Topic: Electromagnetics
A steel pipe is constructed of a material for which $\mu_{r}=200$ and $\sigma=5\times10^{6}\text{mho/m}$. The outer and inner radii are 8 and 6 mm respectively and the length is 80 m. If the total current carried by the pipe is $2\cos(10^{4}\pi t)\text{ A}$, then the skin depth will be
- A. $0.225\times10^{-3}\text{m}$
- B. $0.300\times10^{-3}\text{m}$
- C. $0.352\times10^{-3}\text{m}$
- D. $0.125\times10^{-3}\text{m}$
Answer: A. $0.225\times10^{-3}\text{m}$
Explanation: From the current expression, $\omega = 10^4\pi$, so $f = \frac{10^4\pi}{2\pi} = 5 \times 10^3~\text{Hz}$.
Skin depth $\delta = \frac{1}{\sqrt{\pi f \mu \sigma}} = \frac{1}{\sqrt{\pi \times (5 \times 10^3) \times (200 \times 4\pi \times 10^{-7}) \times (5 \times 10^6)}}$.
$\delta = \frac{1}{\sqrt{20\pi² \times 10^3}} \approx 0.225 \times 10^{-3}~\text{m}$.