Radiation from a Dipole Questions and Answers

Answer: A

Because the field takes time to travel, the potential at a point now depends on the source as it was one propagation time earlier, and writing the source at that retarded time is what turns the static potential integrals into radiating ones. Every radiation result follows from this single modification. Ohmic loss, transient decay and reactive phase lag are all real effects but none of them is what the word retarded refers to here.

Answer: B

The Hertzian or elementary dipole is an idealised element so short that the current may be taken as constant along it, and it serves as the building block from which any wire antenna is assembled by integration. Real short antennas have a current tapering to zero at the ends, which is why their effective length is half the physical length. A half wave dipole, an end fed wire and a crossed pair are all specific practical antennas rather than this abstraction.

Answer: C

The exact solution contains terms in one over r, one over r squared and one over r cubed, and only the first survives to make up the far field. The inverse square term is the induction field and the inverse cube term the electrostatic field, both of which dominate close in and vanish quickly. The distance at which the radiation term overtakes the others is a fraction of a wavelength, which is why the reactive near field is so shallow in wavelengths even for a large antenna.

Answer: D

Only the inverse first power term gives a Poynting vector falling as the inverse square of range, so that the power through a sphere is independent of its radius, which is exactly what radiation means. The other terms carry energy outward and back again within a cycle and transport no net power. That is the cleanest way to distinguish radiation from stored energy, and it explains why the far field is defined by the dominance of a single term, whatever the antenna.

Answer: A

The Poynting vector is the cross product of the electric and magnetic fields and gives the instantaneous power crossing unit area, in watts per square metre. Integrating its time average over a closed surface gives the radiated power, which is how radiation resistance is derived. It points in the direction of propagation, perpendicular to both fields. Energy density and intrinsic impedance are related quantities but are computed differently and mean different things.

Answer: B

Substituting the length ratio directly:

\[ R_{r} = 80\pi^{2}(0.05)^{2} \approx 1.97 \]

giving about 1.97 ohms. Against a conductor and matching loss of even one ohm the efficiency would be barely two thirds, and at a tenth of this length the resistance falls a hundredfold. The square law is the quantitative reason electrically small antennas are inefficient, and it is worth being able to evaluate quickly.

Answer: C

Radiation resistance rises as the square of the length measured in wavelengths, so halving a short antenna quarters its radiation resistance while its loss resistance falls barely at all. That is the whole difficulty of small antennas expressed in one relation. The same square law means that raising the frequency of a fixed physical antenna improves its radiation resistance rapidly, which is why a whip that is hopeless at one frequency can be efficient an octave higher.

Answer: D

The standard thin wire approximation takes a sinusoidal standing wave of current that must vanish at the open ends, since no current can flow into free space there. For a half wave dipole this puts the maximum at the centre, which is why centre feeding gives a convenient impedance. A uniform distribution belongs to the idealised Hertzian element, and the other two shapes describe no physical wire antenna. The assumption is remarkably accurate for thin conductors.

Answer: A

Continuity requires that current vanish where the conductor ends, because there is nowhere for the charge to go. Charge accumulates instead, so the voltage is indeed a maximum there, but that is a consequence rather than the cause. Radiation resistance is a terminal quantity for the whole antenna and not a function of position, and the taper of the wire is irrelevant. The boundary condition at the ends is what forces the standing wave pattern along the whole element.

Answer: B

Setting theta to zero makes the cosine argument pi over two, whose cosine is zero, and although the denominator also vanishes the limit is zero:

\[ \lim_{\theta \to 0} F(\theta) = 0 \]

so there is a null along the wire. The pattern is maximum broadside at ninety degrees, where the function equals one. Every linear current distribution shares this axial null, which is why a vertical dipole radiates nothing straight up.