91Ó°ÊÓ

• Current in all conductors, $\mathrm{ki}=4.50\mathrm{mA}=4.50×{10}^{-3}\mathrm{A}$
• For odd k, the current is out of the page (1,3,5,6)
• For even k, the current is into the page. (2,4,6,8)

Ampere’s law states that,

$\mathbf{âˆ\circledR }\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{â‹\dots }\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{s}}\mathbf{=}{\mathbf{μ}}_{\mathbf{0}}\mathbf{i}$

The line integral in this equation is evaluated around a closed-loop called an Amperian loop.The current ion the right side is the net current encircled by the loop.

By applying Ampere’s law, we will find the value of integral $\mathbf{âˆ\circledR }\mathbf{B}\mathbf{.}\mathbf{ds}$along the closed path.

For odd k, current is out of the page, and it is in the same direction of the outstretched thumb when we curltheright-hand fingers aroundtheAmperian loop. Hencethecurrent through 1,3,5, and 7 conductors is positive.

For even k, current is in to the page, and it is in the opposite direction of the outstretched thumb when we curl the right-hand fingers around the Amperian loop.

Hencethecurrent through 2,4,6, and 8 conductors is negative.

By using Ampere’s law,

$âˆ\circledR \stackrel{→}{\mathrm{B}}â‹\dots \mathrm{d}\stackrel{→}{\mathrm{s}}={\mathrm{μ}}_0{\mathrm{i}}_{\mathrm{enc}}\text{ }$ (i)

${\mathrm{i}}_{\mathrm{enc}}={\mathrm{i}}_1-{\mathrm{i}}_2+{\mathrm{i}}_3-{\mathrm{i}}_4+{\mathrm{i}}_5-{\mathrm{i}}_6+{\mathrm{i}}_7-{\mathrm{i}}_8$

But it is given that the current through each conductor is 4.50 mA.

$\begin{array}{c}∴{\mathrm{i}}_{\mathrm{enc}}=4.50−4.50+4.50−4.50+4.50−4.50+4.50−4.50\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=0\end{array}$

Using in equation (i), we get

$\begin{array}{c}âˆ\circledR \stackrel{→}{\mathrm{B}}â‹\dots \mathrm{d}\stackrel{→}{\mathrm{s}}={\mathrm{μ}}_0×0\\ =0\end{array}$

The value of $âˆ\circledR \stackrel{→}{\mathrm{B}}â‹\dots \mathrm{d}\stackrel{→}{\mathrm{s}}$for the closed path is zero.