91Ó°ÊÓ

• The uniformcurrent density is$j=15A/m^2$
• Three straight-line segments from$\left(x,y,z\right)$ coordinates$\left(4d,0,0\right)$ to$\left(4d,3d,0\right)$ to $\left(0,0,0\right)$to $\left(4d,0,0\right)$where $d=20\text{cm}$.

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 thenetcurrent encircled by the loop.

We have to find the area enclosed by the closed-loop$\mathit{L}$. Using Ampere’s law, we can find the value of$\mathbf{âˆ\circledR }\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{·}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{s}}$.

The area enclosed by the closed loop$L$ is

$$\begin{gathered} A=\frac{1}{2}\left(4d\right)\left(3d\right) \\ =6d^2 \\ =6{\left(0.20\right)}^2 \\ =0.24m^2 \end{gathered}$$

Thus, from equation (i),the value of$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}$is,

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}i$

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}jA$and $i=jA$

$$\begin{gathered} âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}=\left(4\mathrm{Ï€}×{10}^{-7}\right)\left(15.0\right)\left(0.24\right) \\ =4.5×{10}^{-6}T·m \end{gathered}$$

Therefore, the value of $âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}$is $4.5×{10}^{-6}T·m$.