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, \[F=2T\sin \theta F=I(2R\sin \theta )\times B\] |
\[\therefore \] \[2T\sin \theta =I2R\sin \theta \times B\] |
\[T=IRB\] |
You need to login to perform this action.
You will be redirected in
3 sec