Question

Problem 1. An unknown material is characterized by e-80, Nr = 1, and o- 1000 (S/m). At each of the following frequencies, determine if the material may be considered a good conductor, a quasi-conductor, or a low-loss dielectric, and then calculate , B, 2. Hp, ne, and skin depth: (a) 60 Hz (10 pts) (b) 1 GHz (10 pts)

Answer 1

Answer 1-

Given,

$\varepsilon _{r}=80$

  $\mu _{r}=1$

  $\sigma =1000 (S/m)$

(a) At 60 Hz,

Now,

  $\frac{\varepsilon ^{''}}{\varepsilon ^{'}}=\frac{\sigma }{\omega \varepsilon }=\frac{\sigma }{\omega \varepsilon _{r}\varepsilon _{0}}=\frac{1000}{2\pi \times 60\times 80\times (10^{-9}/36\pi )}=3.75\times 10^{9}$

As  

$\frac{\sigma }{\omega \varepsilon }> 10^{2}$

So, the material is considered as a good conductor.

Now,

$\alpha =\sqrt{\pi f\mu \sigma }=\sqrt{\pi f\mu _{0}\sigma }=\sqrt{\pi \times 60\times 4\pi \times 10^{-7}\times 1000}$

$\alpha =0.487(Np/m)$

Now,   $\beta =\alpha =0.487(rad/m)$

Now,

  $u_{p}=\sqrt{\frac{4\pi f}{\mu \sigma }}=\sqrt{\frac{4\pi f}{\mu_{0} \sigma }}=\sqrt{\frac{4\pi \times 60}{4\pi \times 10^{-7} \times 1000}}=774.6(m/s)$

$\lambda =\frac{u_{p}}{f}=\frac{774.6}{60}=12.91(m)$

Now,

  $\eta _{c}=(1+j)\frac{\alpha }{\sigma }=(\sqrt{2}e^{j\pi /4})\frac{0.487}{1000}=6.89\times 10^{-4}e^{j\pi /4}(\Omega )$

Now,

Skin Depth,

  $\delta _{s}=\frac{1}{\alpha }=\frac{1}{0.487}=2.05m$

(b) At 1 GHz = 109 Hz

Now,

$\frac{\varepsilon ^{''}}{\varepsilon ^{'}}=\frac{\sigma }{\omega \varepsilon }=\frac{\sigma }{\omega \varepsilon _{r}\varepsilon _{0}}=\frac{1000}{2\pi \times 10^{9}\times 80\times (10^{-9}/36\pi )}=225$

As  

$\frac{\sigma }{\omega \varepsilon }> 10^{2}$

So, the material is considered as a good conductor.

Now,

$\alpha =\sqrt{\pi f\mu \sigma }=\sqrt{\pi f\mu _{0}\sigma }=\sqrt{\pi \times 10^{9}\times 4\pi \times 10^{-7}\times 1000}$

  $\alpha =1986.9(Np/m)$

  Now, $\beta =\alpha =1986.9(rad/m)$

Now,

  $u_{p}=\sqrt{\frac{4\pi f}{\mu \sigma }}=\sqrt{\frac{4\pi f}{\mu_{0} \sigma }}=\sqrt{\frac{4\pi \times 10^{9}}{4\pi \times 10^{-7} \times 1000}}=3.16\times 10^{6}(m/s)$

$\lambda =\frac{u_{p}}{f}=\frac{3.16\times 10^{6}}{10^{9}}=3.16(mm)$

Now,

$\eta _{c}=(1+j)\frac{\alpha }{\sigma }=(\sqrt{2}e^{j\pi /4})\frac{1986.9}{1000}=2.81e^{j\pi /4}(\Omega )$

Now,

Skin Depth,

  $\delta _{s}=\frac{1}{\alpha }=\frac{1}{1986.9}=0.503(mm)$