question_answer

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 \[\,{{30}^{\text{o}}}\] with the direction of the field, the torque required to keep the coil in stable equilibrium will be

A)  0.15Nm                                              

B)  0.20 Nm

C)                                                         0.24 Nm

D)                         0.12 Nm

Correct Answer: B

Solution :

Given, N = 50 \[B=0.2Wb/{{m}^{2}},I=2A\] \[\theta ={{60}^{{}^\circ }},A=0.12\times 0.1=0.012{{m}^{2}}\] Thus, torque required to keep the coil in stable equilibrium i.e. \[\tau =NIAB\sin \theta =50\times 2\times 0.012\times 0.2\times \sin {{60}^{{}^\circ }}\] \[=50\times 2\times 0.12\times 0.2\times \frac{\sqrt{3}}{2}=0.20Nm\]