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I am currently reading a book about electromagnetic compatibility but I am confused about a section that talks about reducing the coupling between an external magnetic field \$ B_{ext}\$ and a circuit exposed to said field. (It's a problem I am dealing with in reality now).

The section first explains the working principle of a so-called "reduction loop" (see image to the right. It's a shorted wire loop placed next to the circuit to be protected. The external magnetic field \$B_{ext}\$ will induce a voltage in this reduction loop and this voltage will drive a current whose magnetic field is opposed to the external field, essentially lowering the magnetic field that the circuit to be protected "sees".

enter image description here

It is shown that in the presence of the reduction loop, the coupled voltage \$U_{st}\$ is given by:

$$ U_{st} = j\omega \phi_{ext} \frac{R_R+j\omega(L_R-M_{II/R})}{R_R+j\omega L_R} $$

which for high \$\omega\$ and \$L_R\approx M_{II/R}\$ has a low value.

Now the book claims that cable shield shown in the left image essentially acts as a reduction loop and in order to do so, it needs to be grounded on both ends.

Now why is this the case? If the shield wouldn't be grounded on both sides, couldn't there be a circular current flowing alongside the shield counteracting the external magnetic field as shown in the sketch below?

enter image description here

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    \$\begingroup\$ Please provide a link to the book where this rather dubious technique is quoted. In other words (1) context is needed so that I can dispel that feeling of dubiousness and (2) it's a site rule that pictures and extracts embedded need citations. \$\endgroup\$
    – Andy aka
    Commented Dec 17, 2021 at 11:14
  • \$\begingroup\$ Hello, the Picture and section is from the Book "Elektromagnetische Verträglichkeit" (German Book) by A. Schwab. The section is 3.4 (Inductive Coupling), Picture 3.28. I actually kept reading the following chapters and in section 6.1.3. an analytical model is derived for a cylindrical, hollow shield in a transverse, external magnetic field. From this derivation, the shield should actually expel the magnetic field inside of the shield without being grounded on both points if the frequency of the external field is high enough. \$\endgroup\$
    – Mantabit
    Commented Dec 17, 2021 at 12:21
  • \$\begingroup\$ Magnetic fields are ugly to guard against since shielding is somewhat ineffective… the technique is not dubious it's simply how physics work (the Ott book has a whole chapter on it). However the effects varies widely with frequency and there are also field reflection losses to be aware of. AFAIK it's not really needed to ground the shield but it's useful from the electrical field \$\endgroup\$ Commented Dec 17, 2021 at 12:24
  • \$\begingroup\$ Unfortunately with no additional context available (unless you embed the pictures and properly cite within the question itself) there's little anyone can do to uncover your assumptions here. I'm not saying you are wrong of course but, without the concrete facts from the book, it's just an opinion at best you might get AND, answers that seek opinions are normally closed. What can you do to fix this? \$\endgroup\$
    – Andy aka
    Commented Dec 17, 2021 at 12:25
  • \$\begingroup\$ @LorenzoMarcantonio I do think it is dubious because where the shorted loop exists in the prevailing field, the flux through that loop will reduce and have less of an interfering impact directly behind it but, left, right, above and below of the shorted loop, the flux density is increased and, this pushes the problem somewhere else. If proper context was given (as requested above) I might be able to use a better word of course. \$\endgroup\$
    – Andy aka
    Commented Dec 17, 2021 at 12:29

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For shielding high frequency magnetic fields, the shield can be even entirely floating and still block them, due to the induced Eddy currents that you have properly sketched.

For shielding low frequency magnetic fields, the shield will become increasingly useless as frequency drops, because the Eddy currents are damped by the shield resistance and only compensate a fraction of the magnetic field (none at DC1). Again, it doesn't matter whether the shield is grounded at one or both ends, or entirely floating.

What gives you the largest resilience against both DC and AC magnetic fields in the drawn example is the twisted differential pair.

However, it is still important to ground both ends because the oscillating Eddy current will build an oscillating voltage on the cable shield. This voltage couples capacitively onto the differential pair and shifts the common-mode voltages or the two circuits. If both circuits are grounded, the offset common-mode voltage of the signal wires might cause problems for signalling or even damage. You prevent the build-up of the common-mode voltage on the shield ends by clamping both ends to ground.


1 To block DC magnetic fields, you need a material with a permeability of either zero or infinity. Zero permeability means perfect diamagnetism (e.g. type I superconductor) and would expel the magnetic field. Infinite permeability would completely redirect the magnetic field through itself and leak none into the interior. As both methods are impractical, the practical method is to use materials with a very high permeability that redirect almost all the DC magnetic field through themselves around the shielded volume.

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  • \$\begingroup\$ Ok, so the cable shielding will indeed reduce the magnetic field within the shield but I still have to ground on both ends to avoid the capacitive coupling on the ends of the shield that you mentioned. As far as I understand, the ground loop that forms between the devices by grounding both ends of the cable should not have an effect (or at least only a small one) on the differential signal? \$\endgroup\$
    – Mantabit
    Commented Dec 17, 2021 at 12:48
  • \$\begingroup\$ @Mantabit if both circuits are properly designed, the ground loop will be inconsequential. And since the ground voltage can be indeed different at different places (e.g. different buildings or rooms), it is more precise to say that the shield must be attached to the local device ground, because this is what the local circuit is also referenced to. The two cable ends don't need to be at the same CM voltage (this often can't be realized), and some ground current can flow through the cable shield, but it won't affect signalling when it is properly differential. \$\endgroup\$
    – tobalt
    Commented Dec 17, 2021 at 12:58

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