mathrm{Nm} (3) 0.20 \mathrm{Nm} (4) 0.24 \mathrm{Nm}.
Physics | Physics
A rectangular coil of length 0.12 \mathrm{~m} and width 0.1 \mathrm{~m} having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber / \mathrm{m}^{2}. The coil carries a current of 2 \mathrm{~A}. if the plane of the coil is inclined at an angle of 30^{\circ} with the direction of the field, the torque required to keep coil in stable equilibrium will be: (1) 0.12 \mathrm{Nm} (2) 0.15 \mathrm{Nm} (3) 0.20 \mathrm{Nm} (4) 0.24 \mathrm{Nm}.

The correct solution is (3)

\begin{array}{l} \vec{\tau}=\overrightarrow{\mathrm{M}} \times \overline{\mathrm{B}}=\mathrm{MB} \sin 60^{\circ} \\ =\mathrm{Ni} \mathrm{AB} \sin 60^{\circ} \\ 50 \times 2 \times 0.12 \times 0.1 \times 0.2 \times \frac{\sqrt{3}}{2} \\ =12 \sqrt{3} \times 10^{-2} \mathrm{Nm}=0.20748 \mathrm{Nm} \end{array}

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