# Why will we never run into a magnetic field that falls off as $\frac 1 {r^2}$?

For example, Walter Lewin says in many lectures that we will never find a magnetic field $B\propto \frac 1 {r^2}$ - why is this?

I believe it must be related to $\nabla \times E= -\partial_t B$, but I don't see why this would make the previous impossible.

A magnetic field of the form

$$\boldsymbol{B} \propto \frac{\boldsymbol{\hat{r}}}{r^2}$$

is impossible because

$$\nabla \cdot \left( \frac{\boldsymbol{\hat{r}}}{r^2} \right) = 4 \pi \delta(\boldsymbol{r}),$$

so a magnetic field of this form would violate Maxwell's equations, one of which is

$$\nabla \cdot \boldsymbol{B} = 0.$$

It seems that a magnetic monopole might produce a magnetic field like this, but magnetic monopoles are forbidden in classical electromagnetism.

• The question says "falls off as", not "exactly equal to". It also makes no stipulation that the field be radial. – Emilio Pisanty Nov 24 '16 at 22:18
• The question actually says "proportional to" ($\propto$), and the divergence operator is linear. The phrase in the title "falls off as" is vague, but I took it to refer to an inverse-square law. I shall edit my answer to include these caveats. – jc315 Nov 24 '16 at 23:02
• @EmilioPisanty Do you know if it is possible to have a non-radial $\boldsymbol{B}$ field whose magnitude is proportional to $1/r^2$? – jc315 Nov 24 '16 at 23:05
• @EmilioPisanty I strongly suspect that the presence of the term $r^{-2}$ is going to make a delta appear regardless of the direction of the field. Anyway, it would be nice to prove it. – valerio Nov 24 '16 at 23:20
• I see your point, although $\nabla \cdot B$ doesn't exist at $0$, in the regular sense. Are all derivatives in Maxwell's equations to be taken in the distributional sense? – YoTengoUnLCD Nov 25 '16 at 18:00

I believe what he is implying is that there are no magnetic monopoles (that we know of), at least in classical electrodynamics. A magnet has a south and a north pole (a dipole), which produces a field (the vector potential) $$\mathbf{A} (\mathbf{r}) = \frac{\mu_0}{4\pi r^2} \frac{\mathbf{m} \times \mathbf{r}}{r} = \frac{\mu_0}{4\pi} \frac{\mathbf{m} \times \mathbf{r}}{r^3}$$ where $\mathbf{m}$ (vectorial quantities are bolded) is the magnetic moment of this N-S-dipole that is kept constant while the source shrinks to a point. This is the limit of a dipole field. You can read more here on Wikipedia.