# B Does the inverse square rule work with a magnetic field?

1. Jul 12, 2016

### TheAnt

I am interested in mini magnetospheres. How do i calculate the intensity of the field at a certain distance if i already know theits intensity at the source?

2. Jul 12, 2016

### gleem

An elemental current element produces a magnetic field that obeys the inverse square law. However real magnetic fields produced by a concatenation current elements or modeled as such result in fields that do not obey the ISL. In fact magnetic fields are produced by dipoles which have an inverse cube dependence for example a current in a loop of wire. These dipole fields also have an angular dependence with respect to the axis of symmetry of the dipole. For extensive sources the dependences can be more complex.

3. Jul 12, 2016

### Drakkith

Staff Emeritus
What is an elemental current element?

Calculating the strength of a magnetic field from a real source is not trivial. Many people simply end up measuring it instead of calculating it. As gleem said, the strength of the field is not simply a function of distance, but also of direction (strength at distance R from a pole is different than at distance R from the side). So there's not really a simple equation that will tell you the strength at a particular distance.

4. Jul 13, 2016

### vanhees71

The leading-order multipole expansion of the magnetic field is the dipole field which goes like $1/r^3$ for $r \rightarrow \infty$.

5. Jul 13, 2016

### robphy

6. Jul 13, 2016

### TheAnt

Thank you very much for the answers

7. Jul 13, 2016

### wrobel

by the way it is a good task to integrate the problem of planar motion of a particle in the gravity field of the dipole and describe the motion of the particle

8. Jul 19, 2016

### lychette

totally agree.....use the lagrangian transformation

9. Jul 19, 2016

### wrobel

10. Jul 21, 2016

### wrobel

Actually this problem is integrated as follows. In suitable polar coordinates the Hamiltonian is
$$H=\frac{1}{2m}\Big(p^2_r+\frac{p^2_\varphi}{r^2}\Big)+\frac{k\cos\varphi}{r^2}.$$
It is easy to see that the variables are separated:
$$H=\frac{p^2_r}{2m}+\frac{1}{2r^2}F;$$
here $F=p^2_\varphi/m+2k\cos\varphi$ is a first integral: $\{F,H\}=0$

11. Jul 21, 2016

### lychette

Hamiltonian....of course....sorry