A radio wave of wavelength 300 m can be transmitted by a transmission center. A condenser of capacity \( 2.5\,\mu\mathrm{F} \) is available. The inductance of the required coil at resonance is
(Given: \( \pi^{2} = 10 \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.00\times10^{-8}\,\mathrm{H} \)
The circuit is tuned to the radio wave when its resonant frequency equals the frequency of the wave. For an LC circuit the resonant frequency is
\( \displaystyle f = \frac{1}{2\pi\sqrt{LC}} \)
and for the radio wave \( f = c/\lambda \).
\( \displaystyle f = \frac{c}{\lambda} = \frac{3\times10^{8}}{300} = 10^{6}\,\mathrm{Hz} \)
Squaring the resonance condition and solving for \( L \):
\( \displaystyle L = \frac{1}{4\pi^{2} f^{2} C} \)
\( C = 2.5\,\mu\mathrm{F} = 2.5\times10^{-6}\,\mathrm{F} \)
\( 4\pi^{2} = 40 \)
\( \displaystyle L = \frac{1}{40 \times (10^{6})^{2} \times 2.5\times10^{-6}} = \frac{1}{10^{8}} = 1.00\times10^{-8}\,\mathrm{H} \)