How to easily, using standard DIY equipment measure the strength of magnetic field generated by a permanent magnet?

Narrowing down the "loose language" of the above:

strength of magnetic field: either flux density B at given point relative to the magnet or magnetic flux ΦB over area enclosed by a loop made of wire - whichever will be easier to measure, either of those is fine.

standard DIY equipment: commonly found household items, rudimentary tinkering tools. Soldering tools, multimeter, simple electronic parts, or maybe an easy to make spring-based dynamometer - anything of this class of complexity.

The distance of measurement is such that the field is easily noticeable through simplest methods e.g. another magnet held in hand exerts perceptible force - distance of maybe 5cm away at most.

The measurement doesn't need to be very accurate - error of order of 50% is quite acceptable. Simplicity is preferred over accuracy.

Rationale: trying to estimate what coil I need to generate sufficient amount of power to light a LED with a frictionless generator based on that magnet (knowing speed of movement of the magnet and location of the coil relative to the path of the magnet). If you know other simple methods of doing that (without need for measuring the field), they are most welcome them too.

  • $\begingroup$ A magnetic compass will oscillate before aligning itself with a magnetic field. The square of the frequency is proportional to the field strength. Look up the strength and dip of the earths field in your area and use the horizontal component to calibrate your compass. $\endgroup$ – R.W. Bird Nov 22 '19 at 16:04

For what it's worth: http://www.coolmagnetman.com/magmeter.htm - a home-made device based on a Hall effect device - for about $40.

  • $\begingroup$ That's pretty nice - I'll accept if nobody comes up with something even more simple. $\endgroup$ – SF. Dec 8 '12 at 21:08

The easiest method is to put a magnetic compass on one of the magnet's axes of symmetry, and orient the compass and magnet such that the magnet's field is perpendicular to the earth's. Then the tangent of the deflection angle is equal to the ratio of the fields.

  • $\begingroup$ That's nifty, although probably beter suited for weaker fields. Still - to know the absolute value of the field from the ratio I should know the local strength of earth's magnetic field. How would I go about finding it? $\endgroup$ – SF. May 15 '13 at 7:44

The easiest way to make a suitable coil is experimental. After all, what you want is not a precision measurement of the magnet, but all you want is to generate the necessary electricity for your LED and you don't need any math for that. Simply make a coil that satisfies your geometry out of some wire with reasonable resistance. In this case, I would estimate that a coil with around 1-10Ohm resistance will probably get you there. Add turns and see if the LED gets visually brighter. Keep adding until you are happy with the result. If the larger coil produces an inferior result, take some turns away until you are happy...

The last time I did something like this, my intuition got me within 10% of the required result... without any calculation. And how do you get an intuition like that? By grabbing a piece of wire and trying until what you want works the way you want it! After some time you will get quite good at estimating how to go about these things.

Now... that's not what an engineer will do, but that's pretty much what an experimental physicist will do for a non-critical application like this.


Theoretical background

If it's enough to measure magnet strength in relative units - in terms of $g$ "force", then here is a very easy method.
Stick small magnets in a long straight line, put it on a table and slowly push that line forward over table edge, until at some N-th magnet chain integrity will break, and first N magnets will fall down to ground due to gravity.

enter image description here

The main equation which we will be solving here is the fact that near the "breaking point", torque of the COM falling down equals to torque of 1 magnet cube traction force:

$$ NmgL=\frac{d}{2}ma $$ , substituting for $$ L = \frac{dN}{2} $$ and solving for $a$, gives : $$ a=N^2g $$

This means that for evaluating magnetic field traction force in terms of $g$ you just need to count minimum of magnetic cubes required to fall them down (term $N$ in equation) !

Experimental evaluation

For neodymium magnets, i've measured that $N=23$, this means that neodymium magnets has traction force equals to $529g$ force.
According to these fishing neodymium magnets specification:
enter image description here
My predicted theoretical magnet strength error is only $\pm 0.2$%

Hope that helps !


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