question_answer

A uniform ring of mass M and radius R carries a current I (see figure). The ring is suspended using two identical strings OA and OB. There exists a uniform horizontal magnetic field \[{{B}_{0}}\] parallel to the diameter AB of the ring. The ratio of tensions in the two strings \[\frac{{{T}_{AO}}}{{{T}_{BO}}}\] is (Given \[\theta =60{}^\circ \])

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       \[{{T}_{2}}=Mg-\pi IR{{B}_{0}}\]