Skin depth in copper at 1 MHz
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A plane wave at $1\ \text{MHz}$ is incident on copper, for which $\sigma = 5.8 \times 10^{7}\ \text{S/m}$ and $\mu = \mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$.
The skin depth, in micrometres, is ________.
Show the step-by-step derivation
Answer
66 $\mu$m (accept 65 to 67) $\mu$m (accept 65 to 67)
Step-by-step derivation
- Copper is an excellent conductor at $1$ MHz ($\sigma / \omega\varepsilon \ggg 1$), so the good-conductor approximation applies and the skin depth is $$\delta = \frac{1}{\sqrt{\pi f \mu \sigma}}.$$
- Build the product inside the radical one factor at a time: $$\pi f = 3.1416 \times 10^{6},$$ $$\pi f \mu = (3.1416\times 10^{6})(1.2566\times 10^{-6}) = 3.9478.$$
- Multiply by the conductivity: $$\pi f \mu \sigma = 3.9478 \times (5.8 \times 10^{7}) = 2.2897 \times 10^{8}.$$
- Take the square root: $$\sqrt{2.2897 \times 10^{8}} = 1.5132 \times 10^{4}.$$
- Invert: $$\delta = \frac{1}{1.5132 \times 10^{4}} = 6.609 \times 10^{-5}\ \text{m} = \mathbf{66.1\ \mu m}.$$
- The scaling that matters. $\delta \propto 1/\sqrt{f}$, so a hundredfold increase in frequency shrinks the skin depth tenfold. At $100$ MHz copper conducts in a $6.6\ \mu$m shell, which is why RF conductors are plated rather than solid, and why a PCB trace's AC resistance rises with the square root of frequency.
The trap this question is built around
Dropping the $\pi$, or using $\delta = \sqrt{2/(\omega\mu\sigma)}$ without noticing it is the same formula. Both forms are correct and equal, since $\omega = 2\pi f$ makes $2/(\omega\mu\sigma) = 1/(\pi f\mu\sigma)$ - but mixing halves of the two is not.