Question 8 of 8: Magnetic field of a short vertical radiator off the horizontal plane

Question 8: Magnetic field of a short vertical radiator off the horizontal plane (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A short vertical current element radiates into free space. One far-field power density is measured, at a known range and in the plane of maximum radiation, which is enough to calibrate the field everywhere else.

Find. The magnitude and the direction of the magnetic field at the target point.

Approach. Confirm the element is electrically short so that the $\sin\theta$ pattern applies, scale the reference power density by the pattern factor and the inverse-square law, then convert the resulting power density to a magnetic field.

1. Check that the short-element pattern applies. $\lambda = c/f = 15.0\ \text{m}$, so $h/\lambda = 0.5/15.0 = 0.0333 \ll 0.1$: the element is electrically short, its current is essentially uniform in phase, and its far field varies as $\sin\theta$ with $\theta$ measured from the element axis.
2. Convert elevation to polar angle. The axis is vertical, so an elevation $\psi = 30^{\circ}$ above the horizontal plane corresponds to $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$, and the pattern factor is $\sin^{2}\theta = 0.750$.
3. Scale the power density. Power density obeys $S \propto \sin^{2}\theta/r^{2}$, and the reference point lies in the plane of maximum radiation ($\theta = 90^{\circ}$, $\sin^{2}\theta = 1$), so $$S_2 = S_1\left(\frac{r_1}{r_2}\right)^{2}\sin^{2}\theta = 10^{-8}(2)^{2}(0.750) = \boxed{3.00\times10^{-8}\ \text{W/m}^{2}}$$
4. Convert to a magnetic field. In the far field the wave is locally TEM, so $$|H| = \sqrt{\frac{S_2}{\eta_0}} = \sqrt{\frac{3.00\times10^{-8}}{377}} = \boxed{8.921\ \mu\text{A/m}\ (\text{rms})}$$
5. Fix the direction. The far field of a vertical element has $\mathbf{E}$ along $\hat{\theta}$ — in the vertical plane containing the ray — and $\mathbf{H}$ along $\hat{\phi}$, which is horizontal and perpendicular to the ray at every point. For a ray running due east, $$\mathbf{H} \parallel \hat{\phi} = \boxed{\text{horizontal, along the north-south line}}$$

The corresponding electric field is $|E| = \eta_0|H| = 3.363\ \text{mV/m}$, and the two are in phase and mutually perpendicular, as they must be in a far field. Note that the magnetic field is horizontal even though the ray climbs at $30^{\circ}$: the azimuthal unit vector has no vertical component anywhere, so $\mathbf{H}$ stays parallel to the ground at every elevation while $\mathbf{E}$ tilts to remain perpendicular to the ray.

The element length and the frequency are, strictly, decoration. Because the measurement at 10 km already contains every antenna constant — current, effective length, radiation resistance — only the shape of the pattern and the inverse-square law are needed to move to another point. The two data are still worth checking, since they are what license the $\sin\theta$ pattern in the first place; had the element been a half wavelength long the pattern factor would have been different and the scaling would have failed.