An L-C resonant circuit contains a 400 pF ca | vedclass

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(A) Given: Capacitance,$C = 400 \ pF = 400 	imes 10^{-12} \ F$. Inductance,$L = 400 \ \mu H = 400 	imes 10^{-6} \ H$. The frequency of the resonatin

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$754 \ m$

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(A) Given: Capacitance,$C = 400 \ pF = 400 	imes 10^{-12} \ F$. Inductance,$L = 400 \ \mu H = 400 	imes 10^{-6} \ H$. The frequency of the resonating $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$. The wavelength $\lambda$ of the radiated electromagnetic wave is related to the frequency by $\lambda = \frac{c}{f}$,where $c = 3 	imes 10^8 \ m/s$ is the speed of light. Substituting the expression for $f$,we get $\lambda = c 	imes 2 \pi \sqrt{LC}$. Substituting the values: $\lambda = 3 	imes 10^8 	imes 2 	imes 3.14 	imes \sqrt{400 	imes 10^{-12} 	imes 400 	imes 10^{-6}}$. $\lambda = 3 	imes 10^8 	imes 6.28 	imes \sqrt{160000 	imes 10^{-18}}$. $\lambda = 3 	imes 10^8 	imes 6.28 	imes 400 	imes 10^{-9}$. $\lambda = 3 	imes 6.28 	imes 400 	imes 10^{-1} = 753.6 \ m \approx 754 \ m$.

An $L-C$ resonant circuit contains a $400 \ pF$ capacitor and an inductor of $400 \ \mu H$. It is coupled to an antenna. The wavelength of the radiated electromagnetic wave is:

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