Physics - Magnetism Question with Solution | TestHub

Field due to wire$\mathrm{AB}$at the center,

$B_{AB}=\frac{{\mu }_{0}I}{4\pi \left(\frac{L}{2}\right)}\left(\mathit{sin}45°+\mathit{sin}45°\right)$

$B_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}}=4B_{AB}$

$B_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}}=4\frac{{\mu }_{0}I}{4\pi \left(\frac{L}{2}\right)}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)$

$B_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}}=2\sqrt{2}\frac{{\mu }_{0}I}{\pi L}$

$B_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}}\propto \frac{1}{L}$