Topic: Electromagnetics
If the maximum effective aperture of a radiating small loop of constant current is $\frac{3}{(32\pi)}\text{ m}^{2}$, the wavelength of the radiated electromagnetic wave is
- A. 5000 mm
- B. 100 mm
- C. 1000 mm
- D. 500 mm
Answer: D. 500 mm
Explanation: Using $A_e = \frac{\lambda²}{4\pi}G$.
Also, $G = 1.5$ for a small loop antenna.
$\lambda = \sqrt{\frac{4\pi A_e}{G}} = \sqrt{\frac{4\pi \times 3}{32\pi \times 1.5}} = \sqrt{\frac{1}{4}} = 0.5\text{ m} = 500\text{ mm}$.