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
\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) \\<br /> = \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?

 

[Meir Achuz]

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

 

[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)

 

[Meir Achuz]

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) \\<br /> = \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.

 

[Derivator]

Hi,
thanks for your answer.

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?

 

[Meir Achuz]

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}

 

[Meir Achuz]

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

 

[jmwilli25]

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

 

[netheril96]

Hi,
thanks for your answer.

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?

\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.