Magnetic Field of Magnetic Dipole Moment

Morto

I'm seeking some clarification on a topic found in Jackson 'Classical Electrodynamics' Chapter 5. This deals with the magnetic field of a localised current distribution, where the magnetic vector potential is given by
[tex]\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi} \frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}[/tex]

and the magnetic induction [tex]\vec{B}[/tex] is the curl of this [tex]\vec{B} = \nabla \times \vec{A}[/tex].
I know that the end result should be
[tex]\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left[\frac{3\vec{n}\left(\vec{n}\cdot\vec{m} - \vec{m}}{\left|\vec{x}\right|^3}\right][/tex]
Where [tex]\vec{n}[/tex] is the unit vector in the direction of [tex]\vec{x}[/tex]
But I'm struggling to do this calculation for myself.

I know that
[tex]\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left(\nabla \times \left(\frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}\right)\right) \\
= \frac{\mu_0}{4\pi}\left(\vec{m}\left(\nabla\cdot\frac{\vec{x}}{\left|\v ec{x}\right|^3}\right) - \frac{\vec{x}}{\left|\vec{x}\right|^3}\left(\nabla\cdot \vec{m}\right)\right)[/tex]

How does Jackson end up at the result quoted?

clem

Your second term is wrong. It should be
[tex]-({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right][/tex].

Derivator

Hi

I'm new to this forum! Googleing for the questioned problem I found this forum. Having the same problem as the author of this thread, I want to ask if someone has a solution on how Jackson found the formula for [tex]\vec{B}[/tex]?

I tried my luck with the product rule [tex]\nabla\times(\vec{A}\times\vec{B})=(\vec{B}\cdot\nabla)\vec{A}-(\vec{A}\cdot\nabla)\vec{B}+\vec{A}(\nabla\cdot\vec{B})-\vec{B}(\nabla\cdot\vec{A})[/tex] but came to no conclusion.

kind regards,
derivator

(sorry for my english, it's not my mother tongue)

clem

As I said in my post, the curl of the cross product should be
[tex]\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left(\nabla \times \left(\frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}\right)\right) \\
= \frac{\mu_0}{4\pi}\left(\vec{m}\left(\nabla\cdot\frac{\vec{x}}{\left|\v ec{x}\right|^3}\right) - ({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right][/tex].
This follows because m is a constant so only two terms of your four enter.
Given this result, the first term is a delta function is not needed if r>0.
The second term gives the result in the first post if the delta function singularity at the origin is again ignored.
Jackson is a bit obscure on this.

Derivator

Hi,
thanks for your answer.

Could you get a bit more detailed on how to show that

[tex]- ({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right]=\left[\frac{3\vec{n}\left(\vec{n}\cdot\vec{m}\right) - \vec{m}}{\left|\vec{x}\right|^3}\right][/tex]

is correct?

clem

You really need a better textbook.
Sorry I can't seem to find my latex error below. If hyou can read or correct it, it should show you what to do.
- ({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right]=
-{\vec x} ({\vec m}\cdot\nabla)\frac{1}{|{\vec x}|^3}
-\frac{1}{|{\vec x}|^3} ({\vec m}\cdot\nabla){\vec x}

clem

The point is that the m.del acts on only one function at a time.

jmwilli25

[tex]
- (\vec{m}\cdot\nabla)\left[\frac{\vec{x}}{|\vec {x}|^3}\right]=
-{\vec x} ({\vec m}\cdot\nabla)\frac{1}{|{\vec x}|^3}
-\frac{1}{|{\vec x}|^3} ({\vec m}\cdot\nabla){\vec x}
[/tex]

netheril96

[tex]\nabla\frac{\vec{r}}{r^{3}}=\left(\nabla\frac{1}{r^{3}}\right)\vec{r}+\ frac{\nabla\vec{r}}{r^{3}}=-\frac{3\vec{r}\vec{r}}{r^{5}}+\frac{\mathbf{I}}{r^{3}}[/tex]

[tex]\vec{m}\cdot\nabla\frac{\vec{r}}{r^{3}}=\vec{m}\cdot\left(-\frac{3\vec{r}\vec{r}}{r^{5}}+\frac{\mathbf{I}}{r^{3}}\right)=-\frac{3(\vec{m}\cdot\vec{r})\vec{r}}{r^{5}}+\frac{\vec{m}}{r^{3}}[/tex]

I is the unit tensor.