Question 8 of 8: Magnetic field of a monopole above a conducting ground plane

Question 8: Magnetic field of a monopole above a conducting ground 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 monopole standing on a conducting plane, with one measured far-field power density along the plane to calibrate the radiation.

Find. The magnitude and the direction of $\bar{H}$ at the target point.

Approach. Recognise the far field of a short vertical element over a conducting plane: $H_\phi\propto\sin\theta/r$ with $\theta$ measured from the vertical axis. Convert the reference power density to $H$, then scale by the range ratio and the pattern factor — no antenna constants are needed.

1. Field structure of the radiator. With the conducting plane replaced by the in-phase image, the pair radiates into the upper half space as a short dipole: the only far-field components are $E_\theta$ and $H_\phi$, related by $E_\theta=\eta_0H_\phi$, and both vary as $$\left|H_\phi\right|\propto\frac{\sin\theta}{r},\qquad \theta\ \text{measured from the vertical element axis}.$$ At $\lambda=c/f=30\ \text{m}$, a range of 2 km is more than 60 wavelengths, so the far-field form applies at both points.
2. Calibrate at the reference point. On the ground plane $\theta=90^{\circ}$ and $\sin\theta=1$, the pattern maximum. From the given power density, $$H_1=\sqrt{\frac{S_1}{\eta_0}}=\sqrt{\frac{2.62\times10^{-13}}{377}}=2.636\times10^{-8}\ \text{A/m (rms)},$$ equivalently $E_1=\sqrt{S_1\eta_0}=9.94\ \mu\text{V/m}$.
3. Pattern factor at the target. An elevation of $30^{\circ}$ above the plane means $\theta=90^{\circ}-30^{\circ}=60^{\circ}$, so $\sin\theta=0.8660$. The range shrinks from 10 km to 2 km, a factor of 5 increase in field.
4. Scale to the target point. $$H_2=H_1\left(\frac{r_1}{r_2}\right)\sin\theta=\left(2.636\times10^{-8}\right)(5)(0.8660)$$ $$\Rightarrow\;\boxed{\left|\bar{H}\right|=1.142\times10^{-7}\ \text{A/m}=114.2\ \text{nA/m (rms)}} ,$$ i.e. 161.5 nA/m peak, with $E_\theta=\eta_0H_\phi=43.0\ \mu\text{V/m}$ and a power density of $4.91\ \text{pW/m}^{2}$.
5. Direction of the vector. $\bar{H}$ lies along $\hat{\phi}$: it is horizontal, tangent to the horizontal circle of radius $r\sin\theta$ centred on the element's axis, and therefore perpendicular to the vertical plane containing the element and the observation point. Its sense is fixed by $\hat{\theta}\times\hat{\phi}=\hat{r}$, so that $\mathbf{E}\times\mathbf{H}$ points away from the antenna: looking down on the antenna from above, $\bar{H}$ circulates counter-clockwise about the element when the current flows upward.
6. Independent check from first principles. Using the image-doubled short-element result $E_\theta=\eta_0kIh\sin\theta/(2\pi r)$ with $k=2\pi/\lambda=0.2094\ \text{rad/m}$, the reference reading implies $I_{rms}=7.91\ \text{mA}$. Recomputing at $r=2\ \text{km}$, $\theta=60^{\circ}$ returns $43.0\ \mu\text{V/m}$, confirming the scaling. The radiated power follows two ways and agrees: $I_{rms}^{2}R_r$ with $R_r=160\pi^{2}(h/\lambda)^{2}=1.755\ \Omega$ gives $110\ \mu\text{W}$, as does the hemispherical integral $(4\pi/3)S_1r_1^{2}$.

The answer needs nothing about the element beyond its pattern shape, because one measured field value calibrates the whole product of current, length and frequency. That is the practical value of quoting a reference field strength at a reference distance, as broadcast licences do. Note also where the maximum of this antenna lies: a vertical element has a null along its own axis, so the strongest signal grazes along the ground plane at every azimuth and the $30^{\circ}$ elevation point is already 1.25 dB down on the pattern peak. The image is what makes the ground plane useful — it doubles the field above the plane and doubles the radiation resistance, which is why a quarter-wave monopole behaves like a half-wave dipole with half the radiation resistance.