Calculate the power (energy per unit time) transported down the cables of Ex. 7.13 and Prob. 7.62, assuming the two conductors are held at potential difference V , and carry current I (down one and back up the other)

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Example 7.13.

[latex]\left.\begin{array}{l} E =\frac{\lambda}{2 \pi \epsilon_{0}} \frac{1}{s} \hat{ s } \B =\frac{\mu_{0} I}{2 \pi} \frac{1}{s} \hat{\phi}\end{array} ight\} S =\frac{1}{\mu_{0}}( E imes B )=\frac{\lambda I}{4 \pi^{2} \epsilon_{0}} \frac{1}{s^{2}} \hat{ z }[/latex] ;

[latex]P=\int S \cdot d a =\int_{a}^{b} S 2 \pi s d s=\frac{\lambda I}{2 \pi \epsilon_{0}} \int_{a}^{b} \frac{1}{s} d s=\frac{\lambda I}{2 \pi \epsilon_{0}} \ln (b / a)[/latex] .

[latex] ext { But } V=\int_{a}^{b} E \cdot d l =\frac{\lambda}{2 \pi \epsilon_{0}} \int_{a}^{b} \frac{1}{s} d s=\frac{\lambda}{2 \pi \epsilon_{0}} \ln (b / a), ext { so } P=I V ext {. }[/latex]

Problem 7.62.

[latex]\left.\begin{array}{l} E =\frac{\sigma}{\epsilon_{0}} \hat{ z } \B =\mu_{0} K \hat{ x }=\frac{\mu_{0} I}{w} \hat{ x }\end{array} ight\} S =\frac{1}{\mu_{0}}( E imes B )=\frac{\sigma I}{\epsilon_{0} w} \hat{ y }[/latex] ;

[latex]P=\int S \cdot d a =S w h=\frac{\sigma I h}{\epsilon_{0}}, ext { but } V=\int E \cdot d l =\frac{\sigma}{\epsilon_{0}} h, ext { so } P=I V[/latex] .

Question 8.1: Calculate the power (energy per unit time) transported down ...

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