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A coil of inductance $8 \mu \mathrm{H}$ is connected to a capacitor of capacitance $0.02 \mu \mathrm{F}$. To what wavelength is this circuit tuned?
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Verified Answer
The correct answer is:
$7.54 \times 10^2 \mathrm{~m}$
$$
\begin{aligned}
& \text { Here, } L=8 \mu \mathrm{H}=8 \times 10^{-6} \mathrm{H} \text {; } \\
& C=0.02 \mu \mathrm{F}=0.02 \times 10^{-6} \mathrm{~F} \\
& \therefore \text { Resonant frequency, } \\
& f_r=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{8 \times 10^{-6} \times 0.02 \times 10^{-6}}} \\
& =3.98 \times 10^5 \mathrm{~Hz} \\
&
\end{aligned}
$$
If $c\left(=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)$ is the velocity of the electromagnetic wave, then,
Wavelength, $\lambda=\frac{c}{f}=\frac{3 \times 10^8}{3.98 \times 10^5}=7.54 \times 10^2 \mathrm{~m}$
\begin{aligned}
& \text { Here, } L=8 \mu \mathrm{H}=8 \times 10^{-6} \mathrm{H} \text {; } \\
& C=0.02 \mu \mathrm{F}=0.02 \times 10^{-6} \mathrm{~F} \\
& \therefore \text { Resonant frequency, } \\
& f_r=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{8 \times 10^{-6} \times 0.02 \times 10^{-6}}} \\
& =3.98 \times 10^5 \mathrm{~Hz} \\
&
\end{aligned}
$$
If $c\left(=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)$ is the velocity of the electromagnetic wave, then,
Wavelength, $\lambda=\frac{c}{f}=\frac{3 \times 10^8}{3.98 \times 10^5}=7.54 \times 10^2 \mathrm{~m}$
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