Physics - Magnetism Question with Solution | TestHub

Magnetic field due to the circular segment$=\frac{3}{4}.\frac{{\mu }_{0}i}{2a}$

Due to one straight wire segment$=\frac{{\mu }_{0}i}{4\pi b}(\mathrm{ }\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{ }45°+\mathrm{ }\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{ }0°)=\frac{{\mu }_{0}i}{4\sqrt{2}\pi b}$

Net field

$=\frac{3{\mu }_{0}i}{8a}+2 \times \frac{{\mu }_{0}i}{4\sqrt{2}\pi b}=\frac{{\mu }_{0}i}{4\pi }\left(\frac{3\pi }{\text{a}}+\frac{\sqrt{2}}{\text{b}}\right)$