# Describing the strength of a magnetic field

There is one thing I can never get right. If I want to describe the strength of a gravitational field I can use the standard gravitational parameter ($$\mu$$), which does not take into account the distance. For example, if fly on an airplane above the clouds of Saturn I will experience a gravity very similar to what I would experience on Earth, but since I know the standard gravitational parameter of both Saturn and Earth I know that Saturn's gravitational field is much stronger, and if I ever intend to leave the planet I will need much more energy than I would need to leave Earth. This field is measured in cubic metres per second squared (m3/s2).

Similarly I can use $$k_C q$$ to describe the strength of an electric field, using N·m2/C as unit.

But what about magnetic fields? There I always hear “an XX tesla magnet”, but that is an inadequate description, since there I need to apply a distance (XX tesla where? On the surface? One metre away?). For example, saying that Earth's magnetic field is 50 μT is very incomplete; if a build what would normally be called “a 50 μT magnet”, ten metres away its strength will have probably completely faded away, while Earth's magnetic field decreases much much more slowly.

What unit can I use to describe the total strength of a magnetic field?

## 1 Answer

Note that the electric force between two charges is $$F_{\rm elec}=\frac{q_1q_2}{4\pi\epsilon_0 r^2},$$ the gravitational force is $$F_{\rm grav}=G\frac{M_1M_2}{r^2},$$ and finally the (maximal) force for two dipoles is $$F_{\rm mag}= \frac{3\mu_0 |\mathbf m_1|| \mathbf m_2|}{2\pi r^4},$$ where $$\mathbf m_1,\mathbf m_2$$ are the magnetic dipole moments.

On the same grounds as the standard electric and gravitational parameters, can just forget about the distance dependence and you just you define a standard magnetic parameter by multiplying the magnetic field constant by the dipole moment $$\mathbf m$$: $$\mu_{\rm mag}=\frac{\mu_0}{4\pi}|\mathbf m|\;\text{or}\;\frac{3\mu_0}{2\pi}|\mathbf m|$$.

The constant does not matter much, just that it is proportional to its magnetic dipole moment. Some values: Mars has almost none magnetic moment, while Earth's is $$8\times10^{22}$$ A m$${}^2$$ while Jupiter has $$1.5\times10^{27}$$ A m$$^{2}$$. Source.

• Thank you, Mauricio. Finally I can compare the actual strength of Earth's magnetic field (8×10²² A m²) with that of a small neodymium magnet (0.422 A m²). I wonder why people still measure the strength of magnets using teslas (which cannot really capture the strength of magnets)… Commented Mar 21, 2023 at 3:45