question_answer

In the given figure, find magnetic induction at centre O

A) \[\frac{{{\mu }_{0}}I}{4\pi r}\left[ \pi +2+\sqrt{2} \right]\]           

B) \[\frac{{{\mu }_{0}}I}{4\pi r}\left[ \pi +\sqrt{2}+\frac{1}{\sqrt{2}} \right]\]

C) \[\frac{{{\mu }_{0}}I}{4\pi r}\left[ \pi +\frac{1}{\sqrt{2}}+2 \right]\]

D)    \[\frac{{{\mu }_{0}}I}{4\pi r}\left[ \pi +2+2\sqrt{2} \right]\]

Correct Answer: C

Solution :

\[{{B}_{0}}\] (duetoBCD)                 \[\frac{{{\mu }_{0}}I\pi }{4\pi r}=\frac{{{\mu }_{0}}I}{4r}\] \[{{B}_{0}}(deu\,\,to\,\,AB)=\frac{{{\mu }_{0}}I\pi }{4\pi r}(\sin 0+\sin \pi /4]\] \[\frac{{{\mu }_{0}}I\pi }{4\pi r}\frac{1}{\sqrt{2}}\] \[\angle ODP=\pi /4\] \[\angle POD=\angle POE=\pi /4\]           \[OP=r\sin \pi /4=r/\sqrt{2}\] \[{{B}_{0}}(deu\,\,to\,\,DE)\] \[=\frac{{{\mu }_{0}}I\pi }{4\pi r/\sqrt{2}}(\sin {{45}^{o}}+\sin {{45}^{o}})\] \[=\frac{{{\mu }_{0}}I}{4\pi (r/\sqrt{2})}=\frac{{{\mu }_{0}}I}{2\pi r}\] \[{{B}_{total}}=\frac{{{\mu }_{0}}I}{4\pi r}\left[ \pi +\frac{1}{\sqrt{2}}+2 \right]\]