Question 8 of 8: Where the Radiation Field Overtakes the Biot–Savart Field
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2018 — 16-Elec-A7, Electromagnetics. Closed book, three hours, approved Casio or Sharp calculator only. Eight questions, all of equal value; the paper states that any five constitute a complete paper, so each question carries 20 of the 100 marks that are actually graded. All eight are solved here, because this set is a study resource rather than an examination script.
Aids given on the paper. $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, $e = 1.6\times10^{-19}\ \text{C}$. Those two constants fix $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.9986\times10^{8}\ \text{m/s}$ and $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.82\ \Omega$ — which is why Question 1 quotes a 377 Ω line: it is free space itself, rendered as a cable.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, motional EMF, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields and the reactive near-field boundary).
Constants used throughout: the paper's own aid list is taken as authoritative, so $c = 2.9986\times10^{8}$ m/s and $\eta_0 = 376.82\ \Omega$. Question 1 supplies its own propagation velocity ($3\times10^{8}$ m/s) and that stated value is used there. Question 5 is the one place where the difference between $2.9986\times10^{8}$ and $3.00\times10^{8}$ changes the wording of the answer, and both readings are worked through explicitly.
Question 8: Where the Radiation Field Overtakes the Biot–Savart Field (20 marks)
Given. One short element, and two competing expressions for the field it produces.
Given data
Quantity
Symbol
Value
Element length
$\ell$
$\ell \ll \lambda$
Element current
$I$
uniform along $\ell$
Orientation
—
vertical
Observation direction
$\Theta$
horizontal, i.e. $90^\circ$ from the element axis
Supplied aid
$|E_{rad}|$
$Z_0\,\ell I\sin\Theta\,k/4\pi r$
Wavenumber
$k$
$2\pi/\lambda$
Find. The horizontal distance at which the radiation magnetic field equals the element’s Biot–Savart (induction) field.
The element's two field terms plotted against distance. The induction term falls twice as fast on a log scale, so it loses to the radiation term beyond a single crossover point.
Approach. Convert the supplied radiation electric field into a magnetic field by dividing by $Z_0$, write the static Biot–Savart field of the same element, set $\Theta = 90^\circ$ for the horizontal direction, and equate.
Fix the direction first. The element is vertical, so the horizontal direction is broadside to it: $\Theta = 90^\circ$ and $\sin\Theta = 1$. That is the pattern maximum for a short element, and it makes both field expressions take their largest values — which is why the same factor $\sin\Theta$ appears in each and cancels.
Radiation magnetic field. In the far field $E$ and $H$ are related by the free-space impedance, so dividing the supplied aid by $Z_0$ gives $$|H_{\text{rad}}| = \frac{|E_{\text{rad}}|}{Z_0} = \frac{\ell I \sin\Theta\,k}{4\pi r} = \frac{\ell I k}{4\pi r} \quad (\Theta = 90^\circ).$$ It falls as $1/r$.
Biot–Savart field of the same element. Treating the element as a short current filament $I\ell$, the static law gives $$|H_{\text{BS}}| = \frac{\ell I \sin\Theta}{4\pi r^{2}} = \frac{\ell I}{4\pi r^{2}} \quad (\Theta = 90^\circ).$$ It falls as $1/r^{2}$ — twice as fast — so it dominates close in and loses far out.
Equate the two. $$\frac{\ell I k}{4\pi r} = \frac{\ell I}{4\pi r^{2}} \;\Longrightarrow\; \frac{k}{r} = \frac{1}{r^{2}} \;\Longrightarrow\; kr = 1 .$$ Everything about the element — its length, its current, even the constant $Z_0$ — cancels. Only the wavenumber survives.
Solve for the distance. With $k = 2\pi/\lambda$, $$r = \frac{1}{k} \;\Rightarrow\; \boxed{r = \frac{\lambda}{2\pi} = 0.159\,\lambda} .$$ As an illustration, a 100 MHz element ($\lambda = 3.00$ m) has its crossover at $r = 0.477$ m; a 1 GHz element at 4.77 cm.
Interpret the result. $kr = 1$ is the classical boundary of the reactive near field. Inside it the element behaves like a piece of DC wiring: the field is stored, in quadrature with the current, and returns to the source each cycle. Outside it the $1/r$ radiation term takes over and the energy leaves for good. Writing the exact short-dipole result $$H_\phi = \frac{I\ell\sin\Theta}{4\pi}\left(\frac{jk}{r} + \frac{1}{r^{2}}\right)e^{-jkr}$$ shows the two terms side by side and confirms that they are equal in magnitude exactly where $kr = 1$ — and, being in quadrature, that the total there is $\sqrt{2}$ times either one alone.
