• # 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) $\sqrt{6}\,I\,R\,{{B}_{0}}$

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