Discussion
A broadcasting station transmits on a wavelength of 3000m. A capacitor of 0.0005 µF is connected in parallel to a variable inductor to form a closed oscillating circuit. The approximate value of inductor that must be adjusted in order to tune into the particular station is ________.
Answer: B
The inductor must be set to about 5 mH.
The station's frequency follows from its wavelength, and the tuned circuit resonates when \(f = 1/(2\pi\sqrt{LC})\).
\[ \begin{aligned} f &= \frac{c}{\lambda} = \frac{3 \times 10^{8}}{3000} = 10^{5}\ \text{Hz} \\ L &= \frac{1}{4\pi^{2} f^{2} C} \end{aligned} \]
\[ \begin{aligned} L &= \frac{1}{39.48 \times 10^{10} \times 5 \times 10^{-10}} \\ &= \frac{1}{197.4} = 5.07 \times 10^{-3}\ \text{H} \end{aligned} \]
That is about 5 mH. The other values would tune the circuit to different wavelengths: a larger inductance gives a lower frequency and a longer wavelength.
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