Physics Friday 124

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 uJ, 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 Jx 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 ith 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 ith 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 mi and charges qi, in motion with positions ri and velocities vi, then (approximating it as a static current distribution), we have
, and so the magnetic moment integral becomes a sum via the delta functions:
. However, the angular momentum of the ith 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.

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One Response to “Physics Friday 124”

  1. Physics Friday 128 « Twisted One 151's Weblog Says:

    […] our volume, the surface will eventually come to be far from the current distribution. As we noted here, for distant fields, the dipole term dominates, and the vector potential goes as r-2, and the field […]

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