Part 4: Distant Fields

Let us consider the first-order magnetic potential and field far from a localized current density . The first step is to consider the denominator of the integral .

In particular, choosing an origin within the localized distribution, so that , we expand the fraction to first order in **r’**, obtaining

,

giving us the expansion

.

Now, we need to use a bit of vector calculus. For any well-behaved scalar field *u*(**r**) and vector field **v**(**r**), then vector integration by parts over a volume *V* bounded by simple surface *S*

.

Now, letting our vector field be the current density **J**, if our volume *V* completely encloses (and extends beyond) the region to which the current is localized, then **J**, and thus *u***J**, is thus zero on the surface *S*, and so that integral remains zero as the volume is expanded to all space, leaving us

,

or combining,

.

Now, if we let , and use the magnetostatic requirement that , then we see that

,

,

where *J _{x}* is the

*x*component of the current density vector field. Note that this integral is the

*x*component of the integral first integral in our expansion, . Similarly, using and each give us

and

,

respectively. Thus, the initial term in our expansion, the monopole term, is zero. Now, indexing the cartesian components of our vectors with

*i*=1,2,3, we now have

.

Now, we return to our integration by parts formula. We use , and

(where is the unit vector for the

*i*th coordinate); together with our requirement that the current be divergence-free, we obtain

Now, taking our integral term , and expanding the dot product into the sum of coordinate products via index

*j*, we have

.

Now, since , we can subtract half of this integral from each integral in the above sum, and

,

giving us

Now, for

*i*=

*j*, the above integral is zero. Similarly, we note that ; and similar expressions, which allow us to write

,

where is the Levi-Civita symbol. This means

;

but this is the

*i*th component of a cross product (see here), so

we have

.

Now, we define the

*magnetic moment density*by , and it’s integral . Then, in terms of this, we see

.

Taking the curl of this, and using the product rule for curl,

, along with the fact that

**μ**is a constant vector, we see

,

where

*r*=|

**r**| is the distance from the origin, and is the unit vector in the direction of

**r**. Note that the field of an electric dipole of dipole moment

**p**is (see here). Thus, our field is that of a magnetic dipole, and

**μ**is our magnetic moment (thus justifying our terming as magnetic moment density).

Note that if our current is confined to a current

*I*in a closed plane curve

*C*with line element , then our integral for magnetic moment becomes

,

and via Green’s theorem (reversing our use of it here), we see

, where

*n*is the normal to the plane of the loop (with direction via the right-hand rule), and the

*A*over which we integrate is the region enclosed by the loop. Thus, we obtain , where is the area vector. This is the same result that we found here by considering the torque a constant magnetic field exerts on a current loop.

Similarly, if we have point charges with masses

*m*and charges

_{i}*q*, in motion with positions

_{i}**r**

*and velocities*

_{i}**v**

*, then (approximating it as a static current distribution), we have*

_{i}, and so the magnetic moment integral becomes a sum via the delta functions:

. However, the angular momentum of the

*i*th particle is

,

so we obtain

.

If the particles all have the same charge-to-mass ratio , then it factors out of the sum, and

,

where

**L**is the total angular momentum. Note that for electrons, this then gives , the same result we obtained here for a single electron in a circular orbit.