Uniform Plane Waves Questions and Answers

Answer: C

The relation is \(k_{r} = \frac{Z_{L} - Z_{0}}{Z_{0} + Z_{L}}\).

At the load, the voltage and current must satisfy \(V/I = Z_{L}\), while each travelling wave on the line has \(V/I = \pm Z_{0}\). With incident wave \(V^{+}\) and reflected wave \(V^{-}\):

\[ \begin{aligned} Z_{L} &= Z_{0}\frac{V^{+} + V^{-}}{V^{+} - V^{-}} \\ k_{r} &= \frac{V^{-}}{V^{+}} = \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}} \end{aligned} \]

Checks: a matched load (\(Z_{L} = Z_{0}\)) gives no reflection, a short circuit gives -1 and an open circuit +1. The first option becomes infinite for a matched load, and the others do not even have the dimensionless form a reflection coefficient needs.

Answer: C

The key gives \(3.4 \times 10^{-10}\) J/m\(^{3}\), which is the peak total energy density, \(\varepsilon_{0}E_{0}^{2}\).

In a plane wave the electric and magnetic fields carry equal energy, so the total energy density is \(u = \varepsilon_{0}E^{2}\). At the instant the field is at its amplitude:

\[ \begin{aligned} u &= \varepsilon_{0} E_{0}^{2} \\ &= 8.85 \times 10^{-12} \times (6.2)^{2} \\ &= 8.85 \times 10^{-12} \times 38.44 \\ &\approx 3.4 \times 10^{-10}\ \text{J/m}^{3} \end{aligned} \]

The frequency does not enter the calculation. The time-averaged value would be half of this, about \(1.7 \times 10^{-10}\) J/m\(^{3}\), which is not among the options. 1.4, 2.4 and 4.4 do not come from either formula.

Answer: C

The skin depth is 0.5 units.

The skin depth \(\delta\) is the distance over which a wave's amplitude falls to \(1/e\) (about 37%) of its value at the surface. Since the amplitude varies as \(e^{-\alpha z}\), this happens when \(\alpha z = 1\):

\[ \delta = \frac{1}{\alpha} = \frac{1}{2} = 0.5 \]

The answer 2 would follow from mistaking skin depth for the attenuation constant itself, 4 from multiplying instead of dividing, and 1 from ignoring the attenuation altogether. Skin depth and attenuation constant are reciprocals, so a larger \(\alpha\) always means a thinner skin.

Answer: A

The velocity of a wave in a good conductor is \(V = \sqrt{2\omega/\mu\sigma}\).

In a good conductor (\(\sigma \gg \omega\varepsilon\)) the attenuation and phase constants are equal:

\[ \alpha = \beta = \sqrt{\frac{\omega\mu\sigma}{2}} \]

The phase velocity is then

\[ \begin{aligned} v &= \frac{\omega}{\beta} = \omega\sqrt{\frac{2}{\omega\mu\sigma}} \\ &= \sqrt{\frac{2\omega}{\mu\sigma}} \end{aligned} \]

This is very slow compared with light; in copper at 1 MHz it is only about 400 m/s.

Option B has the conductivity and permeability in the numerator, which would make the wave faster in better conductors, the wrong way round. Options C and D drop the square root, so they do not even have the units of velocity.

Answer: D

In a good conductor E leads H by 45 degrees.

The ratio of the tangential electric and magnetic fields is the intrinsic impedance. For a good conductor

\[ \begin{aligned} \eta &= \sqrt{\frac{j\omega\mu}{\sigma}} \\ &= \sqrt{\frac{\omega\mu}{\sigma}}\,\angle 45^{\circ} \end{aligned} \]

because \(\sqrt{j} = e^{j\pi/4}\). Since \(E = \eta H\), the electric field is 45° ahead of the magnetic field.

E and H are in phase only in a lossless dielectric, where \(\eta\) is real. They are never 90° apart in a propagating wave; that limit would need \(\eta\) purely imaginary. 'Out of phase' (180°) does not arise either.

Answer: C

The ratio has the dimension of Resistance.

\(E\) is in V/m and \(H\) is in A/m, so \(\frac{E}{H}\) is in V/A, that is ohms. For a plane wave in free space, \(\frac{E}{H} = \sqrt{\mu_{0}/\varepsilon_{0}} \approx 377\ \Omega\), the impedance of free space.

Inductance is V s/A and capacitance A s/V, and their product LC has dimensions of time squared, so none of these matches.

Answer: C

Skin depth varies inversely with the Half power of frequency.

For a good conductor, the skin depth is:

\[ \begin{aligned} \delta &= \sqrt{\frac{2}{\omega\mu\sigma}} \\ &\propto \frac{1}{\sqrt{f}} \end{aligned} \]

So quadrupling the frequency halves the depth to which the field penetrates. That is why high-frequency currents crowd near the surface of conductors.

First, second or third powers do not follow from this expression.

Answer: B

The SI unit of \(\frac{|E|}{|B|}\) is m/s.

From the Lorentz force, \(qE\) and \(qvB\) are both forces, so \(E\) and \(vB\) have the same units and \(\frac{E}{B}\) has the units of velocity. For an electromagnetic wave in vacuum \(\frac{E}{B} = c\). Explicitly, (V/m)/T = (V/m)/(V s/m²) = m/s.

s/m is the inverse, and the options with \(c^{2}\) put a speed squared where none belongs.

Answer: C

The relation is \(V_{p}V_{g} = c^{2}\).

From \(\omega^{2} = \omega_{0}^{2} + c^{2}k^{2}\):

\[ \begin{aligned} V_{p} &= \frac{\omega}{k} \\ 2\omega\,d\omega &= 2c^{2}k\,dk, \quad V_{g} \\ &= \frac{d\omega}{dk} \\ &= \frac{c^{2}k}{\omega} \\ V_{p}V_{g} &= \frac{\omega}{k}\cdot\frac{c^{2}k}{\omega} \\ &= c^{2} \end{aligned} \]

The phase velocity exceeds \(c\) while the group velocity, which carries energy and information, stays below \(c\). They are equal only without the plasma term, and the square-root relations are dimensionally inconsistent.

Answer: C

In free space, conductivity is zero.

Free space (vacuum) contains no free charges to carry a conduction current, so its conductivity \(\sigma = 0\). This is why electromagnetic waves travel through free space without loss.

Permittivity and permeability of free space are not zero: \(\varepsilon_{0} \approx 8.854 \times 10^{-12}\) F/m and \(\mu_{0} = 4\pi \times 10^{-7}\) H/m, and together they set the speed of light, \(c = 1/\sqrt{\mu_{0}\varepsilon_{0}}\). Resistivity is the reciprocal of conductivity, so in free space it is infinite, not zero.