Magnetic forces vary from gravity and electromagnetic radiation (such as light) in that gravity and radiation diminish by the square of the distance from the source, but magnetism diminishes by the cube of the distance.
Why is this?
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asked Nov 17, 2016 at 22:53
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$\begingroup$ Are you referring to the fact that the magnetic field of a monopole goes as $B\sim r^{-3}$? $\endgroup$– caveracNov 17, 2016 at 23:17
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$\begingroup$ "One characteristic of a dipole field is that the strength of the field falls off inversely with the cube of the distance from the magnet's center." Direct quote from the Wikipedia: en.wikipedia.org/wiki/Force_between_magnets $\endgroup$– Ambrose SwaseyNov 18, 2016 at 0:32
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3 Answers
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It is just a consequence of the fact that the multipole expansion of an arbitrary magnetic field configuration doesn't have a monopole term due to absence of magnetic charges, and the first non-zero term in the expansion is of order of $\frac{1}{r^3}$. Generically, electric and gravitational fields also have analogous terms in their $\frac{1}{r}$ expansions, as well as all the higher-order terms.
If magnetic charges exist, one also has a monopole $\frac{1}{r^2}$ term in the expansion.
answered Nov 18, 2016 at 7:21
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For the case of the gravitational potential, it is true that a point mass (gravitational monopole) generates a field that goes as $\mathcal{G}_{ \rm monopole}\sim 1/r^2$, or equivalently, a potential that decreases as $V_{g,\;\rm monopole}\sim 1/r$. However, when the mass is not point-like, these two behaviors are not valid any longer.
A similar argument applies for the magnetic field, a dipole generates a magnetic field that behaves as $B_{\rm dipole}\sim 1/r^3$, or a magnetic potential that goes as $A_{\rm dipole}\sim 1/r^2$. But again, if the source is not a dipole then this conclusion is not valid either.
You can then ask 'If I were to calculate the gravitational potential of a dipole, does it change as $V_{g,\;\rm monopole}\sim 1/r^2$?' And the answer is: there are no gravitational dipoles, basically because there are not negative masses. But there are negative electric charges, and for an electric dipole its potential indeed drops as $V_{\rm monopole}\sim 1/r^2$, the same behavior as the magnetic dipole $A_{\rm dipole}$.
In general the potential for an arbitrary charge distribution can be calculated as an expansion that include the contribution from the monopole, dipole, quadrupole, $\ldots$ This series is known as multipole expansion
$$ V(\mathbf{r}) = V_{\rm monopole}(\mathbf{r}) + V_{\rm dipole}(\mathbf{r}) + V_{\rm quadrupole}(\mathbf{r}) + \cdots $$
And each term drops as a different power of the distance $1/r$, $1/r^2$, $1/r^3$, $\cdots$ Same for the magnetic field
$$ \mathbf{A}(\mathbf{r}) = \mathbf{A}_{\rm dipole}(\mathbf{r}) + \mathbf{A}_{\rm quadrupole}(\mathbf{r}) + \cdots $$
Note that I intentionally left out the magnetic monopole, because that one is zero
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It behaves as an electric dipole:
https://byjus.com/physics/dipole-electric-field/
"The red line is the actual dipole force, the green line is 1/r^2, and the blue one is 1/r^3, and you can see that the red line starts out following the green line then "moves over" to the blue line. As I said, it's a gradual transition, but the center of sorts is at r=1 on the graph, which represents the length of the dipole. So if the two poles are a distance d apart, roughly speaking, the force is like 1/r^2 if you're closer than d, and like 1/r^3 if you're further away than d." https://www.physicsforums.com/threads/is-1-r-3-descriptive-of-magnetic-force-drop-off.326451/
