Question

Calculate the wavelength of the radiowaves radiated out by a circuit containing $0.02\mu F$ capacitor and $8\mu H$ inductance in series.

• A. 703.8 m
• B. 460 m
• C. 398 m
• D. 753.8 m

Answer 1

Answer: (D)

$C=0.2\mu F=0.02\times 10^{-6}F,L=8\mu H$
$=8\times 10^{-6}H$
Thus, frequency of a circuit i.e., $v=\frac{1}{2\pi \sqrt{LC}}$
$=\frac{1}{2\times 3.14\sqrt{\left(8\times 10^{-6}\right)\left(0.02\times 10^{-6}\right)}}Hz$
$=\frac{1}{2\times 3.14\times 0.4}\times 10^{6}Hz$
$=3.98\times 10^{5}Hz$
Wavelength of electromagnetic wave,
$\lambda =\frac{c}{v}=\frac{3\times 10^8}{3.98\times 10^5}$
$=753.8m$