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How to measure electromagnet by Tesla? Answered

I will be doing an experiment how powerful the magnetic field of an electromagnet, and my guess is the more you twist the wire the more power it generate. But i don't know how to measure the magnetic field by using the unit Tesla.



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Jack A Lopez
Jack A Lopez

4 years ago

It is actually, tesla, with a lowercase "t" when writing about the unit, even if the symbol is an uppercase letter. For example,

0.5 tesla = 0.5 T = 500 mT

More about that, here:

and that concludes the grammar lesson.

There are sensors for measuring magnetic field strength, like pickup coils and Hall-effect sensor ICs, but for some reason it is rare to find a complete instrument for measuring magnetic field strength.

By complete instrument, I mean something analogous to a digital thermometer, which consists of essentially two parts, a small probe on the end of long wire, and a box with a digital display for to translate the signal from the probe into human readable numbers.

For some reason, for measuring magnetic field strength, it is pretty easy to find, or build, just the probe part, but not so easy to find the rest of it; i.e the part that translates the signal from the probe into numbers and displays them.

Regarding sensors, I will describe just two of them that I think would work for measuring the field near a homemade electromagnet, specifically the pickup coil, and Hall-effect IC. There are other methods like moving coil (pickup coil that moves), proton precession magnetometer, and fluxgate magnetometer, but just to keep things simple, I am only going to mention the two sensors I think would work best with a homemade electromagnet, namely pickup coil and Hall-effect IC.

Pickup coil

The main advantage of the pickup coil is its simplicity; essentially it is just a coil, with air core, small area, and a large number of turns.

The main disadvantage of a pickup coil is it only works with AC magnetic fields (also called alternating magnetic field), the kind produced by an electromagnet powered by alternating current.

The physics of the pickup coil is based on Faraday's Law;
i.e. the voltage across the coil is proportional to the number of turns N, and the rate of change of magnetic flux (dPhi/dt)

Vcoil = -N*(dPhi/dt)


By the way, magnetic flux is actually an integral over B*dA, where B is magnetic field (in teslas), and dA is a tiny area, part of the area enclosed by the coil. The approximation
Phi = B*A
is good for the case where B is approximately uniform over the small area A, and A and B are pointed in the same direction.

Assuming A is constant, (dPhi/dt) = A*(dB/dt), so
Vcoil = -N*A*(dB/dt)

which is to say, voltage in the pickup coil, Vcoil, is just due to (dB/dt).

Hall Effect IC

The Hall Effect IC (IC stands for integrated circuit) is a little piece of electronic wizardry that can be used to measure magnetic field strength. Hall Effect ICs will work with DC magnetic fields; i.e the kind of field produced by a permanent magnet, or by an electromagnet powered by DC current.

If you are going to go shopping for a Hall Effect IC, you probably want one described by the word "linear" or the word "ratiometric" which means the same thing. The SS49E is sort of a typical one of these.
It is made to run 5 volts DC, and its output pin produces a voltage proportional to the magnetic field it feels across its little sensor chip;

e.g. Vout= 2.5 V when Bz= 0,
Vout= 4.5 V when Bz= 100 mT,
Vout= 0.5 V when Bz=-100 mT
as a formula,
Vout = 2.5V + K*Bz, where K= 2V/100mT = 20 mV/mT

Here I am using Bz to represent the component of B parallel to sensor. Magnetic field is a vector quantity, and what the chip measures is actually a component of B in the same direction as the chip.

By the way, a lot of the data sheets for these Hall Effect ICs give the sensitivity in millivolts per gauss (mV/G) where 1 gauss = 1 G = 0.0001 T = 0.1 mT, so if you see a quote in mV/G, and you want to convert that to mV/mT, just substitute 1G=0.1 mT. For example the Allegro A1302 has sensitivity of 1.3 mV/G, which is the same as

1.3 mV/G = 1.3mV/(0.1 mT) = 13 mV/mT

Also note, full scale, the highest field one of these 5-volt ratiometric ICs can measure is 2500mV, divided by the sensitivity; e.g.

(2500 mV)/(13 mV/mT) = 192.3 mT ~= 200 mT = 0.2 T

Also BTW, probably about the largest field strength you might expect to present in, or right on top of, a homemade electromagnet is about 1.6T = 1600 mT, because that is the field strength at which ferromagnetic materials saturate. See more about that in the Wiki article on "Saturation(magnetic)":

Also you can expect for the magnetic field, at some fixed point (fixed distance) in the space near an electromagnet, you can expect the magnitude of the magnetic field to be proportional to the current (in amperes) and proportional the number of turns.

Also, and I think -Max- already mentioned this, but magnitude of the field drops pretty fast as a function of distance as you move the probe away from the electromagnet, maybe as (1/r^3)? Something like that...



Reply 2 years ago

May I know what equation you're using to convert the Vout to Tesla? Thanks

Jack A Lopez
Jack A Lopez

Reply 1 year ago

Hey. I am sorry it took me a few months to reply to this. Are you still looking for a reply to this? I am hopeful that you figured something out yourself. However, in the case you are still seeking an answer, I am wondering what kind of sensor you are using. A Hall Effect sensor, or a pickup coil? I thought I tossed in some equations for both those cases.


4 years ago

Your iPhone can run an app that tells you the strength of magnetic fields...


4 years ago

As he left the blazing summer heat outside the Warlock's
cave, the visiting sorcerer sighed with pleasure. "Warlock,
how can you keep this place so cool?... The mana in this
region has decreased to the point where magic is nearly
The Warlock smiled and so did the unnoticeable man young
man who was sorting the Warlock's parchments in a corner
of the cave... The Warlock said "I used a small demon
Harlaz. He was generated by a simple, trivial spell. His
intelligence is low---fortunately, for his task is a dull one.
He sits at the entrance to this cave and prevents the fast-
moving molecules of air from entering and the slow-moving
molecules from leaving. The rest he lets pass. Thus the cave
remains cool"
"That's marvelous, Warlock! I suppose the process can
be reversed in winter?"
"Of course."
"Oh, I didn't think of it," the Warlock said hastily.
"Have you met my clerk? It was his idea."

The Warlock raised his voice. "Oh, Maxwell"
Attributed to the great SF author _ Larry Niven

Anyway 1 Maxwell = 1 Gauss x cm^2

and 1 Tesla x Meter^2 = 100000000 Gauss x cm^2

finally 1 Weber = 1 Tesla x M^2

If you don't have a calibrated hall probe use a compass to compare to a known magnet.


4 years ago

The magnetic field strength will depend on where you do the measurement. It will be stronger (denser) near the wires and inside the loop. As you move away however it will diminish as a inverse(ish) relationship. According to my university physics book, the formula to calculate it is:


Bx = u0*I*a^2 / 2(x^2 + a^2)^3/2 for a single circular loop. Here is a picture that better illustrates how that messy equation is derived:


Hopefully that calculus is not too scary. :P


Yes, adding more turns will proportionally increase the magnetic feild produced. You can roughly measure magnetic feild with a hall effect sensor, although I don't know how accurate that is to actually figure out magnetic feild. You can also use some of the equations and create a loop of wire of known size and use it to measure how fast the change in magnetic field is at a point/area. This is basicly what near field probes do.