Q40E Figure E28.40 shows, in cross se... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

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, $5.03 脳 10^{- 6} T m, 3.14 脳 10^{- 6} T m a n d 5.53 脳 10^{- 6} T m$

Step by step solution

Concept of the current in the loop

The current in the loop is calculated by $鈭� B d l = 渭_{0} l_{enc} where 渭_{0} = 4 蟺脳 10^{鈭� 7} H / m$ 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

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

PART B

role="math" localid="1668250362702" $\begin{aligned} I_{enc} = I_{1} = 4 A \\ 鈭� Bdl = 4 脳渭_{0} = 5.03 脳 10^{鈭� 6} Tm \end{aligned}$

PART C

$\begin{aligned} I_{enc} = I_{2} 鈭� I_{1} \\ = 6.5 鈭� 4 \\ = 2.5 A \\ 鈭� Bdl = 2.5 脳渭_{0} = 3.14 脳 10^{鈭� 8} Tm \end{aligned}$

PART D

$\begin{aligned} I_{enc} = I_{3} + I_{2} 鈭� I_{1} \\ = 6.5 + 1.9 鈭� 4 \\ = 4.4 \\ 鈭� Bdl = 4.4 脳渭_{0} = 5.53 脳 10^{鈭� 6} Tm \end{aligned}$

Therefore, the line integral for each path is $0, 5.03 脳 10^{鈭� 6} Tm, 3.14 脳 10^{鈭� 6} Tm$ and $5.53 脳 10^{鈭� 8} Tm$

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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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