Question

In a particular region there is a uniform current density of $\mathbf{15}\mathbf{}\mathit{A}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$in the positive z direction. What is the value of $\mathbf{\int }\mathit{B}\mathbf{.}\mathit{d}\mathit{s}$when that line integral is calculated along the three straight-line segments from (x, y, z) coordinates (4d, 0, 0) to (4d, 3d, 0) to (0, 0, 0) to (4d, 0, 0), where $\mathit{d}\mathbf{=}\mathbf{20}\mathbf{}\mathit{c}\mathit{m}$?

Short answer

The value of the $\oint \overrightarrow{B}d\overrightarrow{s}$is$4.5\times {10}^{-6}T.m$

Step-by-step solution

Listing the given quantities

• 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 $d=20\text{cm}$.

Understanding the concept of Ampere’s circuital law

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 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$\mathbf{\oint }\overrightarrow{\mathbf{B}}\mathbf{·}\mathit{d}\overrightarrow{\mathbf{s}}$.

Calculating the value of ∮B→·ds→

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$\oint \overrightarrow{B}.d\overrightarrow{s}$is,

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

$\oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_{0}jA$and $i=jA$

$$\begin{gathered} \oint \overrightarrow{B}.d\overrightarrow{s}=\left(4\pi \times {10}^{-7}\right)\left(15.0\right)\left(0.24\right) \\ =4.5\times {10}^{-6}T·m \end{gathered}$$

Therefore, the value of $\oint \overrightarrow{B}.d\overrightarrow{s}$is $4.5\times {10}^{-6}T·m$.