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 place of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle $2 \theta$ at the center of the circle(of which it forms an arch) then the tension in the wire is :

Question

$\begin{array}{1 1} IBR \\ \frac{IBR}{\sin \theta _0} \\ \frac{IBR}{2 \sin \theta_0} \\ \frac{IBR \; \theta_0}{\sin \theta_0} \end{array}$