Question

Consider electromagnetic fields produced by high voltage power lines. Construct a problem in which you calculate the intensity of this electromagnetic radiation in $\mathbf{W}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$ based on the measured magnetic field strength of the radiation in a home near the power lines. Assume these magnetic field strengths are known to average less than a$\mathbf{\mu T}$ .

The intensity is small enough that it is difficult to imagine mechanisms for biological damage due to it. Discuss how much energy may be radiating from a section of power line several hundred meters long and compare this to the power likely to be carried by the lines. An idea of how much power this is can be obtained by calculating the approximate current responsible for$\mathbf{\mu T}$ fields at distances of tens of meters.

Short answer

The intensity of the electromagnetic radiation produced by a high voltage power line is $152.8\mathrm{W}/{\mathrm{m}}^2$. The energy radiated from the power line is $1.53\times {10}^{-4}$ and it is negligible as compared to the power carried by the line.

Step-by-step solution

Definition of Concept

Intensity: The power transferred per unit area is the intensity of radiant energy in physics, where the area is measured on a plane perpendicular to the energy's propagation direction.

Find the intensity of the electromagnetic radiation produced and energy radiated from the power line

Considering the given information,

The magnetic field strength is $\mathrm{B}=0.8\mathrm{\mu T}\left(\frac{{10}^{-6}\mathrm{T}}{1\mathrm{\mu T}}\right)=8.0\times {10}^{-7}\mathrm{T}$.

The length of the power line is $\mathrm{L}=300\mathrm{m}$.

Apply the formula,

The average intensity of electromagnetic radiation is calculated using the formula:

${\mathrm{I}}_{\mathrm{avg}}=\frac{{\mathrm{cB}}^2}{2{\mathrm{\mu }}_0}$

Putting the values in the equation,

$$\begin{gathered} {\mathrm{I}}_{\mathrm{avg}}=\frac{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\times {\left(8.0\times {10}^{-70}\mathrm{T}\right)}^2}{2\times \left(4\mathrm{\pi }\times {10}^{-7}\mathrm{H}/\mathrm{m}\right)} \\ =76.4\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

The peak intensity can be computed as:

$$\begin{gathered} {\mathrm{I}}_0=2{\mathrm{I}}_{\mathrm{avg}} \\ =2\times 76.4\mathrm{W}/{\mathrm{m}}^2 \\ =152.8\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Radiation's energy density can be computed as follows:

${\mathrm{u}}_0=\frac{{\mathrm{I}}_0}{\mathrm{c}}$

The energy radiated from a $300\mathrm{m}$ long segment of the power line can be computed as follows:

At a distance of $20\mathrm{m}$, the magnetic field caused by a power line can be computed as follows:

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}} \\ \mathrm{I}=\frac{2\mathrm{\pi rB}}{{\mathrm{\mu }}_0} \\ \mathrm{I}=\frac{2\mathrm{\pi }\times 20\mathrm{m}\times \left(8.0\times {10}^{-7}\mathrm{T}\right)}{4\mathrm{\pi }\times {10}^{-7}\mathrm{H}/\mathrm{m}} \\ \mathrm{I}=80\mathrm{A} \end{gathered}$$

Assume that the voltage of the electricity line is 500kV .

The transmission line's power can be estimated as follows:

$$\begin{gathered} \mathrm{P}=\mathrm{IV} \\ \mathrm{P}=\left(80\mathrm{A}\right)\times \left(500\times {10}^3\mathrm{V}\right) \\ \mathrm{P}=40\times {10}^6\mathrm{W}\left(\frac{1\mathrm{MW}}{{10}^6\mathrm{W}}\right) \\ \mathrm{P}=40\mathrm{MW} \end{gathered}$$

When the energy emitted from a power line is compared to the power carried by the line, we find that the energy radiated is insignificant.

Therefore, the required intensity of the electromagnetic radiation produced by a high voltage power line is$152.8\mathrm{W}/{\mathrm{m}}^2$ . The energy radiated from the power line is$1.53\times {10}^{-4}\mathrm{J}$ , and it is negligible as compared to the power carried by the line.