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Expert-verified Found in: Page 415 ### Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562 # (a) Show that the skin depth in a poor conductor ${\mathbit{\sigma }}{\mathbf{<}}{\mathbf{<}}{\mathbit{\omega }}{\mathbit{\epsilon }}$ is $\left(\frac{\epsilon }{\sigma }\right)\sqrt{\frac{\mathbf{2}}{\mathbf{\mu }}}$ (independent of frequency). Find the skin depth (in meters) for (pure) water. (Use the static values of ${\mathbit{\epsilon }}{\mathbf{}}{\mathbf{}}{\mathbf{,}}{\mathbf{}}{\mathbf{}}{\mathbit{\mu }}$ and ${\mathbit{\sigma }}$; your answers will be valid, then, only at relatively low frequencies.)(b) Show that the skin depth in a good conductor $\left(\sigma <<\omega \epsilon \right)$ is $\frac{\mathbf{\lambda }}{\mathbf{2}\mathbf{\pi }}$ (where λ is the wavelength in the conductor). Find the skin depth (in nanometers) for a typical metal $\left(\sigma >>\Omega {\left(m{10}^{7}\right)}^{-1}\right)$ in the visible range $\left(\omega \approx {10}^{15}/s\right)$ , assuming ${\mathbit{\epsilon }}{\mathbf{=}}{{\mathbit{\epsilon }}}_{{\mathbf{0}}}$ and ${\mathbit{\mu }}{\mathbf{\approx }}{{\mathbit{\mu }}}_{{\mathbf{0}}}$. Why are metals opaque?(c) Show that in a good conductor the magnetic field lags the electric field by ${{\mathbf{45}}}^{{\mathbf{°}}}$, and find the ratio of their amplitudes. For a numerical example, use the “typical metal” in part (b).

(a) The skin depth in a poor conductor is $d=\frac{2}{\sigma }\sqrt{\frac{\epsilon }{\mu }}$ and the skin depth for pure water is $1.19×{10}^{4}m$.

(b) The skin depth in a god conductor is $d=\frac{\lambda }{2\pi }$ and the skin depth for a typical metal is $12.5nm$.

(c) The ratio of the real amplitude is $1.12×{10}^{-7}S/m$.

See the step by step solution

## Step 1: Expression for the wavenumber and relative permittivity:

Write the expression for the wavenumber for the poor conductor.

${\mathbit{k}}{\mathbf{=}}{\mathbit{k}}{\mathbf{+}}{\mathbit{i}}{\mathbit{K}}$ …… (1)

Here, K is the imaginary part.

Write the expression for the relativity permittivity.

${{\mathbit{\epsilon }}}_{{\mathbf{r}}}{\mathbf{=}}\frac{\mathbf{\epsilon }}{{\mathbf{\epsilon }}_{\mathbf{0}}}\phantom{\rule{0ex}{0ex}}{\mathbit{\epsilon }}{\mathbf{=}}{{\mathbit{\epsilon }}}_{{\mathbf{r}}}{{\mathbit{\epsilon }}}_{{\mathbf{0}}}$ …… (2)

Here, is the Permittivity of the free space.

## Step 2: Determine the skin depth in a poor conductor and for pure water:

(a)

Write the expression for imaginary part K.

$K=\omega \sqrt{\frac{\epsilon \mu }{2}}{\left[\sqrt{1+{\left(\frac{\sigma }{\epsilon \omega }\right)}^{2}}\right]}^{\frac{1}{2}}$

Solve the above expression for the poor conductor.

$K=\omega \sqrt{\frac{\epsilon \mu }{2}}{\left[\sqrt{1+\frac{1}{2}{\left(\frac{\sigma }{\epsilon \omega }\right)}^{2}-1}\right]}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}K=\omega \sqrt{\frac{\epsilon \mu }{2}}\left[\sqrt{\frac{1}{\sqrt{2}}\frac{\sigma }{\epsilon \omega }}\right]\phantom{\rule{0ex}{0ex}}K=\frac{\sigma }{2}\left(\sqrt{\frac{\mu }{\mu }}\right)$

Write the expression for the skin depth.

$d=\frac{1}{K}$

Substitute $K=\frac{\sigma }{2}\left(\sqrt{\frac{\mu }{\epsilon }}\right)$ in the above expression.

Substitute the value of equation (2) in the above expression.

$d=\frac{2}{\sigma }\sqrt{\frac{{\epsilon }_{0}{\epsilon }_{r}}{\mu }}$

Take the value of ${\epsilon }_{r}$ from table 4.2 for pure water.

Substitute $\sigma =2.5×{10}^{5}{\left(\Omega m\right)}^{-1},$role="math" localid="1655714742974" ${\epsilon }_{0}=8.85×{10}^{-12}{C}^{2}/N{m}^{2}$$,{\epsilon }_{r}=80.1$ and $\mu =4\pi ×{10}^{-7}H/m$ in the above expression.

Therefore, the skin depth in a poor conductor is $d=\frac{2}{\sigma }\sqrt{\frac{\epsilon }{\mu }}$ and the skin depth for pure water is $1.19×{10}^{4}m$.

## Step 3: Determine the skin depth in a good conductor and for a typical metal:

(b)

Write the expression for the wavenumber for a good conductor.

${\mathbit{k}}{\mathbf{=}}{\mathbit{K}}$

Here, K is the imaginary part.

Hence, the wavelength of the conductors will be,

$\lambda =\frac{2\mathrm{\pi }}{k}\cong \frac{2\mathrm{\pi }}{K}$

Calculate imaginary part K for a good conductor.

$K=\omega \sqrt{\frac{\epsilon \mu }{2}}{\left[\sqrt{1+{\left(\frac{\sigma }{\epsilon \omega }\right)}^{2}}-1\right]}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}K=\omega \sqrt{\frac{\epsilon \mu }{2}}\sqrt{\left(\frac{\sigma }{\epsilon \omega }\right)}\phantom{\rule{0ex}{0ex}}K=\sqrt{\frac{\omega \mu \sigma }{2}}$

Hence, the skip depth in a good conductor will be,

…… (3)

Substitute $\omega ={10}^{15}{s}^{-1}$, $\mu =4\mathrm{\pi }×{10}^{-7}\mathrm{H}/\mathrm{m}$ and $\sigma ={10}^{7}{\left(\Omega m\right)}^{-1}$ in equation (3).

As the fields do not penetrate far into the metal, they are opaque in nature.

Therefore, the skin depth in a god conductor is $d=\frac{\lambda }{2\pi }$ and the skin depth for a typical metal is $12.5nm$.

## Step 4: Show that in a good conductor, the magnetic field lags the electric field by 45°:

(c)

For a good conductor:

$k=K$

Calculate the phase angle between the magnetic field and the dielectric field.

$\varphi ={\mathrm{tan}}^{-1}\left(\frac{K}{k}\right)\phantom{\rule{0ex}{0ex}}\varphi ={\mathrm{tan}}^{-1}\left(1\right)\phantom{\rule{0ex}{0ex}}\varphi ={45}^{°}$

Take the ratio of the real amplitude of E and B.

$\frac{{B}_{0}}{{E}_{0}}=\frac{K}{\omega }\phantom{\rule{0ex}{0ex}}\frac{{B}_{0}}{{E}_{0}}=\sqrt{\epsilon \mu \sqrt{{\left(1+\frac{\sigma }{\epsilon \omega }\right)}^{2}}}\phantom{\rule{0ex}{0ex}}\frac{{B}_{0}}{{E}_{0}}=\sqrt{\epsilon \mu \frac{\sigma }{\epsilon \omega }}\phantom{\rule{0ex}{0ex}}\frac{{B}_{0}}{{E}_{0}}=\sqrt{\frac{\sigma \mu }{\omega }}$

Substitute $\sigma ={10}^{7}{\left(\Omega m\right)}^{-1}$, $\mu =4\pi ×{10}^{-7}H/m$ and $\omega ={10}^{15}{s}^{-1}$ in the above expression.

$\frac{{B}_{0}}{{E}_{0}}=\sqrt{\frac{\left({10}^{7}{\left(\Omega m\right)}^{-1}\right)\left(4\pi ×{10}^{-7}H/m\right)}{{10}^{15}{s}^{-1}}}\phantom{\rule{0ex}{0ex}}\frac{{B}_{0}}{{E}_{0}}=1.12×{10}^{-7}S/m$

Therefore, the ratio of the real amplitude is $1.12×{10}^{-7}S/m$. ### Want to see more solutions like these? 