Derive the expression for the skin depth $\delta$ of a good conductor and calculate $\delta$ for aluminium (conductivity $\sigma =3.5\times 10^7{\text{S m}}^{-1}$, relative permeability ${\mu }_r=1$) at a frequency of $1\text{MHz}$. Express your answer in metres and in micrometres.

Step-by-step solution

1. Start from the attenuation constant $\alpha =\sqrt{\frac{\pi f\mu \sigma }{2}}$ for a good conductor; the skin depth is defined as $\delta =\frac{1}{\alpha }$.
2. Rewrite $\delta$ as $\delta =\sqrt{\frac{2}{\omega \mu \sigma }}$, where $\omega =2\pi f$.
3. Calculate the angular frequency: $\omega =2\pi \times 1\times 10^6{\text{rad s}}^{-1}=6.283\times 10^6{\text{rad s}}^{-1}$.
4. Compute the absolute permeability: $\mu ={\mu }_0{\mu }_r=(4\pi \times 10^{-7}{\text{H m}}^{-1})\times 1=1.2566\times 10^{-6}{\text{H m}}^{-1}$.
5. Substitute $\omega$, $\mu$, and $\sigma =3.5\times 10^7{\text{S m}}^{-1}$ into the formula: $\delta =\sqrt{\frac{2}{(6.283\times 10^6)(1.2566\times 10^{-6})(3.5\times 10^7)}}$.
6. First evaluate the denominator: $(6.283\times 10^6)(1.2566\times 10^{-6})=7.90$, then multiply by $3.5\times 10^7$ to get $2.76\times 10^8$.
7. Now compute $2/2.76\times 10^8=7.25\times 10^{-9}$.
8. Take the square root: $\sqrt{7.25\times 10^{-9}}=8.5\times 10^{-5}\text{m}$.
9. Convert to micrometres: $8.5\times 10^{-5}\text{m}=8.5\times 10^{-5}\times 10^6\mu \text{m}=85\mu \text{m}$.
10. State the skin depth in both units.

Concept tested: Skin depth in a good conductor

Published 8 September 2026 · How this solution was prepared and checked