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This is a newbie question. I'm pretty sure of the answer, but I would like confirmation. I found this thread in this forum that came close to what I was looking for, but I think my question is different.

I want to shield against ingress of far field electromagnetic radiation. The material I'm working with provides good attenuation of electric fields at the required frequencies, but no magnetic shielding. I've been trying to determine if I need to incorporate magnetic shielding into my design. I'm pretty sure that because there's a fixed relationship between the electric and magnetic fields in a plane wave in free space (E/H = 377 ohm), if the electric field is attenuated, the magnetic field is also equally attenuated, and no magnetic shielding is needed. Do I have this right?

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It depends on the frequency, and the thickness of your electric shield. You need to have several skin-depths thickness of shield.

The skin depth in copper is around 1cm for mains frequency, and 10\$\mu\$m at 50MHz. It varies with the square root of frequency.

So the normal thickness of copper foil on a PCB, 35\$\mu\$m, is thick enough for VHF (100MHz) and higher, marginal at HF (3-30MHz) at RF, and far too thin at mains frequency.

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  • \$\begingroup\$ Are you saying that those thicknesses of copper will attenuate the magnetic field at those frequencies? If so, is there a rule of thumb that predicts by how much? I thought that mu metal was required to shield magnetic fields. \$\endgroup\$ – dcorsello Feb 19 '18 at 15:50
  • \$\begingroup\$ mu metal is required to shield DC magnetic fields. An AC field can be attenuated by a non-magnetic conductor. The attenuation will be small for one skin depth, many skin depths are required for good attenuation. \$\endgroup\$ – Neil_UK Feb 19 '18 at 16:13
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The material I'm working with provides good attenuation of electric fields at the required frequencies, but no magnetic shielding.

If your shield isn't thick enough to attenuate the magnetic component but is useful at strongly attenuating the electric field component, what you are left with is a magnetic field that attenuates with distance squared initially then distance cubed. This is still a useful ploy because in an unattenuated EM field, the components of electric and magnetic field only reduce (or thin out) with distance (no square or cube term).

This is basically why radio waves can transmit vast distances and trying to communicate using a magnetic field can only ever work over short distances.

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  • \$\begingroup\$ My understanding is that a minimum of 50 dB of attenuation is recommended in this application. The magnetic field in a plane wave in air is 1/377 as strong as the electric field, which is a difference of -51.53 dB. So if there's no magnetic shielding, I think I'm okay. That is if the current induced by electric and magnetic fields of equal strength are equal, something I haven't gotten my head around yet. I think the piece that I was missing in my original post is that EMR propagates differently through a conductor than it does through air. \$\endgroup\$ – dcorsello Feb 19 '18 at 16:03
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Each skin depth adds another neper of attenuation, or 8.6dB. Thus 2 skin depths provides 17.2 dB.

At 4MHz, standard copper foil on a standard PCB, with 1 ounce/foot^2 weight of foil producing 1.4mil copper thickness (or 35 microns), the attenuation is 8.6 dB because 35 microns is the skin depth at 4MHz.

At 16 MHz, that same 35 micron foil provides 2 skin depths, or 17.2 dB.

At 36MHz [ 4Mhz * 3 == 4MHz * sqrt(9) ], you'll have 3 * 8.6 = 25.8dB.

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