Q54PP - University Physics with Modern Physics

Question

In the United States, household electrical power is provided at a frequency of 60 Hz, so electromagnetic radiation at that frequency is of particular interest. On the basis of the ICNIRP guidelines, what is the maximum intensity of an electromagnetic wave at this frequency to which the general public should be exposed? (a) $\mathbf{7}\mathbf{.}\mathbf{7}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$; (b) $\mathbf{160}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$; (c) $\mathbf{45}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$; (d) $\mathbf{260}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$.

Step-by-Step Solution

Answer

The maximum intensity of an electromagnetic wave at 60Hz frequency is (c) $\mathbf{45}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$.

Step 1: Calculating the maximum intensity of an electromagnetic wave

The maximum electric field is given by ${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{\left(}\frac{\mathbf{350}}{\mathbf{f}}\mathbf{\right)}\mathit{V}\mathbf{/}\mathit{m}$. So, for frequency 60Hz, this electric field calculated by 

${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{350}\mathbf{*}{\mathbf{10}}^{\mathbf{3}}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{350}\mathbf{*}{\mathbf{10}}^{\mathbf{3}}}{\mathbf{60}\mathbf{H}\mathbf{z}}\mathbf{=}\mathbf{5833}\mathit{V}\mathbf{/}\mathit{m}$ 

When an electromagnetic wave strikes a surface, it exerts radiation pressure, and the maximum intensity of the electric field is given by:

${\mathit{I}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{c}{{\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}}^{\mathbf{2}}$ 

Now, plug the values for ${\mathit{\epsilon }}_{\mathbf{0}}\mathbf{,}\mathit{c}$ and ${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$ into equation (1),

$$\begin{gathered} {\mathit{I}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{c}{{\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}}^{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(8.854*{10}^{-12}\right)\left(3*{10}^8\right)\left(5800\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{45}\mathbf{*}{\mathbf{10}}^{\mathbf{3}}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{45}\mathit{k}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}\mathbf{ } \end{gathered}$$ 

Hence, the maximum intensity of an electromagnetic wave at this frequency is $\mathbf{45}\mathit{k}\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$.