At the surface of a perfect conductor the tangential electric field is
Answer: D
A finite tangential field would drive an infinite current in a perfect conductor, so it must vanish, and continuity then makes it vanish just outside as well. This condition is what fixes the modes of a waveguide and the currents on an antenna.
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The wave equation for the electric field in a source-free region follows from Maxwell's equations by
Answer: A
The curl of a curl expands into a gradient of a divergence minus a Laplacian, and in a source-free region the divergence term vanishes, leaving the Laplacian equal to a second time derivative. That is the wave equation, and its solutions travel at the speed set by the medium.
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The retarded potentials express the fields at a point in terms of the sources
Answer: B
Since no influence travels faster than the wave speed, what happens at a point now depends on what the sources were doing a propagation time ago. The retardation is what turns the static potential expressions into radiating ones and gives rise to antennas.
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A perfectly conducting surface reflects an incident electromagnetic wave because
Answer: C
The boundary condition forces the tangential field to vanish, which can only be satisfied by adding a reflected wave that cancels the incident tangential field at the surface. The surface currents that flow are what carry the reflected wave away.
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A coil of 200 turns links a flux that falls uniformly from 4 mWb to zero in 0.1 s. The induced emf is
Answer: D
The rate of change of flux is \( 4 \times 10^{-3}/0.1 = 0.04 \) Wb/s, and multiplying by two hundred turns gives eight volts. The sign is settled by Lenz's law and describes the sense in which the induced current flows.
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A conductor 0.4 m long moves at 15 m/s perpendicular to a field of 0.5 T. The emf developed is
Answer: A
The product is \( 0.5 \times 0.4 \times 15 = 3 \) V. Reversing either the motion or the field reverses the emf, and reversing both leaves it unchanged, which is worth remembering when tracing the polarity of a generator.
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A capacitor of 1 \( \mu \)F carries an alternating voltage rising at \( 10^{6} \) V/s. The displacement current through its dielectric is
Answer: B
The current is the capacitance times the rate of change of voltage, that is \( 10^{-6} \times 10^{6} = 1 \) A. It is carried entirely by the changing electric flux, no charge crossing the dielectric at all, which is the whole point of the displacement term.
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In copper of conductivity \( 5.8 \times 10^{7} \) S/m at 50 Hz, the displacement current compared with the conduction current is
Answer: C
Correct answer: utterly negligible, smaller by many orders of magnitude
The ratio of displacement to conduction current is \( \omega\epsilon/\sigma \), which for copper at 50 Hz is \[ \frac{2\pi\times50\times8.85\times10^{-12}}{5.8\times10^{7}} \approx 5\times10^{-17}. \] Copper remains a good conductor by this measure to well beyond optical frequencies, which is why the displacement current is dropped without hesitation in any metal.
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The speed of an electromagnetic wave in a medium of relative permittivity 4 and unit relative permeability is about
Answer: D
Correct answer: \( 1.5 \times 10^{8} \) m/s
\[ v = \frac{c}{\sqrt{\epsilon_r\mu_r}} = \frac{3\times10^{8}}{\sqrt{4}} = 1.5\times10^{8}\ \mathrm{m/s}. \] The square root is the step to watch: dividing by 4 gives the \( 7.5\times10^{7} \) offered. The wavelength is halved in the same proportion, which is why a dielectric-loaded antenna or line is physically shorter.
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A square loop of side 0.2 m lies in a field increasing at 0.5 T/s normal to it. The emf induced is
Answer: A
Correct answer: 20 mV
\[ \mathrm{emf} = A\frac{\mathrm{d}B}{\mathrm{d}t} = (0.2)^{2} \times 0.5 = 0.04 \times 0.5 = 0.02\ \mathrm{V} = 20\ \mathrm{mV}. \] The area is the square of the side, and using the side itself gives 100 mV, which is offered here. The field is normal to the loop, so the whole of it links.
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