Question

If a small single-turn loop antenna has a radiation resistance of 0.04 \(\Omega\), how many turns are needed to produce a radiation resistance of 1 \(\Omega\)?

• A. 5
• B. 10
• C. 25
• D. 50

Hint:

For a small loop antenna, the radiation resistance \(R_r\) is proportional to \(N^2 A^2 f^4\), where N is number of turns, A is area of one turn, f is frequency.
If area per turn and frequency are constant, \(R_r \propto N^2\).
So, \(R_{r2}/R_{r1} = (N_2/N_1)^2\).

Answer 1

The Correct Option is A

For a small loop antenna (dimensions much smaller than a wavelength), the radiation resistance \(R_r\) is proportional to the square of the number of turns (N) and the fourth power of the loop area (A) (or fourth power of radius for circular loop, square of area for any shape), and fourth power of frequency. More specifically, for a small loop of N turns with area A, the radiation resistance is given by: \[ R_r \approx K \cdot (NA)^2 \left(\frac{f}{c}\right)^4 \quad \text{or more commonly for fixed area and frequency} \quad R_r \propto N^2 \] If the area A of each turn and the frequency are kept constant, then the radiation resistance \(R_r\) is proportional to \(N^2\): \[ R_r \propto N^2 \] Let \(R_{r1}\) be the radiation resistance for \(N_1\) turns, and \(R_{r2}\) for \(N_2\) turns. Then \(\frac{R_{r2}}{R_{r1}} = \left(\frac{N_2}{N_1}\right)^2\). Given: For a single-turn loop, \(N_1 = 1\), \(R_{r1} = 0.04 \Omega\). We want to find \(N_2\) such that \(R_{r2} = 1 \Omega\). \[ \frac{1 \Omega}{0.04 \Omega} = \left(\frac{N_2}{1}\right)^2 \] \[ \frac{1}{0.04} = N_2^2 \] \[ N_2^2 = \frac{100}{4} = 25 \] \[ N_2 = \sqrt{25} = 5 \] So, 5 turns are needed. This matches option (a). \[ \boxed{5} \]