The skin depth of copper at a frequency of 3 GHz is 1 micron (10^{-6} meter). At 12 GHz, for a non-magnetic conductor whose conductivity is 1/9 times that of copper, the skin depth would be
(a) \sqrt{9 \times 4} microns (b) \sqrt{9 \times 4} microns
(c) \sqrt{4 / 9} microns (d) 1 / \sqrt{9 \times 4} microns
We know that For a good conductor
\begin{aligned}&\text { Skin depth }=\delta=\frac{1}{\sqrt{\pi f \mu \sigma}} \\&\delta \propto \frac{1}{\sqrt{f \sigma}} \\&\frac{\delta_{2}}{\delta_{1}}=\sqrt{\frac{f_{1} \sigma_{1}}{f_{2} \sigma_{2}}}\end{aligned}Given that,
\begin{aligned}&\delta_{1}=1 \text { micron } \\&f_{1}=3 GHz \\&f_{2}=12 GHz \\&\frac{\sigma_{2}}{\sigma_{1}}=\frac{1}{9} \\&\frac{\delta_{2}}{1}=\sqrt{\frac{3}{12} \times \frac{9}{1}}=\sqrt{\frac{9}{4}} \\&\delta_{2}=\sqrt{\frac{9}{4}} \text { microns }\end{aligned}Hence, the correct option is (b)