A transmitting station releases waves of wavelength

Question:

A transmitting station releases waves of wavelength $960 \mathrm{~m}$. A capacitor of $2.56 \mu \mathrm{F}$ is used in the resonant circuit. The self inductance of coil necessary for resonance is__________ $\times 10^{-8} \mathrm{H}$.

Solution:

$\lambda=960 \mathrm{~m}$

$\mathrm{C}=2.56 \mu \mathrm{F}=2.56 \times 10^{-6} \mathrm{~F}$

$\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

$L=?$

Now at resonance, $\omega_{0}=\frac{1}{\sqrt{\mathrm{LC}}}$

[Resoant frequency]

$2 \pi \mathrm{f}_{0}=\frac{1}{\sqrt{\mathrm{LC}}}$

On substituting $f_{0}=\frac{\mathrm{c}}{\lambda}$, we have $2 \pi \frac{\mathrm{c}}{\lambda}=\frac{1}{\sqrt{\mathrm{LC}}}$

Squaring both sides : $4 \pi^{2} \frac{\mathrm{c}^{2}}{\lambda^{2}}=\frac{1}{\mathrm{LC}}$

$=\frac{4 \times 10 \times\left(3 \times 10^{8}\right)^{2}}{(960)^{2}}=\frac{1}{L \times 2.56 \times 10^{-6}}$

$\Rightarrow \frac{1}{L}=\frac{4 \times 10 \times 9 \times 10^{16} \times 2.56 \times 10^{-6}}{960 \times 960}$

$\Rightarrow \mathrm{L}=10 \times 10^{-8} \mathrm{H}$