# Skin effect calculator

Calculate the penetration depth for various materials, and find wire diameter for a given current density.

 frequency: Hz conductivity: $$\mathrm{\frac{S}{m}}$$ Current density: $$\mathrm{\frac{A}{mm^2}}$$ Current: A

Conductivity of typical materials:

 Copper 58 $$\mathrm{\frac{MS}{m}}$$ Brass 16 $$\mathrm{\frac{MS}{m}}$$ Aluminium 37 $$\mathrm{\frac{MS}{m}}$$ Silver 62 $$\mathrm{\frac{MS}{m}}$$ Gold 44 $$\mathrm{\frac{MS}{m}}$$
 skin depth $$\delta$$: $$\mu\mathrm{m}$$ Wire diameter D: mm

# Remarks, and: How does it work?

The calculator assumes nonmagnetic conductors, and in this case, the skin depth $$\delta$$ is calculated as follows:

$$\delta = \sqrt{\frac{2}{\omega\,\mu_0\,\sigma}} = \sqrt{\frac{1}{\pi\,f\,\mu_0\,\sigma}}$$

At a depth of one skin depth, the current density has decreased by 63%; therefore, after approximately 5 skin depths, the current has decreased almost to zero. This is important to find the diameter of the conductor. Assume the conductor has a radius $$r$$. Then, there are two distinct cases:

## Case 1

$$5\,\delta \geq r$$: The current flows in the whole conductor, and therefore, the current density can be found by $$J = \frac{I}{A}$$.

## Case 2

$$5\,\delta \lt r$$: The conductor diameter is large compared to the skin depth, and therefore, the current only flows in an annular ring. The outer diameter $$D$$ is the conductor diameter, while the inner diameter $$d = D – 2\cdot 5\,\delta$$. The surface area of the annular ring is therefore
$$A_\mathrm{ring} = \frac{\pi}{4} \left(D^2 – \left(D-10\,\delta\right)^2\right)$$ and the current density is then $$J=\frac{I}{A_\mathrm{ring}}$$.