Write the Magnetic Field of a dipole in coordinate-free form?

by eyenkay
Tags: curl, dipole, magnetic field, vector potential
eyenkay is offline
Feb20-09, 09:39 AM
P: 7
1. The problem statement, all variables and given/known data
Show that the magnetic field of a dipole can be written in coordinate-free form: B_dip (r)=(μ_o/(4πr^3 ))[3(m*r ̂ ) r ̂-m]

2. Relevant equations
Adip(r)= (μ_o/(4πr^2))(m*sin(theta))

Bdip= curl A = (μ_o*m/(4πr^3))(2cos(theta)(r-direction)+sin(theta)(theta-direction)

3. The attempt at a solution
I figure this must have something to do with the above equations for the vector potential dipole and magnetic field dipole, I just dont have any idea what it means to write in 'coordinate-free form', or how to go about that..
Can anybody point me in the right direction?
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Tom Mattson
Tom Mattson is offline
Feb20-09, 01:32 PM
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"Coordinate free" simply means "in terms of dot products". See how the expression they want you to derive has dot products in it? That's what they want.
drnairb is offline
Feb20-09, 05:59 PM
P: 2
You can [tex]\TeX{}[/tex]-ify your posts. It helps a *lot.* Here's the coordinate-free form:

[tex]\vec{B}_{dip}(\vec{r}) = \frac{\mu_0}{4 \pi r^3}[3(\vec{m} \cdot \hat{r})\hat{r} - \vec{m}][/tex]

where [tex]\vec{m}[/tex] is the dipole moment, right? (I've always used [tex]\vec{p}[/tex].)

Vector potential [tex]\vec{A}[/tex] is

[tex]\vec{A}_{dip}(\vec{r}) = \frac{\mu_0}{4 \pi r^2} (m \sin \theta)[/tex]

Magnetic field [tex]\vec{B}[/tex] is

[tex]\vec{B}_{dip}(\vec{r}) = \vec{\nabla} \times \vec{A} = \frac{\mu_0 m}{4 \pi r^3} (2 \cos \theta \hat{r} + \sin \theta \hat{\theta})[/tex]

To get coordinate-free form, you just need to express [tex]\vec{m}[/tex] in spherical coordinates and manipulate the properties of dot products in that coordinate system. If you assume your dipole is at the origin and points in the [tex]+\hat{z}[/tex] direction, then in spherical it would be [tex]\vec{m} = m \cos \theta \hat{r} - m \sin \theta \hat{\theta}[/tex]. Now use dot products of [tex]\vec{m}[/tex] with the necessary spherical unit vectors in order to eliminate those pesky sine and cosine functions.

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