question_answer

A wire carrying current I is tied between points P and Q and is in the shape of a circular arch of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle \[2{{\theta }_{0}}\] at the centre of the circle (of which it forms an arch) then the tension in the wire is :                                                  [JEE MAIN 11-04-2015]

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 :

[a]

For the area to be in equilibrium, \[F=2T\sin \theta F=I(2R\sin \theta )\times B\]

\[\therefore \] \[2T\sin \theta =I2R\sin \theta \times B\]

\[T=IRB\]