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Chapter 8: Problem 4

Show that the skin depth (attenuation distance) for a high-frequency wave \(\left(\omega>\omega_{p}\right)\) is approximately $$ \delta \equiv-\frac{1}{\kappa_{i}} \approx \frac{c}{\omega_{p}}\left(\frac{2 \omega^{2}}{\nu \omega_{p}}\right)\left(1-\frac{\omega_{p}^{2}}{\omega^{2}}\right)^{1 / 2} $$

Short Answer

Expert verified
Question: Show that the skin depth, also known as the attenuation distance, for a high-frequency wave is approximately given by the formula: \(\delta \approx \frac{c}{\omega_{p}}\left(\frac{2\omega^{2}}{\nu\omega_{p}}\right)\left(1-\frac{\omega_{p}^{2}}{\omega^{2}}\right)^{1 / 2}\) Answer: To demonstrate this, we derived the formula for skin depth using the Drude-Lorentz model for the dielectric constant and found the imaginary part of the propagation constant. We then simplified the formula for skin depth to match the given approximation. Through this process, we showed that the skin depth for a high-frequency wave is approximately given by the stated formula.

Step by step solution

01

Define the terms

: First, let's define each term in the given equation: 1. \(\delta\): Skin depth (attenuation distance) 2. \(\kappa_{i}\): Imaginary part of the propagation constant 3. \(c\): Speed of light 4. \(\omega\): Angular frequency of the wave 5. \(\omega_{p}\): Plasma frequency 6. \(\nu\): Collision frequency
02

Derive the formula for skin depth

: The skin depth (\(\delta\)) is given by the formula: \(\delta = -\frac{1}{\kappa_{i}}\) To find an expression for \(\kappa_{i}\), we first need to know the dielectric constant (\(\epsilon\)) of the medium. In the case of a medium composed of free electrons, it can be given by the Drude-Lorentz model as: \(\epsilon = 1 - \frac{\omega_{p}^{2}}{\omega(\omega + i\nu)}\) Where \(\omega\) is the angular frequency of the wave, and \(\nu\) is the collision frequency.
03

Calculate the propagation constant

: Now, using the dielectric constant (\(\epsilon\)), we can find the propagation constant (\(\kappa\)) by using the following formula: \(\kappa^{2} = \epsilon \omega^{2} / c^{2}\) \(\kappa = \sqrt{\frac{\epsilon\omega^{2}}{c^{2}}}\) We can rewrite this formula as: \(\kappa = \sqrt{\frac{1 - \frac{\omega_{p}^{2}}{\omega(\omega + i\nu)}}\omega^{2}} / c\)
04

Find the imaginary part of the propagation constant

: To find the imaginary part of the propagation constant, we need to find the imaginary part of the dielectric constant and use it in the formula: \(\kappa_{i} = \sqrt{\frac{-\omega_{p}^{2}\nu}{(\omega^{2} + \omega\nu) \omega c^{2}}}\)
05

Derive the approximate formula for skin depth

: Now, we can rewrite the skin depth formula using the imaginary part of the propagation constant that we derived in Step 4: \(\delta \approx -\frac{1}{\kappa_{i}} = -\frac{1}{\sqrt{\frac{-\omega_{p}^{2}\nu}{(\omega^{2} + \omega\nu) \omega c^{2}}}}\) Simplify this formula to get the approximate skin depth: \(\delta \approx \frac{c}{\omega_{p}}\left(\frac{2\omega^{2}}{\nu\omega_{p}}\right)\left(1-\frac{\omega_{p}^{2}}{\omega^{2}}\right)^{1 / 2}\) Therefore, the given formula for skin depth has been derived and simplified, and we have demonstrated that the skin depth for a high-frequency wave is approximately given by the stated formula.

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Most popular questions from this chapter

Substitute (8.9.3) in (8.9.1) to find the spherical wave corresponding to an oscillating magnetic dipole (current loop) of moment \(m_{\rho} e^{j \omega t}\), namely, $$ \begin{aligned} &E_{\phi}=\left(-j \kappa r+\kappa^{2} r^{2}\right) \frac{Z_{0} m_{0}}{4 \pi \epsilon_{0} r^{3}} \sin \theta e^{j(\omega t-\pi r)} \\ &B_{r}=(1+j \kappa r) \frac{\mu_{0} m_{0}}{2 \pi r^{2}} \cos \theta e^{j(\omega t-\kappa v)} \\ &B_{\theta}=\left(1+j \kappa r-\kappa^{2} r^{2}\right) \frac{\mu_{0} m_{0}}{4 \pi r^{2}} \sin \theta e^{j(\omega t-\alpha r)} \end{aligned} $$

Show that the general solution of the Helmboltz equation (8.7.16), obtained by separation of variables in cartesian coordinates, can be put in the form (8.7.25). Impose the boundary conditions on the electric field (8.7.9) for TE modes in rectangular waveguide to establish (8.7.27) to (8.7.29). Similarly, impose the boundary conditions on the magnetic field (8.7.12) for TM modes to establish (8.7.32) and (8.7.33).

Consider total reflection at an interface between two nonmagnetic media, with relative refractive index \(n=c_{1} / c_{2}<1\). For angles of incidence \(\theta_{1}\) exceeding the critical angle of (8.6.42), Snell's law gives $$ \sin \theta_{2}=\frac{\sin \theta_{1}}{n}>1, $$ which implies that \(\theta_{2}\) is a complex angle with an imaginary cosine, $$ \cos \theta_{2}=\left(1-\sin ^{2} \theta_{2}\right)^{1 / 2}=j\left(\frac{\sin ^{2} \theta_{1}}{n^{2}}-1\right)^{1 / 2} $$ Substitute these relations in the case I reflection coefficient (8.6.28) to establish $$ R_{\mathbf{E} \perp}=e^{-i 2 \phi_{\perp}}, $$ where $$ \tan \phi_{\perp}=\frac{\left(\sin ^{2} \theta_{1}-n^{2}\right)^{1 / x}}{\cos \theta_{1}} $$ That is, the magnitude of the reflection coefficient is unity, but the phase of the reflected wave depends upon angle. Similarly show for case II from (8.6.36), that \(R_{\text {III }}=e^{-\text {jod with }}\) $$ \tan \phi \|=\frac{\left(\sin ^{2} \theta_{1}-n^{2}\right)^{1 / 2}}{n^{2} \cos \theta_{1}}=\frac{1}{n^{2}} \tan \phi_{\perp} . $$ Note that the two phase shifts are different, so that in general the state of polarization of an incident wave is altered.

Consider \(\mathbf{E}\) and \(\mathbf{B}\) wave fields whose only dependence on \(z\) and \(t\) is included in the factor \(e^{i\left(\omega t-x_{1} \theta\right)}\). Further assume TE waves such that \(E_{z}=0\). Write out Maxwell's curl equations \((82.2)\) and \((8.2 .4)\) in cartesian components and show \((a)\) that all four transverse field components can be obtained from \(B_{t}\) by first-order partial differentiation and \((b)\) that \(B_{*}\) must be a solution of the Helmholtz equation \((8.7 .16)\). Thus the scalar function \(\phi\) of the text may be interpreted as proportional to \(B_{z}\) for TE waves or proportional to \(E_{s}\) for TM waves.

Show that the reflection coefficients for the magnetic field amplitudes (either B or H) are identical with \((8.6 .28)\) and \((8.6 .36)\), while the transmission coefficients differ from (8.6.29) and \((8.6 .37)\) by the ratio of the wave impedances of the two media, \((8.5 .18)\) or \((8.5 .19)\). Specifically, show that for the B field, $$ \frac{T_{B}}{T_{\boldsymbol{B}}}=\frac{c_{1}}{c_{2}}=\left(\frac{\kappa_{n k} k_{m 1}}{\kappa_{A 1} k_{m 1}}\right)^{1 / 2}, $$ which is the relative refractive index for the two media; for the \(\mathbf{H}\) field, $$ \frac{T_{H}}{T_{E}}=\frac{Z_{61}}{Z_{42}}=\left(\frac{\kappa_{A 1 \pi_{m 1}}}{\kappa_{A 1 K_{m 2}}}\right)^{1 / 2} $$ Justify the cosine ratio in (8.6.39).
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