Simple Nature - Light and Matter

i/The field of a dipole.

j/Example 9.

whereris the distance from the dipole to the point of interest,θ

is the angle between the dipole vector and the line connecting the

dipole to this point, andEzandER are, respectively, the compo-

nents of the field parallel to and perpendicular to the dipole vector.

In the midplane,θequalsπ/2, which producesEz=−kDr−^3

andER= 0. This is the same as the field of amagneticdipole in its

midplane, except that the electric coupling constantkreplaces the

magnetic versionk/c^2 , and the electric dipole momentDis substi-

tuted for the magnetic dipole momentm. It is therefore reasonable

to conjecture that by using the same presto-change-o recipe we can

find the field of a magnetic dipole outside its midplane:

Bz=

km

c^2

(

3 cos^2 θ− 1

)

r−^3

BR=

km

c^2

(3 sinθcosθ)r−^3.

This turns out to be correct.^5

Concentric, counterrotating currents example 9

.Two concentric circular current loops, with radiiaandb, carry

the same amount of currentI, but in opposite directions. What is

the field at the center?

.We can produce these currents by tiling the region between the

circles with square current loops, whose currents all cancel each

other except at the inner and outer edges. The flavor of the calcu-

lation is the same as the one in which we made a line of current

by filling a half-plane with square loops. The main difference is

that this geometry has a different symmetry, so it will make more

sense to use polar coordinates instead ofxandy. The field at

the center is

Bz=

∫

k I

c^2 r^3

dA

=

∫b

r=a

k I

c^2 r^3

· 2 πrdr

=

2 πk I

c^2

(

1

a

−

1

b

)

.

The positive sign indicates that the field is out of the page.

(^5) If you’ve taken a course in differential equations, this won’t seem like a very
surprising assertion. The differential form of Gauss’ law is a differential equation,
and by giving the value of the field in the midplane, we’ve specified a boundary
condition for the differential equation. Normally if you specify the boundary
conditions, then there is a unique solution to the differential equation. In this
particular case, it turns out that to ensure uniqueness, we also need to demand
that the solution satisfy the differential form of Amp`ere’s law, which is discussed
in section 11.4.
694 Chapter 11 Electromagnetism