Physics - Magnetism Question with Solution | TestHub

From figure it is clear that



$B_{AB}=B_{CD}$

$=\frac{{\mu }_{0}i}{4\pi \left(\frac{b}{2}\right)}\left(\mathit{sin}{\varphi }_1+\mathit{sin}{\varphi }_2\right)=\frac{{\mu }_0}{4\pi }\cdot \frac{4i}{b}\cdot \frac{a}{\sqrt{a^2+b^2}}$

$B_{BC}=B_{DA}=\frac{{\mu }_0}{4\pi }\cdot \frac{4i}{a}\cdot \frac{b}{\sqrt{a^2+b^2}}$

$\therefore$$B=B_{AB}+B_{BC}+B_{CD}+B_{DA}$

$B=2\left(\frac{{\mu }_0}{4\pi }\cdot \frac{4i}{b}\cdot \frac{a}{\sqrt{a^2+b^2}}\right)+2\left(\frac{{\mu }_0}{4\pi }\cdot \frac{4i}{a}\cdot \frac{b}{\sqrt{a^2+b^2}}\right)$

$B=\frac{{\mu }_0}{4\pi }\cdot \frac{2\times 4i}{\sqrt{a^2+b^2}}\left[\frac{a}{b}+\frac{b}{a}\right]$

$B=\frac{{\mu }_0}{4\pi }\cdot \frac{8i\sqrt{a^2+b^2}}{ab}$