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. \[\tau =NIAB\sin \theta =50\times 2\times 0.012\times 0.2\times \sin {{60}^{{}^\circ }}\] \[=50\times 2\times 0.12\times 0.2\times \frac{\sqrt{3}}{2}=0.20Nm\]You need to login to perform this action.
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