Time Varying Fields and Maxwells Equations Questions and Answers

Answer: B

The changing flux within the conductor induces currents that add to the current at the surface and subtract from it at the centre, so the current crowds into a layer whose depth falls as the inverse square root of frequency. The effective resistance rises accordingly.

Answer: C

The expression is \( \sqrt{2/(\omega\mu\sigma)} \), so raising the frequency a hundredfold reduces the depth tenfold. In copper it is about nine millimetres at fifty hertz and two hundredths of a millimetre at ten megahertz.

Answer: D

Only a shell of the order of one skin depth carries appreciable current, so the effective cross-section is much smaller than the geometrical one. This is why hollow tubes and stranded conductors are used at radio frequency and why busbars are made flat.

Answer: A

Each strand is thinner than a skin depth and the transposition means no strand stays in the low-current interior, so the whole cross-section is used. It is effective up to a few hundred kilohertz, above which the strand insulation and capacitance become the limiting factors.

Answer: B

Eliminating one field between the two curl equations gives a wave equation for the other, whose solutions travel at the speed fixed by the permittivity and permeability. That prediction, and its numerical agreement with the measured speed of light, is the crowning result of the theory.

Answer: C

A changing magnetic field creates a curling electric field, whose own change creates a curling magnetic field, and the pair regenerate one another indefinitely as they travel. In the static case the time derivatives vanish and the two subjects separate entirely.

Answer: D

Both divergence equations force the fields to have no component along the direction of propagation, and the curl equations make each perpendicular to the other. The wave is therefore transverse in both fields, which is what allows it to be polarised.

Answer: A

If the apparatus is electrically small the retardation across it is negligible and the fields respond as though instantaneously, which is exactly the assumption of circuit theory. Once the dimensions approach a wavelength, transmission line and field methods become necessary.

Answer: B

Dividing the speed of light by fifty gives six million metres. Since no power system approaches that size, lumped circuit theory serves for power engineering and only very long transmission lines need distributed treatment.

Answer: C

Correct answer: 0.1 m

\[ \lambda = \frac{c}{f} = \frac{3\times10^{8}}{3\times10^{9}} = 0.1\ \mathrm{m}. \] Ten centimetres at 3 GHz is a useful anchor, and the scaling is simple: 300 MHz gives a metre, 30 GHz gives a centimetre. Antenna dimensions and waveguide sizes all follow from it.