Question

A dielectric material has relative permittivity of 18 and loss tangent of ${10}^{-3}at100MHz$, then the distance over which the wave amplitude reduces to $\frac{1}{e}$ of its original amplitude is __________ .

• A. 112.5 cm
• B. 112. 5 m
• C. 225.08 m
• D. 32.8 m

Answer 1

The correct option is C 225.08 m

Skin depth $\left(\delta \right)=\frac{1}{\alpha }$

$\alpha =\left[\frac{{\omega }^2\mu ϵ}{2}[\sqrt{1+{\left(\frac{\sigma }{\omega ϵ}\right)}^2}-1]\right]$

$\alpha ={\left[\frac{{\omega }^2\mu ϵ}{2}(1+\frac{1}{2}\frac{{\sigma }^2}{{\omega }^2{ϵ}^2}-1)\right]}^{\frac{1}{2}}$

$\alpha ={\left[\frac{{\omega }^2\mu ϵ}{2}\left(\frac{{\sigma }^2}{2{\omega }^2{ϵ}^2}\right)\right]}^{\frac{1}{2}}$

$\alpha -\frac{\sigma }{2}\sqrt{\frac{{\mu }_c}{{ϵ}_2{ϵ}_2}}-4.44\times {10}^{4}Np/m$

$\delta =\frac{1}{\alpha }=225.08m$