Physics - Magnetism Question with Solution | TestHub

${\text{B}}_1=\frac{{\mu }_0}{4\pi }\frac{I}{a}\left(sin\alpha +sin\beta \right)=\frac{{\mu }_0}{4\pi }\frac{I\times 2sin30^{\circ }}{acos30^{\circ }}$

$\overrightarrow{\text{B}}=\overrightarrow{\text{B}}={\text{B}}_1\widehat{\text{k}}+{\text{B}}_2\widehat{\text{k}}+{\text{B}}_3\widehat{\text{k}}-{\text{B}}_4\widehat{\text{k}}+\text{.....}$

$=\frac{{\mu }_0}{4\pi }\frac{I\times 2sin30^{\circ }}{acos30^{\circ }}\hat{\text{k}}-\frac{{\mu }_0}{4\pi }\frac{I\times 2sin30^{\circ }}{2acos30^{\circ }}\hat{\text{k}}+\frac{{\mu }_0}{4\pi }\frac{I\times 2sin30^{\circ }}{3acos30^{\circ }}\hat{\text{k}}+\text{.....}$

$=\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}a}\hat{\text{k}}-\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}\times 2a}\hat{\text{k}}+\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}\times 3a}\hat{\text{k}}+\text{.....}$

$=\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}a}\left[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\text{...}\right]\hat{\text{k}}=\frac{{\mu }_{0}2I}{4\pi \sqrt{3}a}{\text{log}}_{\text{e}}\left(1+1\right)\hat{\text{k}}=\frac{{\mu }_{0}2I}{4\pi \sqrt{3}a}{\text{log}}_{\text{e}}2\hat{k}$

$=\left(\frac{{\mu }_{0}I}{4\pi \sqrt{3}a}{\text{log}}_{\text{e}}4\right)\hat{\text{k}}$