A) \[1.2\times {{10}^{-4}}T\]
B) \[2.4\times {{10}^{-4}}T\]
C) \[1.2\times {{10}^{-2}}T\]
D) \[2.4\times {{10}^{-2}}T\]
Correct Answer: C
Solution :
(c) \[1.2\times {{10}^{-2}}T\] Here, \[\theta =60{}^\circ ,\text{ }{{B}_{1}}=1.2\times {{10}^{-2}}T\] \[{{\theta }_{1}}=30{}^\circ \] and \[{{\theta }_{2}}=60{}^\circ -30{}^\circ =30{}^\circ \] In stable equilibrium, torques due to two fields must be balanced i.e. \[{{\tau }_{1}}={{\tau }_{2}}\] \[\Rightarrow M{{B}_{1}}\sin {{\theta }_{1}}=M{{B}_{2}}\sin {{\theta }_{2}}\] Or \[{{B}_{2}}={{B}_{1}}\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\] \[={{B}_{2}}\frac{\sin 30{}^\circ }{\sin 30{}^\circ }={{B}_{1}}\] \[=1.2\times {{10}^{-2}}T\]You need to login to perform this action.
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