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This empty outlet of an electric cord plugged in to an outlet can induce 60 Hz powerline distortion in the AD620 and OP490 circuit (isolated with a battery) located 6 inches away.

I can't detect any magnetic field around the outlet, at least not more than 0.2 milligauss (which is the background field of the Trifield meter.)

How do you compute the electric field of this empty outlet 6 inches away? Whatever the value, no matter how miniscule, it can indeed trigger distortion in the 20,000 gain op amps. Is this really possible?

Original message:

If the electromagnetic field from 60 Hz powerline is only say 0.01 milligauss and undetectable by conventional EMF meter, can it still affect op-amps? The gain of an op-amp can be 20,000 times. Does this mean the 0.01 milligauss power line frequency also got amplified, or is it not EMF related but capacitance related or others?

Please share how to compute what happens if 0.01 milligauss got amplified by 20,000 too, if you know how (if this is how it works where the 0.01 milligauss got amplifed by the op-amps.)

  • 1
    \$\begingroup\$ Not sure about the magnetic component, not even sure why would you see anything measurable on certain distance from the wires. But the electric field may really be a menace. Most probably if you see 50hz/60h it's the electric field of mains. \$\endgroup\$
    – TQQQ
    Oct 7, 2022 at 3:07
  • 8
    \$\begingroup\$ If the field has a definite effect on op-amps, then, by definition, it is not "undetectable" \$\endgroup\$ Oct 7, 2022 at 3:20
  • 4
    \$\begingroup\$ Touch the input probe of an oscilloscope and you’ll see a healthy mains signal along with some radio stations if they’re nearby. Electric field, not magnetic. \$\endgroup\$
    – Kartman
    Oct 7, 2022 at 4:59
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    \$\begingroup\$ Your picture shows an extension cord with no load, and hence, no current (except for capacitatively coupled current) flowing through it. Why would you think it is the magnetic field, rather than the electric field which is having an effect? \$\endgroup\$ Oct 7, 2022 at 11:28
  • 1
    \$\begingroup\$ Please do not keep adding requests to a question. Edits to a question should be for clarifications or new information, not new requests. If you have something new to ask, ask it as a separate question. That being said, requests for product recommendations are off topic in this site, so if you ask about available field strength meters, such a question is likely to be quickly closed. \$\endgroup\$ Oct 8, 2022 at 11:18

2 Answers 2


The cord in your image has no significant current through it. (There will be some due to capacitance and leakage conductance, but it will be extremely small. There will however be a significant electric field, as the cord is connected to mains power. If your op-amps is picking up electrical noise from the cord, it is almost certainly almost exclusively from the electric field, not the magnetic field.

And yes, an op-amp can amplify microvolt level signals (or even smaller) that could be induced by the electric (not magnetic) field of a mains powered cord with virtually no current flowing through it.

  • \$\begingroup\$ Yes. its most likely electric field. What is the estimated value in volts per meter (V/m) of the outlet 6 inches away? how do you compute it? \$\endgroup\$
    – Jtl
    Oct 7, 2022 at 12:41
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    \$\begingroup\$ Difficult to compute in reality as it is geometry dependent, but mains E field pickup is an old friend and gets into basically any unshielded high impedance input. Something like a scope probe (1M) will likely show several tens of mV of pickup if you touch the tip. Fortunately it is usually easy to screen out as long as you can get the amp and source you are trying to amplify inside a metal box, failing that you are into the tricks the EEG/EKG crowd use (leg electrodes driven by the CM voltage and suchlike). \$\endgroup\$
    – Dan Mills
    Oct 7, 2022 at 15:57
  • \$\begingroup\$ what you mean by "leg electrodes driven by the CM voltage"? what does CM stand for and how is it driven? \$\endgroup\$
    – Jtl
    Oct 8, 2022 at 12:28
  • \$\begingroup\$ CM is common mode. Basically, instead of removing the noise, they inject it on their common (the ground). Because any amplifier looks at the difference in voltage between its input and its common (often ground), if you have the noise on both the ground and the signal, it will be cancelled out by the common mode rejection ratio of the amplifier. So instead of defeating the noise, you just subtract it by having it on your ground as well. \$\endgroup\$
    – user55924
    Oct 8, 2022 at 12:53

Opamps don't amplify the magnetic field passing through them. A piece of ferromagnet with a relative permeability of 20000 would.

Is 0.01 mG (1 nT) thus inconsequential to electronic circuits ? Depends. If you make tremendous wire loops - something not too unlikely to achieve by accident - even such a tiny field will cause induced voltages in your circuit which are substantial enough.

If you go through some numbers, you can achieve induced voltages at the μV ballpark with only 1 square meter of wire loop. This is very well amplifiable by op-amps.

  • \$\begingroup\$ Please see edit, can a mere empty outlet without any loop induce microvoltage in the op amps? Kindly compute it because there is still effect 6 inches away. \$\endgroup\$
    – Jtl
    Oct 7, 2022 at 11:09
  • \$\begingroup\$ @Jtl Impossible to tell from the image. How is your opamp circuit powered and grounded ? But more likely it can be electric field interference, see answer by MathKeepsMeBusy \$\endgroup\$
    – tobalt
    Oct 7, 2022 at 11:47
  • \$\begingroup\$ this is the exact circuit. electronics.stackexchange.com/questions/636989/… i checked my trifield emf meter carefully and the emf and electric field is near zero near the outlet so if its magnetic field, it is responding to 0.0001 value something. i wrote this new thread bec its about distortion undetected by convensional magnetic and electric meter. so dont say it is duplicate topic \$\endgroup\$
    – Jtl
    Oct 7, 2022 at 12:04

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