How can I solve for the shape of Earth's magnetic field (neglecting the effects of the solar wind which distort it). It looks very similar to the field due to a solenoid, but I can only find solutions for the field inside a solenoid, and I want the field around the Earth. What I ultimately want to know is the angle of the field lines along the planet's surface (angle away from the normal to the planet's surface) as a function of longitude.
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$\begingroup$ Wikipedia has the dipole approximation. $\endgroup$– GhosterCommented Aug 29 at 3:25
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$\begingroup$ @Ghoster Thanks, it does have an expression for the field, but what I want to know is how to derive that expression. $\endgroup$– Random_Astro_StudentCommented Aug 29 at 14:28
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$\begingroup$ The field outside a uniformly magnetized ball is a dipole field. See farside.ph.utexas.edu/teaching/jk1/lectures/node61.html for one derivation using the concept of scalar magnetic potential. $\endgroup$– GhosterCommented Aug 29 at 15:49
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$\begingroup$ In the same way that the electric field outside a uniformly charged ball turns out to be the same as that of an imaginary point charge at the center, the magnetic field outside a uniformly magnetized ball turns out to be the same as that of an imaginary point magnetic dipole at the center. $\endgroup$– GhosterCommented Aug 29 at 15:51
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$\begingroup$ Another derivation you could attempt is to integrate the fields from the actual infinitesimal magnetic dipoles in each infinitesimal volume element inside the ball. This is probably a difficult integral to evaluate analytically, but you could always do it numerically to convince yourself that the outside field is dipolar. Finite-sized magnets are just collections of point dipoles. $\endgroup$– GhosterCommented Aug 29 at 16:00
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1 Answer
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It's not something you solve. It's something you measure:
https://www.ncei.noaa.gov/products/world-magnetic-model
Last time I used it, briefly, it was only out to degree 7 (in contrast to gravity, which is available to degree ~2.5K or higher, depending on your access level).
Edit: So it's traditional with anything around a sphere to do a multipole expansion, which are the irreducible shapes that rotate nicely in spherical coordinates:
$l=0$, monopole = spherically symmetric
$l=1$, dipole = 3 vectors $\leftarrow$
$l=2$, quadrupole = 5 shapes with alignment $\leftrightarrow$ and no direction
...and so on...
Anyway, for uniform magnetized $\vec m$ sphere, that is pure dipole.
For a point $\vec r$ outside the sphere--really for a vector connecting the origin to the point, you can write down the form of the vector potential, $\vec A$:
$$ \vec A = c_1\vec m + c_2 \vec r + c_3(\vec m \times \vec r) $$
since those are the vectors in the problem.
Because $\vec A$ flips sign under parity and $\vec m$ does not:
$$ c_1 = 0$$
Because $\vec A$ flips sign under time reversal and $\vec r$ does not:
$$ c_2 = 0 $$
wile $\vec m \times \vec r$ has the same time/parity behavior as $\vec A$. I'll state w/o derivation:
$$ c_3 = \frac{\mu_0}{4\pi r^3} $$
so
$$ \vec A(\vec r) = \frac{\mu_0}{4\pi r^3} \vec m \times \vec r $$
whence:
$$ \vec B(\vec r) = \vec{\nabla} \times \vec A$$
$$ \vec B(\vec r)= \frac{\mu_0}{4\pi} \Big[ \frac{3\vec r(\vec m \cdot \vec r)}{r^5} - \frac{\vec m}{r^3} \Big] $$
(https://en.wikipedia.org/wiki/Magnetic_dipole)
At the surface, you can--a la Gauss's Law for gravity from a sphere--treat the magnetization as:
$$ \vec m(\vec r) = M\delta(\vec r) $$
Idk what $M$ is for the Earth, and if you want higher order moments: https://en.wikipedia.org/wiki/Multipole_expansion .
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$\begingroup$ Thanks, fair enough, but there are definitely approximate expressions I've seen (for instance in @Ghoster's comment above) which is what I'd like to know how to derive. $\endgroup$ Commented Aug 29 at 14:29
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$\begingroup$ @Random_Astro_Student Are you asking for a dipole (or higher) field out side and at the surface of the sphere of magnetized material? If that's it, I can update my answer. $\endgroup$– JEBCommented Aug 29 at 16:00
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$\begingroup$ Thank you! What is small vector m in your derivation? $\endgroup$ Commented Aug 29 at 20:28
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