Each of the eight conductors in Figure carries $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{A}$of current into or out of the page. Two paths are indicated for the line integral. $\mathbf{\oint }\overrightarrow{\mathbf{B}}\mathbf{}\mathbf{.}\overrightarrow{\mathbf{d}\mathbf{s}}$(a) What is the value of the integral for path 1 and (b) What is the value of the integral for path 2?

Current in all the eight conductors$i=2.0A$

Ampere’s law states that,

$\mathbf{\oint }\overrightarrow{\mathbf{B}}\mathbf{·}\mathit{d}\overrightarrow{\mathbf{s}}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\mathit{i}$ (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 can find value of integral$\mathbf{\oint }\overrightarrow{\mathbf{B}}\mathbf{·}\mathit{d}\overrightarrow{\mathbf{s}}$for both the paths.

In the given diagram, let the conductors be named as:

We can apply the right-hand rule to path. Currents through the conductors $a$and $c$are out of the page and the current through conductor $b$is directed into the page. The conductor $d$is out of the Amperian loop.

Since the path is traversed in the clockwise direction, a current into the page is positive and a current out of the page is negative.

$\therefore i_{enc}=-i_a+i_b-i_c$

But it is given that the current through each conductor is$2.0A$

$\therefore i_{enc}=-2.0A$

Using this inAmpere’s law, we get

$$\begin{gathered} \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_{0}i \\ ={\mu }_0\times -2 \\ =4\times \pi \times {10}^{-7}\times -2 \\ =4\times 3.14\times {10}^{-7}\times -2 \\ =-25.12\times {10}^{-7}T.m \\ =-2.5\times {10}^{-6}T.m \\ =-2.5\mu T.m \end{gathered}$$

Thus, for path 1, value of $\oint B.ds$is $-2.5\mu T.m$.

Similarly, we can apply the right-hand rule to path 2.Currents through the conductors b and d are into the page and currents through conductors a and c are directed out of the page.

Since the path is traversed in the anti-clockwise direction, a current into the page is negative and a current out of the page is positive.

$\therefore i_{enc}=i_a-i_b+i_c-i_d$

But it is given that current through each conductor is 2.0 A

$\therefore i_{enc}=0$

Using this inAmpere’s law, we get

$$\begin{gathered} \oint B.ds={\mu }_0\times 0 \\ \oint B.ds=0 \end{gathered}$$

The value of the$\oint \overrightarrow{B}d\overrightarrow{s}$for path 2 is zero.