Chapter 4: Q40E (page 950)

Figure E28.40 shows, in cross section, several conductors that carry currents through the plane of the figure. The currents have the magnitudes, $l_{1} = 4 A$ , $l_{2} = 6.5 A$ and $l_{3} = 1.9 A$ , 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

Expert verified

The line integral for each path is 0, $5.03 \times 10^{- 6} T m, 3.14 \times 10^{- 6} T m a n d 5.53 \times 10^{- 6} T m$

Step by step solution

Concept of the current in the loop

The current in the loop is calculated by $\int B d l = μ_{0} l_{enc} where μ_{0} = 4 π \times 10^{- 7} H / m$ 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 $I_{1} = 4 A, I_{2} = 6.5 A, I_{3} = 1.9 A$

PART A

$\begin{aligned} I_{enc} = 0 \\ \int Bd l = 0 \end{aligned}$

PART B

role="math" localid="1668250362702" $\begin{aligned} I_{enc} = I_{1} = 4 A \\ \int Bdl = 4 \times μ_{0} = 5.03 \times 10^{- 6} Tm \end{aligned}$

PART C

$\begin{aligned} I_{enc} = I_{2} - I_{1} \\ = 6.5 - 4 \\ = 2.5 A \\ \int Bdl = 2.5 \times μ_{0} = 3.14 \times 10^{- 8} Tm \end{aligned}$

PART D

$\begin{aligned} I_{enc} = I_{3} + I_{2} - I_{1} \\ = 6.5 + 1.9 - 4 \\ = 4.4 \\ \int Bdl = 4.4 \times μ_{0} = 5.53 \times 10^{- 6} Tm \end{aligned}$

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

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Short Answer

Step by step solution

Concept of the current in the loop

STEP 2 Calculate the current in the loop

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