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 :
[b] We know, \[\overset{\to }{\mathop{F}}\,=I(\overset{\to }{\mathop{\ell }}\,\times \overset{\to }{\mathop{B}}\,)\] or \[d\overset{\to }{\mathop{F}}\,=IB\,d\ell \,\sin \,\theta \] \[\therefore \,\,\,\,F=\int_{{{60}^{o}}}^{{{120}^{o}}}{I{{B}_{0}}\,d\ell }\,\sin \,\theta \,d\ell \] 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 \[I{{B}_{0}}R\,\,\hat{k}\] .
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