# Magnetic Field of Magnetic Dipole Moment

1. Oct 11, 2008

### 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
$$\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi} \frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}$$

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

I know that
$$\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left(\nabla \times \left(\frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}\right)\right) \\ = \frac{\mu_0}{4\pi}\left(\vec{m}\left(\nabla\cdot\frac{\vec{x}}{\left|\vec{x}\right|^3}\right) - \frac{\vec{x}}{\left|\vec{x}\right|^3}\left(\nabla\cdot \vec{m}\right)\right)$$

How does Jackson end up at the result quoted?

2. Oct 11, 2008

### clem

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

3. Apr 4, 2009

### 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 $$\vec{B}$$?

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

kind regards,
derivator

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

Last edited: Apr 4, 2009
4. Apr 4, 2009

### clem

As I said in my post, the curl of the cross product should be
$$\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left(\nabla \times \left(\frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}\right)\right) \\ = \frac{\mu_0}{4\pi}\left(\vec{m}\left(\nabla\cdot\frac{\vec{x}}{\left|\vec{x}\right|^3}\right) - ({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right]$$.
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.

5. Apr 4, 2009

### Derivator

Hi,

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

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

is correct?

6. Apr 5, 2009

### 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}

Last edited: Apr 5, 2009
7. Apr 5, 2009

### clem

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

8. Mar 22, 2011

### jmwilli25

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

9. Mar 22, 2011

### netheril96

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

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

I is the unit tensor.