A) \[IBR\]
B) \[\frac{IBR}{\sin {{\theta }_{0}}}\]
C) \[\frac{IBR}{2\sin {{\theta }_{0}}}\]
D) \[\frac{IBR{{\theta }_{0}}}{\sin {{\theta }_{0}}}\]
Correct Answer: A
Solution :
For the area to be in equilibrium, \[F=2T\sin \theta F=I(2R\sin \theta )\times B\] \[\therefore \] \[2T\sin \theta =I2R\sin \theta \times B\] \[T=IRB\]You need to login to perform this action.
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