Physics - EMI/AC Question with Solution | TestHub

The frequency given is $0.5$ kilo cycle per second $\Rightarrow f=500 \mathrm{Hz}\Rightarrow \omega =2\pi f=1000\pi \mathrm{rad} {\mathrm{s}}^{-1}$.

Total capacitance of the wire is $C=0.01\times {10}^{-6}\times 100 \mathrm{F}$.

Impedance is minimum at resonance. Therefore, for minimum impedance

$X_L=X_C$

$$\begin{gathered} \Rightarrow \omega L=\frac{1}{\omega C} \\ \Rightarrow L=\frac{1}{{\omega }^{2}C}=\frac{1}{{1000}^2\times {\pi }^2\times 0.01\times {10}^{-6}\times 100}={10}^{-1} \mathrm{H}=100 \mathrm{mH} \end{gathered}$$