Write the Magnetic Field of a dipole in coordinatefree form?by eyenkay Tags: curl, dipole, magnetic field, vector potential 

#1
Feb2009, 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 coordinatefree form: B_dip (r)=(μ_o/(4πr^3 ))[3(m*r ̂ ) r ̂m] 2. Relevant equations A_{dip}(r)= (μ_o/(4πr^2))(m*sin(theta)) B_{dip}= curl A = (μ_o*m/(4πr^3))(2cos(theta)(rdirection)+sin(theta)(thetadirection) 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 'coordinatefree form', or how to go about that.. Can anybody point me in the right direction? 



#2
Feb2009, 01:32 PM

Emeritus
Sci Advisor
PF Gold
P: 5,540

"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.




#3
Feb2009, 05:59 PM

P: 2

You can [tex]\TeX{}[/tex]ify your posts. It helps a *lot.* Here's the coordinatefree 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 coordinatefree 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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