# Homework Help: Magnetic potential and inverse square law

1. Jan 9, 2006

### hotel

Hi
I am in doubt about the lack of the squared distance r in the expression below:
v(r)= constant* int_vol div M /|r_0 - r| dV
M: magnetization vector
r : displacement vectors
Is it correct to say that magnetic potential V(r) obeys the inverse square law ? Or it is more correct to say that the inverse square law can be "infered" from the expression above?

2. Jan 11, 2006

### Tom Mattson

Staff Emeritus
First, let me retype your expression in LaTeX:

[tex]v(r)=k\int_V\frac{\vec{\nabla}\cdot\vec{M}}{|\vec{r}_0-\vec{r}|}dV[/itex]

In the above, $k$ is a constant.

OK, now what does $v(r)$ mean? You say that it is the magnetic potential, but the magnetic potential is a vector, while your expression above is a scalar.

3. Jan 11, 2006

### LeoAlpha

The magnetic potential can be defined as a scalar or a vector, the magnetic potential (as is its electric counterpart) is used classically as a tool to compute the magnetic field, so depending on the problems as long as one still gets the correct B-field, it can be defined in a non-unique way. So the expression written is correct.

The scalar magnetic potential is mainly used for magnetostatic problems (refer to the corresponding chapter in Jackson), and has a form you typed. This is analogous to the electric potential (which is a scalar), and the -DivM corresponds to a "magnetic charge" coming from the magnetization of the material.

Back to the original question by hotel, this is equivalent to the inverse square law. The equation you've written is formally equivalent to the electric potential case.

Hope this helps,

4. Jan 13, 2006

### hotel

why the magnetic potential is a vector?

And the other thing is could you (anyone) explain how the equation above is equivalent to the inverse square law?

Because I cannot see the indication of 'square of the distance' in the fraction inside the integral !

for example what is the problem with:
[tex]v(r)=k\int_V\frac{\vec{\nabla}\cdot\vec{M}}{|\vec{r}_0-\vec{r}|^2}dV[/itex]

?