91影视

Write the expression for the wavenumber for the poor conductor.

$\mathit{k}\mathbf{=}\mathit{k}\mathbf{+}\mathit{i}\mathit{K}$ 鈥︹�� (1)

Here,Kis the imaginary part.

Write the expression for the relativity permittivity.

$$\begin{gathered} {\mathit{蔚}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{蔚}}{{\mathbf{蔚}}_{\mathbf{0}}} \\ \mathit{蔚}\mathbf{=}{\mathit{蔚}}_{\mathbf{r}}{\mathit{蔚}}_{\mathbf{0}} \end{gathered}$$ 鈥︹�� (2)

Here, is the Permittivity of the free space.

(a)

Write the expression for imaginary part K.

$K=蝇\sqrt{\frac{蔚渭}{2}}{\left[\sqrt{1+{\left(\frac{蟽}{蔚蝇}\right)}^2}\right]}^{\frac{1}{2}}$

Solve the above expression for the poor conductor.

$$\begin{gathered} K=蝇\sqrt{\frac{蔚渭}{2}}{\left[\sqrt{1+\frac{1}{2}{\left(\frac{蟽}{蔚蝇}\right)}^2-1}\right]}^{\frac{1}{2}} \\ K=蝇\sqrt{\frac{蔚渭}{2}}\left[\sqrt{\frac{1}{\sqrt{2}}\frac{蟽}{蔚蝇}}\right] \\ K=\frac{蟽}{2}\left(\sqrt{\frac{渭}{渭}}\right) \end{gathered}$$

Write the expression for the skin depth.

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

Substitute $K=\frac{蟽}{2}\left(\sqrt{\frac{渭}{蔚}}\right)$in the above expression.

$$\begin{gathered} d=\frac{1}{{\displaystyle \frac{蟽}{2}}\left(\sqrt{{\displaystyle \frac{渭}{蔚}}}\right)} \\ d=\frac{2}{蟽}\sqrt{\frac{蔚}{渭}} \end{gathered}$$

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

$d=\frac{2}{蟽}\sqrt{\frac{{蔚}_0{蔚}_r}{渭}}$

Take the value of ${蔚}_r$from table 4.2 for pure water.

Substitute $蟽=2.5脳{10}^5{\left(惟m\right)}^{-1},$role="math" localid="1655714742974" ${蔚}_0=8.85脳{10}^{-12}{C}^2/N{m}^2$$,{蔚}_r=80.1$and $渭=4蟺脳{10}^{-7}H/m$in the above expression.

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

Therefore, the skin depth in a poor conductor is $d=\frac{2}{蟽}\sqrt{\frac{蔚}{渭}}$and the skin depth for pure water is $1.19脳{10}^{4}m$.

(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,

$位=\frac{2\mathrm{蟺}}{k}鈮�\frac{2\mathrm{蟺}}{K}$

Calculate imaginary part K for a good conductor.

$$\begin{gathered} K=蝇\sqrt{\frac{蔚渭}{2}}{\left[\sqrt{1+{\left(\frac{蟽}{蔚蝇}\right)}^2}-1\right]}^{\frac{1}{2}} \\ K=蝇\sqrt{\frac{蔚渭}{2}}\sqrt{\left(\frac{蟽}{蔚蝇}\right)} \\ K=\sqrt{\frac{蝇渭蟽}{2}} \end{gathered}$$

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

$d=\frac{1}{\sqrt{{\displaystyle \frac{蝇渭蟽}{2}}}}$ 鈥︹�� (3)

Substitute $蝇={10}^{15}{s}^{-1}$, $渭=4\mathrm{蟺}脳{10}^{-7}\mathrm{H}/\mathrm{m}$and $蟽={10}^7{\left(惟m\right)}^{-1}$in equation (3).

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

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

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

(c)

For a good conductor:

$k=K$

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

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

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

$$\begin{gathered} \frac{B_0}{E_0}=\frac{K}{蝇} \\ \frac{B_0}{E_0}=\sqrt{蔚渭\sqrt{{\left(1+\frac{蟽}{蔚蝇}\right)}^2}} \\ \frac{B_0}{E_0}=\sqrt{蔚渭\frac{蟽}{蔚蝇}} \\ \frac{B_0}{E_0}=\sqrt{\frac{蟽渭}{蝇}} \end{gathered}$$

Substitute $蟽={10}^7{\left(惟m\right)}^{-1}$, $渭=4蟺脳{10}^{-7}H/m$and $蝇={10}^{15}{s}^{-1}$in the above expression.

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

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