I weighed a Neodymium disc magnet and found that it weighs about $1$ mg more when the north pole is upwards than when the south pole is upwards.
Can this be explained by "conventional" physics?
Apparatus
Scales: Kern EMB 100-3 Max 100g d=0.001g
Cardboard tower: 455 mm high
Magnet:
- Neodymium N52 disk 30 mm diameter 10 mm thickness
- South pole marked "0" north pole marked "1"
- Dipole moment: $8.32$ A m$^2$
- Magnetic field at surface: $4105$ gauss ($10^{-4}$T)
- Magnetic field at 455 mm from surface (height of cardboard tower): $0.1708$ gauss
- Earth's magnetic field $0.25$ to $0.65$ gauss
- See https://www.kjmagnetics.com/calculator.asp?calcType=disc
Control
I first suspended the magnet over the scales while weighing a steel $100$g test weight. It was at a height that was slightly less than the height of the cardboard tower.
I flipped the magnet at random and got the following results for the test weight. "0" is south pole upwards and "1" is north pole upwards. The weight of the test weight was $99.9+n/1000$ grams where "n" is tabulated below.
I performed a Point-Biserial Correlation calculation to check whether there was a correlation between:
Magnet orientation
0,0,0,0,0,1,1,1,0,1,1,1,1,1,0,0,0,0,1,1,1,0,1,0,1,1,1,0,1,1,0,1,1,0,1,1,1,1,1,0,0,1,1,0,0,0,1,1,0,0,0,0,1,1,1,0,0,0,1,0,1,1,1,0,1,0,0,1,0,0,0,0,1,1,0,1,1,0,0,0,0,0,1,1,0,0,1,0,1,1,1,0,1,0,1,1,1,0,0,1
Test weight
101,99,101,101,97,97,98,99,98,99,98,99,99,100,100,97,98,98,98,97,98,98,99,98,99,99,98,101,99,98,100,99,99,100,100,99,98,99,99,101,99,98,98,97,99,97,100,96,100,98,99,98,97,98,99,101,99,99,101,99,100,100,99,100,100,99,97,100,100,99,99,98,99,98,99,98,97,101,97,99,98,99,99,98,100,98,99,100,98,98,100,97,98,99,101,99,96,100,97,97
The result was:
r-value: $-0.13424$
p-value: $0.183$
Therefore the correlation was not significant at $p<0.05$
Experiment
I put a cardboard tower on the scales to keep the magnetism from affecting it. I zeroed the scales and then weighed the magnet with different orientations on the tower.
I flipped the magnet at random and got the following results for its weight. "0" is south pole upwards and "1" is north pole upwards. The weight of the magnet was $52.6(\rm{dd})$ grams where "dd" is tabulated below.
I performed a Point-Biserial Correlation calculation to check whether there was a correlation between:
Magnet orientation
0,0,0,0,0,1,1,1,0,1,1,1,1,1,0,0,0,0,1,1,1,0,1,0,1,1,1,0,1,1,0,1,1,0,1,1,1,1,1,0,0,1,1,0,0,0,1,1,0,0,0,0,1,1,1,0,0,0,1,0,1,1,1,0,1,0,0,1,0,0,0,0,1,1,0,1,1,0,0,0,0,0,1,1,0,0,1,0,1,1,1,0,1,0,1,1,1,0,0,1
Magnet weight
63,62,62,60,63,63,60,63,61,65,62,62,62,62,63,61,60,59,61,62,61,61,64,60,62,62,64,62,62,62,59,62,61,61,63,62,63,62,63,59,60,63,63,60,60,62,62,62,60,62,61,62,63,62,62,60,61,61,62,61,62,62,63,63,63,61,61,62,62,61,61,61,63,62,60,61,65,60,62,60,63,63,62,63,59,61,61,62,63,63,61,61,62,61,62,61,62,61,61,62
The result was:
r-value: $0.51659$
p-value: < $0.00001$
Therefore the correlation was significant at $p<0.05$.
I performed a two-tailed T-Test for 2 Independent Means to find the mean weight difference between:
south pole upwards (0)
63,62,62,60,63,61,63,61,60,59,61,60,62,59,61,59,60,60,60,62,60,62,61,62,60,61,61,61,63,61,61,62,61,61,61,60,60,62,60,63,63,59,61,62,61,61,61,61
Mean: $61.04$ Standard deviation: $1.12$ Sample size: $48$
north pole upwards (1)
63,60,63,65,62,62,62,62,61,62,61,64,62,62,64,62,62,62,61,63,62,63,62,63,63,63,62,62,63,62,62,62,62,62,63,63,62,63,62,61,65,62,63,61,63,63,61,62,62,61,62,62
Mean: $62.29$ Standard deviation: $0.96$ Sample size: $52$
The result was:
t-value: $-5.97273$
p-value: < $0.00001$
The result was significant at $p<0.05$.
I used this calculator to find the effect size $g$.
Hedges' g: $1.2$ mg
Conventional explanation in terms of interaction with Earth's magnetic field
For the sake of simplicity let us assume that the experiment is conducted at the Earth's magnetic south pole. If we assume the Earth has a magnetic dipole moment of $m_1$ then the difference in magnetic force $\Delta F$ when one flips a magnet with a dipole moment of $m_2$ is given by: $$\Delta F = \frac{3 \mu_0 m_1 m_2}{\pi r^4}$$ Taking: \begin{eqnarray} \mu_0 &=& 4\pi \times 10^{-7}\hbox{ N A$^{-2}$}\\ m_1 &=& 8.0\times 10^{22}\hbox{ A m$^2$}\\ m_2 &=& 8.32\hbox{ A m$^2$}\\ r &=& 6.371\times 10^6\hbox{ m} \end{eqnarray} $$\Delta F = 5\times 10^{-10}\hbox{ N}$$ This is equivalent to a weight difference of just $50$ nanograms!
Correction
The Earth's magnetic field is far from a dipole field - especially in a house. Using an Ipad I measured $dB_z/dz=-2 \mu$T$/$m. See David Bailey's answer.