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$$
\begin{array}{l}{\text { Figure } \mathrm{E} 28.42 \text { shows, in }} \\ {\text { cross section, several conductors }} \\ {\text { that carry currents through the }} \\ {\text { plane of the figure. The currents }} \\ {\text { have the magnitudes } I_{1}=4.0 \mathrm{A}} \\ {I_{2}=6.0 \mathrm{A}, \text { and } I_{3}=2.0 \mathrm{A}, \text { and }}\end{array}
$$

$$
\begin{array}{l}{\text { the directions shown. Four paths, }} \\ {\text { labeled a through } d \text { , are shown. }} \\ {\text { What is the line integral } \oint \vec{B} \cdot d \vec{l}} \\ {\text { for each path? Each integral in- }} \\ {\text { volves going around the path in }} \\ {\text { the counterclockwise direction. Ex- }} \\ {\text { plain your answers. }}\end{array}
$$

$$
\oint \vec{B} \cdot \vec{d}=M_{0} I_\text { encl }
$$

(a) $$
I_{\text {encl }}=0
$$$$
\longrightarrow \oint \vec{B} \cdot d \vec{l}=0
$$

(b) $$
I_\text { encl }=-I_{1}
$$$$
=-4 A \longrightarrow \oint \vec{B} \cdot d \vec{L}
$$$$
=M_{0}(-4)
$$$$
=-5.03 * 10^{-6} \mathrm{T.m}
$$

(c) $$
I_\text { encl }=-I_{1}+I_{2}=
$$$$
-4+6=2 A
$$$$
\rightarrow \oint \vec{B} \cdot \vec{d} L
$$$$
=M_{0}(2)=2.51*10^{-6} \mathrm{T.m}
$$

(d) $$
I_{encl} d=
$$$$
-I_{1}+I_{2}+I_{3}
=$$$$
-4+6+2=4 A \rightarrow \oint \vec{B} d \vec{L}=M_{0}(4)
$$

\(∴ \quad=5.03 * 10^{-6} \mathrm{T.m} \)

$$
\begin{array}{l}{\text { A closed curve encircles several conductors. The line }} \\ {\text { integral } \oint \vec{B} \cdot d \vec{l} \text { around this curve is } 3.83 \times 10^{-4} \mathrm{T} \cdot \mathrm{m} . \text { (a) What }} \\ {\text { is the net current in the conductors? (b) If you were to integrate }} \\ {\text { around the curve in the opposite direction, what would be the value }} \\ {\text { of the line integral? Explain. }}\end{array}
$$

$$
\oint \vec{B} \cdot d \vec{l}
$$$$
=M_0 I _{encl}
$$

(a) $$
\oint \vec B \cdot d \vec{l}
$$$$
=M_{0} I_{encl}
$$

$$
3.83 * 10^{-4}=4 \pi * 10^{-7} I_\text {encl } \longrightarrow I_\text { encl }=305 \mathrm{A}
$$

(b) $$
-3.83 * 10^{-4} \mathrm{T} \cdot \mathrm{m}
$$

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