Wire bent as ABOCD as shown, carries current I entering at A and leaving at D. Three uniform magnetic fields each \[{{B}_{0}}\] exist in the region as shown. The force on the wire is |
A) \[\sqrt{3}I\,R\,{{B}_{0}}\]
B) \[\sqrt{5}I\,R\,{{B}_{0}}\]
C) \[\sqrt{8}I\,R\,{{B}_{0}}\]
D) \[\sqrt{6}I\,R\,{{B}_{0}}\]
Correct Answer: D
Solution :
\[\vec{F}=\vec{F}=I\vec{\ell }\times \vec{B}\] |
\[\vec{\ell }=\overrightarrow{AD}=R(\vec{i}-\vec{j})\] |
\[\vec{B}={{B}_{0}}(\hat{i}+\hat{j}+\hat{k})\] |
\[\therefore \vec{F}=IR{{B}_{0}}(\hat{i}-\hat{j})\times (\hat{i}+\hat{j}-\hat{k})\] |
\[=IR{{B}_{0}}\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -1 & 0 \\ 1 & 1 & -1 \\ \end{matrix} \right|=IR{{B}_{0}}(\hat{i}+\hat{j}+2\hat{k})\] |
\[F=IR{{B}_{0}}\sqrt{6}\] |
Aliter: |
\[\vec{B}={{B}_{0}}(\hat{i}+\hat{j}-\hat{k})\,\,:\,\,\,\,\,\,\vec{\ell }=R(\hat{i}-\hat{j})\] |
\[\vec{B}\,\vec{\ell }=0\,\,\,\,\,\,\Rightarrow Angle\,\,=90{}^\circ \,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,F=BI\ell \] |
\[=\sqrt{3}\,{{B}_{0}}\,I\,\sqrt{2}\,R=\sqrt{6}\,{{B}_{0}}\,IR\] |
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