Does the inverse square rule work with a magnetic field?

[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?

 

[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.

 

[Drakkith]

An elemental current element

What is an elemental current element?

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?

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.

 

[vanhees71]

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

 

[robphy]

A possibly useful starting point:
https://en.wikipedia.org/wiki/Biot–Savart_law
followed by
https://en.wikipedia.org/wiki/Gauss's_law_for_magnetism
https://en.wikipedia.org/wiki/Magnetic_monopole

 

[TheAnt]

Thank you very much for the answers

 

[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

 

[lychette]

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

totally agree...use the lagrangian transformation

 

[wrobel]

did not understand your suggestion

 

[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: $\{F,H\}=0$

 

[lychette]

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: $\{F,H\}=0$

Hamiltonian...of course...sorry