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 × × ×) in the vicinity of the wire. If the wire subtends an angle 2θ0 at the centre of the circle (of which it forms an arc) then the tension in the wire is

(A) \frac{{IBR}}{{2\,\sin \,{\theta _0}}}

(B) \frac{{IBR{\theta _0}}}{{\sin \,{\theta _0}}}

(D) \frac{{IBR}}{{\,\sin \,{\theta _0}}}

JEE Mains · Physics · STD 12 - Moving charges and magnetism

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

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Verified Solution

Correct Answer

(C) \(IBR\)

Detailed explanation

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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