Question

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).

Short answer

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

Step-by-step solution

Expression for the Poynting vector:

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.

Determine the power transported through the cable of wire:

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.

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

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

Apply Gauss’s law for the closed surface:

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)

Determine the strength of the magnetic field on the surface of the conductor:

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)

Determine the power transported down the cables:

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).

$P=IV$

Determine the power transported through the thin sheet:

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).

$P=IV$

Final Solution:

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.