Question E9F06:
What is the approximate physical length of an air-insulated, parallel conductor transmission line that is electrically 1/2 wavelength long at 14.10 MHz?  

Choices:  
A: 7.0 meters  
B: 8.5 meters  
C: 10.6 meters  
D: 13.3 meters    

Correct Answer is  C: 10.6 meters  

To find the physical length of a line that is electrically 1/2 wavelength, use the line’s velocity factor (VF). For an air-insulated parallel conductor line, VF ≈ 1.0 (wave travels at nearly the speed of light).

Useful formulas:
- Free-space wavelength: $$\lambda = \frac{300}{f_{\text{MHz}}}\ \text{meters}$$
- Half-wave physical length in a line: $$L_{\frac{1}{2}} = \frac{150 \cdot \text{VF}}{f_{\text{MHz}}}\ \text{meters}$$

Compute at 14.10 MHz with VF ≈ 1:
$$L_{\frac{1}{2}} = \frac{150 \cdot 1}{14.10} \approx 10.64\ \text{m} \Rightarrow 10.6\ \text{m}$$

Why the correct answer (C: 10.6 meters) is correct:
- It matches the half-wavelength in an air-insulated line where VF ≈ 1.0, so the physical length is essentially the free-space half-wavelength.

Why the other choices are incorrect (implied VF shown):
- A: 7.0 m → VF ≈ 7.0 / 10.64 ≈ 0.66 (typical of solid polyethylene coax, not air-insulated open wire)
- B: 8.5 m → VF ≈ 8.5 / 10.64 ≈ 0.80 (typical of plastic-insulated twinlead, not air)
- D: 13.3 m → VF ≈ 13.3 / 10.64 ≈ 1.25 (impossible; VF cannot exceed 1)

Rule of thumb:
- In meters: half-wave length ≈ 150/f(MHz), then multiply by VF.
- For true air dielectric lines, use VF ≈ 1.0.