From my understanding, correct me if I'm wrong, the electromagnetic field outside a toroid is 0, but it has a EM field within the shell itself. So my question is can a magnetic sensor pick up on this field still? or would it only be picking up the weak field around it since real life toroids are not ideal? I'm assuming no?

I ask because I've been trying to use a 3D magnetic sensor for a contactless position system for quite some time now, but if the same pole is not facing the sensor at the exact same angle, or the the magnet is flipped relative to the sensor, the data becomes completely scrambled. I was thinking that because a toroid magnetic field, if detectable, since in theory has infinite/no poles, the data from the sensor would be constant or at least more stable and not easily affected by the rotation/orientation of the toroid in relation to the sensor?

I wasn't terrific in Physics, so if I'm completely off here, please let me know.

  • \$\begingroup\$ A toroid does tend to cancel inside the diameter due to cancellation of the stray field inside, I imagine it is quite hard to use a stray torroidal field as a linear proximity sensor. WHy is this necessary? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 7 '18 at 0:40
  • \$\begingroup\$ @TonyEErocketscientist I'm not using a linear proximity sensor, I'm using a 3 axis magnetometer in an effort to essentially reproduce this - youtube.com/watch?v=fHhbM_2UMiI - but asking if toroids would be better than solenoids for the problems I described above? \$\endgroup\$ – Dante Biase Dec 7 '18 at 0:53

Toroids won't emit detecable fileds, you need to use bar electromagnets.

The Finexis people are feeding the magnets with AC and resolving both the location and orientation of the electromagnets.

That's 6 degrees of freedom computed from the 9 inputs from their 3 3-axis magnetometers.


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.