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


by eyenkay
Tags: curl, dipole, magnetic field, vector potential
eyenkay
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#1
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
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#2
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
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#3
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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