Physics - Magnetism Question with Solution | TestHub

Here, $a=\frac{r}{\sqrt{2}}$ Magnetic field at point $O$ due to $AB$ is $B_1=\frac{{\mu }_0}{4\pi }\frac{I}{a}=\frac{{\mu }_{0}I}{4\pi (r/\sqrt{2})}$ Magnetic field at point $O$ due to $BCD$ is $B_2=\frac{{\mu }_{0}I}{4\pi r}\left(\frac{\pi }{2}\right)$ Magnetic field at point $O$ due to $DE$ is $B_3=\frac{{\mu }_{0}I}{4\pi a}=\frac{{\mu }_{0}I}{4\pi (r/\sqrt{2})}$ Resultant magnetic field at point $O$ is $\begin{array}{rl}B& =B_1+B_2+B_3\\ & =\frac{{\mu }_{0}I}{4\pi (r/\sqrt{2})}+\frac{{\mu }_{0}I}{4\pi r}\left(\frac{\pi }{2}\right)+\frac{{\mu }_{0}I}{4\pi (r/\sqrt{2})}\\ & =\frac{{\mu }_{0}I}{4\pi r}\left(\sqrt{2}+\frac{\pi }{2}+\sqrt{2}\right)=\frac{{\mu }_{0}I}{4\pi r}\left(2\sqrt{2}+\frac{\pi }{2}\right)\\ & =\frac{{\mu }_{0}II}{4\pi r}\left(\sqrt{2}+\frac{\pi }{4}\right)\\ & =\frac{10^{-7}\times 2I}{r}\left(\sqrt{2}+\frac{\pi }{4}\right)\left[\because \frac{{\mu }_0}{4\pi }=10^{-7}\right]\end{array}$