(a) Show that the skin depth \(\delta\) can be put in the form $$ \delta=\left(\frac{\lambda_{0}}{\pi Z_{\mu} \pi_{m} g}\right)^{1 / 2} $$ where \(Z_{0}\) is the free-space wave impedance \((8.3,10)\) and \(\lambda_{0}=2 \pi c / \omega\) is the vacuum wavelength. (b) Evaluate s for copper, for waves having wavelengths in vacuum of \(5,000 \mathrm{~km}(60-\mathrm{Hz}\) power line); \(100 \mathrm{~m}\) ( \(\sim\) AM broadcast band); \(1 \mathrm{~m}\) ( \(\sim\) television and FM broadcast); \(3 \mathrm{~cm}\) ( \(\sim\) radar); \(500 \mathrm{~nm}\) ( \(\sim\) visible light). How does the size of the skin depth affect the technology of these various applications" (c) The electrical conductivity of sea water is about \(4 \mathrm{mhos} / \mathrm{m}\). How would you communicate by radio with a submarine \(100 \mathrm{~m}\) below the surface?

We start with the definition of the skin depth, \(\delta\), as $$ \delta = \sqrt{\frac{2}{\mu \sigma \omega}}, $$ where \(\mu\) is the magnetic permeability, \(\sigma\) is the electrical conductivity, and \(\omega\) is the angular frequency in radians per second. Now, the magnetic impedance, \(Z_{\mu}\), is defined as $$ Z_{\mu} = \sqrt{\frac{\mu \omega}{2 \sigma}}. $$ Squaring and dividing both sides by \(\mu \omega\), we get $$ \frac{1}{\pi_m^2} = \frac{2}{\mu^2 \omega^2 \sigma^2} \Rightarrow \frac{1}{\pi_m} = \frac{2}{\mu \sigma \omega}. $$ We can then write the skin depth, \(\delta\), as $$ \delta = \sqrt{\frac{1}{\pi_m \mu \sigma \omega}}. $$ Substituting the vacuum wavelength, \(\lambda_0\), as \(\frac{2 \pi c}{\omega}\), where \(c\) is the speed of light, we find $$ \delta= \sqrt{\frac{\lambda_{0}}{\pi Z_{\mu} \pi_{m} g}}. $$

Given the equation for \(\delta\) derived in the previous step, we need to compute the skin depth for various wavelengths in vacuum: \(5,000 \,\mathrm{km}\), \(100\, \mathrm{m}\), \(1\, \mathrm{m}\), \(3\,\mathrm{cm}\), and \(500\, \mathrm{nm}\). For copper, \(Z_{0} = 8.3\times10^3 \, \Omega m\) and \(g = 5.8\times10^7 \, \mathrm{S/m}\). We calculate the skin depth for each wavelength, $$ \delta=\sqrt{\frac{\lambda_{0}}{\pi Z_{\mu} \pi_{m} g}}. $$ Values for each application: 1. Power line (\(\lambda_0 = 5,000\, \mathrm{km}\)): $ \delta_{1} = \sqrt{\frac{5000 \times 10^3}{\pi (8.3\times10^3) \pi (5.8\times10^7)}} \approx 1.32 \times 10^{-5}\, \mathrm{m}$. 2. AM Broadcast (\(\lambda_0 = 100\, \mathrm{m}\)): $ \delta_{2} = \sqrt{\frac{100}{\pi (8.3\times10^3) \pi (5.8\times10^7)}} \approx 3.57\times 10^{-7}\, \mathrm{m}$. 3. Television and FM Broadcast (\(\lambda_0 =1\, \mathrm{m}\)): $ \delta_{3}= \sqrt{\frac{1}{\pi (8.3\times10^3) \pi (5.8\times10^7)}} \approx 1.13 \times 10^{-8}\, \mathrm{m}$. 4. Radar (\(\lambda_0 = 3\,\mathrm{cm}\)): $ \delta_{4}= \sqrt{\frac{0.03}{\pi (8.3\times10^3) \pi (5.8\times10^7)}} \approx 6.49 \times 10^{-9}\, \mathrm{m}$. 5. Visible light (\(\lambda_0 = 500\, \mathrm{nm}\)): $ \delta_{5}= \sqrt{\frac{500\times 10^{-9}}{\pi (8.3\times10^3) \pi (5.8\times10^7)}} \approx 2.26 \times 10^{-12}\, \mathrm{m}$. With increasing frequency, the skin depth becomes smaller. So the size of the skin depth influences the technology in various ways. For applications like power lines and broadcasting, larger skin depth allows for the use of thicker cables. However, for applications like radar and visible light, the skin depth is very small, making them more suitable for thin layers.

The electrical conductivity of seawater is given as \(4\,\mathrm{S/m}\). We need to find a suitable frequency to communicate with a submarine \(100\,\mathrm{m}\) below the surface. We can use the expression for skin depth \(\delta\): $$ \delta = \sqrt{\frac{2}{\mu \sigma \omega}}. $$ For sea water, \(\sigma = 4\, \mathrm{S/m}\). By rearranging the above formula, we can find the angular frequency (\(\omega\)) required to communicate with the submarine, $$ \omega = \frac{2}{\mu\sigma \delta^2}. $$ As the submarine is \(100\, \mathrm{m}\) below the surface, we set \(\delta = 100\, \mathrm{m}\). Using the permeability of free space \(\mu = 4\pi \times 10^{-7}\, \mathrm{Tm/A}\), we calculate the angular frequency, $$ \omega = \frac{2}{(4\pi \times 10^{-7})(4)(100)^2}\, \mathrm{rad/s} \approx 1.59\times 10^2\, \mathrm{rad/s}. $$ Converting this to frequency, we have $$ f = \frac{\omega}{2 \pi} = \frac{1.59\times 10^2}{2 \pi}\, \mathrm{Hz} \approx 25\, \mathrm{Hz}. $$ Therefore, communication with a submarine at \(100\, \mathrm{m}\) below the surface would require the use of ultra low-frequency radio signals with a frequency of approximately \(25\, \mathrm{Hz}\).