A) \[2I{{B}_{0}}R\hat{k}\]
B) \[I{{B}_{0}}R\hat{k}\]
C) \[-2I{{B}_{0}}R\hat{k}\]
D) \[-I{{B}_{0}}R\hat{k}\]
Correct Answer: B
Solution :
We know, \[\vec{F}=I(\vec{\ell }\times \vec{B})\]or \[d\,\vec{F}=IBd\ell \sin \theta \] \[\therefore \]\[F=\int_{{{60}^{o}}}^{{{120}^{o}}}{I{{B}_{0}}d\ell \sin \theta d\ell \,\text{where,}d\ell =rd\theta }\] \[F=\int_{{{60}^{o}}}^{{{120}^{o}}}{I{{B}_{0}}R\sin \theta d\theta =IBR\left[ -\cos \theta \right]_{{{60}^{o}}}^{{{120}^{o}}}}\] \[=I{{B}_{0}}R\left[ -\left( -\frac{1}{2} \right)+\left( -\frac{1}{2} \right) \right]=I{{B}_{0}}R\] Hence, force acting on the arc is IB R k\[I{{B}_{0}}R\,\hat{k}\] Hence, the correction option is [b].
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