Physics - Magnetism Question with Solution | TestHub

Magnetic field due to wire $\mathrm{AC}$,

${\mathrm{B}}_{\mathrm{AC}}=\frac{{\mathrm{\mu }}_0\text{I}}{4\mathrm{\pi }\left(2Rsin30\right)}\left(cos30-cos90\right)=\frac{\sqrt{3}{\mathrm{\mu }}_0\text{I}}{8\mathrm{\pi }\text{R}}$

Magnetic field due to wire $\mathrm{BC}$,

${\mathrm{B}}_{\mathrm{BC}}=\frac{{\mathrm{\mu }}_0\text{I}}{4\mathrm{\pi }\left(2Rsin60\right)}\left(cos60-cos90\right)=\frac{{\mathrm{\mu }}_0\text{I}}{8\sqrt{3}\mathrm{\pi }\text{R}}=\frac{\sqrt{3}{\mathrm{\mu }}_0\text{I}}{24\mathrm{\pi }\text{R}}$

${\overrightarrow{\text{B}}}_{\text{net}}={\text{B}}_{\text{A}}-{\text{B}}_{\text{B}}$

       $=\frac{\sqrt{3}{\mathrm{\mu }}_0\text{I}}{8\mathrm{\pi }\text{R}}\left(1-\frac{1}{3}\right)$

       $=\frac{2\sqrt{3}{\mu }_0\text{I}}{24\pi \text{R}}=\frac{{\mu }_0\text{I}}{4\sqrt{3}\pi \text{R}}$