Topic: Electromagnetics
A Gold cavity resonator resonating at 10 GHz is fully coated using $2~\mu m$ YBCO (Yttrium Barium Copper Oxide) material and operating at boiling temperature of liquid nitrogen. What is the Skin Depth of RF at this frequency? Assume $\sigma_{Au}=4e7~\text{Siemens/meter}$ at cavity operating temperature.
- A. $2~\mu m$
- B. $(2.5/\pi)\mu m$
- C. $0~\mu m$
- D. None of the above
Answer: B. $(2.5/\pi)\mu m$
Explanation: The skin depth is given by $\delta = \sqrt{\frac{2}{\omega \mu \sigma}}$. For Gold, $\omega = 2\pi \times 10^{10}$, $\mu = 4\pi \times 10^{-7}$, and $\sigma = 4 \times 10^7$. Plugging these in gives $\delta = \sqrt{\frac{2}{2\pi \times 10^{10} \times 4\pi \times 10^{-7} \times 4 \times 10^7}} = \frac{10^{-5}}{4\pi} = \frac{2.5}{\pi} \mu m$.