A rectangular coil of length 0.12m and width 0.1m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/m2. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

τ=BiAnsin⁡θ=0.2×2×0.12×0.1×50×32=0.12×1.732=0.207N−m

A rectangular coil of length 0.12m and width 0.1m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/m2. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

τ=BiAnsin⁡θ=0.2×2×0.12×0.1×50×32=0.12×1.732=0.207N−m

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A rectangular coil of length 0.12m and width 0.1m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/m². The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

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0.20 Nm

0.24 Nm

0.15 Nm

0.12 Nm

answer is C.

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Detailed Solution

$\begin{array}{l} τ = BiAnsin θ = 0 .2 \times 2 \times 0 .12 \times 0 .1 \times 50 \times \frac{\sqrt{3}}{2} \\ = 0 .12 \times 1 .732 = 0 .207 N - m \end{array}$

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