Eight wires cut the page perpendicularly at the points shown in Figure 29-70 wire labeled with the integer $\mathbf{k}\mathbf{}\mathbf{(}\mathbf{k}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{.}\mathbf{}\mathbf{.}\mathbf{}\mathbf{.}\mathbf{}\mathbf{,}\mathbf{}\mathbf{8}\mathbf{)}$carries the current $\mathbf{ki}$, where $\mathbf{i}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{mA}$. For those wires with odd $\mathbf{k}$, the current is out of the page; for those with even $\mathbf{k}$, it is into the page. Evaluate$\mathbf{\oint }\overrightarrow{\mathbf{B}}\mathbf{}\mathbf{.}\overrightarrow{\mathbf{ds}}\mathbf{}$ along the closed path in the direction shown.

• Current in all conductors, $\mathrm{ki}=4.50\mathrm{mA}=4.50\times {10}^{-3}\mathrm{A}$
• 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,

$\mathbf{\oint }\overrightarrow{\mathbf{B}}\mathbf{\cdot }\mathbf{d}\overrightarrow{\mathbf{s}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}$

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 $\mathbf{\oint }\mathbf{B}\mathbf{.}\mathbf{ds}$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,

$\oint \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{s}}={\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{enc}}$ (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.

$\begin{array}{c}\therefore {\mathrm{i}}_{\mathrm{enc}}=4.50-4.50+4.50-4.50+4.50-4.50+4.50-4.50\\ =0\end{array}$

Using in equation (i), we get

$\begin{array}{c}\oint \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{s}}={\mathrm{\mu }}_0\times 0\\ =0\end{array}$

The value of $\oint \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{s}}$for the closed path is zero.