# Astrophysics: Simple magnetic field issues

1. Apr 1, 2014

### qftqed

Hello! I have an assignment question that is a proving a real stone in my shoe, mostly because it is reminding how little I can recall about electromagnetism. It is part of a series of questions leading to the calculation the Alfven radius, where the solar wind decouples from the Sun's magnetic field.

1. The problem statement, all variables and given/known data
Write down the equation for magnetic energy density. How does this change with radius [from the sun], assuming that the magnetic B field is purely radial?

2. Relevant equations
Magnetic energy density u=B2/2μ

3. The attempt at a solution

I really feel like this ought to decrease with radius, but I don't know how. The real issue is that I've not the slightest idea of where to start on working this out. The geometry of the field is not sitting well with me, when I read 'purely radial' I think field lines directed from the center outwards, which seems like something Gauss's law for magnetism would have a fit over. The magnetic field strength should be inversely proportional to the radius, I suspect. It seems my lack of familiarity with electromagnetism has finally come to collect! I would be overjoyed if someone could lend me a hand with this (=

Last edited: Apr 1, 2014
2. Apr 2, 2014

### Simon Bridge

Welcome to PF;
Do you know what the letters in your equation stand for?
Do you have notes for what the magnetic field of the Sun looks like?

3. Apr 2, 2014

### qftqed

Hello there, thanks for the speedy reply! We've not been given details on the geometry of the magnetic field, only the words 'purely radial'. I do know what the equation means, and in fact I think (hope) I'm making some progress. I've decided B2 is better off being written as a dot product of the vector B with itself, and from there I've thought I should try to solve div(B)=0 in spherical coordinates, interpreting 'purely radial' to mean that only the r component has a non-zero derivative or something to that effect. Having found the vector B in that way I can substitute that into the expression for energy density and hopefully get something meaningful. Would I be on the right track at all?

4. Apr 2, 2014

### Simon Bridge

"purely radial" will mean that B is a function of r alone.
So you can write: $\vec B = B(r)\hat r$.

5. Apr 2, 2014

### qftqed

Excellent! Thanks, I'm quite sure I can work with this.

6. Apr 2, 2014

### Simon Bridge

Actually - looking at how a dipole works, it should be $\vec B = B(r)\hat k$ choosing the z-axis to point the same way as the dipole, and the origin at the center, this will work for points in the x-y plane.