• # question_answer 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) $\sqrt{6}I\,R\,{{B}_{0}}$

 $\vec{F}=\vec{F}=I\vec{\ell }\times \vec{B}$ $\vec{\ell }=\overrightarrow{AD}=R(\vec{i}-\vec{j})$ $\vec{B}={{B}_{0}}(\hat{i}+\hat{j}+\hat{k})$ $\therefore \vec{F}=IR{{B}_{0}}(\hat{i}-\hat{j})\times (\hat{i}+\hat{j}-\hat{k})$ $=IR{{B}_{0}}\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -1 & 0 \\ 1 & 1 & -1 \\ \end{matrix} \right|=IR{{B}_{0}}(\hat{i}+\hat{j}+2\hat{k})$ $F=IR{{B}_{0}}\sqrt{6}$ Aliter: $\vec{B}={{B}_{0}}(\hat{i}+\hat{j}-\hat{k})\,\,:\,\,\,\,\,\,\vec{\ell }=R(\hat{i}-\hat{j})$ $\vec{B}\,\vec{\ell }=0\,\,\,\,\,\,\Rightarrow Angle\,\,=90{}^\circ \,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,F=BI\ell$ $=\sqrt{3}\,{{B}_{0}}\,I\,\sqrt{2}\,R=\sqrt{6}\,{{B}_{0}}\,IR$