Magnetic Field of Magnetic Dipole Moment

by Morto
Tags: dipole, field, magnetic, moment
 P: 12 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 $$\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left(\nabla \times \left(\frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}\right)\right) \\ = \frac{\mu_0}{4\pi}\left(\vec{m}\left(\nabla\cdot\frac{\vec{x}}{\left|\v ec{x}\right|^3}\right) - \frac{\vec{x}}{\left|\vec{x}\right|^3}\left(\nabla\cdot \vec{m}\right)\right)$$ How does Jackson end up at the result quoted?
 Sci Advisor P: 1,250 Your second term is wrong. It should be $$-({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right]$$.
 P: 144 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)
P: 1,250

Magnetic Field of Magnetic Dipole Moment

As I said in my post, the curl of the cross product should be
$$\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left(\nabla \times \left(\frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}\right)\right) \\ = \frac{\mu_0}{4\pi}\left(\vec{m}\left(\nabla\cdot\frac{\vec{x}}{\left|\v ec{x}\right|^3}\right) - ({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right]$$.
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.
 P: 144 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?
 Sci Advisor P: 1,250 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}
 Sci Advisor P: 1,250 The point is that the m.del acts on only one function at a time.
 P: 5 $$- (\vec{m}\cdot\nabla)\left[\frac{\vec{x}}{|\vec {x}|^3}\right]= -{\vec x} ({\vec m}\cdot\nabla)\frac{1}{|{\vec x}|^3} -\frac{1}{|{\vec x}|^3} ({\vec m}\cdot\nabla){\vec x}$$
P: 193
 Quote by 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?
$$\nabla\frac{\vec{r}}{r^{3}}=\left(\nabla\frac{1}{r^{3}}\right)\vec{r}+\ frac{\nabla\vec{r}}{r^{3}}=-\frac{3\vec{r}\vec{r}}{r^{5}}+\frac{\mathbf{I}}{r^{3}}$$

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

I is the unit tensor.

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