Question 8 of 8: Magnetic Field of a Short Vertical Radiating Element

Question 8: Magnetic Field of a Short Vertical Radiating Element (equal value)

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. Short (Hertzian) vertical dipole, $f=20$ MHz; power density $S(r_1{=}10\text{ km},\,\theta{=}90^\circ)=10^{-8}$ W/m² in the horizontal plane through the element (i.e. broadside, $\theta=90^\circ$ from the vertical axis); observation point at $r_2=5$ km, $30^\circ$ elevation above the horizontal (so $\theta=60^\circ$ from the vertical axis), due East of the element.

Find. Magnitude and direction of $H$ at the observation point.

Approach. A short vertical dipole's far-field power density follows $S(r,\theta)=S_{max}(r_1/r)^2\sin^2\theta$ referenced to the given broadside value; scale from $(r_1,90^\circ)$ to $(r_2,60^\circ)$, then get $H$ from the plane-wave relation $S=\eta_0H^2$ in the far field. The field direction follows the dipole's radiation pattern: $\mathbf H$ is purely azimuthal ($\hat\phi$, circling the vertical axis).

1. Scale the power density to the new point. For a Hertzian dipole, $S\propto\sin^2\theta/r^2$, so $$S_2=S_1\left(\frac{r_1}{r_2}\right)^2\frac{\sin^2(60^\circ)}{\sin^2(90^\circ)}=(10^{-8})\left(\frac{10}{5}\right)^2(0.75)=\boxed{3\times10^{-8}\ \text{W/m}^2}.$$
2. Magnetic field magnitude. In the radiation (far) zone, $E$ and $H$ are in phase and related by the free-space wave impedance, so $S=\eta_0 H_{rms}^2$: $$H_{rms}=\sqrt{\frac{S_2}{\eta_0}}=\sqrt{\frac{3\times10^{-8}}{377}}=\boxed{8.92\ \mu\text{A/m}}.$$
3. Field direction. A vertical current element's far-zone magnetic field has only a $\hat\phi$ component (it circles the antenna axis, by symmetry of a current confined to the $\hat z$ direction); at a point due East of the element, $\hat\phi$ is horizontal and points due North (for current assumed flowing in $+\hat z$, by the right-hand rule) — i.e. $\mathbf H$ is horizontal, perpendicular to the vertical plane containing the antenna and the observation point, tangent to the circle of constant $\theta$ around the antenna axis.