# Weight of magnet depends on its N-S orientation?

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:

1. Neodymium N52 disk 30 mm diameter 10 mm thickness
2. South pole marked "0" north pole marked "1"
3. Dipole moment: $$8.32$$ A m$$^2$$
4. Magnetic field at surface: $$4105$$ gauss ($$10^{-4}$$T)
5. Magnetic field at 455 mm from surface (height of cardboard tower): $$0.1708$$ gauss
6. Earth's magnetic field $$0.25$$ to $$0.65$$ gauss
7. 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.

Belatedly I have noticed that the scales itself strongly effects the Earth’s magnetic field in its vicinity.

• How about a histogram showing the two distributions of measurements?, or a scattering plot? We're physicists, not medical researchers. Some units/decimals on the measurements would help.
– JEB
Commented Mar 11, 2023 at 16:15
• You might want to work out how much force you'd expect on that magnet from the Earth's magnetic field at your location. Or, if you happen to be in a science lab or other "science-y" place, make sure you're not one office over from some strong magnet. I think your results are qualitatively correct if you live in the Southern hemisphere, but this is exactly the sort of problem where I get my signs reversed, and I have no clue what the actual quantitative result would be. Commented Mar 11, 2023 at 17:04
• I don't understand the close votes on this question. It is an on-topic request for help interpreting a well-done home experiment.
– rob
Commented Mar 12, 2023 at 2:41

Local magnetic field gradients in a lab can easily be orders of magnitude larger that predicted by a naive Earth dipole model. Local Earth surface magnetic fields are typically $$\sim 50\,\mu\mathrm{T}$$ , and this field is shaped, concentrated, and redirected by any nearby ferrous material - steel chair or table frames, rebar in concrete floors or ceilings, galvanized ventilation and electrical conduits in walls, apparatus with steel cases or frames, …. This is an issue for many sensitive physics experiments. I have more than once had to rearrange the lab furniture to smooth out the local gradient sufficiently to make an experiment work.
Many modern cell phones have quite good magnetometers, which you can use to measure the local magnetic field with an app like Phyphox or Physics Toolbox. In the centre of a relatively empty room in my wood frame house, my iPhone tells me that the magnetic field changes by about $$2\,\mu\mathrm{T/m}$$. That would give a weight force, $$F=\nabla \left( \vec{m}\cdot \vec{B}\right)$$, of about $$2$$ mg on a magnetic dipole with $$\mathrm{m}=8.32\, \mathrm{A\,m^2}$$, completely consistent with your observation.
• You're right. I measured the gradient in the z-component of the magnetic field using an Ipad. It came out to $dB/dz=-2 \mu$T$/$m. Commented Apr 7, 2023 at 17:18