The skin depth effect in metals such as copper is usually thought of as a high frequency effect, but of course it depends what you mean by 'high'. What matters is the refractive index of the metal. Well above a certain frequency known as the 'plasma frequency' the refractive index is real, and the metal is transparent. Below that frequency the index has an imaginary component, and because of this an e-m wave will be attenuated as it travels through the metal. For copper the plasma frequency is about 1012Hz, so for transmission line problems we are only concerned with the 'low frequency' behaviour.
In the case of a twin wire transmission line carrying a signal there will be resistive losses in the wires, and to replace this loss energy must be supplied from the surrounding field. To accelerate the conduction electrons there must be a component of the electric field, E, parallel to the wire, and the magnetic field, B, is in closed loops round the cable. The direction of flow of energy is given by the 'Poynting Vector' which is proportional to the vector cross-product ExB and this has a component into the wire at its surface. To maintain a current distributed throughout the wire the energy must reach the centre of the wire with little attenuation, and it is the refractive index of the metal which determines the rate of attenuation.
A detailed classical analysis of the refractive index and its effect can be found in 'The Feynman Lectures on Physics, vol.2' by R.P.Feynman, (Addison-Wesley 1967) Chapter 32.
To summarise the effect, the refractive index at low frequencies is:
Refractive index, n = ( 1-i ).sqrt( k/f )
'sqrt' means the square root, k is a constant, f is frequency in Hz, and i is the square root of -1.
For copper the value of k is about 5 x 1017 sec.-1.
The electric field E in the conductor is given by E = Eo exp (iw(t - nx/c))
where w is angular frequency, Eo is the peak electric field component along the wire at its surface, t is time, and x is the distance into the metal.
The imaginary component of n multiplied by i gives a real exponential term, and so the field level falls exponentially inside the metal, and the skin depth is conventionally defined as the depth at which the field has fallen to e-1, about 0.37, of its original level. The value of n is a function of the frequency, and putting this into the equation for the electric field gives a value for the skin depth inversely proportional to the square root of the frequency. In other words, increasing the frequency by a factor of 4 reduces the skin depth by half.
The skin depth of copper at 20kHz is about 0.47mm, and for the resistance of the wire at 20kHz to be double the d.c. value the radius must be 1.4mm. (This applies to solid core wires, and for lower radius wires the skin depth gives a lower relative increase in resistance, and so is less significant.) The d.c. resistance per metre is about 0.0028 ohms, so the 20kHz resistance is 0.0056 ohms, half of this contributed by the skin effect.
To take an extreme example, suppose 10 metres of this wire is used in a speaker cable to drive a speaker with impedance as low as 2 ohms at 20kHz ( fortunately very rare ). The cable will have two conductors giving total added skin effect resistance of 0.056 ohms. In series with the 2 ohm speaker resistance this gives attenuation by a factor 0.97, i.e. -0.24 dB. It has been claimed that response changes around 0.2dB are just audible in direct comparison at lower frequencies, but a variation of 2dB or more may be needed at 20kHz, and so 0.24dB is of no significance. (Few speakers are flat to within even 3dB up to this frequency.) For most speakers with impedance well above 2 ohms at 20kHz, and using cable lengths much less than 10 metres, even 1mm diameter is usually more than adequate as regards added skin depth resistance, though the d.c. resistance may be a little excessive for long cables, being around 0.04 ohms per metre total for the two conductors.