Earth's Magnetism

To leading order the Earth behaves like a giant magnetic dipole, tilted by about $11.3^\circ$ relative to the rotation axis. The compass works because near the surface, the Earth's field has a horizontal component pointing roughly toward geographic north. To completely specify the field at a location, three numbers — the elements of Earth's magnetism — are needed.

Concept

At any point on Earth, the local geomagnetic field $\vec B_E$ has magnitude $B_E$ and direction characterised by:

1. Declination $D$ — the angle (East or West of true north) between the geographic meridian and the magnetic meridian (the vertical plane containing $\vec B_E$).
2. Inclination or dip $I$ — the angle that $\vec B_E$ makes with the horizontal plane.
3. Horizontal component $H_E = B_E\cos I$ — the projection on the horizontal plane.

The vertical component is $V_E = B_E\sin I$, so

$B_E = \sqrt{H_E^2 + V_E^2},\qquad \tan I = \frac{V_E}{H_E}.$

Important convention. The Earth's magnetic south pole lies near the geographic north (in the Canadian Arctic) and vice versa. That is why the "north-seeking" end of a compass points to geographic north — it is being attracted to a magnetic south.

Derivation

Treat the Earth as a dipole of moment $\vec m_E$ at the centre, tilted by an angle small enough that the field at the surface is approximately that of a point dipole. At magnetic latitude $\lambda$ (measured from the magnetic equator), the field strength has components

$B_r = \frac{\mu_0}{4\pi}\,\frac{2 m_E \sin\lambda}{R_E^3},\qquad B_\theta = \frac{\mu_0}{4\pi}\,\frac{m_E \cos\lambda}{R_E^3},$

where $B_r$ is radial (vertical at the surface) and $B_\theta$ is tangential (horizontal). Hence

$\tan I = \frac{B_r}{B_\theta} = \frac{2\sin\lambda}{\cos\lambda} = 2\tan\lambda.$

At the magnetic equator $\lambda = 0$, dip $I = 0$ — field is horizontal. At the magnetic poles $\lambda = 90^\circ$, dip $I = 90^\circ$ — field is vertical and a compass dips straight down.

Worked Example

At a station, $H_E = 0.25\,G$ and dip $I = 60^\circ$. Find total field and vertical component.

$B_E = \frac{H_E}{\cos I} = \frac{0.25}{0.5} = 0.50\,G,$ $V_E = B_E\sin I = 0.50 \times \frac{\sqrt 3}{2} = 0.433\,G.$

The same station's magnetic latitude can be estimated from $\tan I = 2\tan\lambda \Rightarrow \tan\lambda = \tan 60^\circ /2 = 0.866 \Rightarrow \lambda \approx 41^\circ$.

Common Confusions

• Magnetic and geographic poles are not the same and the magnetic poles drift over time.
• Declination is measured at the local point; it varies hugely with longitude.
• Dip is measured from the horizontal, not from the vertical.
• The horizontal component is what deflects a freely pivoted compass needle; the vertical component points it down (only seen with a dip needle).

Key Takeaways

• Three elements: declination $D$, dip $I$, horizontal component $H_E$.
• $H_E = B_E\cos I$, $V_E = B_E\sin I$, $\tan I = V_E/H_E$.
• $\tan I = 2\tan\lambda$ at magnetic latitude $\lambda$ (dipole model).
• At magnetic equator $I=0$; at magnetic poles $I=90^\circ$.