# Magnetism: Same Pole Repulsion & Heteropolar Attraction

• I
• Hir
In summary, the Earth's magnetic field has a force on the compass that is difficult to determine mathematically.f

#### Hir

We know that the two magnets have same-pole repulsion and heteropolar attraction; In most parts of the planet, a magnet is hoisted with a thin wire, one side of the magnet will point to the guide, and the other side will point to the north; If you come to the magnetic pole of the earth and place the magnet horizontally, will the magnet still follow the same pole repulsion and heteropolar attraction? If I may, I would like to know how much force there is when they attract each other, and how much force they have when they repel each other.

Can you clarify what you want to know? Are you wanting to know if would be dangerous to be near the Earth's magnetic pole (as the warning message on your pictured magnet)?

Can you clarify what you want to know? Are you wanting to know if would be dangerous to be near the Earth's magnetic pole (as the warning message on your pictured magnet)?
The post did not give the size of the magnet. The picture (from the Internet) is a reference. I have watched a scientific expedition documentary. The compass points to the ground at the two poles of the geomagnetic field. I am not sure whether the geomagnetic field is still attractive to the magnet, or Repulsion, and its size.

I am not sure whether the geomagnetic field is still attractive to the magnet, or Repulsion, and its size.
What a strange remark. The field is there, period. Post #3 gets you all you can possibly desire !

What a strange remark. The field is there, period. Post #3 gets you all you can possibly desire !
Things are not that simple. Some say that a uniform magnetic field has only torque, and some say that the attractive force of a magnet is related to the gradient of the magnetic field.

A force is a measure of interaction between two objects. "Force of a magnet" is meaningless.

A force is a measure of interaction between two objects. "Force of a magnet" is meaningless.

Things are not that simple. Some say that a uniform magnetic field has only torque, and some say that the attractive force of a magnet is related to the gradient of the magnetic field.
Who is "some"? And where do they say this?

vanhees71
We know that the two magnets have same-pole repulsion and heteropolar attraction; In most parts of the planet, a magnet is hoisted with a thin wire, one side of the magnet will point to the guide, and the other side will point to the north; If you come to the magnetic pole of the earth and place the magnet horizontally, will the magnet still follow the same pole repulsion and heteropolar attraction? If I may, I would like to know how much force there is when they attract each other, and how much force they have when they repel each other.
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These questions remind me a bit of an old thread where someone asked how DC ammeters work and ended up with me doing all manner of science experiments to determine the strength/directions of the earth's magnetic field where I live.

Your first question seems a bit silly, though I may be misinterpreting what you were asking. I know in the past I've had dreadfully difficult days trying to formulate a coherent question. In any event, I'll ignore it for now.

From another thread, I seem to recall that unless magnets are identical, which obviously your Earth geo-magnet and fancy man-made magnet, are not, the is force is extremely difficult to determine mathematically. I believe this is why I went the experimental/empirical route.

Pay particular attention to the data provided by the link posted by @dlgoff ,https://www.ngdc.noaa.gov/geomag/calculators/magcalc.shtml , as it was crucial in determining when and where my experiment had gone astray.

scottdave
No matter how big it is, a magnet does not have a force. Size does not matter in this case.

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There's nothing wrong with my question, it can be confirmed with a super simple experiment, but extremely costly due to location.

No matter how big it is, a magnet does not have a force. Size does not matter in this case.
It takes more force to turn the compass! Are you sure you understand my question?

"The Earth's magnetic field intensity is roughly between 25,000 - 65,000 nanoTeslas (.25 - .65 gauss)."

Does that help?

There's nothing wrong with my question
Yes there is. The force on the tip of a compass needle has one answer. A piece of chalk? Another answer, An electron in the upper atmosphere? Yet another answer,

"The Earth's magnetic field intensity is roughly between 25,000 - 65,000 nanoTeslas (.25 - .65 gauss)."

Does that help?
Knowing the intensity of magnetic induction is useless, only experiments can prove all conjectures in the future

davenn
It takes more force to turn the compass! Are you sure you understand my question?
The force depends on both object interacting through the magnetic field. It is not an intrinsic property of the earth alone. Same is true for any other magnet.
Also, it is true that a dipole (like a compass needle or any other magnet) experience no net force in an uniform magnetic field. The earth's field is not uniform at large scale but for the size of a compass needle the gradient is probably not significant. The needle experiences a torque that orients it along the local field line. This torque will depend on the local value of the field and the properties of the needle (or other magnet).
The strength of the magnetic field near poles is about 60 microtesla versus the minimum value of about 30 microtesla and average of about 50. So there is nothing special about the poles in terms of the size of the torque acting on the compass. What is unusual is the direction of the torque. The compass tends to take a direction closer to vertical than horizontal. But there is a vertical component of the field most everywhere else (magnetic inclination).

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Yes there is. The force on the tip of a compass needle has olne answer. A piece of chalk? Another answer, An electron in the upper atmosphere? Yet another answer,
There are two forces between the magnets, one is torque (torque) and the other is the attractive force that moves closer to each other

davenn, weirdoguy and malawi_glenn
The torque is not a type of force.

The torque is not a type of force.
Let me put it another way, put iron and non-magnetic materials of the same mass on both sides of the balance, and put the balance on the south pole of the earth's magnetic field. Due to the relationship between the earth's magnetic field, will the balance remain parallel?

Between the magnetic field and what?

It is not the magnetic field itself that causes the attraction, it is the gradient of the magnetic field.

https://physics.stackexchange.com/q...s-more-heavy-due-to-the-earths-magnetic-field
In a multipole expansion of the Earth's magnetic field ##\vec{B_{E}}##, the dipole contribution constitutes about 90% of the total field. Hence, to a first approximation, the ##\left(r,\theta\right)## components of ##\vec{B_{E}}## are (https://en.wikipedia.org/wiki/Dipole_model_of_the_Earth's_magnetic_field): $$B_{Er}=-2B_{0}\left(\frac{R_{E}}{r}\right)^{3}\cos\theta,\;B_{E\theta}=-B_{0}\left(\frac{R_{E}}{r}\right)^{3}\sin\theta\tag{1}$$where ##B_{0}=3.12\times10^{-5}\text{T}## and ##R_{E}=6.37\times10^{3}\text{m}## is the mean radius of the Earth. We can obtain the force ##\vec{F}## exerted on a magnet located on the surface of the Earth by treating it as a magnetic dipole with moment ##\vec{m}##:$$\vec{F}=\vec{\nabla}\left(\vec{m}\cdot\vec{B_{E}}\right)\tag{2}$$From this, the maximum possible forces are:$$\left|\vec{F}_{max}\right|_{Equator}=\frac{3B_{0}\left|\vec{m}\right|}{R_{E}},\;\left|\vec{F}_{max}\right|_{Poles}=\frac{6B_{0}\left|\vec{m}\right|}{R_{E}}\tag{3a,b}$$ Note that the gradient operator in eq.(2) has introduced a net factor of ##R_{E}## into the denominators of eqs.(3), so we expect these forces to be tiny. Indeed, consider the case of a neodymium disk magnet about the size of a US penny with a surface field of ##0.3\text{T}##, resulting in a dipole-moment ##\left|\vec{m}\right|## of ##0.422\text{ N-m/T}## (https://www.kjmagnetics.com/blog.asp?p=dipole). The maximum possible force (3b) on the magnet evaluates to ##1.24\times10^{-8}\text{N}## at the poles; i.e., a negligible 12 nanoNewtons. In contrast to this tiny force, the K & J Magnetics site shows that, while the torque tending to align the magnet to the Earth's field also seems small at ##2\times10^{-5}\text{N-m}##, it's still enough that when placed edge-on atop a low friction surface, the magnet promptly rotates like a compass needle to point north.

Hornbein and nasu
For a non-magnetized piece of ferromagnetic material the force depends on both the gradient of the field and the absolute field value itself. This will make the force even smaller, by several orders of magnitude. (compared with post 23).

hutchphd
Knowing the intensity of magnetic induction is useless, only experiments can prove all conjectures in the future
Thing is, I can't really tell exactly what you're asking for. Like others, I'm taking shots in the dark.

If the earth is a large neodymium magnet, and the large magnet attracts the small magnet, the attractive force between them is so small, which is very surprising;
Special thanks to #23
Thanks to all the people above.

If the earth is a large neodymium magnet, and the large magnet attracts the small magnet, the attractive force between them is so small, which is very surprising

You have to remember Newtons third law. Let's say there is an attractive force between body 1 and body 2. That means that they are both attracted by each other. The sun is attracthed by the earth too.

Large is spatial size means nothing, as has been pointed out several times. Anyway, back to the gravity analogy, the force from the earth on a small pebble is pretty small. Do you think this is surprising as well?

which is very surprising;
It is not at all surprising to me, and simple if you understand how magnets produce force. The force is small because the earth field is small at the surface and the field gradient is also very small accross a magnet in your pocket.

The geomagetic field is generated by its iron core.

Compare with being at a large distance from your large neodymium magnet...

a negligible 12 nanoNewtons.
Confirm again, 12 nanonewtons is the attraction between the magnet and the earth's magnetic field?
A new question is added here, if the earth's magnetic induction increases by 10,000 times, find the attraction between the magnet and the earth's magnetic field at this time.

If nothing saturates ( probably true in this situation) then it will increase by ten thousand. We are assuming that you know how to turn up the Earth field.....an unlikely circumstance.

This thread seems to have run its course, so it will be closed now.

Hir is correct that the magnetic force on a magnet is caused by the gradient of the magnetic field (its non-uniformity), and without a gradient you only get a torque (which is why compass needles align with the magnetic field).

BvU