A) \[3W\]
B) \[\sqrt{3}\,W\]
C) \[\frac{W}{3}\]
D) \[\frac{W}{\sqrt{3}}\]
Correct Answer: B
Solution :
The instantaneous moment of the deflecting couple or torque acting on the needle is \[\tau =\] force \[\times \] perpendicular distance = work done When axis of needle makes an angle \[\theta \] with the magnetic field, then for magnetic moment M and magnetic field B, we have \[\tau =MB\,\,\sin \,\theta \] ...(i) \[W=MB\,\,\cos \,\,B\] ...(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{\tau }{W}=\frac{MB\,\sin \,\theta }{MB\,\cos \,\theta }\] Given, \[\theta ={{60}^{o}}\] \[\therefore \] \[\tau =W\,\,\frac{\sin \,{{60}^{o}}}{\cos \,{{60}^{o}}}=W\sqrt{3}\]
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