====== Magnetic Dipole Field ======

Near the Earth's surface, the magnetic field can be considered dipolar, directed from the south pole to the north pole. Many charged particles gyrate and oscillate around the field lines in the north-south direction and drift in the east-west direction. The Earth's inner core, outer core, convection currents, and magnetic field lines are shown below.

{{https://upload.wikimedia.org/wikipedia/commons/thumb/9/91/Dynamo_Theory_-_Outer_core_convection_and_magnetic_field_generation.svg/1024px-Dynamo_Theory_-_Outer_core_convection_and_magnetic_field_generation.svg.png?nolink&650}}

The magnetic field is generated due to the motion of liquid metal in the Earth's outer core. Convection occurs in the liquid outer core due to the heat from the solid inner core. If the inner core is like a stove, the outer core is like a pot of water on that stove; just as the heat from the stove causes bubbles in boiling water, the heat from the inner core causes convection in the liquid metal of the outer core. Due to the [[coriolis-force|Coriolis force]] generated by the Earth's rotation, this convection occurs in a circular path. The circular motion of the liquid metal creates current loops, resulting in a magnetic field in the direction of the loop's area vector (\(\mathbf{A}\)). If \(I\) is the current flowing through the loop, then the magnetic dipole moment of this field is:

$$ \vec{\mu}_E = I\mathbf{A} $$

The average magnitude, according to our calculations, is \(\mu_E = 8.05\times 10^{22}\) A m\(^2\). Using the [[Biot-Savart law]], it is straightforward to calculate the magnetic field created along the area vector at the center of a current loop.

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Using the magnetic moment, the radial component \(B_r\) and azimuthal component \(B_\lambda\) of the Earth's magnetic field in spherical coordinates can be written together as a vector:

$$ \mathbf{B} = \frac{\mu_0\mu_E}{4\pi r^3} (-\hat{r}2\sin\lambda+\hat{\lambda}\cos\lambda) $$

where \(\hat{r}\) and \(\hat{\lambda}\) are unit vectors in the radial and azimuthal directions, respectively; these directions are marked in the figure above. The azimuthal direction's coordinate can be referred to as magnetic latitude, which differs from Earth's geographic latitude. Since our magnetic field is axisymmetric, the third coordinate in the spherical system does not need to be used here, as the field magnitude remains the same across 0 to 360 degrees along a specific latitude. The magnitude of this field is:

$$ B = \sqrt{B_r^2+B_\lambda^2} = \frac{\mu_0 \mu_E}{4\pi r^3} \sqrt{1+3\sin^2\lambda} $$

where the two component values have been substituted for calculation. The magnetic field at any point on the field line is tangent to that line. Therefore, if \(d\mathbf{s}\) is an infinitesimal element of the field line:

$$ d\mathbf{s}\times \mathbf{B} = 0 $$

Solving this through determinants yields:

$$ \frac{dr}{r} = \frac{B_r}{B_\lambda} d\lambda = -d\lambda\frac{2\sin\lambda}{\cos\lambda} $$

Integrating this finally gives the **field line equation**:

$$ r = r_{eq} \cos^2\lambda $$

where \(r_{eq}\) is an integration constant equal to the equatorial radius of a field line, which is the maximum distance of the field line from Earth. This is evident because when \(\lambda=0\), \(r=r_{eq}\). If we recall:

$$ ds^2 = dr^2 + r^2 d\lambda^2 $$

where the arc element along the circle's circumference is \(r d\lambda\), then the relationship between the total length of a field line and its magnetic latitude can be written as:

$$ \frac{ds}{d\lambda} = r_{eq} \cos\lambda \sqrt{1+3\sin^2\lambda} $$

Integrating with respect to latitude gives the total length \(s\) of the field line.

The magnetic field is often expressed in terms of Earth's equatorial radius \(R_E\), but it is more convenient to use the \(L\)-shell parameter or \(L\)-value \(L=r_{eq}/R_E\). The process is straightforward. Along Earth's equator (\(\lambda=0\)), the magnetic field is only \(B_E = \mu_0 \mu_E / (4\pi R_E^3)\). Dividing the general magnetic field equation by this gives:

$$ \frac{B}{B_E} = \frac{R_E^3}{r^3} \sqrt{1+3\sin^2\lambda} $$

Substituting \( r = r_{eq} \cos^2\lambda \) and \(L=r_{eq}/R_E\) yields:

$$ B = \frac{B_E}{L^3} \frac{\sqrt{1+3\sin^2\lambda}}{\cos^6\lambda} $$

which is useful in many cases. Using the \(L\)-value, the field line equation can also be written as:

$$ \cos^2\lambda_E = \frac{R_E}{r_{eq}} = L^{-1} $$

where \(\lambda_E\) is the magnetic latitude at which a specific \(L\)-value field line touches Earth's surface.