Uniform Plane Waves Questions and Answers

Answer: A

The surfaces of constant phase are parallel planes, and over each of them the field is uniform. It is an idealisation, since a real source produces spherical waves, but far from the source a small region of a spherical wave is very nearly plane.

Answer: B

The intrinsic impedance of a lossless medium is purely real, so the two fields rise and fall together. In a lossy medium the impedance acquires a phase angle and the magnetic field lags the electric, reaching forty-five degrees in a good conductor.

Answer: C

It is \( \sqrt{\mu/\epsilon} \) for a lossless medium, about 377 ohms in free space, and it governs how much of a wave is reflected at a boundary. Being a ratio of a volt per metre to an ampere per metre, it has the dimensions of a resistance.

Answer: D

Dividing the angular frequency by the phase constant gives this result, which in free space is the speed of light. In a dielectric of relative permittivity four and unit permeability it is half that value, so the refractive index is two.

Answer: A

Correct answer: \( \beta = 2\pi/\lambda \)

The phase constant is the phase advance per unit distance, and a full wavelength corresponds to \( 2\pi \) radians, so \( \beta = 2\pi/\lambda \) with units of radians per metre. It is the spatial counterpart of the angular frequency \( \omega = 2\pi/T \), and the phase velocity follows as \( v = \omega/\beta \).

Answer: B

It measures how much phase is accumulated per unit distance travelled, so its unit is an angle over a length. Multiplying it by a distance gives the phase shift over that distance, which is how a transmission line calculation proceeds.

Answer: C

The cross product is the Poynting vector, whose direction is that of energy flow and hence of propagation. The three vectors form a right-handed set, which fixes the orientation of the magnetic field once the electric field and the direction are known.

Answer: D

Applying the divergence conditions to a field varying only along the direction of propagation forces the component in that direction to be constant, and it must therefore be zero for a wave. This is why an electromagnetic wave can be polarised while a sound wave cannot.

Answer: A

Correct answer: the attenuation constant, in nepers per metre

In \( \gamma = \alpha + j\beta \), the real part governs how the amplitude decays and the imaginary part how the phase advances: the wave varies as \( e^{-\alpha z}e^{-j\beta z} \). The neper is the natural-logarithm unit, so one neper is a fall to \( 1/e \) - and one neper equals 8.686 decibels.

Answer: B

Both come to \( \sqrt{\pi f\mu\sigma} \), so a wave in a conductor loses one neper of amplitude in the same distance as it gains one radian of phase. That distance is the skin depth, and it is why fields scarcely penetrate metal.