Q56P - University Physics with Modern Physics

Question

The ICNIRP also has guidelines for magnetic-field exposure for the general public. In the frequency range of \(25Hz\) to\(3kHz\), this guidelines states that the maximum allowed magnetic field amplitude is \(\frac{5}{f}T\), where\(f\) is the frequency is kHZ. Which is a more stringent limit on allowable electromagnetic-wave intensity in this frequency range: the electric-field guideline or the magnetic-field guideline? (a) The magnetic-field guideline, because at a given frequency the allowed magnetic field is smaller than the allowed electric field. (b) The electric field guideline, because at a given frequency, the allowed intensity calculated from the electric-field guideline is smaller. (c) It depends on the particular frequency chosen (both guidelines are frequency dependent). (d) Neither – for any given frequency, the guidelines represent the same electromagnetic-wave intensity.

Step-by-Step Solution

Answer

The electric field guideline is more stringent limit on allowable electromagnetic-wave intensity in the given frequency range.

Step 1: Formula for the maximum electric field and allowed magnetic field

${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{350}}{\mathbf{f}}$

${\mathit{B}}_{\mathbf{max}}\frac{\mathbf{5}}{\mathbf{f}}$

Step 2: Determine the intensity

The maximum electric field is given as: ${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{350}}{\mathbf{f}}$ 

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

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

Substitute ${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$ to get the intensity for the electric field,

${\mathit{l}}_{\mathbf{E}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{c}{\mathbf{\left(}\frac{\mathbf{350}}{\mathbf{f}}\mathbf{\right)}}^{\mathbf{2}}$                                                   (1)

The allowed magnetic field is 

${\mathit{B}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{5}}{\mathbf{f}}$ 

Also, ${\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathit{c}{\mathit{B}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$. Substitute to get the intensity for the magnetic field.

${\mathit{l}}_{\mathbf{B}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{c}{\left(c{B}_{max}\right)}^{\mathbf{2}}$ 

$\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{c}}^{\mathbf{3}}{{\mathit{B}}_{\mathbf{m}\mathbf{a}\mathbf{x}}}^{\mathbf{2}}$                                                             

$\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{c}}^{\mathbf{3}}{\mathbf{\left(}\frac{\mathbf{5}}{\mathbf{f}}\mathbf{\right)}}^{\mathbf{2}}$                                                                                                                    (1)

Compare equation (1) and (2), the intensity of the magnetic field is larger than the electric field. 

Hence, the correct answer is (b) The electric field guideline, because at a given frequency, the allowed intensity calculated from the electric-field guideline is smaller.