Generalization of the magnetic moment

In summary, the conversation revolves around the equation for magnetic moments (\vec\mu=I\vec{A}) and torque (\vec\tau=\vec\mu\times\vec{B}), and the attempt to prove it in a general case. It is clarified that the torque is indeed \vec\mu\times\vec{B}, and the second term in the integral is shown to be zero. The general definition of magnetic moment for a current loop is also provided.
  • #1
tim_lou
682
1
I have searched various websites... but all they have about magnetic moments is:

[tex]\vec\mu=I\vec{A}[/tex]
and
[tex]\vec\tau=\vec\mu\times\vec{B}[/tex]
but all the the sources only showed that the equation is true for rectangular loop of current.

but I need a more satisfactory answer. So I try to prove this in a general case:

I want to prove that the torque is indeed [tex]\vec\mu\times\vec{B}[/tex], when magnetic moment is defined as:
[tex]\vec\mu=I\int d\vec{A}[/tex]
(after proving that, i will try to prove the energy relation: [tex]E=-\vec\mu\cdot{\vec{B}}[/tex])

So, I forge ahead, let a closed current loop be the curve r, and a constant magnetic field B exists (I is current in the loop, tau is torque):
[tex]d\vec\tau=\vec{r}\times d\vec{F}=I \vec{r}\times (d\vec{r}\times\vec{B})[/tex]
where I is current in the wire, assume that it is constant.

[tex]\vec\tau=I\oint\vec{r}\times(d\vec{r}\times\vec{B})[/tex]

I used the triple product formula and some manipulations, then I got:
[tex]\vec\tau=I\oint\vec{r}\times{d\vec{r}}\times{\vec{B}}+I\oint\vec{r}(\vec{B}\cdot{d\vec{p})[/tex]

the first term looks like [tex]\vec\mu\times\vec{B}[/tex] but then I can't tell if the second term is zero...well, i checked the manipulation numerous times... i basically used:
[tex]\vec{r}\times{d\vec{r}}\times\vec{B}=d\vec{r}(\vec{B}\cdot\vec{r})-\vec{r}(\vec{B}\cdot{d\vec{r}})[/tex]

and
[tex]\vec{r}\times{}(d\vec{r}\times\vec{B})=d\vec{r}(\vec{r}\cdot\vec{B})-\vec{B}(\vec{r}\cdot{d\vec{r}})[/tex]


when integrating, B(r * dr) goes to zero since it is a closed loop.
Hence,
[tex]\vec\tau=I\oint\vec{r}\times{d\vec{r}}\times{\vec{B}}+I\oint\vec{r}(\vec{B}\cdot{d\vec{p})=\vec\mu\times\vec{B}+I\oint\vec{r}(\vec{B}\cdot{d\vec{p})[/tex].

I just changed the integrand.

now I'm stuck...supposedly, torque=mu cross magnetic field. but what is the second term of the integral doing there? perhaps the formula for the torque on any closed current loop is not simply mu cross B?

Questions: what frame should I use to calculate the torque? does it matter? should I choose the center of mass frame? Is the torque always the same regardless of what reference frame one is in?

edited: clarified my intentions.
 
Last edited:
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  • #2
Hi, I don't know what you're trying to do here but if as the title of the thread suggest, you're looking for the general definition of the magnetic moment of a current loop, it is the following thing:

Given a current loop in which circulates a current I, the magnetic moment is the vector

[tex]\vec{\mu}=I\int\hat{n}dA[/tex]

where the integral is taken over any surface whose boundaries are the current loop itself (the result is independant of the choice of the surface). [itex]\hat{n}[/itex] denotes the vector normal to the surface.
 
  • #3
I'm trying to prove that the torque is indeed mu cross A. I should've specified that. sorry about that.
 
  • #4
[itex]\tau =I\oint{\bf r\times(dr\times B)}\\
=I\oint{\bf dr(r\cdot B)}
-I\oint{\bf B(r\cdot dr)}.
[/itex]
The second integral vanishes for constant B.
 
  • #5
For the first integral:
[itex]
\oint{\bf (r\cdot B)dr}=\int_S{\bf dS\times\nabla(r\cdot B)}
={\bf S}\times{\bf B}.
[/itex]
 
  • #6
Thx, Meir, it looks rather simple... Can you tell me what theorem you used? I am currently learning vector calculus and I'm not that familiar with all those different theorems (stokes, greens...).
 
  • #7
What Meir used in his last post is a corollary of Stokes theorem. It's a nice exercise proving it.
 

1. What is the definition of magnetic moment?

Magnetic moment refers to the measure of the strength and orientation of a magnetic field generated by a magnet or a current-carrying wire.

2. How is magnetic moment calculated?

The magnetic moment of a system can be calculated by multiplying the strength of the magnetic field by the area of the loop perpendicular to the field and the number of turns in the loop.

3. What is the significance of generalizing magnetic moment?

Generalizing magnetic moment allows us to understand and predict the behavior of different magnetic materials and systems, including atoms, molecules, and larger structures.

4. What factors affect the generalization of magnetic moment?

The generalization of magnetic moment can be affected by factors such as temperature, external magnetic fields, and the material's atomic and molecular structure.

5. How is generalization of magnetic moment used in practical applications?

The generalization of magnetic moment has various applications in industries such as electronics, telecommunications, and medicine. It is also used in scientific research to study magnetic materials and their properties.

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