Physics - EMI/AC Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Here, }L=8\mu \mathrm{H}=8\times 10^{-6}\mathrm{H}\text{; }\\ & C=0.02\mu \mathrm{F}=0.02\times 10^{-6}\mathrm{F}\\ & \therefore \text{ Resonant frequency, }\\ & f_r=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{8\times 10^{-6}\times 0.02\times 10^{-6}}}\\ & =3.98\times 10^5\mathrm{Hz}\\ & \end{array}$$ If $c\left(=3\times 10^8\mathrm{m}{\mathrm{s}}^{-1}\right)$ is the velocity of the electromagnetic wave, then, Wavelength, $\lambda =\frac{c}{f}=\frac{3\times 10^8}{3.98\times 10^5}=7.54\times 10^2\mathrm{m}$