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Why does magnetism appear to decay much faster than gravity with distance?

A clear indication of this is the fact that a magnet that in short distance able to overcome gravity and pick up some object, is quickly overpowered by gravity as any short distance is put between the magnet and said object.

Gravity, as we all know, is governed by the inverse square law. Shouldn't they both decay at the same rate?

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    $\begingroup$ Why should they be the same rate ? You can find in any book of electromagnetism (or wikipedia), the analytical solution to the magnetic dipole problem. The magnetic field decreases as 1/r^3. why should it be 1/r^2 ? $\endgroup$
    – Barbaud Julien
    Oct 30, 2018 at 3:02
  • $\begingroup$ And even then, just because two things drop off like $1/r^2$ does not mean they have the same strength. As an example, I will hear much more sound intensity standing two meters away from a helicopter than standing two meters away from a person talking to me. $\endgroup$
    – BioPhysicist
    Oct 30, 2018 at 3:09
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    $\begingroup$ @AaronStevens The problem would arise if walking two steps away from both the helicopter and the person, the person suddenly became louder than the chopper. $\endgroup$
    – uKER
    Oct 30, 2018 at 3:25
  • $\begingroup$ @BarbaudJulien I figured it would, basically because that's the rate at which every energy in the universe falls off. I'm guessing it being an exception to this is probably due to the fact that magnetism isn't uniformly radiated in every direction, but a structured phenomenon. $\endgroup$
    – uKER
    Oct 30, 2018 at 3:27
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    $\begingroup$ Possible duplicate of Why does magnetic force vary in proportion to the cube of the distance instead of the square? $\endgroup$
    – John Rennie
    Oct 30, 2018 at 7:04

3 Answers 3

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Magnetic dipole forces fall off with $r^{-3}$ when relatively close to the magnet and transition to $r^{-2}$ at greater distances.

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  • $\begingroup$ the gravitational force is f=mg in the case described above. It is the constants in front that determine the size of the forces in neutons $\endgroup$
    – anna v
    Oct 30, 2018 at 6:32
  • $\begingroup$ I'm sure it's the other way around. $r^-3$ dominates near the dipole and $r^-2$ dominates at greater distances $\endgroup$
    – Lewis Kelsey
    Oct 17, 2022 at 10:09
  • $\begingroup$ @LewisKelsey, my bad, will edit. -NN $\endgroup$
    – niels nielsen
    Oct 18, 2022 at 1:51
  • $\begingroup$ I am confused myself now – if you just look at the x axis and have 2 point charges at 0 and - 4, you get $\frac{1}{x^2}-\frac{1}{(x+4)^2}$ = $\frac{8}{x^3+ 8 x^2 + 16x} + \frac{16}{x^4 + 8 x^3 + 16 x^2} $ so in this case $x^3$ is the main term at a distance and $x^2$ is the main term near to the dipole. But originally I was thinking of the hertzian dipole equations where E is claimed to be $r^3$ in the near field phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/… $\endgroup$
    – Lewis Kelsey
    Oct 18, 2022 at 14:50
  • $\begingroup$ $\frac{1}{x^2}-\frac{1}{(x+4)^2}$ isn't quite right because the fields sum in between the charges but it is right for the field outside the dipole. I've always been confused about the equations exactly tie in to electrostatics $\endgroup$
    – Lewis Kelsey
    Oct 18, 2022 at 15:01
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You are confusing two issues here, the mathematical format of the force, and the magnitude/strength of the force acting on a body.

The format of the force felt by a pole of a magnet does fall as $1/r^2$, as seen here, which is an approximation, as magnetic monopoles do not exist in nature the way electric ones do. $$ F=\frac{\mu \, q_{\text{m}1} \, q_{\text{m}1}}{4 \pi r^2} $$

There are constants and functions entering the problem if one wants to compute relative strengths acting on a body.

F is force (SI unit: newton)

$q_{m1}$ and $q_{m2}$ are the magnitudes of magnetic poles (SI unit: ampere-meter)

$μ$ is the permeability of the intervening medium (SI unit: tesla meter per ampere, henry per meter or newton per ampere squared)

r is the separation (SI unit: meter).

The gravitational force which also falls as $1/r^2$ has different constants in front of it and depends on the masses.

The earth is a huge ensemble of masses and at the distances you describe the gravitational force is constant,$$ F=mg \,,$$ where $m$ is the mass of the particle and $g$ the gravitational acceleration.

The force generated by the magnetic field has to be computed for the particular situation, but the constants are such that the magnetic force cannot overcome the gravitational force. It is the magnitude of forces that has to be compared for induced motion.

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Mangentism is inherently a dipole force. The force on a point charge exerted by an electric dipole falls of as 1/r³. The magnetic force could be formulated in terms of hypothetical magnetic monopoles - the use of 'hypothetical' here having no bearing on whether they exist in an ultimate sense. According to some models of the universe there is probably one magnetic monopole in the entire universe - there might be none - there might be two - there might (according to a poisson or poisson-like distribution) be more; but the most probable number of magnetic monopoles in the entire universe is literally one!. And there are other theories relating to their existence. But whether they exist or not, magnetism could be formulated in terms of hypothetical ones: and then the monopole magnetic force would have the 1/r² dependence.

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