91Ó°ÊÓ

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

Ampere’s law states that,

$\mathbf{âˆ\circledR }\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{·}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{s}}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}\mathit{i}$ (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$\mathbf{âˆ\circledR }\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{·}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{s}}$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.

$∴i_{enc}=-i_a+i_b-i_c$

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

$∴i_{enc}=-2.0A$

Using this inAmpere’s law, we get

$$\begin{gathered} âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}i \\ ={\mathrm{μ}}_0×-2 \\ =4×\mathrm{Ï€}×{10}^{-7}×-2 \\ =4×3.14×{10}^{-7}×-2 \\ =-25.12×{10}^{-7}T.m \\ =-2.5×{10}^{-6}T.m \\ =-2.5\mathrm{μ}T.m \end{gathered}$$

Thus, for path 1, value of $âˆ\circledR B.ds$is $-2.5\mathrm{μ}T.m$.

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.

$∴i_{enc}=i_a-i_b+i_c-i_d$

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

$∴i_{enc}=0$

Using this inAmpere’s law, we get

$$\begin{gathered} âˆ\circledR B.ds={\mathrm{μ}}_0×0 \\ âˆ\circledR B.ds=0 \end{gathered}$$

The value of the$âˆ\circledR \stackrel{→}{B}d\stackrel{→}{s}$for path 2 is zero.