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 \[{{30}^{o}}\] with the direction of the field, the torque required to keep the coil in stable equilibrium will be [NEET (Re) 2015]

A)                                0.15 Nm                                  

B)  0.20 Nm

C)  0.24 Nm

D)                   0.12 Nm

Correct Answer: B

Solution :

[b] 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.

\[\tau =NIAB\sin \theta =50\times 2\times 0.012\times 0.2\times \sin {{60}^{{}^\circ }}\]

\[=50\times 2\times 0.12\times 0.2\times \frac{\sqrt{3}}{2}=0.20Nm\]