# Measuring the Magnetic Field Strength on the Surface of a Neodymium Magnet (Without Gaussmeter)?

As the title states, I need to measure the magnetic field strength on the surface of a neodymium magnet but I do not have access to a Gaussmeter.

Initially, I tried placing the magnet next to a compass so that the magnet's magnetic field is acting on the compass needle in a perpendicular direction to earth's magnetic field. Then, using the angle the needle is displaced from pointing north (and earth's known magnetic field), the magnet's magnetic field can be worked out. Of course, however, the magnet's magnetic field is much stronger than that of the earth, so, the compass needle pointed directly at the magnet.

To solve this, I decided to back the magnet up far enough that the compass needle is displaced by a lesser amount. As I was not sure exactly how the distance between the magnet and the compass needle changes the magnetic field strength (perhaps someone can clear this up?), I took a number of measurements of the angle displacement at different distances.

Making a plot revealed that the magnet's magnetic field strength is inversely proportional to the cube of the distance from the compass needle. This concerned me because at distance 0 (the surface of the magnet), the magnetic field strength is "infinite." Of course, this is incorrect but I am not sure where I have gone wrong. Could someone point out the flaw in my understanding and suggest how the experiment could be modified to find the magnetic field strength on the surface of the magnet (or an alternate better experiment)? Thanks.

Your magnet is essentially equivalent to a loop of wire with a given current (correct me if wrong). As such it would be expected that the Lorentz force will drop off $$\propto \frac{1}{r}$$. This is still likely imperfect, and I'm entirely unsure as to whether or not a compass could provide data without a laughably bad error.
My go-to for this would be to use induction. I would suspect (again correct me in the comments), that this could be derived from dropping the magnet through a solenoid and using an ammeter. If the coil was around the same size as the magnet, then you could probably calculate $$\frac{\partial \vec{B}}{\partial t}$$ and work from there.