A wire carrying current $I$ is tied between points $P$ and $Q$ and is in the shape of a circular arc of radius $R$ due to a uniform magnetic field $B$ (perpendicular to the plane of the paper, shown by $\times \times \times $) 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 arc) then the tension in the wire is
JEE MAIN 2015, Medium
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For small are length
$2 T \sin \theta=\operatorname{BIR} 2 \theta$
$\text { (As }\mathrm{F}=\mathrm{BIL} \text { and } L=\mathrm{RZ} \theta)$
$T=\mathrm{BIR}$
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