Question

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

• A. $\frac{{\mu }_{o}I}{8\pi }\frac{ln2}{\sqrt{3}a}\hat{k}$
• B. $\frac{{\mu }_{o}I}{4\pi }\frac{ln4}{\sqrt{3}a}\hat{k}$
• C. $\frac{{\mu }_{o}I}{4\pi }\frac{ln4}{\sqrt{3}a}(-\hat{k})$
• D. $zero$

Answer 1

The correct option is B $\frac{{\mu }_{o}I}{4\pi }\frac{ln4}{\sqrt{3}a}\hat{k}$

Magnetic field at $P$ is equal to magnetic field due to $AB,CD,EF$ and so on.

Magnetic field due to $AB$ is $+Z$ direction.

and $CD$ is negative $-Z$ direction nad $EP$ is positive $+Z$drection and so on.

$B_P=\frac{{\mu }_{0}I}{4\pi d_1}[\sin 30°-\sin (-30°)\hat{k}]+\frac{{\mu }_{0}I}{4\pi d_2}[\sin 30°-\sin (-30°)l-\hat{k}]+\frac{{\mu }_{0}I}{4\pi d_3}[\sin 30°-\sin (-30°)\hat{k}]+...\infty$

$B_P=2\sin 30°\times \frac{{\mu }_{0}I}{4\pi }[\frac{1}{d_1}-\frac{1}{d_2}+\frac{1}{d_3}-\frac{1}{d_4}+\frac{1}{d_5}....\infty ]\hat{k}$

$=\frac{{\mu }_{0}I}{4\pi }[\frac{1}{a\cos 30°}-\frac{1}{2a\cos 30°}+\frac{1}{3a\cos 30°}-....\infty ]\hat{k}$

$=\frac{{\mu }_{0}I}{4\pi \left(a\cos 30°\right)}[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}....\infty ]\hat{k}$

$=\frac{{\mu }_{0}I}{4\pi }\frac{2}{\sqrt{3}a}\ln (1+1)\hat{k}=\frac{{\mu }_{0}I}{4\pi \sqrt{3}a}\ln 2^2\hat{k}=\frac{{\mu }_{0}I}{4\pi \sqrt{3}a}\ln 4\hat{k}$