Derive the skin depth $\delta$ for a good conductor and use it to find $\delta$ for copper (conductivity $\sigma =5.96\times 10^7{\text{S m}}^{-1}$) at a frequency of $50\text{MHz}$. Express your answer in metres and micrometres.

Step-by-step solution

1. Start from the general expression for skin depth in a conductor: $\delta =\sqrt{\frac{2}{\omega \mu \sigma }}$.
2. For a good conductor $\mu \approx {\mu }_0=4\pi \times 10^{-7}{\text{H m}}^{-1}$.
3. Compute angular frequency $\omega =2\pi f=2\pi (50\times 10^6)=3.14\times 10^8{\text{rad s}}^{-1}$.
4. Insert the values into the formula: $\delta =\sqrt{\frac{2}{(3.14\times 10^8)(4\pi \times 10^{-7})(5.96\times 10^7)}}$.
5. Evaluate the denominator: $(3.14\times 10^8)(4\pi \times 10^{-7})(5.96\times 10^7)\approx 2.35\times 10^{10}$.
6. Thus $2/\text{denominator}\approx 8.5\times 10^{-11}$.
7. Take the square root: $\delta \approx \sqrt{8.5\times 10^{-11}}=9.2\times 10^{-6}\text{m}$.
8. Convert to micrometres: $9.2\times 10^{-6}\text{m}=9.2\mu \text{m}$.

Concept tested: Skin depth derivation and calculation

Published 11 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked