Question

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current In this case, the wave equation for an electric field $\overrightarrow{\mathbf{E}}\left(x,t\right)\mathbf{=}{\mathbf{E}}_{\mathbf{y}}\left(x,t\right)\stackrel{\mathbf{⏜}}{\mathbf{j}}$propagating in the +x-direction within a conductor is

$\frac{{\mathbf{\partial }}^{\mathbf{2}}{\mathbf{E}}_{\mathbf{y}}\left(x,t\right)}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{\mu }}{\mathbf{\rho }}\frac{\mathbf{\partial }{\mathbf{E}}_{\mathbf{y}}\left(x,t\right)}{\mathbf{\partial }\mathbf{t}}$

where $\mathit{\mu }$is the permeability of the conductor and $\mathit{\rho }$is its resistivity. (a) A solution to this wave equation is ${\mathbf{E}}_{\mathbf{y}}\left(x,t\right)\mathbf{=}{\mathbf{E}}_{\mathbf{max}}{\mathbf{e}}^{\mathbf{-}{\mathbf{k}}_{\mathbf{c}}\mathbf{x}}\mathbf{cos}\left(k_{c}x-\mathrm{\varpi t}\right)$where ${\mathit{k}}_{\mathbf{c}}\mathbf{=}\sqrt{\frac{\mathbf{\omega }\mathbf{\mu }}{\mathbf{2}\mathbf{\rho }}}$. Verify this by substituting ${\mathit{E}}_{\mathbf{y}}\left(x,t\right)$into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propogates. Explain why this happens. (Hint: The field does work to move charges within a conductor. The current of these moving charges causes ${\mathit{i}}^{\mathbf{2}}\mathit{R}$heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric field amplitude decreases by a factor of $\frac{\mathbf{1}}{\mathbf{e}}$in a distance $\frac{\mathbf{1}}{{\mathbf{k}}_{\mathbf{C}}}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{\rho }}{\mathbf{\varpi }\mathbf{\mu }}}$, and calculate this distance for a radio wave with frequency $\mathit{f}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathit{M}\mathit{H}\mathfrak{L}$in copper (resistivity $\mathbf{1}\mathbf{.}\mathbf{72}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{\Omega }\mathbf{}\mathbf{m}$; permeability) $\mathbf{}\mathbf{\mu }\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{R}$. Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

Short answer

(a)$\frac{{\partial }^2{E}_y}{\partial x^2}=\frac{\mu }{\rho }\frac{\partial E_y}{\partial t}$

(b) The energy comes from the electromagnetic waves that vibrate inside the conductor.

(c) $d=66\mu m$

Step-by-step solution

Formula for Power dissipated by the moving charges

$\mathit{P}\mathbf{=}{\mathit{i}}^{\mathbf{2}}\mathit{R}$

Verify the solution of the wave equation by taking the first differentiation

Given the second differentiation of the electric field by

$\frac{{\partial }^2{E}_y\left(x,t\right)}{\partial x^2}=\frac{\mu }{\rho }\frac{\partial E_y\left(x,t\right)}{\partial t}$ (1)

The electric field function is given by

$E_y\left(x,t\right)=E_{max}{e}^{-k_{c}x}\cos \left(k_{c}x-\varpi t\right)$

Also, Given an expression for$k_c$by

$$\begin{gathered} k_c=\sqrt{\frac{\omega \mu }{2\rho }} \\ {k}_c^2=\frac{\varpi \mu }{2\rho } \\ \frac{2k_c^2}{\omega }=\frac{\mu }{\rho } \end{gathered}$$

DifferentiateE for t, we get

$\frac{\partial E_y}{\partial t}=-E_{max}{e}^{-k_{c}x}\omega \cos \left(k_{c}x-\omega t\right)$ (2)

Take the first differentiation for $E_y=\left(x,t\right)$for x

localid="1664082486614" $$\begin{gathered} \frac{\partial E_y}{\partial x}=\frac{\partial }{\partial x}\left(E_{max}{e}^{-k_{c}x}\cos \left(k_{c}x-\omega t\right)\right) \\ =E_{max}\left(-k_c\right)e^{-k_{c}x}\cos \left(k_{c}x-\omega t\right)+E_{max}\left(-k_c\right)e^{-k_{c}x}\sin \left(k_{c}x-\omega t\right) \end{gathered}$$

Take the second differentiation

$$\begin{gathered} \frac{{\partial }^2{\mathrm{E}}_{\mathrm{y}}}{\partial {\mathrm{x}}^2}={\mathrm{E}}_{\max }{\mathrm{k}}_{\mathrm{c}}\left[{\mathrm{k}}_{\mathrm{c}}{\mathrm{e}}^{-\mathrm{kx}}\sin \left({\mathrm{k}}_{\mathrm{c}}\mathrm{x}-\mathrm{\omega t}\right)\right]-{\mathrm{e}}^{-\mathrm{kx}}\left({\mathrm{k}}_{\mathrm{c}}\right)\cos \left({\mathrm{k}}_{\mathrm{c}}\mathrm{x}-\mathrm{\omega t}\right)-{\mathrm{e}}^{-\mathrm{kx}}\left(\mathrm{kx}\right)\sin \left({\mathrm{k}}_{\mathrm{c}}\mathrm{x}-\mathrm{\omega t}\right) \\ =-2{\mathrm{E}}_{\max }{\mathrm{k}}^2{\mathrm{ce}}^{-{\mathrm{k}}_{\mathrm{c}}\mathrm{x}}\cos \left({\mathrm{k}}_{\mathrm{c}}\mathrm{x}-\mathrm{\omega t}\right) \end{gathered}$$

(3)

Using, equation (2), replace the term $E_{max}{e}^{-k_{c}x}\omega \cos \left(k_{c}x-\omega t\right)$into equation (3) by role="math" localid="1664082068443" $\frac{1}{-\omega }\frac{\partial E_y}{\partial t}$, so we can get

$$\begin{gathered} \frac{{\partial }^2{E}_y}{\partial x^2}=\left(\frac{2k_c^2}{\omega }\right)\frac{\partial E_y}{\partial t} \\ =\frac{\mu }{\rho }\frac{\partial E_y}{\partial t} \end{gathered}$$

Hence, proved.

Explanation for why the electric field decreases in amplitude as it propagates.

The energy comes from the electromagnetic waves that vibrate inside the conductor and creates a magnetic field that induces a current that dissipates power by

$P=i^{2}R$

Determine the distance for the radio wave

The distance is given by:

$d=\frac{1}{k_c}=\sqrt{\frac{2\rho }{\omega \mu }}$

Substitute the values,

$d=\sqrt{\frac{2\rho }{\omega \mu }}=\sqrt{\frac{2\left(1.72*10\right)}{2\mathrm{\pi }\left(1*{10}^6\right)\left(4\mathrm{\pi }*{10}^{-7}\right)}}=66*{10}^{-6}m=66\mu m$

Thus, the distance for the radio wave is $66\mu m$.