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Localized Current Distribution
Consider the magnetic field generated by a current distribution that is localized in some relatively small region of space centered on the origin.
From Equation (621), we have
![$\displaystyle {\bf A}({\bf r}) = \frac{\mu_0}{4\pi}\int\frac{{\bf j}({\bf r}')}{\vert{\bf r}-{\bf r}'\vert}\,dV'.$](img1310.png) |
(661) |
Assuming that
, so that our observation point lies well outside the distribution, we can write
![$\displaystyle \frac{1}{\vert{\bf r}-{\bf r}'\vert}= \frac{1}{\vert{\bf r}\vert}+\frac{{\bf r}\cdot{\bf r}'}{\vert{\bf r}\vert^{\,3}}+\cdots.$](img1380.png) |
(662) |
Thus, the
th Cartesian component of the vector potential has the expansion
![$\displaystyle A_i({\bf r})= \frac{\mu_0}{4\pi}\,\frac{1}{\vert{\bf r}\vert}\int...
...{{\bf r}}{\vert{\bf r}\vert^{\,3}}\cdot\int j_i({\bf r}')\,{\bf r}'\,dV'+\cdots$](img1381.png) |
(663) |
Consider the integral
![$\displaystyle K = \int (f\,{\bf j}\cdot \nabla' g+g\,{\bf j}\cdot\nabla' f)\,dV',$](img1382.png) |
(664) |
where
is a divergence-free [see Equation (618)] localized current distribution, and
and
are
two well-behaved functions.
Integrating the first term by parts, making use of the fact that
as
(because the current distribution
is localized), we obtain
![$\displaystyle K=\int\left[-g\,\nabla'\cdot(f\,{\bf j})+g\,{\bf j}\cdot\nabla' f\right]dV'$](img1388.png) |
(665) |
Hence,
![$\displaystyle K = \int\left[-g\,{\bf j}\cdot\nabla' f -g\,f\,\nabla'\cdot{\bf j}+g\,{\bf j}\cdot\nabla' f\right]dV'=0,$](img1389.png) |
(666) |
because
. Thus, we have proved that
![$\displaystyle \int (f\,{\bf j}\cdot \nabla' g+g\,{\bf j}\cdot\nabla' f)\,dV'=0.$](img1391.png) |
(667) |
Let
and
(where
is the
th component of
). It immediately follows from Equation (668) that
![$\displaystyle \int j_i({\bf r}')\,dV'= 0.$](img1395.png) |
(668) |
Likewise, if
and
then Equation (668) implies that
![$\displaystyle \int \left(x_i'\,j_j + x_j'\,j_i\right)\,dV' = 0.$](img1398.png) |
(669) |
According to Equations (664) and (669),
![$\displaystyle A_i({\bf r})= \frac{\mu_0}{4\pi}\,\frac{{\bf r}}{\vert{\bf r}\vert^{\,3}}\cdot\int j_i({\bf r}')\,{\bf r}'\,dV'+\cdots.$](img1399.png) |
(670) |
Now,
![$\displaystyle {\bf r}\cdot\int j_i({\bf r}')\,{\bf r}'\,dV' = x_j\int x'_j\,j_i\,dV'= -\frac{1}{2}x_j\int(x_i'\,j_j-x_j'\,j_i)\,dV',$](img1400.png) |
(671) |
where use has been made of Equation (670), as well as the Einstein summation convention. Thus,
![$\displaystyle {\bf r}\cdot\int j_i\,{\bf r}'\,dV' = -\frac{1}{2}\int \left[({\b...
... = -\frac{1}{2}\left[{\bf r}\times \int ({\bf r}'\times {\bf j})\,dV'\right]_i.$](img1401.png) |
(672) |
Hence, we obtain
![$\displaystyle {\bf A}({\bf r})=-\frac{\mu_0}{8\pi}\,\frac{{\bf r}}{\vert{\bf r}\vert^{\,3}}\times \int ({\bf r}'\times {\bf j})\,dV'+\cdots.$](img1402.png) |
(673) |
It is conventional to define the magnetization, or magnetic moment density, as
![$\displaystyle {\bf M}({\bf r}) = \frac{1}{2}\,{\bf r}\times {\bf j}({\bf r}).$](img1403.png) |
(674) |
The integral of this quantity is known as the magnetic moment:
![$\displaystyle {\bf m} = \frac{1}{2}\int{\bf r}'\times {\bf j}'({\bf r}')\,dV'.$](img1404.png) |
(675) |
It immediately follows from Equation (674) that the vector potential a long way from a localized current distribution takes the form
![$\displaystyle {\bf A}({\bf r}) = \frac{\mu_0}{4\pi}\,\frac{{\bf m}\times {\bf r}}{r^{\,3}}.$](img1405.png) |
(676) |
The corresponding magnetic field is
![$\displaystyle {\bf B}({\bf r}) = \nabla\times {\bf A} = \frac{\mu_0}{4\pi}\left[\frac{3\,({\bf m}\cdot{\bf r})\,{\bf r} -r^{\,2}\,{\bf m}}{r^{\,5}}\right].$](img1406.png) |
(677) |
Thus, we have demonstrated that the magnetic field far from any localized current distribution takes the form of a magnetic
dipole field whose moment is given by the integral (676).
Consider a localized current distribution that consists of a closed planar loop carrying the current
. If
is a
line element of the loop then Equation (676) reduces to
![$\displaystyle {\bf m} = I\oint\frac{1}{2}\,{\bf r}\times d{\bf r}.$](img1408.png) |
(678) |
However,
, where
is a triangular element of vector area defined by the two ends of
and the origin.
Thus, the loop integral gives the total vector area,
, of the loop. It follows that
![$\displaystyle {\bf m} = I\,A\,{\bf n},$](img1411.png) |
(679) |
where
is a unit normal to the loop in the sense determined by the right-hand circulation rule (with the current determining the sense
of circulation). Of course, Equation (680) is identical to Equation (659).
Next: Exercises
Up: Magnetostatic Fields
Previous: Circular Current Loop
Richard Fitzpatrick
2014-06-27