A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Wb m⁻². 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…

Question

A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Wb m⁻². 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

Accepted Answer

(4) 0.20 N m. Given: N=50, l=0.12 m, b=0.1 m, I=2 A, B=0.2 Wb m⁻², plane at 30° to the field. Area: A=0.12×0.1=0.012 m². Idea: the torque formula uses the angle between the normal to the coil and the field, not the angle the plane makes. If the plane is at 30° to ⃗B, the normal is at 90°-30°=60° to ⃗B. NIAB=50×2×0.012×0.2=0.24 N m. τ=NIAB sin 60°=0.24×(√3)/2=0.208 N m. To the two figures the options carry, that is 0.20 N m.

Step-by-step solution

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