Question

(a) Show that the skin depth in a poor conductor $\mathit{\sigma }\mathbf{<}\mathbf{<}\mathit{\omega }\mathit{\epsilon }$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 $\mathit{\epsilon }\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathit{\mu }$and $\mathit{\sigma }$; 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 $\mathit{\epsilon }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}$and $\mathit{\mu }\mathbf{\approx }{\mathit{\mu }}_{\mathbf{0}}$. 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).

Short answer

Answer

(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 $1.19\times {10}^{4}m$.

(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 $1.12\times {10}^{-7}S/m$.

Step-by-step solution

Expression for the wavenumber and relative permittivity:

Write the expression for the wavenumber for the poor conductor.

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

Here,Kis the imaginary part.

Write the expression for the relativity permittivity.

$$\begin{gathered} {\mathit{\epsilon }}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{\epsilon }}{{\mathbf{\epsilon }}_{\mathbf{0}}} \\ \mathit{\epsilon }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{r}}{\mathit{\epsilon }}_{\mathbf{0}} \end{gathered}$$ …… (2)

Here, is the Permittivity of the free space.

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.

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

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.

$$\begin{gathered} d=\frac{1}{{\displaystyle \frac{\sigma }{2}}\left(\sqrt{{\displaystyle \frac{\mu }{\epsilon }}}\right)} \\ d=\frac{2}{\sigma }\sqrt{\frac{\epsilon }{\mu }} \end{gathered}$$

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 $\sigma =2.5\times {10}^5{\left(\Omega m\right)}^{-1},$role="math" localid="1655714742974" ${\epsilon }_0=8.85\times {10}^{-12}{C}^2/N{m}^2$$,{\epsilon }_r=80.1$and $\mu =4\pi \times {10}^{-7}H/m$in the above expression.

$$\begin{gathered} d=\frac{2}{\left({\displaystyle \frac{1}{2.5\times {10}^5}}{\left(\Omega m\right)}^{-1}\right)}\sqrt{\frac{\left(8.85\times {10}^{-12}{C}^2/N{m}^2\right)\left(80.1\right)}{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{H}/\mathrm{m}\right)}} \\ d=1.19\times {10}^{4}m \end{gathered}$$

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 $1.19\times {10}^{4}m$.

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

(b)

Write the expression for the wavenumber for a good conductor.

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

Here,Kis 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.

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

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

$d=\frac{1}{\sqrt{{\displaystyle \frac{\omega \mu \sigma }{2}}}}$ …… (3)

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

$$\begin{gathered} d=\frac{1}{\sqrt{{\displaystyle \frac{\left({10}^{15}{s}^{-1}\right)\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{H}/\mathrm{m}\right)\left({10}^7{\left(\Omega m\right)}^{-1}\right)}{2}}}} \\ d=1.25\times {10}^{-8}m\times \left(\frac{{10}^{9}nm}{1m}\right) \\ d=12.5nm \end{gathered}$$

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 $12.5nm$.

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.

$$\begin{gathered} \varphi ={\tan }^{-1}\left(\frac{K}{k}\right) \\ \varphi ={\tan }^{-1}\left(1\right) \\ \varphi ={45}^{°} \end{gathered}$$

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

$$\begin{gathered} \frac{B_0}{E_0}=\frac{K}{\omega } \\ \frac{B_0}{E_0}=\sqrt{\epsilon \mu \sqrt{{\left(1+\frac{\sigma }{\epsilon \omega }\right)}^2}} \\ \frac{B_0}{E_0}=\sqrt{\epsilon \mu \frac{\sigma }{\epsilon \omega }} \\ \frac{B_0}{E_0}=\sqrt{\frac{\sigma \mu }{\omega }} \end{gathered}$$

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

$$\begin{gathered} \frac{B_0}{E_0}=\sqrt{\frac{\left({10}^7{\left(\Omega m\right)}^{-1}\right)\left(4\pi \times {10}^{-7}H/m\right)}{{10}^{15}{s}^{-1}}} \\ \frac{B_0}{E_0}=1.12\times {10}^{-7}S/m \end{gathered}$$

Therefore, the ratio of the real amplitude is $1.12\times {10}^{-7}S/m$.