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JEE Mains · Physics · STD 12 - 4. 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

  1. A \(\frac{{IBR}}{{2\,\sin \,{\theta _0}}}\)
  2. B \(\frac{{IBR{\theta _0}}}{{\sin \,{\theta _0}}}\)
  3. C \(IBR\)
  4. D \(\frac{{IBR}}{{\,\sin \,{\theta _0}}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(IBR\)

Step-by-step Solution

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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