Question 89 of 276

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, as shown in the figure) in the vicinity of the wire. If the wire subtends an angle $2{\theta }_{o}$ at the center of the circle (of which it forms an arch) then the tension in the wire is:

A$IBR$B$\dfrac{IBR}{sin{\theta }_{0}}$C$\dfrac{IBR}{2sin{\theta }_{0}}$D$\dfrac{IBR{\theta }_{0}}{sin{\theta }_{0}}$

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Answer: (A) $IBR$

Consider any small element <mml:math xmlns:m="http://schemas.openxmlformats.org/officeDocument/2006/math" xmlns:mml="http://www.w3.org/1998/Math/MathML"><mml:mi mathvariant="normal">dl</mml:mi></mml:math> of arc of angle $d\theta$ . Then magnetic force $dF=Idl.B$ outwards



For equilibrium of element

$Idl B=2T sin\dfrac{d\theta }{2}$

For small angle $sin\theta ≃\theta$

$IdlB=2T\dfrac{d\theta }{2}$

$Idl B=T\dfrac{dl}{R}$

$T=IBR$