Infinite number of straight wires each carryi | vedclass

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Question

$\vec{\mathrm{B}}=\left(\mathrm{B}_{1}-\mathrm{B}_{2}+\mathrm{B}_{3}-\mathrm{B}_{4} \ldots \ldots+\infty\right) \hat{\mathrm{k}}$

Here $\quad \math

Vedclass · Accepted Answer

$\frac{{{\mu _0}I}}{{4\pi }}\,\,\frac{{\ln \,4}}{{\sqrt 3 \,a}}\hat k$

Vedclass · Answer

$\vec{\mathrm{B}}=\left(\mathrm{B}_{1}-\mathrm{B}_{2}+\mathrm{B}_{3}-\mathrm{B}_{4} \ldots \ldots+\inftyight) \hat{\mathrm{k}}$

Here $\quad \mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left(\sin 30^{\circ}+\sin 30^{\circ}ight)$

$B_{1}=\frac{\mu_{0} I}{4 \pi a \cos 30^{\circ}}\left(\frac{1}{2}+\frac{1}{2}ight)$

$B_{1}=\frac{2 \mu_{0} I}{4 \sqrt{3} \pi a}$

Similarly, $\mathrm{B}_{2}=\frac{2 \mu_{0} \mathrm{I}}{8 \sqrt{3} \pi \mathrm{a}}$ and so on

$	herefore \quad \vec{\mathrm{B}}=\frac{2 \mu_{0} \mathrm{I}}{4 \sqrt{3} \pi \mathrm{a}}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \ldots \ldots+\inftyight)$

$\vec{\mathrm{B}}=\frac{2 \mu_{0} \mathrm{I}}{4 \sqrt{3} \pi \mathrm{a}} \ln 2 \hat{\mathrm{k}}$

$\vec{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{4 \sqrt{3} \pi \mathrm{a}} \ln 4 \hat{\mathrm{k}}$

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

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