| 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: |
| \[\vec{B}={{B}_{0}}(\hat{i}+\hat{j}-\hat{k})\,\,:\,\,\,\,\,\,\vec{\ell }=R(\hat{i}-\hat{j})\] |
| \[\vec{B}\,\vec{\ell }=0\,\,\,\,\,\,\Rightarrow Angle\,\,=90{}^\circ \,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,F=BI\ell \] |
| \[=\sqrt{3}\,{{B}_{0}}\,I\,\sqrt{2}\,R=\sqrt{6}\,{{B}_{0}}\,IR\] |
You need to login to perform this action.
You will be redirected in
3 sec
