Question

Figure E28.40 shows, in cross section, several conductors that carry currents through the plane of the figure. The currents have the magnitudes,${\mathit{l}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathit{A}$ ,${\mathit{l}}_{\mathbf{2}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{5}\mathit{A}$and${\mathit{l}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{9}\mathit{A}$, and the directions shown. Four paths, labelled a through d, are shown. What is the line integral Ab of each path? Each integral involves going around the path in the counter clockwise direction. Explain your

Short answer

The line integral for each path is 0,$5.03\times {10}^{-6}Tm,3.14\times {10}^{-6}Tmand5.53\times {10}^{-6}Tm$

Step-by-step solution

Concept of the current in the loop

The current in the loop is calculated by$\mathbf{\int }\mathit{B}\mathit{d}\mathit{l}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}{\mathit{l}}_{\mathbf{\text{enc}}}\mathbf{\text{where}}{\mathit{\mu }}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathit{\pi }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{H}\mathbf{/}\mathbf{m}$is permeability of free space.

 STEP 2 Calculate the current in the loop

The integral involves going around the path in the counter clockwise direction for path A, B C and D. Given that$I_1=4\mathrm{A},I_2=6.5\mathrm{A},I_3=1.9\mathrm{A}$

PART A

$\begin{array}{r}{\mathrm{I}}_{\text{enc}}=0\\ \int \mathrm{Bd}l=0\end{array}$

PART B

role="math" localid="1668250362702" $\begin{array}{r}{\mathrm{I}}_{\text{enc}}={\mathrm{I}}_1=4\mathrm{A}\\ \int \mathrm{Bdl}=4\times {\mu }_0=5.03\times {10}^{-6}\mathrm{Tm}\end{array}$

PART C

$\begin{array}{r}{\mathrm{I}}_{\mathrm{enc}}={\mathrm{I}}_2-{\mathrm{I}}_1\\ =6.5-4\\ =2.5\mathrm{A}\\ \int \mathrm{Bdl}=2.5\times {\mu }_0=3.14\times {10}^{-8}\mathrm{Tm}\end{array}$

PART D

$\begin{array}{r}{\mathrm{I}}_{\mathrm{enc}}={\mathrm{I}}_3+{\mathrm{I}}_2-{\mathrm{I}}_1\\ =6.5+1.9-4\\ =4.4\\ \int \mathrm{Bdl}=4.4\times {\mu }_0=5.53\times {10}^{-6}\mathrm{Tm}\end{array}$

Therefore, the line integral for each path is $0,5.03\times {10}^{-6}\mathrm{Tm},3.14\times {10}^{-6}\mathrm{Tm}$and$5.53\times {10}^{-8}\mathrm{Tm}$