Question

An infinite number of straight wires each carrying current $I$ are equally placed as shown in the figure. Adjacent wires have current in the opposite direction. The net magnetic field at point $P$ is

• A. $\frac{{\mu }_{0}I}{4\pi }\frac{l\text{n}2}{\sqrt{3}a}\hat{\text{k}}$
• B. $\frac{{\mu }_{0}I}{4\pi }\frac{\text{ln}4}{\sqrt{3}a}\hat{\text{k}}$
• C. $\frac{{\mu }_{0}I\ln 4}{4\pi \sqrt{3}a}(-\hat{\mathrm{k}})$
• D. Zero

Answer 1

Answer: (B)

${\mathrm{B}}_1=\frac{{\mu }_{0}I}{4{\pi }^a}(\sin \alpha +\sin \beta )=\frac{{\mu }_0}{4\pi }\frac{I\times 2\sin 30^{\circ }}{a\cos 30^{\circ }}$
$\overrightarrow{B}=\overrightarrow{B}={\text{B}}_1\hat{\text{k}}+{\text{B}}_2\hat{\text{k}}+{\text{B}}_3\hat{\text{k}}-{\text{B}}_4\hat{\text{k}}+\text{.....}$
$=\frac{{\mu }_0}{4\pi }\frac{I\times 2sin30^{{}^{\circ }}}{acos30^{{}^{\circ }}}\hat{\text{k}}-\frac{{\mu }_0}{4\pi }\frac{I\times 2sin30^{{}^{\circ }}}{2acos30^{{}^{\circ }}}\hat{\text{k}}+\frac{{\mu }_0}{4\pi }\frac{I\times 2sin30^{{}^{\circ }}}{3acos30^{{}^{\circ }}}\hat{\text{k}}+\text{.....}$
$=\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}a}\hat{\text{k}}-\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}\times 2a}\hat{\text{k}}+\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}\times 3a}\hat{\text{k}}+\text{.....}$
$=\frac{{\mu }_{0}2I}{4\pi \sqrt{3}a}\left[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots \right]\hat{\mathrm{k}}=\frac{{\mu }_{0}2I}{4\pi \sqrt{3}a}{\log }_{\mathrm{e}}(1+1)\hat{\mathrm{k}}=\frac{{\mu }_{0}2I}{4\pi \sqrt{3}a}{\log }_{\mathrm{e}}2\hat{\mathrm{k}}$
$=\left(\frac{{\mu }_{0}I}{4\pi \sqrt{3}a}{\log }_{\mathrm{e}}4\right)\hat{\mathrm{k}}$