A) \[\frac{Mg-\pi IR{{B}_{0}}}{Mg+\pi IR{{B}_{0}}}\]
B) \[\frac{Mg+\pi IR{{B}_{0}}}{Mg-\pi IR{{B}_{0}}}\]
C) \[\frac{Mg-\left( {}^{IR{{B}_{0}}}/{}_{\pi } \right)}{Mg+\left( {}^{IR{{B}_{0}}}/{}_{\pi } \right)}\]
D) \[\frac{Mg+\left( {}^{IR{{B}_{0}}}/{}_{\pi } \right)}{Mg-\left( {}^{IR{{B}_{0}}}/{}_{\pi } \right)}\]
Correct Answer: B
Solution :
[b] Magnetic dipole moment \[\vec{\mu }=I.\pi {{R}^{2}}\hat{k}\] Magnetic torque \[{{\vec{\tau }}_{B}}=\vec{\mu }\times \vec{B}\] \[=I\pi {{R}^{2}}{{B}_{0}}(\hat{k}\times (-\hat{j}))=I\pi {{R}^{2}}{{B}_{0}}(\hat{i})\] To counterbalance this torque we must have \[{{T}_{1}}>{{T}_{2}}\]Torque about centre C \[-({{T}_{1}}\cos \theta .R-{{T}_{2}}\cos \theta R)\hat{i}+I\pi {{R}^{2}}{{B}_{0}}\hat{i}=0\] \[\therefore \,\,\,\,\,\,\,\,\,{{T}_{1}}-{{T}_{2}}=\frac{I\pi R{{B}_{0}}}{\cos \theta }\] ?..(i) And \[({{T}_{1}}+{{T}_{2}})\cos \theta =Mg\] \[\Rightarrow \,\,\,\,\,\,\,\,{{T}_{1}}+{{T}_{2}}=\frac{Mg}{\cos \theta }\] ?...(ii) Solving (i) and (ii), \[{{T}_{1}}=\frac{1}{2\cos \theta }(Mg+\pi IR{{B}_{0}})=Mg+\pi IR{{B}_{0}}\] and \[{{T}_{2}}=Mg-\pi IR{{B}_{0}}\]You need to login to perform this action.
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