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 Online Paper (Held On 11 April 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 :

For the area to be in equilibrium, \[F=2T\sin \theta F=I(2R\sin \theta )\times B\] \[\therefore \] \[2T\sin \theta =I2R\sin \theta \times B\] \[T=IRB\]