0
\$\begingroup\$

Why, using a EMF tester on a smartphone in airplane mode, do I get a very low measure (\$< 1 mG\$), whereas using some applications which exploit the phone magnetometer, I get a very high value, something as \$35 \mu T\$? I get also a significant difference when I switch airplane mode off.

\$\endgroup\$
  • \$\begingroup\$ I'm guessing it's something to do with the cellular antennas being switched off in airplane mode. \$\endgroup\$ – AlfroJang80 May 27 at 19:39
  • 1
    \$\begingroup\$ Somehow I doubt that these applications are any accurate. \$\endgroup\$ – Eugene Sh. May 27 at 19:40
  • 1
    \$\begingroup\$ Can you please provide context as to what this "EMF tester" instrument is, preferably with a link to a datasheet? Electromagnetic compatibility is a complicated field (pardon the pun). \$\endgroup\$ – Peter May 27 at 19:40
  • 1
    \$\begingroup\$ @Bento - the EMF tester almost certainly only measures AC fields. The magnetometer only DC fields of which the Earth's magnetic field is the most significant. \$\endgroup\$ – Kevin White May 27 at 20:12
  • 1
    \$\begingroup\$ @Bento What sort of radiation? You have a device which is (presumably) giving you output in Gauss. Is it an AC gaussmeter? If so, over what frequency range does it measure? \$\endgroup\$ – Peter May 27 at 20:27
1
\$\begingroup\$

The Earth's magnetic field is about 30uG at the equator and 60 uG at the poles.
A small solar flare can cause 200uG.

For RF measurements you must also consider the power and Bandwidth of your magnetic fields to make reasoned comparisons.

1 T = 10,000 G

So 1 mG= 0.1 uT

\$\endgroup\$
  • \$\begingroup\$ Do you mean that the measure by the phone might be falsified by geographic mf? But the app says that to overcome this inconvenience it tries to take calibrate field from the system.. and anyway 35 uT seems to be an exaggerate value. Moreover, why does the external tester says almost null field? Is it perhaps calibrate in order not to consider Earth emf? \$\endgroup\$ – Bento May 27 at 20:30
  • \$\begingroup\$ sorry, I was wrong: It says to overcome the inconvenience of interferences by fields of the internal electronic circuits \$\endgroup\$ – Bento May 27 at 21:07
  • \$\begingroup\$ Okay, after a little search, that makes sense, but the correct data are 30-60 uT, not Gauss. I guess the problem is so solved \$\endgroup\$ – Bento May 27 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.