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Magnetic Field of Magnetic Dipole Moment

by Morto
Tags: dipole, field, magnetic, moment
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Morto
#1
Oct11-08, 10:51 AM
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
[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?
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clem
#2
Oct11-08, 01:48 PM
Sci Advisor
P: 1,256
Your second term is wrong. It should be
[tex]-({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right][/tex].
Derivator
#3
Apr4-09, 07:45 AM
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 [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
#4
Apr4-09, 04:25 PM
Sci Advisor
P: 1,256
Magnetic Field of Magnetic Dipole Moment

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
#5
Apr4-09, 05:14 PM
P: 144
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
#6
Apr5-09, 10:58 AM
Sci Advisor
P: 1,256
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
#7
Apr5-09, 11:10 AM
Sci Advisor
P: 1,256
The point is that the m.del acts on only one function at a time.
jmwilli25
#8
Mar22-11, 03:58 AM
P: 5
[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
#9
Mar22-11, 05:38 AM
P: 193
Quote Quote by Derivator View Post
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?
[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.


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