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).

The term Poynting vector is defined as the instantaneous energy flux of the wave, and its direction is in the electromagnetic wave propagation.

Write the expression for the Poynting vector.

$\mathit{S}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\left(E\times B\right)\mathbf{}$

Here, E is the electric field, and B is the magnetic field.

Write the expression for power transported through the wire cables.

$P=-\frac{dU}{dt}+\int S\cdot da$

As E and B are independent of time, the term $\frac{dU}{dt}$will be zero. Hence,

$\begin{array}{c}P=-0+\int S\cdot da\\ P=\int S\cdot da\end{array}$

$\begin{array}{c}P=-0+\int S\cdot da\\ P=\int S\cdot da\end{array}$

Substitute the value of the poynting vector (S) in the above equation.

$\begin{array}{c}P=\int \frac{1}{{\mu }_0}\left(E\times B\right)\cdot da\\ P=\frac{1}{{\mu }_0}\int \left(E\times B\right)\cdot da\end{array}$

$\begin{array}{c}P=\int \frac{1}{{\mu }_0}\left(E\times B\right)\cdot da\\ P=\frac{1}{{\mu }_0}\int \left(E\times B\right)\cdot da\end{array}$

Write the formula for the linear charge density.

$\lambda =\frac{q}{l}$ …….. (2)

Here, q is the charge, and l is the Gaussian surface of length.

Apply Gauss’s law for the closed surface.

$\begin{array}{c}\oint \overline{E}\cdot \overline{da}=\frac{q}{{\epsilon }_0}\\ E_r\left(2\pi sl\right)=\frac{q}{{\epsilon }_0}\end{array}$ …….. (3)

Substitute the value of equation (2) in equation (3).

$\begin{array}{l}E\left(2\pi sl\right)=\frac{\lambda l}{{\epsilon }_0}\\ \overrightarrow{E}=\frac{\lambda }{2\pi s{\epsilon }_0}\widehat{s}\end{array}$ …… (4)

Write the expression for Ampere’s law.

$\oint B\cdot ds={\mu }_{0}I\mathrm{......}\left(5\right)$

$\oint B\cdot ds={\mu }_{0}I\mathrm{......}\left(5\right)$

Here, B is the magnetic field, I is the current in the wire and localid="1653733166875" $\oint ds$is the circumference of the conductor.

Write the formula for the circumference of the conductor.

localid="1653733172530" $\oint ds=2\pi s$ …… (6)

Substitute the value of equation (6) in equation (5).

localid="1653733214471" role="math" $\begin{array}{c}B\left(2\pi s\right)={\mu }_{0}I\\ B=\frac{{\mu }_{0}I}{2\pi s}\end{array}$

Now, use the right-hand rule to find the direction of the magnetic field around the direction of the current in a straight line.

localid="1653733219733" $\overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi s}\widehat{\varphi }$ …… (7)

Write the expression for power transported down the cables.

$P=\int \overrightarrow{S}\cdot \overrightarrow{da}$ …… (8)

As the current is opposite to each other, the Poynting vector in equation (1) becomes,

$\overrightarrow{S}=\frac{1}{{\mu }_0}\left(\overrightarrow{E}\times \overrightarrow{B}\right)$ …… (9)

Substitute the value of equations (4) and (7) in equation (9).

$\begin{array}{c}\overrightarrow{S}=\frac{1}{{\mu }_0}\left(\frac{\lambda }{2\pi s{\epsilon }_0}\widehat{s}\right)\left(\frac{{\mu }_{0}I}{2\pi s}\widehat{\varphi }\right)\\ \overrightarrow{S}=\frac{\lambda I}{4{\pi }^2{\epsilon }_0{s}^2}\widehat{z}\mathrm{......}\left(10\right)\end{array}$

Substitute the value of equation (10) in equation (8).

$\begin{array}{c}P=\int \left(\frac{\lambda I}{4{\pi }^2{\epsilon }_0{s}^2}\widehat{z}\right)\cdot \left(2\pi sds\right)\\ P=\frac{\lambda I}{2\pi {\epsilon }_0}{\int }_a^b\frac{1}{s}ds\\ P=\frac{\lambda I}{2\pi {\epsilon }_0}{\left(\ln s\right)}_a^b\\ P=\frac{\lambda I}{2\pi {\epsilon }_0}\ln \left(\frac{b}{a}\right)\end{array}$ ……. (11)

Write the potential difference along the cable.

$V=\frac{\lambda }{2\pi {\epsilon }_0}\ln \left(\frac{b}{a}\right)\mathrm{......}\left(12\right)$

Substitute the value of equation (12) in equation (11).

$P=IV$

Write the expression for the electric field of an infinite plane sheet.

$\overrightarrow{E}=\frac{\sigma }{{\epsilon }_0}\widehat{z}$

Here,$\sigma$is the surface charge density of the sheet and${\epsilon }_0$is the permittivity of the medium.

Write the expression for the magnetic field of an infinite plane sheet.

$\begin{array}{c}\overrightarrow{B}={\mu }_{0}k\widehat{x}\\ \overrightarrow{B}={\mu }_0\left(\frac{I}{\omega }\right)\widehat{x}\\ \overrightarrow{B}=\frac{{\mu }_{0}I}{\omega }\widehat{x}\end{array}$

Here,${\mu }_0$is the magnetic permeability of the medium and$\omega$is the angular frequency of the charge.

Substitute the values in equation (9).

$\begin{array}{c}\overrightarrow{S}=\frac{1}{{\mu }_0}\left(\frac{\sigma }{{\epsilon }_0}\widehat{z}\times \frac{{\mu }_{0}I}{\omega }\widehat{x}\right)\\ \overrightarrow{S}=\frac{1}{{\mu }_0}\left(\frac{\sigma }{{\epsilon }_0}\widehat{z}\right)\times \left(\frac{{\mu }_{0}I}{\omega }\widehat{x}\right)\\ \overrightarrow{S}=\frac{\sigma I}{{\epsilon }_0\omega }\widehat{y}\end{array}$

Substitute the values in equation (8).

$\begin{array}{c}P=\int \left(\frac{\sigma I}{{\epsilon }_0\omega }\widehat{y}\right)\cdot \left(\omega dh\right)\\ P=\frac{\sigma Ih}{{\epsilon }_0}\mathrm{......}\left(13\right)\end{array}$

Write the potential difference across the cable.

$V={\int }_a^{b}E\cdot dl$

Substitute the values in the above equation.

$V=\frac{\sigma }{{\epsilon }_0}h$

Substitute the values in equation (13).

$P=IV$

The power transported down the cables and through the plane sheets is equal to$P=IV$ where I is the carry current, and V is the potential difference.