GATE-2018 (24) :Wyllie – Time average equation

 22. Skin Depth in homogeneous media of resistivity $\rho_{1}$ and $\rho_{2}$ are 100 m and 200 m respectively, at 1000Hz frequency. The ratio $\frac{\rho_{1}}{\rho_{2}}$ will be __________.

 (Thanks to Pragnath, AU)

Solution:

Skin depth formula  $D = 503.8 * \sqrt{(\frac{\rho}{f})}$

Given that,

 $D_{1}= 100m~,D_{2}= 200 m$;         $f=1000~Hz$

$\frac{D_{1} = 503.8 * \sqrt{(\frac{\rho_{1}}{f})}}{D_{2} = 503.8 * \sqrt{(\frac{\rho_{2}}{f})}}$

$\frac{100}{200}=\frac{503.8 * \sqrt{(\frac{\rho_{1}}{1000})}}{ 503.8 * \sqrt{(\frac{\rho_{2}}{1000})}}$

$\frac{1}{2}=\frac{\sqrt{(\frac{\rho_{1}}{1000})}}{\sqrt{(\frac{\rho_{2}}{1000})}}$

$\frac{1}{2}=\frac{\sqrt{(\rho_{1}}}{\sqrt{(\rho_{2}})}$

Squaring on both sides we get

$\frac{1}{4}=\frac{{\rho_{1}}}{\rho_{2}}$

$\frac{{\rho_{1}}}{\rho_{2}}=0.25$